organic chemistry 118a - ericpdotutoring.com€¦ · organic chemistry, with its focus on compound...

Dr. S. Lievens (2016)

Organic Chemistry 118A

Supplementary Materials for Discussion Sections Spring 2016

Department of Chemistry University of California, Davis

118A Discussion Manual Spring 2016 University of California, Davis

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Table of Contents Introduction – Learning Organic Chemistry ……….......................................... 2 Week 1 – Structure of Organic Molecules I: Structural Formulae and Resonance Drawing Organic Structures ……………………………………………….. 3 Resonance ……………………………………………………….………….. 14 Practice ………………………………………………………………………. 20 Week 2 – Structure of Organic Molecules II: Hybridization Hybridization ………………………………………………………………… 22

Assigning Hybridization ……………………………………………………. 26 Practice ………………………………………………………………………. 28 In Class Worksheet …………………………………………………………. 28

Week 3 – Nomenclature of Organic Molecules: Alkanes, Cycloalkanes, and Haloalkanes Alkyl Groups ……………………………………………….………………… 34

Nomenclature of Alkanes ……………………………...…………………… 37 Nomenclature of Cycloalkanes ……………..……………………………… 39 Structures from names ……………………………………………………… 41

Practice ……………………………………………..………………………… 42 Week 4 – Stereoisomerism What are stereoisomers? …………………………………………………… 44 Assigning R/S ………………………………………………………………… 52 Practice ………………………………………………………………………... 56

In Class Worksheet …………………………………………………………. 59 Week 5 – Introduction to Spectroscopy I: Infrared Spectroscopy Principles of IR ………………………………………………………………. 65

Interpreting IR ………………………………………………………………... 70 Practice ……………………………………………………………………….. 71 Week 6 – Introduction to Spectroscopy II: Basic Principles of NMR Nuclear Magnetic Resonance ……………………………………………… 73 H NMR and Chemical Shifts ……………………………………………….. 75 Chemical Equivalence ……………………………………………………… 78 Practice ………………………………………………………………………. 80 Week 7 – Introduction to Spectroscopy III: Further Principles of NMR Integration …………………………………………………………………… 81 Coupling ……………………………………………………………………… 82 Practice ………………………………………………………………………. 86 Week 8 – Introduction to Spectroscopy IV: Approaching Spectroscopy Problems Approaches to Spectroscopy ……………………………………………… 89 Practice ………………………………………………………………………. 93 Week 9 – Approaching Synthesis Problems Approaches to Synthesis …………………………………………………... 99 Practice ………………………………………………………………………. 107 Appendix A – Other Useful Things ……………………………………………….. 109 Appendix B – Practice Problems Key ……………………………………………. 113

© 2016 Sarah Lievens and Department of Chemistry, University of California, Davis. All rights reserved.


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Introduction – Learning Organic Chemistry

Organic chemistry is often met with trepidation by college students, who have heard all sorts of stories about how difficult, abstract, and confusing it can be. But organic chemistry is all around us and is a key component to understanding myriad subjects including biochemistry, metabolism, nutrition, materials and plastics, food science and safety, pharmacy, human and animal medicine, and sustainable technology.

The study of organic chemistry is the study of molecules that are based on carbon with attendant hydrogen, oxygen, nitrogen and other non-metallic atoms adding in for variety. Organic chemistry, with its focus on compound structure, chemical reactions, and reaction mechanisms, relies on analysis of molecular structures and shapes and a lot of ‘picture’ interpretation and drawing. So methods of studying in organic chemistry are perceived by many to be distinctly different from general chemistry, which often focus on mathematical relationships and calculations. However, the basic principles of bonding, electrostatic attraction, electronegativity, stability, and kinetics do carry through. It’s still chemistry! Positive charges attract negative charges, negative charges repel negative charges and everything wants to lower its overall energy. It’s how we determine what’s low energy and how molecules change in the process of lowering their energy that becomes the focus of organic chemistry. The goal of discussion in 118A is to provide a solid grounding in principles and practice for the basic skills of reading, writing, and describing organic molecules, visualizing molecules in three-dimensions, identifying organic compounds using real world methods, and building larger molecules from smaller pieces. These skills are foundational to a complete understanding of organic molecules and organic chemistry and will be expanded upon in 118B and 118C as new structures and properties are explored. Each week in discussion topics will be covered that expand upon topics covered in lecture, answer ‘what about when… ?’, and provide an opportunity for extensive practice of key skills. Discussion sections provide a unique opportunity work through chemical problems in real time and get immediate feedback from both peers and teaching assistants in a small group setting.

It is tempting in the study of organic chemistry to try memorize patterns and structures, however it’s more productive to understand the principles behind how reactions work and problem solving processes that allow chemists to derive answers to chemical problems that are not exact copies of textbook problems. The methods and practice outlined in this manual endeavor to provide students beginning with their study of organic chemistry with a basic set of tools to understand how chemical questions are asked and methods to approach complex chemical questions in a step-wise and logical manner. The first four weeks focus on methods of representing organic molecules in two and three dimensions expanding on lecture material. The weeks five through eight focus on spectroscopy and how chemists begin to identify the chemical structure of unknown compounds. The final week of discussion introduces concepts of organic synthesis where individual chemical reactions learned in lecture can be combined in various orders to produce complex molecules for specific uses such as pharmaceuticals, food additives, fabrics, plastics, and technology.

Welcome to the fascinating world of carbon, organic molecules, and moving electrons!


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Week 1 – Structure of Organic Molecules I Structural Formulae and Resonance

Learning Outcomes – • Reading and writing structural formula of organic structures and translation between

Kekule structures, condensed formula, bond-line structures, and dash-wedge structures. • Recognizing and writing resonance structures and resonance hybrids in various organic

structures and interpretation of relative stabilities. • Assigning formal charges.

Additional Reading –

K. P. C. Vollhardt, N. E. Schore; Organic Chemistry: Structure and Function Chapters -1.4, 1.5, 1.9

Introduction – Organic compounds are extensive networks based on covalent bonds. These bonds are formed primarily between carbon and hydrogen, with occasional oxygen, nitrogen, sulfur, fluorine, chlorine, bromine, and iodine atoms. Bonds are formed by sharing electrons either equally between atoms in non-polar covalent bonds (e.g. C-H, C-C or C=C) or unequally between atoms of significantly different electronegativity in polar covalent bonds (e.g. C-O, O-H, C-N, N-H, or C-Cl). In general chemistry compounds are written by chemical formula such as H2O, NH3 or PCl5. Here compounds with single central atoms or simple networks are represented by simply stating the number of each kind of atom per molecule or formula unit. However organic molecules are much more complex: organic compounds contain extended chains of central atoms, rings, and multiple pairs of electrons shared between the same two atoms in multiple bonds.

Because of the complexity of how atoms can bond in organic compounds, substances with the same formula may have completely different bonds. Substances that have the same formula but have different structures are isomers and will have distinct chemical and physical properties. Generally isomers are broken into two general categories: constitutional isomers have the same formula but different connections (bonds) between atoms. Stereoisomers have the same connections, but different three-dimensional orientations.

A simple chemical formula isn’t adequate to indicate the complexity of these molecules, a structural formula that indicates the exact position, bonds, and potentially orientation is required to distinguish between isomers with the same chemical formula but distinct shapes and properties.

Organic chemists use several different types of structural formula to represent molecules on paper, each of which has advantages and disadvantages. The four common ways of writing organic molecules are Kekule structures, condensed formula, bond-line structures and dash-wedge structures. Structural Formula – Kekule (Straight-line) Structures: Atoms are written with proper connections, but there is no angle information, all bonds are written as 90° angles. Functionally the same as Lewis Dot structures, but all bonds are written as lines representing electron pairs rather than dots and lone pairs may or may not be explicitly shown.


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Pros: • Atoms all explicitly shown

Cons: • No shape information. • No stereochemistry information • Difficult to write branches • Takes a long time to write • Cannot easily be typed

Condensed Structures (or Condensed Formula: Bonds are left out and each central atom is written with its attached atoms directly to the right. Generally each central atom is followed first by any attached H, then all other elements in alphabetical order. No bonds or attachments are explicitly shown. Condensed structures are meant to be typed using a typewriter so are organized to maximize clarity of connections while minimizing keystrokes.

Pros: • Can easily be typed • Atoms are all explicitly shown

Cons: • No shape information • No stereochemistry information • Can be difficult to see multiple bonds • Can be difficult to interpret branches. • Difficult to show rings.

Bond-Line Structures: Bond-line structures are the most commonly used structural formula in organic chemistry. Here the C and H atoms are not explicitly written. Carbons are implied by bends, ends, and transitions to multiple bonds, while hydrogens are left out entirely unless part of a functional group. The number of hydrogens is assumed to be sufficient to give each carbon exactly 4 bonds. Angles for bonds are correct for sp and sp2 hybridized atoms, but extra information is needed for sp3 hybridized central atoms with substituents. Non C and H atoms are written explicitly, as well as H bonded to non-carbon atoms.

C C C C C C

C

C

HH H

H

H

HH

H HH

CH

C C C C C CH

Br

Br

Cl

H

H

H

C

OO C C

H

H

H

C

CC C C CH

CH

Cl

HCH

CH

C

CH

C

C

H HHH

Compound A Compound B Compound C

H

H OH

H

HH

HH H

COH

H HH

HH

HH

HCH HH

H

CH

H

H

HH

H HH

(CH3)3CCHCHCHCHCHOCompound A

Compound B

Compound C

CH3CH(OH)CH2CCCHClCH2CH(CH2CH3)2

CH3CBrClCHBrCH2CH(CH2CH3)COOCH2CH(CH3)CH2CH3


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Note: All stereochemical information given must be determined independently of the Kekule or condensed structures. So the above alkenes have been arbitrarily assigned as trans/E, but could be either in the Kekule/condensed structure Pros: • Easy to write • Branches and multiple bonds are clearly

shown • Functional groups emphasized. • Shape information is clear

Cons: • Carbons and hydrogens not explicitly

shown. • Cannot be easily typed without special

programs • Stereochemistry requires additional labels

Wedge/Dash Structures: Can be written as bond line structures or with the atoms explicitly stated. Gives a full 3D representation of the molecular structure. Wedges represent atoms pointed out of the page (towards the reader) while dashes represent atoms pointed into the page (away from the reader). Essentially each central atom is shown by its VSEPR structure.

Note:

1) All stereochemical information given is determined independently of the Kekule or condensed structures.

2) Compound A and B are shown with all atoms explicitly shown, compound C uses a mix of bondline and wedge/dash structure to show stereochemistry without excessive writing.

Pros: • Straight forward stereochemical

information • Quick to write when using bond-line

structures

Cons: • Cannot easily be typed without special

programs

H

OOH

ClO

Br

Cl Br

O


CC

CC

HO CC

CC

CC

H HH

HCH2CH3

HH

H

HHH

HH

H

H

HCl

CC

CC

CC

CH

O

H

H

HH

H

HH

H

HO

Br

OBr Cl



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Applications - Writing Condensed Structures: There are several tricks and pitfalls to writing condensed formula. Like any written language there are conventions and patterns that can be difficult to distinguish for a beginning reader. Basics:

Condensed formula list central atoms in order left to right. Each central atom is directly followed by the atoms bonded to that central atom. It is important to be aware of how many bonds each central atom can make in a neutral compound to know when it is time to move on to the next central atom.

Carbon – 4 Bonds Nitrogen – 3 Bonds Oxygen – 2 bonds Hydrogen – 1 bond (never a central atom) Halogens – 1 bond (never a central atom) Once a central atom is ‘full’ it is time to move to the next central atom, if an atom is not at it’s proper number of bonds and there are no more atoms attached to it, then multiple bonds must be used to complete the necessary number of bonds. Atom Order: In condensed formula atoms are listed in a fixed order. First the central atom, then any H attached directly to that central atom, and then all other elements attached to that central atom in alphabetical order. This is the same atomic order used in writing molecular formula for organic molecules (e.g. C6H12O6 (glucose), C2HBrClF3 (halothane), and C3H4O2 (acetic acid)). If the central atom is not carbon the same pattern still applies.

Multiple Bonds: In neutral molecules carbon always makes four bonds, nitrogen makes three, oxygen makes two, halogens make one, and hydrogen makes one. Any exceptions to this pattern in a compound will be indicated in the formal charges. So we can interpret multiple bonds by finding the number of attached atoms and extrapolating any ‘empty’ spaces as double or triple bonds between appropriate atoms. When converting to condensed formula from written structures we can simply look at each carbon and it’s attachments separately.

When we’re drawing structures from condensed formula, multiple bonds show up as gaps where atoms have insufficient bonds to make them neutral (give them octets).

Cl

Br HN

1-bromo-1-chloropropane N-methyl-1-propanamineCHBrClCH2CH3 CH3CH2CH2NHCH3

OCl

2-chloro-1-propoxybutane

CH3CH2CH2OCH2CHClCH2CH3

H C C C CH

H

H

H

H

H

H

HH

butane

CH3CH2CH2CH3

trans-2-butene

H C C C CH

H

H

H

H

HH H C C C C

H

H

H

HH

CH3CHCHCH3 CH3CCCH3

2-butyne


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Double bonds: Two adjacent central atoms (C, N or O) are both missing one bond. They will be double bonded to each other. This can be carbon/carbon, carbon/oxygen, carbon/nitrogen or even nitrogen/oxygen and nitrogen/nitrogen in advanced substances. Halogens and hydrogens do not make double bonds. So if you see adjacent CHCH with no branches or other substitutents you likely have a carbon-carbon double bond.

Carbonyls: Carbonyls (the basic C=O structure) are so commonly found in organic chemistry they deserve a special comment. They can be found in many functional groups by following the same rules as any double bond, but are often best recognized by seeing adjacent CO followed by the continuing chain. The exception to this pattern being aldehydes (CHO) which are listed as first the central carbon, then attached H, and lastly the double bonded oxygen due to atom order rules. The atom order keeps the H attached to C rather than confusing the issue by implying an O-H bond. Writing –COH would require a C/O triple bond and a +2 formal charge on oxygen to give four bonds to carbon!

Triple bonds: Two adjacent central atoms (C or N) are both missing two bonds. They will be triple bonded to each other. This can be carbon/carbon or carbon/nitrogen. It’s possible to find nitrogen/nitrogen triple bonds, but that results in N2 not an organic molecule. Oxygen, halogens and hydrogen do not make triple bonds. So if you see adjacent CC with no other substituents you likely have a carbon-carbon triple bond.

Parenthesis: There are three common uses for parenthesis in condensed formula: branches containing central atoms, extended repeating sections within a chain, and symmetrical ends of a chain.

O

H H C C C CH

H

H

H

H

HCO

H CH3CHCHCH2CHO

trans-3-pentenal

CH3CH2CH2CHO CH3COOCH2CH3CH3COCH3

C CH

HCO

H H CH

HCO

C H CH

HCO

O CH

HC

H

HH

H

O

butanal

CHH

H

H

H

O

acetone

HH

H

O

O

ethyl acetate

CHON(CH3)2

H CO

NC

C

H N

O

N,N-dimethylformamide (DMF)

HH

HHH

H

CNC C C C C

H

H

H

HCH

H

HC

H

CN

H HH

CH3CH2CCCH2CH(CH3)CN

2-methylhept-4-ynenitrile


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Branches: When an alkane chain branches parenthesis are placed around the branch to distinguish

it from the main chain. Parenthesis are always used when the branch contains C’s and H’s (e.g. -CH(CH3)2-, -C(CH2CH2CH3)2-, or –CH(CH2OH)-), can be used for non-carbon branches (e.g. NH2, OH, =O), but are generally not used for terminal atoms (e.g. H, Cl, Br, F, I). Halogens are generally not considered branches as they are single atoms that merely replace a hydrogen in the formula. Parenthesis of this kind are found within a chain and end with a terminal group on the right side of the branch.

Repeating subunits:

When the same structure is repeated parenthesis may be used to shorten the name. Generally parenthesis are used when there are 3 or more of the same unit repeating (eg. (CH2)-4), or if the repeating unit is large. The goal is to save space/typing. Parenthesis of this kind are found within a chain and do not terminate, they have bonds to other central atoms on both the left and right sides. Be particularly aware of similar carbons that are not the same –CCC(CH3)3 is correct, -(C)3(CH3)3 is not, also –CHCHCHO is correct, -(CH)3O is not. In both of these cases (triple bond plus quaternary carbon and alkene plus aldehyde) the three ‘identical’ carbons in a row are not in fact the same as they belong to different functional groups.

Symmetrical ends:

This is a combination of repeating units and branches. If the end of a molecule has two or three identical branches, parenthesis should be used to show that all of the branches are the same and that there is no ‘main-chain’ vs. ‘branch’ distinction. This also usually ends up with less typing overall. This type of parenthesis is found at the ends of a chain and contain a terminal group. If the chain begins with a symmetrical end there will be a terminal group at the left of the enclosed group, if the chain ends with a symmetrical end there will be a terminal group at the right of the enclosed group.

HO

H C C C CH

H

H

HCH

CH HC HHC

CH

HCH

HC

CH HOH

CH

H

HCO

CH

HH

H

H

C

C

H HH

H HH

H HH

CH3CH2C(CH3)2CH2CH(CH2CH2CH3)CH2CH2CH(CH2OH)CH2COCH3

4-(hydroxymethyl)-9,9-dimethyl-7-propylundecan-2-one

O

O

O

butyl (2E,4E,6E)-octa-2,4,6-trienoate

CH3(CH)6COO(CH2)3CH3

H C CH

HC

H

HCH

CH

CH

CH

CO

O CH

HCH

HCH

HCH

HH


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Example: Kekule to Condensed Structure

Looking at example A left to right the first non-hydrogen we encounter is a carbon. Carbon always makes four bonds and this carbon has 3 bonds to H and one bond to the next central atom. So we write the first central atom followed by the atoms connected only to that central atom: CH3- . The second central atom is a C with one bond back to the previous atom, two bonds to H, and one bond to the next central atom, a -CH2- , which we write directly after our first atom: CH3CH2-. The third central atom is a C with one bond back to the previous atom, one H, one Br, and on bond to the next atom. There are two different atoms attached to the C, so we list the H first and all other atoms in alphabetical order, a –CHBr-, which we attach to our growing chain: CH-3CH2CHBr-. The fourth central atom is a C with two Cl, -CCl2-, which gives: CH3CH2CHBrCCl2-. Then comes the triple bond, since there are no atoms other than the two central C we can just add –CC-, to give: CH3CH2CHBrCCl2CC-. We can then finish the molecule with the last central atom and it’s 3 bonds to H and one bond to the previous atom, a –CH3. So our finalized condensed formula is: CH3CH2CHBrCCl2CCCH3.

Applications - Writing Bond-Line Structures: Basics:

Carbon always forms four bonds, but since most organic molecules are backbones of carbon surrounded by mostly hydrogen with occasional other atoms, chemists have simplified writing organic structures by using bond line structures.

In bond-line structures the carbons are not explicitly stated, but the ends of a chain, bends in a chain, and occasionally transitions in bond number are used to indicate the position

O

Br(CH3CH2)2CHCH2CHBrCOCH2C(CH3)3H C C C

H

H

H

H

HC

C HHCH HH

H

HCH

BrCO

CH

HC C

CH HH

CH HH

H

HH

5-bromo-7-ethyl-2,2-dimethylnonan-4-one

H CH

HCH

HCBr

HCCl

ClC C C

H

HHExample A:

H CH

HCH

HCBr

HCCl

ClC C C

H

HHExample A: CH3CH2CHBrCCl2CCCH3.


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of carbon atoms. The bonds between carbons are written as lines hence bond-line structures and the main chain becomes a zig-zag line. To simplify the structure further hydrogen atoms that are not part of functional groups are left out and only implied by the number of ‘missing’ bonds for each carbon.

Functional groups and non C/H atoms are written out explicitly. Functional groups are where most reactions happen and are the focus of most names, reactions, mechanisms, and physical/chemical properties. As such they’re written out explicitly.

The angles in bond-line structures match the angles in the molecule as much as possible sp3 hybridized atoms are tetrahedral, sp2 hybridized atoms are trigonal planer, and sp hybridized atoms are linear. If stereochemistry is not known around an sp3 atom, then the bonds are shown as lines. If stereochemistry is known, dash/wedge bonds can be included to indicate the full 3D shape of the molecule. Writing Bond-Line Structures

1) Find the longest chain of continuous carbons and write that as a zig-zag line. (It is common, but not required to start with an upward line.) This chain (the base name) is particularly important if starting with a name.

2) a. The angles in this line should be 120°. Bond angles around triple bonds should

be 180°. Angles too far from these angles will be counted as incorrect. b. Ether or ester oxygens and amine or amide nitrogens may be included in the

original base chain or added with their substituents in step 3.

3) Add any requisite multiple bonds and straighten the chain if you find a triple bond. e.g. a ten carbon zig-zag chain

an eight carbon zig-zag chain with a triple bond between carbons 3 and 4.

4) Counting from left to right (or clockwise around a ring) add lines for any carbon substituents and lines and atoms for any functional group. A general rule of writing substitutents is to follow the shape of the molecule. If there isn’t space for the substituent to be written, then there probably isn’t space for it there in the molecule!

a. H in functional groups are EXPLICITLY written out. i. The H in OH, COOH, NH, and CONH are written out since they are

attached to non-carbon atoms. ii. The hydrogen of an aldehyde (CHO) is written out by convention it

doesn’t act like a normal H so we show it as unique. iii. The H on a terminal alkyne MAY be written out for clarity.

b. If only one substituent is present the group will extend straight off the point of the bend (either straight up or straight down). Single substituents will not go in the ‘cup’ of the zig-zag.

correctincorrect

e.g. 3-ethyl-2-methylheptane


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Example: Kekule to Bond-Line Structure

C C C C C C

C

C

H Cl

H

H HH

H

HC

H HH

Cl

HC

C

HCH

C

C C C O CH

HC

HH H

O H

H HHH

HC

OCC

HCH

H

H HH

H

H

H

Example A:O

H

1) The longest chain is 13 carbons followed by an O then 3 carbons for a 17 atom zig-zag chain.

3) working left to right the substituents are added.

O

O

2) Add multiple bonds...

O

2a) and adjust the bond angles.

OO

H

OO

H

OO

HCl

Cl

OO

HCl

Cl

OH

OO

HCl

Cl

OH

O

OO

HCl

Cl

OH

O4/5) No stereochemical information was given so no dash/wedge are drawn and either E/Z orientation (118B) around the alkene would be considered correct.


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c. If two substitutents are present they can be drawn as two angled groups or off the point and into the ‘cup’ if one group is small or just zig-zags. If a group goes into the ‘cup’ it will be the smaller group.

5) If stereochemical information is given check the stereocenters are correctly indicated

using dash/wedge bonds. This is only applicable if stereochemistry was originally given in the name/structure. If no stereochemical information is given leave the bonds as plain lines.

a. Stereochemistry cannot be determined from Kekule structures or condensed structures without additional information.

b. Stereochemistry cannot be determined from the name if no R/S (or cis/trans on rings) information is given.

c. (118B) Check that all double bonds have the correct orientation if E/Z or cis/trans information is given.

NOTE: If starting from a Kekule structure rather than a name (and with practice) steps 1-3 can be combined and each central atom in the chain can be written independently in sequence left to right or clockwise around a ring. NOTE: There are many other ways to approach bond-line structures, if a different approach works, use it! Applications - Writing Kekule Structures from Bond-Line Structures Converting a bond-line structure to a Kekule structure is in many ways more straight-forward than writing the original bond-line structure. Most chemists never use Kekule structures directly, we simply mentally follow the process below when necessary and with practice the translation becomes automatic.

1) For each end or bend or multiple bond transition in the molecule add a carbon. a. The two carbons in a carbon/carbon triple bond are the most common carbons to

miss! b. Remember the bonds to non-carbon atoms already have atoms they don’t get an

extra C. 2) For each carbon, count how many bonds it has. Draw in enough bonds to H to make

exactly 4 bonds to carbon. a. This can be anywhere from 3 H to 0 H. b. Heteroatoms should already have the proper number of H.

3) Adjust the angles to 90° to make it a proper Kekule structure. 4) HINT: While learning to look at bond-line structures the above process can be very

helpful in seeing the actual molecule, such as when approaching nomenclature questions or reaction products. You don’t need to make the formal Kekule structure, but drawing the atoms can help prevent mistakes. Eventually it will become too time consuming, but by then you’ll have had enough practice not to need the extra translation step.

correct correct incorrect

e.g. 2,2,5-trimethy-5-(1-methylethyl)loctane


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Example: Bond-Line to Kekule Structure

Applications – Constitutional Isomers: Basics: Organic molecules often have multiple constitutional isomers, multiple ways of connecting the atoms that produce different bonds, different structures, and different properties. It’s difficult to know how many isomers are possible for a particular formula, but there are methods to help write out all of the possibilities. Alkanes:

1) Finding the longest chain. a. List all of the carbon atoms as a single line. This is the longest possible

isomer (the straight chain alkane). b. Fill in the hydrogen around the chain.

2) Adding a single branch a. Remove one carbon from the chain. b. Attach the single carbon back to the main chain at a middle carbon.

i. Adding the single carbon back to an end carbon will reform the original straight chain.

c. Repeat for each different kind of middle atom. d. Fill in the hydrogen around the chain for all the isomers.

3) Adding a second branch a. Remove two carbons from the original chain. b. Attach the single carbons back to the main chain at middle carbons.

i. Adding the single carbon back to an end carbon will reform the original straight chain.

ii. Carbons can be added as two branches on the same carbon iii. Carbons can be added as two branches on different carbons

NH

O

OH

Br

C C C C C C NH

O

CC

C C C CCC

C

OH

Br C C C C C C NH

O

CC

C C C CCC

C

OH

Br

HH

H

H H

HHH

H H

H H

H

H H

H H

H H HH

HHH

H C CH

HC C

H

HCBr

CC

H HCH HH

H

HCH

OH

CO

NH

CH

HCH

CH

CH

CH HHH

HH

1)2)

3)

Example B:

Adding carbons Adding hydrogens

Flattened to a Kekule structure


14

iv. If the chain is long enough carbons can be added as a single two carbon (ethyl) branch. This will not work if it makes the branch into the longest chain.

4) Repeat until you cannot add anymore branches. This is all the possible isomers for that alkane.

a. If molecules have different names, they will be different isomers. b. If molecules can be bent into the same shape they are the same isomer.

Molecules with Functional Groups: 1) Use the method for alkanes to determine the number of possible carbon backbones. 2) Fill in the functional group in all the possible places on each backbone.

Example: Constitutional Isomers of Hexane (C6H14)

Resonance – All structural formula are based off of Lewis Dot structure; where two atoms share pairs of electrons to form single (one shared pair), double (two shared pairs), or triple (three shared pairs) bonds. Some molecules can keep all the same atom connections (same isomer), but still have multiple Lewis structures that vary in the positions of the electrons. When Lewis structures vary only in the positions of the electrons they are called resonance structures. We can convert between resonance structures by moving pi bonds, lone pairs, and charges around the framework of single bonds. Any movement of the atoms will change the bonds and create a new constitutional isomer, only the movement of electrons is permitted in resonance.

Resonance structures can be drawn using any of the types of structural formula, but while resonance structures can be drawn as proper Lewis structures they’re not accurate representations of the real structures.

1) C C C C C C C C C C C CH H H H H

HH

HHHHHH

H

2) C C C C C + CC C C C C

C

C C C C CC

3) C C C + C + CC C C C

C

C C C CC

C C

C

C C C + C + C + C C C CC

C+ C no possible isomers

that are not above. Done!!

hexane

2-methylpentane

3-methylpentane

2,2-dimethylbutane

2,3-dimethylbutane

longest chain

one branch

two branches

three branches?


15

Resonance Structures for Some Common Organic Compounds

In reality, the molecule does not undergo rapid exchanges between the possible resonance structures. The reality of the molecule is a blending of the resonance structures where electrons are spread out over multiple central atoms (delocalized) rather than shared only between two atoms as in a normal bond (localized) or as a lone pair on a single atom (localized). This blended sharing is called the resonance hybrid. Unfortunately resonance hybrids are difficult to draw accurately so we often still write the individual resonance structures when writing reactions or mechanisms.

Resonance Hybrids for Some Common Organic Compounds

Resonance hybrids typically show partial bonds where the bond order between the

atoms the number of electron pairs shared divided by the number of connections (bonds) they’re shared over. Resonance hybrids also often show partial charges, with the formal charge of spread out over several atoms.

Bond order = number of electrons shared Number of atoms

Formal Charge = # of valence e- in neutral atom - (# e- in lone pairs + ½ x # e- in bonds)

The relationship between resonance structures and resonance hybrids can be thought of like a labradoodle puppy. A labradoodle puppy has some traits of its labrador parent and some traits of its poodle parent all the time. The puppy is not a labrador on Monday, Tuesday, Wednesday and every other Sunday and a poodle on Thursday, Friday, Saturday, and every other Sunday (alternating between resonance structures). The puppy is always itself (the resonance hybrid).

For some molecules not all of the resonance structures are equivalent, for these molecules we can determine which resonance structures are the most relevant (major contributor to the hybrid) by looking at formal charges. The hybrid is the average of all the resonance structures with greater weight going to the more significant structures.

H N

OCH3

CH3

H N

OCH3

CH3

O O

O

O

O

O

both equivalent

major minormajor minor

both equivalent

slightly major

O O O

major minor minor

toluene acetate ion

N,N-dimethylformamide (DMF) acetone

pentadienyl cation

cyclohex-2-enone

both equivalentallyl cation

O

O

δ−

δ−δ+δ+

H N

OCH3

CH3

δ+

δ−O δ−

δ+

Oδ−

δ+δ+

δ+ δ+ δ+


16

Writing resonance in organic molecules is complicated by the fact that we often use resonance to explain why a particular ion or intermediate is formed, or preferences for certain reactions or mechanisms. So while a particular resonance structure may only be a minor contributor to the molecule as a whole, it may still be relevant to the discussion at hand. Additionally, if one resonance structure is major, often only that major contributor is written in most contexts (e.g. amides or ketones), but the minor contributors may become relevant when considering hybridization, mechanisms, reactivity, or other properties. It is important to consider whether resonance is possible whenever looking at an organic molecule even if no resonance is explicitly shown.

Applications – Writing Resonance Structures: Basics:

Resonance structures contain electrons that are able to move around the molecule without moving the atoms. These mobile electrons are delocalized and are not part of the basic single bonds that hold the molecule together (sigma bond network). Writing delocalized electrons as localized on particular atoms in resonance structures as per typical Lewis dot structures (and Kekule, condensed and bond-line structures) is misleading since those particular structures are extreme possibilities, but not the whole picture. Delocalized electrons are better shown as resonance hybrids or full molecular orbital diagrams. However these better pictures are difficult to write, and so chemists use resonance as a tool to describe possible electron positions in a molecule. It’s important to note that resonance structures are a model and not the ‘real’ structure.

In order to have delocalized electrons structures that display resonance will have some combination of multiple bonds, lone pairs, and charges in p-orbitals. When these ‘extra’ electrons are positioned adjacent to each other, sharing of electrons between several atoms is possible and several resonance structures can be written. However, not all resonance contributors are equal. Some are more or less significant for a variety of reasons.

Individual resonance structures are written separated by double-headed resonance arrows and all the structures that contribute to the hybrid are enclosed in brackets. Typically the electron flow to covert between structures is written by using double-pronged electron pushing arrows (single pronged fish-hook arrows if the resonance involves radicals). The resonance hybrids are written by converting any bond that has delocalization to a dotted ‘partial’ bond. Looking for Resonance in Bond-Line Structures:

1) Identify any pi bonds. Are any of these pi bonds: a. Next to another pi bond? b. Next to a lone pair? c. Next to a charge (radical, anion or cation)? d. Polar due to being a C=X bond?

2) Identify any lone pairs. Are any lone pairs: a. Adjacent to a pi bond (above) b. Adjacent to a cation or radical?

3) If any of these is yes there is probably at least one resonance structure. Though it may be more or less relevant to the hybrid.

4) Write the resonance structure a. Write the basic the bond connections from the original molecule b. Move the electrons from the position with more electrons (anion, lone pair, pi

bond) to the position with fewer electrons (cation, polar bond, pi bond). c. Make sure the new structure follows the octet rule.


17

Prioritizing Resonance Structures: Not all resonance structures within a molecule are identical. Structures with octets tend to be more significant contributors, as do those without formal charges. However, in organic chemistry even minor structures can become relevant when dealing with reaction intermediates or extreme conditions. Typically the following rules can be applied in order to determine the major and minor contributors to a resonance hybrid.

1) Atoms in organic molecules generally prefer octets and resonance structures with full octets are generally most significant.

a. Hydrogen will have a duet not an octet. b. Central atoms of row 3 or higher may expand their octets if it helps minimize

formal charges and in these cases the expanded octet version is preferred. 2) All atoms should have minimal (ideally 0) formal charge.

a. Total formal charges should equal the charge on the compound/polyatomic ion.

b. Any resonance structure that has a formal charge of +/- 2 or larger will not be a significant contributor.

c. Structures with no formal charge will nearly always be most significant except when it violates rule 1.

3) If formal charges are necessary: a. Opposite charges should be near each other and similar charges should be

separate. b. Negative formal charges prefer more electronegative atoms and positive

formal charges prefer less electronegative atoms. c. It is possible to get non-preferable formal charges if the molecule is charged

(e.g. RCOHOH+ will have a positive formal charge on oxygen in it’s two major structures).

Examples - Pi bonds: Molecules with adjacent pi bonds will show resonance. This resonance may result in these pi bonds acting as a unit in reactions. When alkenes are adjacent in a structure, the resonance structures without formal charges are the major contributors. However, when polar bonds are included minor structures with formal charges may become significant portions of the resonance hybrid.

Benzene – Electrons flow evenly around the ring with no significant localized charges.

Conjugated carbonyl – Electronegative atom pulls electrons towards itself creating an electron deficient carbon. The electron rich alkene pushes electrons towards the electron deficient carbon creating a different electron deficient carbon.

equivalent not significant

O OO

major minor minor


18

Examples - Lone Pairs and Anions: Lone pair electrons and anions can be treated as concentrations of electrons (piles of dirt). These electrons can flow onto other atoms if there is space or a way for other electrons to get out of the way. (Move the pile of dirt from one place to another). Lone pairs and anions tend to resonate when they are adjacent to a multiple bonds.

Amide – Electronegative oxygen in the carbonyl pulls electrons towards itself creating an electron deficient carbon. The lone pair on nitrogen donates to the electron deficient carbon so all atoms have an octet, but nitrogen gains a positive formal charge.

Carbanion – Lone pair electrons push into pi bond forcing the pi electrons out of the way and forming a lone pair on the farther alkene carbon.

Examples – Cations: Carbocations can be treated as empty spaces (holes in the ground). These spaces pull electrons towards themselves if there are any available to attract. (Fill in the hole and create a new one).

Carbocation – Electrons from the pi bond push into the adjacent empty orbital, leaving an empty orbital on the farther alkene carbon.

Cations formed from atoms with greater than standard number of bonds can also accept

delocalized electrons when they are part of a polar pi bond or adjacent to a pi bond.

Protonated Carbonyl – Lone pair electrons have acted as a Bronsted-Lowry base and produced a cation. The oxygen has too many bonds and pulls electrons away from the pi bond. This other oxygen shares into the ‘hole’ left behind.

N

O

H

HN

O

H

HN

O

H

H

major minorminimal

approximately equivalent

approximately equivalent

O

OH

equivalent equivalentnot significant

H

O

OH

H

O

OH

H


19

Practice – 1) Draw all the constitutional isomers of the following formulae. Ignore consequences of

stereochemistry. A) Butanes – C4H10, C4H10Cl, and C4H10O

B) Butenes – C4H8, C4H7Cl, and C4H8O

C) Pentanes – C5H12, C5H11Br, and C5H12O

D) Pentene – C5H10

E) Larger alkanes – C6H14, C7H16, and C8H18

2) A. Write the molecular formula for each of the given compounds.

B. Convert the given bond-line structures to condensed formula.

3) A. Convert the given condensed formula to bond-line structures B. Fill in all relevant lone pairs on the structures in questions 3. A) (CH3)3CCH2CH2CHBrCOCH3

B) (CH3CH2)2CHCHCHCH2CH(CH2CH3)2

C) CH3(CH2)4COOCH2CH(CH3)CH2CH3

D) CH3CCCCl2CH2CHO

E) (CH3)2CHCH2C(CH3)2CONHCH2CH3

F) CH2CHCHCClCH2CHBrC(CH3)3

O

O

O

HN

O

Cl

Br

O O

OH

OH OH

O

O

Cl

N

O

H

OH

O

O

OH

HN


Compound D Compound E Compound F

Compound G Compound H Compound I

Compound J Compound K Compound L


20

G) CH3(CH2)5CHBrCO(CH2)6CF3

H) CH3CHOHCH2CH2OCH2CH3

I) CH3CH2CHCHCO(CH2)4CN

J) NH2CH2CH2NHC(CH2CH3)2CHCHCH2OH

K) (CH3)2CH(CH)4COOC(OCH2CH3)3

L) CH3CHBrCCl2COCH2COOH

4) A. Fill in any relevant lone pair electrons on each compound below. B. Draw significant resonance structures of the given compounds with appropriate electron movement arrows and resonance arrows. C. Assign which are major, minor or equivalent structures. D. Draw the resonance hybrid of each structure.

O

NH

OO

H O

O

OH

O H

NH2 Cl OH

OH Br

O OHO

NH

NH2

Compound A Compound B Compound C Compound D Compound E

Compound F Compound G Compound H Compound I Compound J

Compound K Compound L Compound M Compound N Compound O

Compound P Compound Q Compound R Compound S Compound T


21

Week 2 – Structure of Organic Molecules II Hybridization

Learning Outcomes – • Identify, label, and compare sigma and pi bonds and molecular geometries • Assign hybridization to atoms in organic molecules • Analyze consequences of molecular geometry in three-dimensions • Describe and compare energies and steric hindrances of three-dimensional conformers.


K. P. C. Vollhardt, N. E. Schore; Organic Chemistry: Structure and Function Chapters -1.8 (hybridization), and 2.5 - 2.7 (conformations)

Introduction – Valence Shell Electron Pair Repulsion (VSEPR) theory produces structures, which match with the observed three-dimensional structures of most covalently bonded molecules. However, there is no correlation between the observed three-dimensional structure, VSEPR, and the position of electrons in atomic orbitals. The need to position electrons into molecules led to Valence Bond Theory. Valence Bond Theory describes bonds as overlaps (constructive interference) between the wavefunctions of atomic orbitals. The overlaps can form either sigma or pi bonds.

Sigma bonds (σ bonds) – Involve the end-to-end overlap of two atomic orbitals. Sigma bonds have electron density directly between the two bonded atoms and are stronger than pi bonds on a per bond basis. The first bond between two atoms is a sigma bond.

Pi bonds (π-bonds) – Involve side-to-side overlap of two atomic orbitals. Pi bonds have electron density above and below (or in front and behind) the plane of the two bonded atoms and have a node directly on the plane of the two bonded atoms. Pi bonds are weaker than sigma bonds on a per bond basis. The second and third bonds of any multiple bond are pi bonds.

s - s sigma bond

s - psigma bond

p - psigma bond

sp* - s sigma bond

sp* - psigma bond

sp* - sp*

sigma bond

p - p pi bond


22

Hybrid Orbitals and Hybridization: While the typical atomic orbitals as shown by Schrodinger’s equations give the shapes predicted by VSEPR and observation for most two atom molecules (e.g. HCl, H2, I2, IBr etc.), the atomic orbitals do not match the shapes of most molecules that have a central atom with multiple connections. Valence Bond Theory (VBT) reconciles the tetrahedral shape (109.5° bond angles) around carbon in a molecule with the 2s22p2 electron configuration and a spherical s-orbital and three perpendicular (90°) p orbitals. VBT or hybrid theory uses the formation of hybrid orbitals to match reality to theory. Hybrid orbitals are localized blends of atomic orbitals. Meaning that while the electrons are still in standing waves around the central atom, the shape and energy of those orbitals is different in molecules than in single atoms. Atomic orbitals are the lowest possible energy states for the electrons when the atom is a single atom. However when the central atom begins to form bonds, other possible standing waves, which we now call hybrid orbitals, are possible

It’s simplest to envision these hybrid orbitals as blends of the contributing atomic orbitals (add and average), which changes the energy and shape of the standing wave. The process of blending atomic orbitals is called hybridization and the resulting blended orbitals are the hybrid orbitals. The number of hybrid orbitals that result is always exactly equal to the number of atomic orbitals used: # hybrid orbitals = # atomic orbitals. Energetically the total energy of the hybrid orbitals is the same as the total energy of the atomic orbitals, but the energy of the individual orbitals is different, thus the different shape and standing wave.

Which atomic orbitals are involved in hybridization depends on how many electron groups are around the central atom. More electrons around the central atom requires more spaces for electrons to be, and so involves more orbitals. The number of electron groups around a central atom will be the number of atomic orbitals blended into hybrid orbitals.

Hybridization and VBT are still model and do not perfectly reflect the reality of some molecules. Hybrid orbitals are fixed to their original central atom (localized electrons) and in cases where resonance spreads electrons out over several atoms (delocalized electrons), hybrid orbitals no longer tell the whole picture. The full delocalization of electrons by resonance is better described by Molecular Orbital (MO) theory. An in depth coverage of MO theory is beyond the scope of this course, but it can be simplified as hybrid orbitals making up the single bonds (sigma bonds) and unhybridized p-orbitals making up any delocalized charges, lone pairs or pi bonds. The unhybridized p-orbitals can then be considered as localized (single pi bonds, VBT) or delocalized (extended pi systems, limited MO theory) depending on the molecule.

Shapes of Hybrid Orbitals –

In general hybrid orbitals look like lopsided p orbitals with a node at the nucleus, but with greater electron density on the side that forms the bond increasing overlap and strengthening the bond. They are an average of the energy of their parent atomic orbitals and are built by constructive interference between atomic orbitals. Typically the orientations of the hybrid orbitals mimic the bond angles found by VSEPR.

2 electron groups – sp hybridization: s + p = sp To form sp hybrid orbitals an s orbital and a p orbital are blended to form two sp orbitals. There are two possibilities with this blend a) the left* side of the p orbital has constructive interference and most of the electron density ends up on the left or b) the right side of the p orbital has constructive interference and most of the electron density ends up on the right. The overall shape is linear.


23

3 electron groups – sp2 hybridization: s + p + p = sp2

In sp2 hybrid orbitals a second p orbital is added to the blend resulting in three sp2

orbitals. The first p orbital gives left/right1 character as in sp hybridization (above), and the second perpendicular p orbital gives up/down* character where one orbital is pointed up, the second down and leftward, and one down and rightward. The overall up character is cancelled by the down of the other two, and the overall leftward character of the second orbital is cancelled by the rightward character of the third. The overall shape is trigonal planar.

1 NOTE: The terms left/right, up/down, and front/back are used below to conveniently distinguish between the three perpendicular p orbitals (px, py, pz) from the unhybridized atom. In reality the atom doesn’t care which p orbital(s) hybridize, but blends along 1 (linear), 2 (planar), or 3 (spatial) perpendicular axes to give the final hybrid orbitals.

Ener

gy

2s

2p

+ +s p sp x 2

Ener

gy

2psp

BeCl Cl

empty p orb

sp in sigma bond

bonding p orbCl Be Cl

A) Blending of energy from atomic to sp hybrid orbitals. B) Blending of shapes from atomic to sp hybrid orbitalsC) Orbital overlap and bonding in an sp hybridized molecule containing only sigma bonds. D) Structure of C.

A

B

C D

Ener

gy

2s

2p

+ +s p

sp2 x 3

Ener

gy

2psp2

empty p orb

sp2 in sigma bond

bonding s orb

+

p

BH

H

H HB

H

H

A) Blending of energy from atomic to sp2 hybrid orbitals. B) Blending of shapes from atomic to sp2 hybrid orbitalsC) Orbital overlap and bonding in an sp2 hybridized molecule containing only sigma bonds. D) Structure of C.

A

B

C D


24

4 electron groups – sp3 hybridization: s + p + p + p = sp3

In sp3 hybrid orbitals a third p orbital with front/back* character is blended into the mix resulting in four sp3 orbitals. One has greater electron density pointed up, one down and left, one down, right and forward, and one down, right and backwards. The overall up character of the first orbital is cancelled by the down character of the other three, the left character of the second orbital by the right character of the third and fourth, and the front character of the third orbital by the back character of the fourth orbital. The overall shape is tetrahedral.

Multiple bonds: Multiple bonds occur when additional pairs of electrons are shared between two atoms. The first bond between any two atoms is a sigma bond with the electron density directly between the two nucleii, but the second and third bonds are pi bonds with electron density above/below or in front/behind the sigma bond. This prevents overlap between electrons in different bonds. Pi bonds require side to side overlap of unhybridized p-orbitals, and so hybridization of atoms with octets is not always sp3. Atoms can use blends of hybrid and unhybridized orbitals in order to find the lowest energy state for the number of atoms and electrons involved. In organic molecules atoms with all sigma bonds/lone pairs will be sp3 hybridized with approximately 109.5° bond angles and tetrahedral electron pair geometry.

Adding a pi bond changes the hybridization. Atoms with a single pi bond must be sp2

with an unhybridized p-orbital available to form the pi bond, producing a trigonal planar

Cl

Ener

gy

2s

2p

+ +

s p sp3 x 4

Ener

gy sp3

sp3 in sigma bond

bonding s orb

+

p

C

H

p+

H

H

bonding p orb

C

H

Cl HH

A) Blending of energy from atomic to sp3 hybrid orbitals. B) Blending of shapes from atomic to sp3 hybrid orbitalsC) Orbital overlap and bonding in an sp3 hybridized molecule containing only sigma bonds. D) Structure of C.

A

B

C D

C

H

HH

C

H

H

H

C CH

HH

HH

Hsp3

4 σ bondssp3 hybridizationbond angles = 109.5o


25

molecules with 120° bond angles and a pi bond above and below the plane of the atoms. Though the pi bond appears to have two parts, it’s just one orbital with a node along the plane of the atoms where the node in the p orbitals is.

Atoms with two pi bonds will be sp hybridized with two unhybridized p-orbitals producing

a linear molecule with 180° bond angles and two pi bonds, one above and below and one in front and behind the sigma bond. Again the pi bonds have nodes along the lines of the atoms where the sigma bond is strongest.

A way to look at the sigma vs. pi bond networks is a triple bond is like a sigma bond hot dog surrounded by a pi bond bun, a double bond is thus more like a hamburger with the pi bond bun only above and below. Atoms without pi bonds are like chunks of tri-tip steak with no bun at all. The position and presence or absence of pi bonds will significantly affect both the chemical and physical properties of molecules so being able to envision the positions of orbitals is important to understanding how organic chemistry works. Applications – Assigning Hybridization: There are two common methods for assigning hybridization. Counting electron groups and looking at pi bonds.

Counting electron groups works for any central atom, but is easiest when the full Lewis structure is shown. This works well with Kekule structures but not as well with bond-line structures. So counting groups is typically used with inorganic chemistry, where the central atom is explicitly shown and we commonly get electron-deficient atoms or expanded octets.

Looking at pi bonds only works for central atoms that have exactly 8 valence electrons so is less common in inorganic chemistry, but is simpler for molecules that are shown using organic bond-line structures where hydrogens are left out. Since neutral organic molecules all have octets this method is primary in organic chemistry.

Several cases require special attention even in organic molecules. If the atom is charged it is important to consider electron deficiency (carbocations and carbon radicals are sp2 while carbanions are sp3). In cases where atoms show resonance, hybridization is assigned by looking at the resonance hybrid rather than any particular structure. Alternatively, each atom can be analyzed using the resonance structure that has the LEAST electron groups (most pi bonds). Additionally atoms that only have one bond are not central atoms and are best considered as unhybridized, since hybridization is a model and in reality the electron configuration of the outer atoms is much closer to the free atomic orbitals than hybrid orbitals.

CH

HC

H

HC

HC

H

H

H

HC C

H

HH

H 3 σ bonds, 1 π bondsp2 hybridizationbond angles = 120o

sp2

p π above and below

σ between atoms

C CH H C CH H H C C H2 σ bonds, 2 π bondsp hybridizationbond angles = 180oπ above and below

π front and backσ between atoms

sp

p


26

Method 1 - Counting Electron Groups

1) When looking at the central atom count how many electron groups it has. (Assign AXE) Each lone pair or bonded atom counts as one electron group.

2) The number of electron groups is the number of hybrid orbitals needed to space all the groups around the atom. Count orbitals from low E to high E (s, then p, then d)

a. 2 electron groups – s + p = sp b. 3 electron groups – s + p + p = sp2 c. 4 electron groups – s + p + p + p = sp3 d. 5 electron groups – s + p + p + p + d = sp3d e. 6 electron groups – s + p + p + p + d + d = sp3d2

i. NOTE: This shows why only atoms in row 3 or higher can expand octets. They must have a d orbital in the valence to form sp3d or sp3d2 orbitals.

3) When drawing the orbitals they will follow the shape of the VSEPR prediction for that many electron groups.

a. 2 electron groups – Linear b. 3 electron groups – Trigonal Planar c. 4 electron groups – Tetrahedral d. 5 electron groups – Trigonal Bipyramidal e. 6 electron groups – Octahedral

Method 2 - Counting pi bonds

1) First check that the atom has exactly 8 valence electrons. If yes continue, if no use the counting electron groups method above.

2) In the bond line structure count the number of pi bonds around the central atom. 3) Each pi bond will require an unhybridized p orbital so subtract one p orbital from the

hybrid per pi bond. a. Zero pi bonds, the atom has no free p orbitals

i. Hybridization = sp3 b. One pi bond, the molecule requires a free p orbital

i. Hybridization= sp2 (+ p).

c. Two pi bonds, the atom requires two free p orbitals i. Hybridization = sp (+ p + p)


27

Practice: 1) Assign hybridization to each atom in the following molecules. Remember to consider all

resonance structures.

In Class Exercise – sp3 Hybridization - Ethane:

1) Build the model of ethane (CH3CH3). a. What is the angle between H-C-H? ________________ b. What is the molecular shape around the C? ________________

2) Look at the molecule so that the two carbon atoms are one behind the other. Rotate it

until one H on the front carbon points straight up and one H on the rear carbon points straight down. This is a staggered conformer or staggered rotamer. Using a ruler measure how far apart the H on C1 and C2 are in your model. ___________ cm

3) Rotate the rear carbon 60°. The C-H bonds should be directly behind each other when you look down the C-C bond. This as an eclipsed conformer or eclipsed rotamer. Using a ruler measure how far apart the H on C1 and C2 are in your model. ___________ cm

4) Eclipsed conformers are higher in energy than staggered. Explain why using what you’ve observed from your models.

NH2

O

Cl OH

OH

OCN

NH2

H OH

ON

N

OO

HO

Br

Compound A Compound B Compound C Compound D

Compound E Compound F Compound G Compound H

Compound I Compound J Compound K Compound L


28

5) Rotate the rear carbon 360° in 60° increments. Sketch what you see below. What happens to the energy as the molecule rotates? Sketch an energy plot to go with the conformers. Label each conformer as staggered or eclipsed.

Conformers:

0° 60° 120° 180° 240° 300° 360°

sp3 Hybridization - Butane:

6) Build the model of butane (CH3CH2CH2CH3). Look at the molecule so that carbon 2 and 3 are one behind the other. Rotate it until the CH3 on the front carbon points straight up and the CH3 on the rear carbon points straight down. This is a staggered conformer in the anti configuration.

7) Using a ruler measure how far apart the H on C1 and C3 are in your model. Does rotating the C1-C2 bond affect this distance? What are the maximum (___________ cm) and minimum (__________ cm) distances?

8) Rotate the rear carbon 60°. The C-H bonds should be directly behind each other when

you look down the C-C bond. This as an eclipsed conformer. Using a ruler measure how far apart the H on C1 and C3 are in your model. Does rotating the C1-C2 bond affect this distance? What are the maximum (___________ cm) and minimum (__________ cm) distances?

9) Rotate the rear carbon 60°. The two CH3 should be closer to each other. This as an

staggered conformer in the gauche configuration. Using a ruler measure how far apart the H on C1 and C3 are in your model. Does rotating the C1-C2 bond affect this distance? What are the maximum (___________ cm) and minimum (__________ cm) distances?

10) Rotate the rear carbon 60°. The two CH3 should be directly behind each other when you

look down the C-C bond. This as an eclipsed conformer. Using a ruler measure how far apart the H on C1 and C3 are in your model. Does rotating the C1-C2 bond affect this distance? What are the maximum (___________ cm) and minimum (__________ cm) distances?


29

11) Gauche conformers are higher in energy than anti. Explain why using what you’ve observed from your models.

12) Rotate the rear carbon 360° in 60° increments. Sketch what you see below. What happens to the energy as the molecule rotates? Sketch an energy plot to go with the conformers. Label each rotamer as anti, gauche or eclipsed.

Conformers:

0° 60° 120° 180° 240° 300° 360°

13) One eclipsed conformer is higher in energy than the others. Label it above and explain why.


30

sp2 Hybridization - Ethene: 1) Build the model of ethene (CH2CH2).

a. What is the angle between H-C-H? ___________________ b. What is the molecular shape around C? ___________________

2) Attempt to rotate ethene around the C=C bond. Why does ethane show rotational

conformers in nature while ethene does not? sp2 Hybridization – 2-Butene

3) Build the model of 2-butene (CH3CHCHCH3). Set up the molecule so that C1 and C4 are as far apart as possible. This is the trans isomer. Sketch it here:

4) Build a second model of 2-butene so that C1 and C4 are on the same side of the alkene. This is the cis isomer. Sketch it here:

5) What is the distance between the H on C1 and the H on C3 in the trans isomer?

(_____________ cm) What is the distance between the H on C1 and the H on C4 in the cis isomer? (_____________ cm).

6) Which isomer do you think is more stable and why?


31

7) Do you think trans-2-butene and cis-2-butene will interconvert like anti and gauche rotational conformers? Why or why not?

8) Do you think trans-2-butene and cis-2-butene will have the same physical properties? Why or why not?

sp Hybridization - Ethyne: 1) Build the model of ethyne (CHCH).

a. What is the angle between H-C-C? ___________________ b. What is the molecular shape around C? ___________________

2) Attempt to rotate ethyne around the CC bond. Does ethyne act more like ethane or

ethane? Why?

3) Build the model of 2-butyne (CH3CCCH3) and sketch the structure below. Will you expect

2-butyne to have alternative rotational conformers as per butane, or isomers as per 2-butene, or something else?


32

Week 3 – Nomenclature of Organic Molecules: Alkanes, Cycloalkanes, and Haloalkanes

Learning Outcomes – • Identify alkyl groups by systematic and common names. • Identify main chain of a molecule and distinguish substituents from main chain. • Systematic nomenclature of simple alkanes, cycloalkanes, and haloalkanes. • Draw structures of organic compounds from a given systematic name.


K. P. C. Vollhardt, N. E. Schore; Organic Chemistry: Structure and Function Chapters – 2.2, 2.3

Introduction – The Chemical Abstract Service Registry, which catalogs known chemical substance, contains more than 91 million unique substances and adds an estimated 15,000 substances daily. It would be impossible to memorize the common names of all of these unique substances if there was no link between the compound and the name. Think of how complicated it would be to know everyone in California by their nickname (population 38 million in 2013)! So in 1919 the International Union of Pure and Applied Chemistry was founded to standardize chemical nomenclature with a set of descriptive rules. The rules for organic nomenclature are listed in the “Blue Book”, the IUPAC Nomenclature of Organic Compounds.

Following the rules outlined by IUPAC each organic compound will have a unique name characterized by it’s bonds and functional groups, and chemists in different labs can use the same name to draw the same structure and vice versa. Even with these rules nomenclature can be complex and many compounds are still referred to by common names for simplicity, we may not know nicknames for all the people in California, but we know the nicknames of our friends.

It is important to understand the basics of IUPAC nomenclature, because it allows chemists to talk about the same molecules at the same time, to describe molecules without writing them down, and to refer to molecules and portions of molecules succinctly both in writing and in speaking. For now, we’ll be exploring the basics of nomenclature of alkanes and additional functional groups will be added as they are introduced. Alkanes – Alkanes are hydrocarbons, they contain only carbon and hydrogen and come in straight chains, branched chains, and rings. All alkanes contain carbons that are sp3 hybridized and are named for the number of carbons in the chain. All alkanes have the ending –ane in IUPAC nomenclature to show that the only functional group present is alkane. Straight-Chain Alkanes:

Straight chain alkanes have exactly two ends and all the carbons are in single linear sequence. The formula for all straight chain alkanes is CnH2n+2. Straight chain alkanes are named for the number of carbons in the chain. Typically alkanes with 5 or more carbons count using Greek counting prefixes similar to those used for binary covalent molecules. The names of straight chain alkanes are also used as base names for more complex molecules as well, so you’ll want to remember the names for chains up to 20 carbons.


33

Basic Straight Chain Alkanes C1 – C20

Branched Chain Alkanes and Alkyl Groups: Branched alkanes have more than two ends, but still have the formula CnH2n+2. They no longer have just a linear sequence of carbons, but have places where the chain forks into more complex structures. Nomenclature of branched chain alkanes distinguishes the longest chain of continuous carbons (aka the main chain), from the carbons that make up the branches (aka the substituents or alkyl groups). Alkyl groups come from basic alkanes by removing a hydrogen atom and attaching the main chain (or other funcational group) to the ‘open’ spot. Alkyl groups of straight chains are named by dropping the ending –ane and adding the ending –yl to the base name of the chain.

CH4

CH3CH3

CH3CH2CH3

CH3(CH2)2CH3

CH3(CH2)3CH3

CH3(CH2)4CH3

CH3(CH2)5CH3

CH3(CH2)6CH3

CH3(CH2)7CH3

CH3(CH2)8CH3

CH3(CH2)9CH3

CH3(CH2)10CH3

CH3(CH2)11CH3

CH3(CH2)12CH3

CH3(CH2)13CH3

CH3(CH2)14CH3

CH3(CH2)15CH3

CH3(CH2)16CH3

CH3(CH2)17CH3

CH3(CH2)18CH3

CH4C1 methane

C2 ethane

C3 propane

C4 butane

C5 pentane

C6 hexane

C7 heptane

C8 octane

C9 nonane

C10 decane

C11 undecane

C12 dodecane

C13 tridecane

C14 tetradecane

C15 pentadecane

C16 hexadecane

C17 heptadecane

C18 octadecane

C19 nonadecane

C20 icosane


34

However, once we get to three or more carbons there are choices. We can form straight

chain alkyl groups or alkyl groups with further branches.

Many smaller branched alkyl groups get common names. These names are old fashioned, but are still often used. The following common names are in general use and should be remembered.

Common Names of Branched Alkyl Groups

Branched alkyl groups can also be named by IUPAC by treating each branch as a substituent off the main chain to form complex substituents. Larger alkyl groups with five or more carbons are nearly always named using the full IUPAC nomenclature. When naming branched alkyl groups using IUPAC the carbon attached to the point of interest (e.g. main chain) is carbon one. Branches off of the main branch are then named as alkyl groups along that new main chain to give alkylalkyl- type groups.

It is useful to note a few things about alkyl groups before looking at larger molecules: 1) The variable –R represents any alkyl group. 2) iso- groups generally have the form (CH3)2CH-X. 3) sec- groups attach via a secondary carbon 4) tert- groups attach via a tertiary carbon

HCH

H HHCH

H

a methane molecule a methyl group

HCH

H CHCH

H CH

HH

H

Han ethane molecule an ethyl group

HCH

C CH

HH

a propane molecule

H

HH

HCC C

HH

Han isopropyl group

H

HH

HCH

C CH

Ha propyl group

H

HH

a straight chain orn-alkyl substituent

a branched chain substituent

isopropylHCC C

HH

H

H

HH

Xthe iso- prefix is included in alphabetization

isobutylHCC

C CH

H

H

HH the iso- prefix is included in alphabetization

H HH

X

sec-butylHCC C

H

H

H

HH the sec- prefix is italicized and separate

from the name and is NOT included in alphabetization.

CH

HH

X

tert-butylCCC

CH

HH the tert- prefix is italicized and separate

from the name and is NOT included in alphabetization.H HH

H HH

X


35

IUPAC Naming of Branched Alkyl Groups

Cycloalkanes:

Cycloalkanes have only carbon and hydrogen atoms and have no ends in the simplest variations. Their formulas are CnH(2n). In more complex cases cycloalkanes can be combined with branched alkanes to give molecules with rings and branches (CnH2n) or halogens to give haloalkanes with rings (CnH2n-yXy). The presence of the ring does not significantly affect the properties of the alkane, but it must be included in the name.

isopropylHCC C

HH

H

H

HH

X1-methylethyl

isobutylHCC

C CH

H

H

HH

H HH

X

sec-butylHCC C

H

H

H

HH 1-methylpropylC

H

HH

X

tert-butylCCC

CH

HH 1,1-dimethylethyl*

H HH

H HH

X

2-methylpropyl

Common IUPAC1

2

12

3

1 32

12

methyl

methyl

methyl

methyl

* Unlike most prefixes di-, tri-, tetra- etc prefixes in branched substituents count towards alphabetical order. e.g., 1,1-Dimethylethyl rather than 1,1-diMethylethyl.

N/AHCC C

H

H

H

HH 1-methylbutylC

H

HC

X

1 32

methyl

H

HH

4

N/ACCC

CH

HH 2,2-dimethylpropyl*

H HH

H HH

C1

2

methyl

H

H

X3

N/AHC C

HH

H

Cl

X1-chloroethyl

12

chloro

Cl

FC F

F F

X1,1,1-trifluoromethyl*1

fluoro x 3

F

F


36

Applications – Writing IUPAC Names of Branched Alkanes and Haloalkanes IUPAC Nomenclature of branched alkanes follows these steps: 1) Find the longest chain of continuous carbons. This is now the main chain. Name this

chain as if it were a straight chain alkane, this is the base name. a. If two chains are the same length choose the one with more branches.

2) Count the carbons in the main chain left to right and right to left. The direction you

first run into a substituent is the direction use to number the chain. a. If the numbers are the same go to the next nearest substituent b. If the numbers are exactly the same regardless of direction the lower

numbers go to the first substituent in alphabetical order.

3) Give each substituent a number according to which carbon it’s attached to.

4) List the substituents in alphabetical order in front of the main chain. a. Branched substituents get parenthesis around them b. Numbers are separated from numbers by commas c. Numbers are separated from words by hyphens

5) Multiple straight chain substituents of the same kind are combined and given a prefix

to indicate the number (di-, tri-, tetra-). These prefixes do not count towards alphabetical order. If the substituent is branched (bis-, tris-, tetrakis-) are used.

Notes: • Sometimes it helps to think of each carbon in the main chain as a house on a street. The

substituents that live there each get an address (number) according to which house (carbon) they live at.

• Once you’ve found your main chain circling or labeling it can show you what you still have left to name as substituents

• To find the longest chain pick an end and follow it to the first branch. Make sure you have the longest end on that branch. Follow the chain to the next branch selecting the longest continuing chain each time you find a branch until you run out of molecule.

• Haloalkanes are included in normal alkane nomenclature as substituents. F- is Fluoro, Cl- is Chloro, Br- is Bromo, I- is Iodo. They get no special priority and are treated similarly to alkyl groups.


37

Examples:

Cl

Br

Br

1)Cl

Br

Br

We could have started in the lower left or ended in the upper right and gotten the same length, but this chain has 5 substituents rather than 3 or 4.

We could start at the lower left methyl and the result would be exactly the same.

2)Cl

Br

Br

13

2 4 65

7

8 9

123

456

78

Left to right we get substituents at positions 2,3,5,7,and 8Right to left we also get substituents at postions 2,3,5,7, and 8.The first substituent L-->R is Chloro, the first substituent R-->L is Methyl.C comes first so L-->R is correct.

Left to right we get substituents at positions 2, 5, 6, and 7.Right to left we also get substituents at postions 2,3,4, and 7.2 = 2 so we go on to the next substituent and 3 < 5 so we use R--> L

3) Cl

Br

Br

13

2 4 65

7

8 9

123

456

78

2-chloro3-ethyl5-methyl7-ethyl8-methyl

2-bromo3-bromo4-isopropyl or 4-(1-methylethyl)7-methyl

4,5)

2-chloro-3,7-diethyl-5,8-methylnonane

a nonane an octane

2,3-dibromo-4-isopropyl-7-methyloctaneor2,3-dibromo-7-methyl-4-(1-methylethyl)octane

Compound A Compound B


38

Applications – Writing IUPAC Names of Cyclic Alkanes and Haloalkanes IUPAC Nomenclature of cycloalkanes follows these steps:

1) Find the ring. This is now the main chain. Count the carbons and name chain as if it were a straight chain alkane. Add the prefix cyclo- to the name to get the full base name of the ring.

2) If there are substituents: a. 1 substituent ! the carbon the substituent is attached to is carbon-1. There is

no need to number the substituent in the name. b. 2 substituents ! the substituents will have the same numbers, so the first in

alphabetical order will be at carbon-1. Substituents should be numbered in the name to indicate relative position.

c. 3 or more substituents ! Number substituents so that the lowest possible numbers are used. If there is a tie use alphabetical order.

3) Give each substituent a number according to which carbon it’s attached to.

4) List the substituents in alphabetical order in front of the main chain. a. Branched substituents get parenthesis around them b. Numbers are separated from numbers by commas c. Numbers are separated from words by hyphens

5) Multiple straight chain substituents of the same kind are combined and given a prefix

to indicate the number (di-, tri-, tetra-). These prefixes do not count towards alphabetical order. If the substituent is branched (bis-, tris-, tetrakis-) are used.

6) Indicate stereochemistry using cis- or trans- prefixes for disubstituted rings or R,S labels for multiply substituted rings.

Notes: • Cycloalkanes can become cycloalkyl substituents

• Two substituents on the same carbon are almost always 1,1-disubstituted.

• Cis-substituents are on the same side, which sounds similar, trans- substituents are on

opposite sides and you have to go across the ring.

• Cis- and trans- work when there are exactly two substituents, rings with more substituents must be named using R/S (see Week 4).


39

Examples:

Cl Cl

1)Cl Cl

a cyclohexane a cyclopentane

2) Cl Cl1

23

12

3 4

5

butyl comes before propyl and 1,3 is lower than 1, 5

1, 1, 3, 4 is lower than 1, 2, 4, 4(1,1 always wins)ethyl comes before methyl so we go counterclockwise

3) Cl Cl1

23

12

3 4

5

1-sec-butyl or 1-methylpropyl3-propyl

1-chloro1-chloro3-ethyl4-methyl

4,5)

1-sec-butyl-3-propylcyclohexaneor 1-(1'-methylpropyl)-3-propylcyclohexane

1,1-dichloro-3-ethyl-4-methylcyclopentane

(S)(R)

(R)

(S) (S)

Cl Cl6)

same side = cis ethyl = #1 CH, #2 CH2CCl2, #3 CH2CH3, #4 H = Smethyl = #1 CH, #2 CH2CCl2, #3 CH3, #4 H = S

cis-1-sec-butyl-3-propylcyclohexaneorcis-1-(1'-methylpropyl)-3-propylcyclohexane

(3S,4S)-1,1-dichloro-3-ethyl-4-methylcyclopentane

Compound C Compound D


40

Applications – Writing Structures from IUPAC Names 1) Find the base name and write a chain or ring with that many carbons.

2) Starting from the left of a chain or top of a ring number the main chain left to right or

clockwise for a ring.

3) Add substituents at the indicated carbons. a. After chapter 8 when additional functional groups are present be sure to add any

substituents that are in suffixes b. (118B) When alkenes/alkynes have been added rewrite the structure to give the

proper bond-line structure around the pi bonds.

4) Add dash/wedge to properly indicate stereochemistry. Examples:

Compound E

3-ethyl-4-fluoro-2,3,6-trimethyloctane

Compound F

trans-1-(1,1-dimethylethyl)-3-propylcyclohexane

1+2) 3-ethyl-4-fluoro-2,3,6-trimethyloctane trans-1-(1,1-dimethylethyl)-3-propylcyclohexane

1

12

34

56

78

2

3

4

5

6

3)

3-ethyl4-fluoro2-methyl3-methyl6-methyl

2 3 4 6

F 1-(1,1-dimethylethyl)3-propyl

1

3

3-ethyl-4-fluoro-2,3,6-trimethyloctane trans-1-(1,1-dimethylethyl)-3-propylcyclohexane

4)

No stereochemistry. Structure is done.

2 3 4 6

F

1

3

3-ethyl-4-fluoro-2,3,6-trimethyloctane trans-1-(1,1-dimethylethyl)-3-propylcyclohexane

1

3

trans = one wedge and one dash

or


41

Practice – 1) Give the IUPAC names of the following substances:

Br

F

Cl

Cl

FBr

F

F

F Br

Br

Cl

BrCl

I

F

Br

ClBr Cl

Cl

Br Br

Br Br

BrBr

Br

Br

Br




Compound M Compound N Compound O Compound P

Compound Q Compound R Compound S Compound T

Compound U Compound V Compound W Compound X

Compound Y Compound Z


42

2) Write the structure of the following IUPAC names:

AA) 3,3-trifluoro-2-methylhexane

BB) 1-ethyl-4-(1-methylethyl)cyclohexane

CC) 3-bromo-4,7-diethylnonane

DD) 2,3-dichloro-5-cyclopentyl-6,6-dimethyloctane

EE) cis-1-(1,1-dimethylethyl)-4-trifluoromethylcycloheptane

FF) 3-bromo-2,2,6,6-tetramethylheptane

GG) 1-chloro-4-(1-chloroethyl)-5,8-diethylundecane

HH) trans-1-fluoro-2-isopropylcyclopropane

II) 4-ethyl-2,4-dimethylheptane

JJ)1-chloro-2,4-diethyl-1-methylcyclohexane

KK) 1-cyclopentyl-3-(1-methylpropyl)cyclopentane

LL) 1,1,1-trifluoro-4,6-dimethyloctane

MM) 1,2-diethyl-4-isobutylcyclooctane

NN) 1,4-bis-(1,1-dimethylethyl)cyclohexane

OO) trans-1-methyl-3-(1-methylethyl)cyclobutane

PP) 7-bromo-3,3-diethyl-6-methylnonane

QQ) 8,8,9,9-tetrachloro-4,5-diethyltridecane

RR) 5-cyclohexyl-2-methyl-4-(1-methylpropyl)undecane


43

Week 4 – Stereoisomerism Learning Outcomes –

• Identify chiral molecules, stereocenters / chiral centers, and mirror planes. Explain principles of [α]D.

• Assign R/S configurations to stereocenters. • Determine relationships between organic molecules e.g. enantiomers, diastereomers,

meso, identical, constitutional isomers. • Describe relative vs. absolute stereochemistry • Properly include stereochemistry in systematic nomenclature.


K. P. C. Vollhardt, N. E. Schore; Organic Chemistry: Structure and Function Chapters – 5.1-5.7.

Introduction – As organic molecules get larger (more than three central atoms) there are often multiple ways to bond the atoms in a given formula. The resulting substances, called isomers, are entirely different molecules, and can have widely different properties. Isomers are broadly separated into two categories depending on how the structures vary: constitutional isomers and stereoisomers.

Constitutional isomers are molecules with the same atoms (chemical formula), but distinctly different bonds. Constitutional isomers have different shapes, occasionally different functional groups, and distinct physical and chemical properties. They have completely different IUPAC names and they and are distinctly different substances.

C5H12 – 3 constitutional isomers

C4H10O – 7 constitutional isomers

Stereoiosmers are molecules with the same atoms (chemical formula) and same bonds/connectivity, but different shapes. Stereoisomers are not superimposable, meaning if they were laid on top of one another they would not be the same. Stereoisomers have different shapes and different physical properties, but will have the same functional groups and the same basic name in IUPAC nomenclature.

C5H12pentane

C5H122-methylbutane

C5H122,2-dimethylpropane

OHHO OH

OH

H3CO O

C4H10O1-butanol

C4H10O2-butanol

C4H10O2-methyl-1-propanol

C4H10O2-methyl-2-propanol

C4H10O1-methoxypropane

C4H10Oethoxyethane

OCH3

C4H10O2-methoxypropane


44

C6H12O6 – 4 of the stereoisomers of glucose

Stereoisomers – What causes a Stereoisomer?

Molecules with stereoisomerism must have some sort of asymmetry to provide a point of difference. There are three common points of asymmetry: a stereocenter, an alkene, or an overall asymmetry. A stereocenter (aka chiral center or chiral carbon) is an sp3-hybridized carbon that has four different substituents attached. Chiral centers have no planes of symmetry. To find stereocenters look for sp3 CH or quaternary carbons. CH3 and CH2 cannot be stereocenters as they have at least two of the same group (hydrogen) attached. Carbons with sp2 and sp hybridization cannot be stereocenters; they cannot have four different groups attached. In looking at a molecule to determine if it has stereoisomers check each of carbons to see if it has four different groups attached. It doesn’t matter how far away the difference is, any difference counts.

When the two carbons that make up the alkene each have two different substituents attached, the molecule will have stereoisomers. An in depth discussion of the stereoisomerism of alkenes will wait until 118B, but it’s helpful to be aware that alkenes can be cis or trans like rings.

Molecules where steric hindrance causes a permanent twist may display stereoisomerism without a stereocenter or asymmetrical alkene. These molecules are overall asymmetrical without any particular stereocenter or alkene. Inherently asymmetric molecules are relatively few and do not show up in the 118 series, but they do exist. (e.g. helicene)

H

O

OH

OH

OH

OHHO H

O

OH

OH

OH

OHHO

(2R,3S,4R,5R)-2,3,4,5,6-pentahydroxyhexanal

glucose - formed by photosynthesis

H

O

OH

OH

OH

OHHO

H

O

OH

OH

OH

OHHO

(2R,3S,4S,5R)-2,3,4,5,6-pentahydroxyhexanal

galactose - found in milk

(2S,3S,4R,5R)-2,3,4,5,6-pentahydroxyhexanal

mannose - common protein modifier

(2S,3R,4S,5R)-2,3,4,5,6-pentahydroxyhexanal

idose -unstable and not found in nature

H

CHH

H

H

CClCl

ClH

CH

ClHO

CH3C

H

achiraloverall symmetrical4 identical groups

achiral3-fold axis

2 different groups

120o

achiralmirror plane

2 different pairs

achiralmirror plane

3 different groups

Cl H

F

CCH3

ClH

chiraloverall asymmetrical

4 different groups

F

CCH3

ClH

a stereocenter an alkene a helicene (overall asymmetric)


45

Molecules with One Stereocenter: Substances with exactly one stereocenter are chiral they are not superimposable on

their mirror images. The pair of non-identical mirror images are described as enantiomers.

Enantiomers are considered chiral (handed), are optically active (see below), and have identical physical properties except for rotation of plane-polarized light or when interacting with other chiral molecules or chiral environments. The terms chiral and optically active are used interchangeably to describe molecules that have enantiomers and are not superimposable on their mirror images. A chiral molecule will have an enantiomer and will be overall asymmetrical. One enantiomer of the pair will be assignable as R stereochemistry and the other will be S.

Individually chiral molecules demonstrate optical activity and act differently in a chiral environment such as a protein, living systems, or interacting with another chiral molecule. But a mixture of enantiomers has it’s own distinct properties. An equal mixture of both substances in a pair of enantiomers is a racemic mixture. Racemic mixtures have the same properties as the individual enantiomers, but the mixture is achiral, does not rotate plane-polarized light, and does not have any special properties in chiral situations. Polarimetry / Optical Rotation:

Enantiomers have nearly identical physical properties; a major exception is the rotation of plane-polarized light, where one enantiomer will rotate light clockwise and the other will rotate light an equal amount counter-clockwise. Molecules that rotate plane-polarized light are said to be optically active. The process of measuring the rotation of plane-polarized light is called polarimetry.

F

CCH3

ClH

F

CH3C Cl

H

Compound A Compound A'

mirror plane

rotate Compound A'F

CCH3

ClH

Compound A

F

CHCl

CH3

Compound A'

mirror reflection of a chiral molecule produces two different substances

Compound A and A' are not superimposable,they are enantiomers.

Br Br

(R)-(-)-2-bromopentane (S)-(+)-2-bromopentane

Br

2-bromopentane(a racemic mix of both enantiomers)

achirala chiral moleculea chiral molecule

Light PolarizingFilter

Plane-Polarized Light

Chiral Sample

(typically 10 cm path)

Rotated Plane-Polarized Light

Rotating Slit/Polarizing Filter

Observer/Detector

Polarimeter Schematic


46

• All chiral molecules (molecules with a nonsuperimposable mirror image) are optically active and all optically active molecules are chiral. Thus the terms chiral and optically active are functionally synonymous can be used interchangeably.

• The amount a compound turns plane-polarized light measured as an angle α. The [α]D is a physical constant that can be calculated from α, pathlength, and concentration of the sample.

• Molecules that rotate plane polarized light clockwise (rightwards) from the point of view of an observer looking at the light through the sample are said to be dextrorotatory (D) and are given a positive (+) sign.

• Molecules that rotate plane polarized light counterclockwise (leftwards) are said to be levorotatory (L) and are given a negative (-) sign.

• For any given pair of enantiomers one molecule will be levorotatory (-) and the other will be an equal amount dextrorotatory (+), however which is which cannot be predicted from just observing the structure or R/S assignments, and the (+) or (-) assignment must be determined experimentally

• A racemic mixture is a 1:1 mixture of enantiomers and will have an optical rotation of 0 degrees (equal amounts levorotatory and dextrorotatory).

Drawing Chiral Molecules: Chiral molecules are commonly written as bond-line structures that use dashes or wedged bonds to show the configuration at the stereocenter if the configuration is known. However, occasionally other methods are used. Fischer projections are another way of representing stereochemistry that is often used for sugars or amino-acids, but less commonly in organic chemistry. Typically a Fischer projection involves representing the four different groups on a stereocenter on a cross shaped framework. Bonds that are pointing vertically are assumed to be ‘back’ or dashed and bonds that are pointed horizontally are assumed to be ‘forward’ or wedges. We won’t be working much with Fischer Projections in 118A, but it’s helpful to recognize them when working in the greater realm of biology, chemistry, and biochemistry (and MCAT’s!).

It’s also useful to note that when the stereochemistry of a center is unknown, it’s written with normal bond-line structures with no dashes/wedges. This is the method we’ve been using since the beginning of class, before we brought up the concept of stereoisomers. So molecules with stereocenters that are represented as lines are assumed to be racemic/unknown and all stereochemical considerations are ignored.

α

initial plane of light

observed plane of light

+_Dextrorotatory

(clockwise)Levorotatory

(counter-clockwise)


47

Various Methods of Drawing Chiral Molecules

Relative vs. Absolute Stereochemistry: When a molecule has a stereoisomer it is possible to know all, part or nothing about which stereoisomer is present in a particular case. We can know the absolute stereochemistry, or the relative stereochemistry, or only the constitutional isomer not which stereoisomer is present. Absolute stereochemistry describes the exact stereoisomer with ALL relationships between atoms determined for molecules with either single or multiple stereocenters. Determining absolute stereochemistry requires the assignment of R vs. S for stereocenters and E vs. Z for alkenes. A molecule where absolute stereochemistry is known is a particular enantiomer and will rotate plane polarized light (have an assignable [α]D). Relative stereochemistry describes the relative positioning of multiple stereocenters within a molecule. Giving relative stereochemistry determines the diastereomer, but not which of the two possible enantiomers. Relative stereochemistry requires the assignment cis vs. trans on rings or alkenes or anti/syn in a mechanism, but does not require exact R/S assignments. A molecule where only relative stereochemistry is known can be differentiated from its diastereomers, but is usually a racemic mix with an [α]D of 0. Molecules with Multiple Stereocenters – Enantiomers, Diastereomers, and Meso

As the number of stereocenters increases a molecule will get more stereoisomers. The number of stereoisomers is 2n, where n is the number of stereocenters. So larger chiral molecules will have exactly one enantiomer, but may have many other diastereomers or related molecules. Stereoisomers can be compared to other isomers of the same structure and be related by their stereochemistry. Most relationships require comparison between two or more molecules.

• Enantiomers – Non-superimposable and exact mirror images. Molecules that are enantiomers are chiral (have at least one stereocenter) and are exact reflections. They are nearly identical in properties, but have opposite [α]D and different properties in chiral environments. All stereocenters are exactly reversed when comparing enantiomers. (Left and right hands)

H BrH Cl

CH3

CH3

=H BrH Cl

CH3

CH3

CH3HBr

CH2CH3

=CH3

HBrCH2CH3

=

=

Br(2R)-2-bromobutane

Fischer ProjectionDash-Wedge

(Bowtie) Structure Bond-line Structure IUPAC Name

=

Cl

Br

= (2S,3R)-2-bromo-3-chlorobutane

CHOOHHHHOOHHOHH

CH2OH

= = =H

O

OH

OH

OH

OHHO

(2R,3S,4R,5R)-2,3,4,5,6-pentahydroxyhexanalor

(D)-glucose

OHHCHO

HHOOHHOHH

CH2OH


48

• Identical (chiral) molecules – Superimposable and not mirror images. Molecules that are

chiral and identical (have at least one stereocenter) and are identical in all ways including [α]D. They are exactly the same and have exactly the same properties in all cases. This relationship does not have a fancy name, but it’s important to realize when two stereoisomers really are the same thing. All stereocenters are exactly the same when comparing identical chiral molecules. (Two left hands)

• Diastereomers – Non-superimposable and not mirror images. Molecules that are

diastereomers must have multiple chiral centers and will have completely different properties. They will be no more similar than any two constitutional isomers and will have different boiling or melting points, solubility, reactivity, [α]D etc. Some stereocenters are the same and some stereocenters are reversed when comparing diastereomers. (Left hand vs. right foot or left hand vs. left foot)

• Meso – Superimposable and exact mirror images. The property of being meso is

inherent to the molecule and doesn’t depend on a comparison. Meso molecules are not chiral! They do have chiral centers, but will have an internal mirror plane in at least one rotamer. Meso molecules must have multiple stereocenters and each stereocenter will have a paired center with the same four groups attached in the reverse stereochemistry (R/S pairs). All stereocenters are both the same and reversed (R,S). (Faces are meso, eyes are different from ears or mouths, but the whole has a mirror place)

• Constitutional Isomers – Same formula but different bonds. Not stereoisomers at all.

(person vs. chimpanzee, same general stuff but put together differently)


49

Example: Two Stereocenters with Different Substituents

(S) (S)

Br

OH(R) (R)

Br

OH

(S) (R)

Br

OH

(R) (S)

Br

OH

CH3(S)H Br(S)

CH3

HO H

CH3(R)Br H(R)

CH3

H OH

CH3(S)H Br(R)

CH3

H OH

CH3(R)Br H(S)

CH3

HO H

A B

C D

enantiomers

enantiomers

diasteromers diastereomers

diastereomers

If we had a mixture of the above compounds and picked out two random molecules we would get three different possibilities.

A/B and C/D are non-superimposible exact mirror images: ENANTIOMERS

A/C, B/C, A/D, B/D are non-superimposable but are not mirror images: DIASTEREOMERS

A/A, B/B, C/C, D/D are superimposable but not mirror images:IDENTICAL CHIRAL MOLECULES

H3C (S)(S)CH3

OHHBr

H

eclipsed formof A

H3C (R)(R)CH3

HBrH

OH

eclipsed formof B

H3C (S)(R)CH3

HHBr

OH

eclipsed formof C

H3C (R)(S)CH3

OHBrH

H

eclipsed formof DEnantiomers

Diastereomers

Meso

Identical

Superimposable Mirror Images

YesNoNo No

Yes YesYes No

Optically Active*YesYes

No

Yes

*as pure compounds


50

Example: Two Stereocenters with the Same Substituents

(S)(S)(S)(S)

Br

Br(R)(R)

(R)(R)

Br

Br

(S)(S)(R)(R)

Br

Br

(R)(R)(S)(S)

Br

Br

CH3(S)(S)H Br(S)(S)

CH3

Br H

CH3(R)(R)Br H(R)(R)

CH3

H Br

CH3(S)(S)H Br(R)(R)

CH3

H Br

CH3(R)(R)Br H(S)(S)

CH3

Br H

E F

G G'

enantiomers

meso

diasteromers diastereomers

diastereomers

Enantiomers

Diastereomers

Meso

Identical

Superimposable Mirror Images

YesNoNo No

Yes YesYes No

If we had a mixture of the above compounds and picked out two random molecules we would get four different possibilities.

E/F are non-superimposible exact mirror images: ENANTIOMERS

E/G, E/G', F/G, F/G' are non-superimposable but are not mirror images: DIASTEREOMERS

E/E, F/F are superimposable but not mirror images: IDENTICAL CHIRAL MOLECULES

G/G', G/G, G'/G' are superimposable and are mirror images, this is a special case where two molecules are exactly the same and have an internal mirror plane and are MESO.

extra internalmirror plane

H3C (S)(S)(S)(S)CH3

BrHBr

H

H3C (S)(S)(R)(R)CH3

HHBr

Br

eclipsed formof E

eclipsed form of G

H3C (R)(R)(S)(S)CH3

BrBrH

H

eclipsed form of G'

H3C (R)(R)(R)(R)CH3

HBrH

Br

eclipsed formof F

Optically Active*YesYes

No

Yes

*as pure compounds


51

Applications – Assigning R vs. S to Stereocenters: Basics: To describe the orientation of substituents around a stereocenter or an alkene each substituent is assigned priority through the Cahn-Ingold-Prelog rules. The relationship between these groups in three-dimensional space can then be evaluated to assign R vs. S to stereocenters. The same prioritizations (but different relationships) are used later for alkenes too. Cahn-Ingold-Prelog Rules

1) Priority is primarily assigned by atomic number, higher atomic number get first priority

a. In order 1st Br then Cl then C and lastly H 4th i. H always has lowest priority. ii. CH3 is typically next lowest after H. iii. Heteroatoms typically get high priority.

2) If isotopes are encountered the heavier isotope gets priority a. 14C has priority over 13C, which has priority over 12C

3) If two of the same atom are encountered priority is assigned by the first difference in bonded atoms

a. –CH(CH3)CH2CH2CH3 has priority over –CH2CH2CH2Br 4) Multiple bonds count as multiple single bonds to the same atom

Assigning Priority

1) Look at the atoms attached directly to the stereocenter/alkene. 2) If any have clearly different atomic numbers this is the first difference, use the atomic

number to assign priority. 3) If any have the same atomic number (typically C), determine what atoms are

attached (2 bonds from the stereocenter/alkene. a. List the attached atoms in order of atomic number and look for a higher atomic

number. i. A C attached to C, C, and H is higher priority than a C attached to C, H,

and H. 4) Continue following the attached atoms away from the stereocenter/alkene until a

difference is found. a. The first difference is the only one that matters. b. If no difference is found then the two substituents are the same and the carbon is

not a stereocenter or the alkene is symmetrical. Assigning R vs. S Stereochemistry:

1) Assign priority to all 4 groups around the stereocenter as per the Cahn-Ingold-Prelog Rules.

a. If no difference is found between two groups the atom is not a stereocenter. 2) Rotate the molecule so the lowest priority group is pointing straight backwards.

a. If 1 ! 2 ! 3 proceeds clockwise the center is R (right turning) b. If 1 ! 2 ! 3 proceeds counterclockwise the center is S (left/sinister turning)

C N CN

NN

C CC

CCC

CO

HCO

HO C

C


52

3) Special notes: There are many ‘short cuts’ out there, some work better than others.

When in doubt make sure the lowest priority is in back, this will ALWAYS work. It is worth practicing visualizing molecules starting from different orientations as Newman projections where the stereocenter is the front carbon and the lowest priority group is the rear carbon.

Applications – Stereochemistry in IUPAC nomenclature When stereochemistry is known, this should be reflected in the name. The typical IUPAC name shows the constitutional isomer. By adding R/S to the front of the name we can clearly show A) that the stereochemistry is known and B) exactly which enantiomer is present.

1) Name the connections (constitutional isomer) as per IUPAC nomenclature. 2) Assign R/S to each stereocenter with explicit stereochemistry 3) Write stereochemistry in parenthesis at beginning of name. Separate

stereochemistry from name with a hyphen. a. If more than one stereocenter list in numerical order with number indicators b. Each stereocenter is separated by commas. c. Stereochemistry of alkenes can be included in with R/S.

2 3

11

3 2

44

Counterclockwise(Leftward turning)

S

Clockwise(Rightward turning)

R


53

Cahn-Ingold-Prelog Examples:

R CH

HCH

HH R C

H

HO H

The C is attached toC, H, H

The C is attached toO, H, H

C < O

<

R CH

HCH

HH R C

H

HC C



H

H

H

HH so we go to the next C

R CH

HCH

HH R C

H

HC C

The C is attached toH, H, H


<H

H

H

HH

H < C

vs.

R CH

HCH

HOH R C

H

HC C

The C is attached toO, H, H


>H

H

H

HBr

O > CThe Br is too far away, it is not the first difference. Closer differences take priority not the ultimately largest atom.

R CC

HCH

HH R C

H

HC H

The C is attached toC, C, H


>

Oxygen is larger than carbon so the alcohol gets priority, functional groups usually get priority as they tend to have larger atoms.

H HHH

H

C > HThe first C attached to each is the same, but the second C is larger than H. Branched chains will usually get priority.

Longer chains will get priority as any atom will win over H.

R CN CN

NNC

C

C

C

> CH

HNH

HR C NH2

H NC

C

The C is attached toN, N, N

The C is attached toN, H, H

N > H

The triple bond counts as 3 bonds to N so N is larger than H.

Example 1: Larger Atomic Number gets Priority

Example 2: Longer Chains Get Priority

Example 3: Branched Chains get Priority Example 4: First Difference gets Priority

Example 5: Multiple Bonds Count Multiple Times

R C CH3

OCO

OCH

HH

C

C

The carbonyl counts as two C-O bonds from the carbons perspective so we need to go on to the next atom. The O is attached to H in the diol and back through the second bond to the C in the carbonyl.

CO

OCH

HH

H

H

R C CH3

HO OH

Example 6: Mulitple Bonds Count Extra for Both Atoms

The C is attached toO, O, C

The C is attached toO, O, C

=R C CH3

OCO

OCH

HH

C

C

> CO

OCH

HH

H

H

R C CH3

HO OH

C > HThe O is attached to C

The O is attached toH

next atom


54

Assigning R vs. S Examples:

CH2CH2BrCOH

H3CH

FC

H COOHCH2OH

Br

NC CHO

Compound A:

CH2CH2BrCOH

H3CH

1

2

3

4CH2CH2Br

COH

H3CH

1

2

3

4clockwise = R

(2R)-4-bromo-2-butanol

assign priority low priority in back

Compound B:

assign priority FC

H COOHCH2OH

1

2

3

4

FC

HOOC CH2OHH

1

2

3

4

counterclockwise = S

(2S)-2-fluoro-3-hydroxypropanoic acid

low priority in back

Compound C:

assign priorityBr H

1

23

4 low priority in back

Br

H CH3

1 2

43


(2S)-2-bromo-3-methylbutane

Compound D:

assign prioritylow priority

in backNC CHONC

OHC CH3

12

34

1

2 3

4


(2S)-2-formyl-2-methylbutanenitrileCompound E:

assign prioritylow priority

in back

1

2 3

4OH HO H

1

23

4HHO

flip over clockwise = R

(1R,2Z)-cyclohex-2-en-1-ol


55

Practice: 1) Circle the optically active molecules. Draw the mirror plane for all meso substances.

2) Assign R/S to the stereocenters in each molecule.

Br

Br

OH

OH

OO

Cl

OH

OH

Br

F

OHCl Br

Cl

Br

CN

Cl

H3CO

CN CN

OCH3

OCH3

H

Br

Cl

H

H

CH3

HO CH2CH3

COOHF

Br

H HSH

CHOCH3

HOOCH

BrH2CH3

BrCOOH

H

H3CO CH3

H OHCH3

H3C BrH

OH

Br OCH3

OH

OH

CN

Br

SCH3

CNNC

O

O

O

OH

O

O

BrOH

NC O O

O H

OHBr

H CH2CH3CF3

COOHCl

OH





Compound U Compound V Compound W Compound X Compound Y


56

3) Determine if each pair of molecules are enantiomers (E), diastereomers (D), identical chiral molecules (I), the same meso molecule (M), or constitutional isomers (C).

H BrCH3

CH2CH3HOH

H3C FH

HH3COCH3

Br HCH2CH3

CNHCH3

H CNCH2CH3

CNHCH2CH3

H OHCH3

BrHCH3

Cl

OCH3

Br

Cl

Cl

OH

OH

Cl

Br

Cl

Br

NC Br

Br

HO

OH OH

Br HCH3

HHOCH2CH3

Cl

OCH3

HOCN Brvs.

vs.

vs.

vs.

F CH3

H

OCH3H3CH

HO OH

F

F

vs.

vs. vs.

vs.

H OHCH3

CH3BrH

Cl

OH

Cl

vs.

vs.

vs.

vs.

NC CH2CH3

H

HH3CH2CCN

Cl

Br

Cl

Br

OHOH

OH

vs.

vs.

vs.

vs.

NC HCH2CH3

BrHCH3

Br

Br

Br

Br

vs.

vs.

vs.

vs.


57

4) Give the IUPAC name with proper stereochemistry for the given compounds.

Cl

Br

ClCl

Br

F

CF3

Br

Compound A Compound BCompound C Compound D



58

In Class Exercises – Finding Stereocenters 1) Using the tetrahedral central atoms build two CH4 molecules.

a. Are these molecules superimposable? Yes or No b. Are these molecules mirror images? Yes or No c. How many mirror planes can you find? _______ d. Pick the molecule up by one of the H. How many axis of rotation can you find?

_______ i. These are _____-fold axis.

e. Are these molecules chiral? Yes or No f. Sketch the molecules below using dash/wedge configurations:

mirror plane 2) Using the tetrahedral central atoms build two CH3Cl molecules

a. Are these molecules superimposable? Yes or No b. Are these molecules mirror images? Yes or No c. How many mirror planes can you find? _______ d. Pick the molecule up by the Cl and find the axis of rotation. This is a _____ -fold axis. e. Pick the molecule up by the H. How many axis of rotation can you find? _______ f. Are these molecules chiral? Yes or No g. Sketch the molecules below using dash/wedge configurations:

mirror plane

3) Using the tetrahedral central atoms build two CH2Cl2 molecules. a. Are these molecules superimposable? Yes or No b. Are these molecules mirror images? Yes or No c. How many mirror planes can you find? _______ d. Are these molecules chiral? Yes or No e. Sketch the molecules below using dash/wedge configurations:

mirror plane


59

4) Using the tetrahedral central atoms build two CH2BrCl molecules. a. Are these molecules superimposable? Yes or No b. Are these molecules mirror images? Yes or No c. How many mirror planes can you find? _______ d. Are these molecules chiral? Yes or No e. Sketch the molecules below using dash/wedge configurations:

mirror plane

5) Using the tetrahedral central atoms build two CHBrClFmolecules that are mirror images. a. Are these molecules superimposable? Yes or No b. How many mirror planes can you find? _______ c. Are these molecules chiral? Yes or No d. Sketch the molecules below using dash/wedge configurations, show the mirror image

relationship and assign R/S to each.

mirror plane

Multiple Stereocenters and Stereocenters in Rings 6) Build C6H12 (Cyclohexane).

a. Gently flatten the molecule so that the C atoms are all in the same plane. i. What is the angle of the C-C bonds? ________ ii. Look down one of the C-C bonds so one C is directly behind the other. What

is the angle between the C-H on the front carbon and the back carbon?

Eclipsed / staggered / skew

iii. Draw a sketch of the molecule with dash/wedge configuration below.


60

iv. Would you expect the molecule to be stable in this shape? Explain why or why not.

Yes / No

b. Gently pull the left-most carbon upwards until the bonds are 109.5°. This is a half-chair configuration.

c. Gently pull the right-most carbon downwards until the bonds are 109.5°. This is a chair configuration.

i. What is the angle of the C-C bonds? ________ ii. Look down one of the C-C bonds so one C is directly behind the other. What

is the angle between the C-H on the front carbon and the back carbon?

Eclipsed / staggered / skew

iii. Draw a sketch of the molecule with dash/wedge configuration below.

iv. Would you expect the molecule to be stable in this shape? Explain why or why not.

Yes / No

7) Orient the molecule so that one carbon is pointing towards 12’o clock and the axial H is

pointed up. Replace the axial H on the top (12 o’clock) carbon with Br. On the carbon just to the RIGHT (2 o’clock) of the Br replace the equatorial H with a Cl. This forms 1-bromo-2-chlorocyclohexane.

a. Is this molecule cis or trans? b. When you flip the chair what happens to the Br? Axial / equatorial c. When you flip the chair what happens to the Cl? Axial / equatorial


61

8) Make a second model cyclohexane. Orient the molecule so that one carbon is pointing towards 12’o clock and the axial H is pointed up. Replace the axial H on the top (12 o’clock) carbon with Br. On the carbon just to the LEFT (10 o’clock) of the Br, replace the equatorial H on with a Cl.

a. Are the chair forms of these molecules superimposable? Yes or No mirror images? Yes or No

b. Are the planar forms of these molecules superimposable? Yes or No mirror images? Yes or No

c. Are these molecules chiral? Yes or No d. What is the relationship between these two molecules? ___________ e. Sketch the molecules below using dash/wedge configurations, show the mirror image


mirror plane 9) For both of the 1-bromo-2-chlorocyclohexanes, find the carbon bonded to the Br. Switch the

H and Br bonds on this atom without changing any other atoms. a. Is this molecule cis or trans? b. Are the chair forms of these molecules superimposable? Yes or No mirror images?

Yes or No c. Are the planar forms of these molecules superimposable? Yes or No mirror images?

Yes or No d. Are these molecules chiral? Yes or No e. What is the relationship between these two molecules? ____________ f. What is the relationship between these molecules and the molecules from step 8?

______________________________ g. Sketch the molecules below using dash/wedge configurations, show the mirror image


mirror plane


62

10) For both of the 1-bromo-2-chlorocyclohexanes, find the carbon bonded to the Cl. Remove

the Cl and replace it with a Br to form 1,2-dibromocyclohexane. a. Is this molecule cis or trans? b. Are the chair forms of these molecules superimposable? Yes or No mirror images?

Yes or No c. Are the planar forms of these molecules superimposable? Yes or No mirror images?

Yes or No d. Are these molecules chiral? Yes or No e. What is the relationship between these two molecules? ____________ f. Sketch the molecules below using dash/wedge configurations, show the mirror image


mirror plane 11) For each of the 1,2-dibromocyclohexanes pick on carbon with a Br attached and switch the

H and Br bonds on that carbon. a. Is this molecule cis or trans? b. Are the chair forms of these molecules superimposable? Yes or No c. Are the planar forms of these molecules superimposable? Yes or No d. Are these molecules chiral? Yes or No e. What is the relationship between these two molecules? ____________ f. What is the relationship between these molecules and the molecules from step 10?

______________________________ g. Sketch the molecules below using dash/wedge configurations, show the mirror image


mirror plane


63

12) Open the ring on both models to make 3,4-dibromohexane. a. Rotate the C3-C4 bond until the two Br are eclipsed. Sketch the molecule below and

show the internal mirror plane. Rotate the chain into a staggered, anti configuration. Sketch the molecule below.

eclipsed staggered

b. Are these molecules chiral? Yes or No 13) Pick one of the two models and switch the H and Br on one of the carbons.

a. Rotate the C3-C4 bond until the two Br are eclipsed. Sketch the molecule below and show the lack of an internal mirror plane. Rotate the chain into a staggered, anti configuration. Sketch the molecule below.

eclipsed staggered

b. Is this molecule chiral? Yes or No


64

Week 5 - Introduction to Spectroscopy I: Infrared Spectroscopy

Learning Outcomes – • Describe general principles of IR spectroscopy. • Identify bond types and functional groups in IR spectra • Predict IR signals for given structures • Analyzing IR spectra or spectral data.


K. P. C. Vollhardt, N. E. Schore; Organic Chemistry: Structure and Function Chapters – 10.2, 11.5

Introduction – Covalent bonds exist as sharing of electrons between non-metal atoms. These bonds act in many ways like springs with the atoms as weights that can vibrate back and forth along the bond length or even wag off the bond axis. These vibrations can be symmetrical, asymmetrical, they can stay in a plane or twist out of a plane, and they can be between two bonded atoms or multiple bonded atoms to give more complex harmonic oscillations.

At ground state the atoms are at the lowest possible energy (with the spring in a neutral state) and are at a particular distance apart (the bond length or the length of the spring). If the bond absorbs energy the atoms can move a small distance closer together or further apart as the bond (spring) compresses or expands. The more energy that is absorbed the larger the compression and expansion. If sufficient energy is absorbed the bond will break. Breaking bonds generally requires absorption of ultraviolet light, but infrared is sufficient energy to get bonds vibrating back and forth.

The frequency of light absorbed will be the natural oscillation frequency of the bond as it expands and contracts. The frequencies associated with Infrared (IR) Spectroscopy are the mid-range infrared (~2.5-25 µm). IR absorptions are usually measured in wavenumbers or inverse wavelength (υ = 1/λ), which have the units of cm-1 and the range of wavenumbers we observe for IR spectroscopy is 4000 to 400 cm-1. Since the wavenumber of a vibration is the resonance frequency of the oscillation of the bond, it follows Hooke’s Law of elasticity: where c = speed of light, k = spring constant/bond strength, m1 = mass of atom 1, m2 = mass of atom 2, and υ = the wavenumber of the associated frequency.

symmetric stretchstretching waggingasymmetric stretch

rocking scissoring wagging (out of plane) twisting (out of plane)


65

Hooke’s Law shows that the stronger the bond (large k), the larger the frequency, the shorter the wavelength and the larger the wavenumber associated with that resonance frequency. Decreasing the mass of the atoms (small m) will also give a larger resonance frequency, a shorter wavelength and a larger associated wavenumber. Small atoms (e.g. H) and strong bonds (e.g. triple bonds, sp C-H), will tend to have large wavenumbers (υ), while larger atoms (e.g. Br, Cl) and weaker bonds (e.g. single bonds) will tend to have smaller wavenumbers. The difference means that different types of bonds will have different resonance frequencies. The different resonance frequencies associated with each kind of bond allow us to identify what kinds of bonds are present and so what kinds of functional groups may be present.

In order to be visible by IR spectroscopy a vibration must show a change in dipole. An N2 stretch will not show up in the IR as the electrons are evenly distributed, but a CO stretch will show up as there is an overall dipole and the distance between the areas of positive and negative charge changes as the bond compresses/expands. In practice even most C-C single bonds have some difference in polarity due to neighboring atoms, shapes, rotamers etc. so are visible in the IR spectrum. The increased intensity of the resonance frequency for strongly polar bonds means that functional groups with C=O, or O-H, N-H etc. appear very clearly in the IR spectrum.

0

Ener

gy

Distance between Nucleii

Bond Energy

Bond Length

Excited Vibration States lowest energy

(ground state)

vibration state - A

Atoms can move back and forth between bond distances when there is sufficient energy. The higher the energy the greater the degree of freedom.

vibration state - B

vibration state - C

Covalent Bond Diagram

m1m2

m1 + m2

k1

2πcxν =

sam

ple

original intensity

(I0)

I0

transmitted intensity

(IT)

detectorIR source

refe

renc

e

transmitted intensity

(IR)

detector

computer display

ITIR

= T

T x 100 = %T


66

Key Points: • IR spectroscopy measures the vibration of covalent bonds.

• IR spectroscopy is typically measured as percent transmittance with a scale of

wavenumbers decreasing from left to right.

• Frequencies of light that are associated with a particular molecular vibration are absorbed and show a low percent transmittance.

• Frequencies of light that are not absorbed pass through the sample and show 100%

transmittance of the light.

• Resonance frequencies are associated with bond strength and atomic mass therefore each particular kind of bond will have its own resonance frequencies. O-H is different from C-H which is different from C-C orC=C.

• IR spectroscopy allows identification of bond type (functional groups), but not number of

bonds or relative positioning of functional groups. Interpreting IR –

Each absorption peak (peak of minimum transmittance) in the IR corresponds to a particular bond type. Knowledge of the bond type(s) in a molecule can lead identification of what functional groups are present, but NOT the exact identity of the molecule unless a comparison is made to a known compound.

The IR absorption for two similar functional groups (e.g. two ketones) is likely to be very similar, so it can be difficult to tell if there is more than one of a similar bond type. This makes it difficult to tell how many of a particular bond type are present. Also there is only minimal information in IR about what groups are nearby, so it is nearly impossible to place groups in particular positions relative to each other based only on IR. This makes assignment of exact identity a challenge.

However, given the large number and variety of shapes around C-C bonds in most organic molecules the single bond region is very complex. The very complexity of these peaks can be used to match molecules, as no two different C-C structures will have exactly the same shapes. It is possible to find an exact identity using IR if the compound can be matched to a known compound and the IR are found to be identical.

IR is of particular use when coupled with other types of spectroscopy, in particular for functional groups that do not have H or C present. Wavenumbers:

NOTE: The scale on a typical IR spectrum isn’t a normal linear x-axis, a peak halfway between 2000 cm-1 and 1500 cm-1 isn’t exactly 1750 cm-1, but is closer to 1700 cm-1.

3000 2000 1500 1000 5004000ν (wavenumbers) cm-1Large ν"

smaller λhigher energy

smaller νlarger λ

lower energy

X-H C X C X C X(fingerprint region)


67

Table of Common IR Wavenumbers1 Functional Group Bond Wavenumber Range

Zone 1 - X-H bond Region Alcohols O-H 3400-3650 cm-1 (broad, strong) Amines N-H 3250-3500 cm-1 (broad, strong)

(NH2 doubled) Acids COO-H 2500-3100 cm-1 (broad, strong) Alkynes CC-H 3260-3330 cm-1

(sharp) Alkenes =C-H 3050-3150 cm-1

Aromatics Benzene-H ~3030 cm-1

Alkanes C-H 2850-3000 cm-1

Aldehydes O=C-H 2850 and 2750 cm-1

Zone 2 – Triple bond Region2 Alkynes CC 2100-2260 cm-1

Nitriles CN 2220-2260 cm-1

Zone 3 – Double Bond Region Carbonyls (generic) C=O 1650-1850 cm-1 (strong) Carboxylic acid O=COH 1710-1760 cm-1

Ester O=COR 1735-1750 cm-1

Aldehyde O=CH 1690-1750 cm-1 Ketone O=CR 1690-1750 cm-1

Amide O=CNR2 1650-1690 cm-1

Alkenes (generic) C=C 1620-1680 cm-1

Zone 4 – Single bonds/Fingerprint region3

Oxygen C-O 1000-1260 cm-1

Nitrogen C-N 1030-1230 cm-1

Chlorine C-Cl 600-800 cm-1

Bromine C-Br 500-600 cm-1

1 Wavenumber ranges taken from Organic Chemistry: Structure and Function 6th Ed.; Vollhardt, P.; Schore, N.; W. H. Freeman and Co; New York; 2011 and Spectrophotometric Identification of Organic Compounds 6th Ed.; Silverstein, R. M.; Webster, F.X.; John Wiley&Sons, New York, 1998. 2 Alkynes and nitriles are difficult to distinguish in IR. Check the 13C NMR or formula to distinguish between them. 3 Frequencies given in the fingerprint region are for information purposes only. The fingerprint region is not generally used for structure elucidation, but can be used to match to known compounds. There are 4 major regions of wavenumbers in an IR spectrum:

• 4000 to 2700 cm-1: The X-H region. H atoms are much lighter than other atoms in organic molecules so have higher wavenumber oscillating frequencies.

o CC-H, C=C-H, C-C-H, O-H, N-H, and O=C-H

• 2300 to 2150 cm-1: The triple bond region. Triple bonds are stronger so have a larger k and thus higher wavenumbers than other bond types between the same atoms.

o CC and CN


68

• 1820 to 1550 cm-1: The double bond region. Double bonds are stronger than single

bonds, but weaker than triple bonds so are their oscillation frequency is in between the ranges of triple and single bonds.

o C=O, C=N, and C=C

• 1500 to 500 cm-1: The single bond region a.k.a. the fingerprint region. Single bonds are much weaker than multiple bonds so have smaller wavenumber oscillating frequencies. Since most organic molecules have many C-C bonds this region is generally not easily interpreted, but can be used to ‘match’ a known compound as a fingerprint.

o C-C, C-N, C-O, and C-X NOTE: Know the peaks in Table 11.5 of Vollhardt and Schore for basic IR peaks used in 118A. Your instructor may give additional tables or information about what functional group ranges you should know. Additional peak ranges will be added as we learn new functional groups! Peak Shape and intensity:

Intensity is also a factor in interpreting IR spectra. Peaks may be strong (low %T) or weak (high %T). A C=O is typically a strong absorption (low %T) while C-H is weak to medium. If the C=O peak is weaker than the C-H peak it may imply there are many C-H, while if the C-H peak is weak compared to other peaks, it may imply that there are relatively few C-H bonds in the molecule. This will not allow assignment of the exact structure, but can along with other information help narrow the possibilities.

3000 20004000

ν (wavenumbers) cm-1

C HC H C C

(sharp)

% T

rans

mitt

ance

0%3000 20004000ν (wavenumbers) cm-1

C H

% T

rans

mitt

ance

100%

0%1500

OH

3000 20004000ν (wavenumbers) cm-1

C H

% T

rans

mitt

ance

100%

0%1500

OC O

O

O H

OC H

OC

(broad)

(broad)


69

Peak shape can also be relevant to assignment of functional groups. Peaks may be sharp (narrow) or broad. Some peaks (NH2 groups) are doubled. The exact peak shape can help differentiate between an alkyne C-H (a short, sharp peak at 3300 cm-1) and an alcohol O-H (a broad deep peak at 3300 cm-1) or a carboxylic acid O-H (a very broad, shallower peak, at 3150 cm-1). Alcohols tend to be distinct from the alkyl C-H, while acids tend to blur into the alkyl C-H.

The broadening of NH and OH peaks is commonly seen in IR. This is due to extensive hydrogen bonding within the sample, which changes the bond length of the X-H bond and so gives a greater range of resonance frequencies. Samples with minimal hydrogen bonding will show less broadening depending on how the sample is measured. Applications – Determining Structure in IR Spectroscopy There are many ways to analyze IR spectra, this method gives a general overview of deciding between possible functional groups.

1) Find the alkane C-H peak near 2950 cm-1. Nearly all IR spectra will have some sp3 C-H to show up here.

a. Look left. Is there a peak that is distinct from the mass of C-H? i. Yes, and it’s relatively sharp and narrow and near 3100 cm-1 ! sp2 C-H

look for an alkene C=C near 1650-1600 cm-1 ii. Yes, and it’s relatively broad and centered near 3300 cm-1 ! alcohol or

secondary amine OH or NH iii. Yes, and it’s relatively broad and centered near 3100 cm-1 and

overlapping the alkane C-H peak ! acid OH, look for a C=O near 1700-1750 cm-1.

iv. Yes, and it’s a narrow peak well away from the alkane C-H near 3300 cm-

1. ! terminal alkyne C-H, look for an alkyne CC near 2200 cm-1 v. Yes, two that are both rather broad and connected centered near 3300

cm-1 ! primary amine –NH2 vi. No. The molecule may have alkenes/alkynes, but they have no H

attached. b. Look right. Is there a peak that is distinct from the mass of C-H but closely

associated? i. Yes, aldehyde C-H. Look for an aldehyde C=O between 1700-1750 cm-1.

2) Look between 2100-2300 cm-1, is there a peak? a. Yes, with alkyne C-H (above). ! Terminal alkyne b. Yes, without alkyne C-H ! internal alkyne or nitrile. Check the formula for N. c. No, then no triple bonds are present in the molecule.

3) Look between 1820-1600 cm-1, is there a peak? a. Yes, between 1820-1680 cm-1 ! a carbonyl C=O

i. With acid OH ! carboxylic acid ii. With aldehyde C-H ! aldehyde iii. No other associated peaks ! ketone or ester.

b. Yes between 1650-1600 cm-1 ! an alkene C=C i. Most alkenes will show sp2

C-H, but some do not have C-H bonds c. No, then no double bonds are present in the molecule.

NOTE: One of the most common difficulties is distinguishing C=O from C=C bonds. This is a problem even for experienced spectroscopists in cases where conjugation, ring strain, or other groups blurs the difference between them. Look for other clues aldehyde C-H, alkene C-H, acid OH, distance from fingerprint region etc.


70

Practice: 1) What bonds are present in the following IR. What functional groups may be present?

Label all the significant peaks in each of the given spectra.

2) Write a reasonable chemical structure with 6 carbons that could have produced spectra

A – D.

3000 20004000

% T

rans

mitt

ance

100%

0%1500 1000

(cm-1)

3000 20004000

% T

rans

mitt

ance

100%

0%1500 1000

(cm-1)

3000 20004000

% T

rans

mitt

ance

100%

0%1500 1000

3000 20004000

% T

rans

mitt

ance

100%

0%1500 1000

(cm-1)

Compound A

Compound B

Compound C

Compound D


71

3) Match the given IR with the possible compounds listed below. Some IR may have more than one possible match.

3000 20004000

% T

rans

mitt

ance

100%

0%1500 1000

(cm-1)

3000 20004000

% T

rans

mitt

ance

100%

0%1500 1000

(cm-1)

3000 20004000

% T

rans

mitt

ance

100%

0%1500 1000

(cm-1)

Compound E

Compound F

Compound G

H

OO

H

OO OO

H

OO

H

OO

CN

OO

H

OO

OH OCH3 OH OCH3

OH NOCH3

NOH OCH3

O

O N

O

O

O

H

O

OH

O O OOH


72

Week 6 - Introduction to Spectroscopy II: Basic Principles of NMR

Learning Outcomes – • Describe basic principles of NMR. • Properly define and utilize spectroscopic terms such as shielded, deshielded, upfield,

downfield, chemical shift, and chemical equivalence. • Predict chemical shifts of H associated with various functional groups in H NMR • Predict chemical equivalence for H NMR


K. P. C. Vollhardt, N. E. Schore; Organic Chemistry: Structure and Function Chapters -10.2 – 10.5.

Introduction – NMR or Nuclear Magnetic Resonance allows a chemist to use radio waves to look at the chemical environment around active nuclei when the nucleus is placed in a magnetic field. It is one of the most useful techniques in modern chemistry to find the structure of novel compounds as small variations in bonding, neighboring nuclei, and functional group produced distinct predictable signals. Physics of NMR – Magnets and Nuclear Spins:

The nuclei of atoms contain protons and sometimes neutrons and are overall positively charged. Because they are rotating charges they have a magnetic moment, which is randomly oriented under normal conditions, but will align either parallel (α) or anti-parallel (β) to a magnetic field. Only nuclei that produce a magnetic moment will be NMR active, so not every isotope of every atom is detectable.

The parallel spins are very slightly lower energy than the anti-parallel spins, so there are slightly more parallel spins than anti-parallel spins in a sample. The difference in energy (and the difference in number of spins) depends on the strength of the magnet, the larger the magnetic field, the larger the energy difference and the fewer anti-parallel spins. Even with very large magnets the difference in energy (and spin population) is very small. A 11.7 Tesla magnet will produce an energy difference in a 1H nucleus of about 500 MHz (3.31x10-25 J) and the

B

A charged nucleus spins and produces the magnetic field B. Nuclei are often represented by an arrow in the direction of the magnetic field.

B

The magnetic moments of nuclei under normal conditions point in random directions.

B0In a magnetic field (B0) the nuclear magnetic moments will line up parallel or anti-parallel to the magnetic field.


73

largest magnet used in a commercial NMR to date (June 2009) is the Bruker AVANCE 1000, which has a 23.5 Tesla magnet and produces an energy difference of 1000 MHz (6.63x10-25 J) in 1H nuclei. NMR Active Nuclei:

To be visible using NMR a nucleus must have a quality called spin (ℓ).

• Nuclei with an EVEN mass number and an EVEN number of neutrons have ℓ= 0 so cannot be observed with NMR.

• Nuclei with an EVEN mass number and ODD number of neutrons have ℓ = 1, 2, 3 etc and can be observed with NMR.

• Nuclei with an ODD mass number will have ℓ = 1/2, 3/2, 5/2 etc and can be observed with NMR.

Common NMR active nuclei in organic chemistry are 1H and 13C. Other NMR active nuclei include: 15N, 19F, 29Si, and 31P. Most high-resolution NMR works with nuclei that are spin ½, as larger spins have

additional complexities that will not be discussed here. Both 1H and 13C are spin ½ nuclei. What the Radio Waves do: The magnet produces two different positions for the nuclear magnetic spin (parallel and anti-parallel). Because the difference in energy is on the order of the energy in radio waves, the nuclei can be put into an excited state by irradiating them with radio waves, just as an electron can be excited into a different orbital using ultraviolet light (absorption spectra). We can then observe the radio waves that are given off as the excited state relaxes back to the ground state (just as in atomic emission spectra). When the sample is irradiated at the ΔE for a particular nucleus we see continuous absorption/emission, this is called the resonance frequency.

The identity of the active nucleus and the magnet strength are the major factors in the

ΔE between the parallel and anti-parallel spins. The operating frequency is the frequency in MHz that correlates to the ΔE for that particular nucleus and magnet combination. Most NMR

B0

ΔE hv

(radio wave absorption at ΔE)

B0Ground state in a magnetic field

Excited state in a magnetic field

hv

(radio wave emisson at ΔE)

(detected)

B0

ΔE

Relaxation back to ground state by radio wave emission


74

spectrophotometers are referred to by their 1H operating frequency i.e. 300 MHz, 400 MHz or 500 MHz spectrophotometers. 13C NMR operating frequencies are about ¼ of 1H operating frequencies for the same magnet. However, the exact environment of the nucleus also affects ΔE to give a particular resonance frequency for that particular nucleus and those differences are what allow us to use NMR as a diagnostic tool in organic molecules. High-resolution NMR can differentiate between the 1H or 13C nuclei in different chemical environments, which allows us to make conclusions about structure.

NOTE: A consequence of increasing the magnet strength is with ΔE, small changes become more significant, a .0001% difference is a larger ΔE with a larger magnet so subtle differences become more distinct. What is a blur of signals a 60 MHz, may resolve into a half dozen peaks at 300 MHz, and clearly distinct peaks at 600 MHz.

Chemical Shift and TMS: Since the magnet strength affects the ΔE of a nucleus each individual NMR spectrophotometer has a very slightly different operating frequency for 1H nuclei, as each individual magnet is slightly different. This means that if the resonance frequency of a particular 1H nucleus is .001% off the operating frequency it will resonate at 300.003 MHz in a 300 MHz NMR and at 500.005 MHz in a 500 MHz NMR. This causes different signals for different people using different physical NMR spectrophotometers for the same nucleus. To avoid confusion chemists use a ratio unit instead of individual resonance frequencies: chemical shift. Chemical shift (δ) is measured in parts per million (ppm). This is the ratio of the resonance frequency of the particular nucleus to the operating frequency of the spectrophotometer times 106.

Chemical shift is the same for all spectrophotometers so chemists can easily compare

data from lab to lab and spectrophotometer to spectrophotometer. To further standardize chemical shift, tetramethylsilane (TMS, (CH3)4Si) is defined as a standard 0.00 ppm in both 1H and 13C NMR. The typical chemical shift ranges compared to TMS are 0-10 ppm for 1H NMR and 0-200 ppm for 13C NMR.

NOTE: As the range of ΔE (resonance frequencies) increases with larger magnets the

ppm do not change, but the appearance of the signals becomes sharper and narrower allowing much clearer differentiation of different signals. This makes interpretation of spectra where many signals are similar or highly coupled much easier. NMR in Organic Chemistry – Shielded vs. Deshielded: In organic chemistry different chemical environments in a molecule will result in different resonance frequencies. The general pattern is signals near TMS (low ppm) are upfield. Upfield signals are shielded and have relatively high electron densities around the observed atom’s nucleus. The large electron densities buffer the nucleus from effects of the magnetic field (shield the nucleus). The movement of electrons (also charges in a magnetic field) around the nucleus produces local magnetic field (Blocal) anti-parallel to the applied field B0. The Blocal reduces the

δ (ppm) = resonance frequency of nucleus (Hz)operating frequency of NMR (MHz)

x 106


75

effective strength of B0, thus the nucleus feels less B0, and the resonance frequency has a smaller ΔE. Signals further from TMS (high ppm) are downfield. Downfield signals are deshielded and have relatively low electron densities around the observed atom’s nucleus. The small electron densities allow the nucleus to feel the magnetic field (B0) more effectively (less Blocal) so it has a larger ΔE.

Typically nuclei are deshielded by induction due to nearby electronegative atoms or by movement of pi electrons (electrons are charges too) to produce a parallel Blocal increasing the applied magnetic field (B0). Chemical Shift: The chemical shift of 1H NMR signals ranges from 0 ppm to 10 ppm for most functional groups. The acidic -OH of carboxylic acids can be found as far downfield as 13 ppm, but this is the only common exception. Tables of typical chemical shift ranges can be found in many sources (Table 10.4 Vollhardt + Schore) and these numbers should be memorized in order to facilitate interpretation of spectra. While these numbers may seem random there are patterns based on chemical principles of organic chemistry.

While there are exceptions, H attached to sp3 hybridized atoms tend to show up between 0-5 ppm. And while there are exceptions, H attached to sp2 hybridized atoms tend to show up between 5-10 ppm. The presence of the pi bonds sets up a consistent effect on the chemical shift accounting for about 5 ppm of downfield shift. Alkyne C-H are the only common H bonded to sp hybridized atoms and tend to show up between 1.7-3.1 ppm due to the electronic effects of two pi bonds.

Within those ranges hydrocarbons are most upfield (most shielded), then C-H adjacent

to pi-bonds, and C-H attached to electronegative atoms are furthest downfield, this pattern repeats in the sp3 and sp2 regions. The downfield shift is caused by to increased deshielding of the H due to increasing electron-withdrawing by the attached functional group. Alkyl groups are not electron-withdrawing, pi bonds are only slightly electron withdrawing, and electronegative atoms are more so. Oxygen is more electronegative and so more electron withdrawing than nitrogen. Halogens cover a broad range depending on which halogen is present.

downfield O-C-H > N-C-H > =C-C-H > C-C-H upfield

0 ppm

TMS

upfield,high field,shielded,small ΔE

downfield,low field,deshielded,large ΔE

10 ppm

TMS10 9 8 7 6 5 4 3 2 1 01112

sp2 sp3

sp(ppm)


76

Also when looking within a range for a particular functional group a CH (methine proton)

will be downfield/left of a CH2 (methylene protons), which will be downfield/left of a CH3 (methyl protons). This is a subtle effect, but if a peak is at the borderline between two ranges (e.g. 3.3 ppm could be O-C-H or N-C-H) then if it’s a CH3 and it should be to the upfield side of the range (3.3 ppm makes sense for an OCH3, but an NCH3 would be closer to 2.5 ppm), and if it’s a CH then it should be to the downfield side of the range (3.3 ppm makes sense for a -NCHR2 but 3.8 ppm would be more reasonable for an -OCHR2).

Downfield CH > CH2 > CH3 upfield

Similarly to IR spectroscopy H attached to OH and NH will have a broadened signal with

a variable chemical shift due to varied degrees of hydrogen bonding from molecule to molecule. However, alcohols and amines (sp3) will tend to be between 0.5-5 ppm, amides and phenols (sp2) will tend to be between 6-8 ppm and carboxylic acids (sp2) will tend to be between 10-13 ppm.

TMS10 9 8 7 6 5 4 3 2 1 01112

H

OH

H

H(adjacent to any pi bonds)

H

X

H

(ppm)

TMS6 5 4 3 2 1 0

RCH3RCH2R

R3CHCH2H

(ppm)

OH or NH CH


77

Table of 1H NMR Chemical Shifts: Functional Group Shape Chemical Shift Sp3 hybridized Alkyl (1°) -CH3 0.7-1.3 ppm Alkyl (2°) -CH2- 1.2-1.6 ppm Alkyl (3°) -CH- 1.4-1.8 ppm Near Pi bonds Allylic (next to alkene) C=C-CH 1.8-2.2 ppm Next to C=O O=C-CH 2.0-2.4 ppm Next to aromatic Benzene-CH 2.4-2.7 ppm Alkyne CC-H 1.7-3.1 ppm Near Heteroatoms Alkyl Halide CH-X 2.5-4.0 ppm fluoride CH-F chloride CH-Cl bromide CH-Br iodide CH-I Alkoxy CH-O 3.3-4.5 ppm Sp2 Hybridized Alkene CH=CH 4.5-6.5 ppm (terminal alkene) C=CH2 4.5-5.2 ppm (internal alkene) C=CHR 5.2-6.5 ppm Aromatic Benzene 6.5-8.5 ppm Formyl XCHO 8-8.5 ppm Aldehyde RCHO 9.7-10.0 ppm H-bonded Amine (sp3) -NH 0.5-5.0 ppm (broad) Alcohol (sp3) -OH 0.5-5.0 ppm (broad) Amide (sp2) -CONH 6-8 ppm (broad) Phenol (sp2) Benzene-OH 6-8 ppm (broad) Acids COOH 11.0-12.0 ppm 1 Wavenumber ranges taken from Organic Chemistry: Structure and Function 6th Ed.; Vollhardt, P.; Schore, N.; W. H. Freeman and Co; New York; 2011 and Spectrophotometric Identification of Organic Compounds 6th Ed.; Silverstein, R. M.; Webster, F.X.; John Wiley&Sons, New York, 1998. Chemical Equivalence: Differences in chemical environment due to bonds and functional groups will cause different atoms in a molecule to have different chemical shifts (resonance frequencies). NMR is capable of differentiating very small differences in chemical environment. The following discussion will refer directly to 1H NMR, but the principles apply to all nuclei.

• Homotopic protons are exactly identical. If replaced with a bromine atom, they would give exactly the same molecule. Homotopic protons are always the same chemical shift. The 3H in any CH3 are homotopic. The 2H on carbon-2 of propane are homotopic.


78

• Enantiotopic protons are the same carbon, but if replaced with a bromine atom, they would give a pair of enantiomers. Enantiotopic protons usually give the same signal, but may be different if a chiral solvent is used. (Remember the physical properties of enantiomers are the same except in a chiral environment). The 2H on carbon-2 of butane are enantiotopic.

• Diastereotopic protons are attached to the same carbon, but if replaced with a bromine

atom, they would give a pair of diastereomers. Diastereotopic protons usually give different signals, but may be the same if the two diastereotopic environments are similar enough. The 2H on carbon-3 of 2-chlorobutane are diastereotopic.

Generally linear and branched alkane systems have free rotation around the C-C bonds so

even diastereotopic protons can average out and have the same signal. However, systems without free rotation (rings and alkenes) will have distinctly different signals for diastereotopic protons. For purposes of 118 A/B/C consider diastereotopic protons on linear and branched alkanes and rings without stereochemistry as the same and diastereotopic protons on rings with explicit stereochemistry or alkenes as different, this is an over-generalization, but makes things simpler to start.

Examples of Chemical Equivalence

Br

OH Br

O

Br

OHBr

O

A

A

B

D

C

F

EG

7 Signalsall homotopic

7 Signals

4 Signalsall homotopic

9 Signalsdiastereotopic ring

7 Signalsdiastereotopic alkene

4 Signalssymmetric

7 Signalssymmetric

diastereotopic ring

6 Signalsdiastereotopic alkene

A CB

D

F

EG

A CB B

A

D

AB

C/DC/D

E/F E/FG

A

B

C D

A/B

C/DE F G

HI

H3CO

H3CO

A

B

C

D

E

F

G

A

A

C

BD

EF


79

Practice – 1) How many signals would you expect in the 1H NMR spectrum of each of the following

compounds?

2) Circle the H in each of the above compounds that is most deshielded.

3) Look at the given NMR scale. Label the ppm ranges where you would expect to find the following kinds of H in a 1H NMR. a. Alkene (C=C-H) b. Aldehyde (CHO) c. Alcohol (OH) d. α to an amine (CH-N)

e. α to a ketone (CH-COR) f. Aromatic (benzene C-H) g. Alkane (C-H) h. Ether (CH-OR)

O

Br

OHO

O

OCH3 O

H

OHO OH

OOHCl

Compound A = _______ Compound B = _______ Compound C = _______ Compound D = _______

Compound E = _______ Compound F = _______ Compound G = _______ Compound H = _______

Compound I = _______ Compound J = _______ Compound K = _______ Compound L = _______

10 9 8 7 6 5 4 3 1 02(ppm)


80

Week 7 – Introduction to Spectroscopy III: Further Principles of NMR

Learning Outcomes – • Properly define and utilize spectroscopic terms such as integration, coupling, doublet,

triplet, multiplet, and coupling constant. • Predict integration ratios for given compounds • Utilize n+1 rule to predict coupling or interpret number of neighboring H.


K. P. C. Vollhardt, N. E. Schore; Organic Chemistry: Structure and Function Chapters -10.6 – 10.8

Introduction – NMR also provides additional information beyond just the individual signals for each type of active nucleus. It is possible to tell not only what the chemical environment (chemical shift) of a nucleus is, but also how many nuclei are in that exact environment (integration) and even how many other active nuclei are nearby (coupling or spin-spin splitting). When combined with information from IR and chemical formula, the full molecular structure of nearly any organic molecule can be determined. With additional types of spectroscopy it’s even possible to determine exactly what diasteromer or even enantiomer is present. Integration – Since the signals in NMR are related to the relaxation emission they are also directly related to the population of active nuclei. Since a molecule has a fixed whole number of atoms in fixed positions, the ratio of active nuclei at each resonance frequency (chemically equivalent nuclei) depends on the formula. Thus the intensity of the signal is related back to the number of Integration Examples:

6 5 4 3 2 1 0(ppm)

O

OCH3

1H

3H 3H

9H

8 7 6 5 4 3 2(ppm)

1H

3H

O

O

6 5 4 3 2 1 0(ppm)

2H

3HOH

Br(H3C)2N

8 7 6 5 4 3 2(ppm)

1H1H

6H

1H 1H


81

atoms with the same chemical equivalence within the molecule. And the signal intensity can be calculated by integration of the area under the curve of the peak for each resonance frequency. Classically the integral is charted and the heights of the integrals are measured by a ruler, and then compared to give whole number ratios. Or if no integral was measured, the peaks can be cut out of the chart paper and weighed to find the mass ratio! More modern spectrophotometers can automatically calculate peak area ratios. However, the integral only shows the ratios of signal intensity, so some care is needed in determining the exact number of H per peak.

Generally the smallest peak in the spectra integrates to 1H, however, in highly symmetrical molecules it may integrate to 2H or even 3 or more H!. When interpreting integrations if the number of protons in the spectrum sum to the correct number then the integrations are correct, if they only sum to less than the total then the molecule is symmetrical. If the sum is half the number of protons then all the ratios need to be multiplied by a factor of 2, if sum to a third of the number of protons then all the ratios need to be multiplied by a factor of 3 etc.

Coupling / Spin-spin splitting – Basic Principles: Since nuclei are spinning charges that create magnetic fields and the measurement in NMR is the ΔE caused by the total magnetic field, nearby nuclei will effect each other’s chemical environment. This effect is coupling, also known as spin-spin splitting. Picture each carbon in a carbon chain as a house along a street and the hydrogens attached to those carbons as the people living in the house. Neighboring hydrogen atoms can ‘talk’ to each easily other over the fence, but atoms that are further away cannot ‘talk’ easily because there is too much distance in between them. The ‘talking’ between hydrogens on adjacent carbons is called 3-bond coupling and it’s the most common type of coupling. Typically the H on the same carbon (2-bonds) are the same so do not ‘talk’ or couple with each other, and while H further away (4-bonds) may couple it’s typically too small an effect to be measured in alkanes. (We’ll look into this further in 118B with alkenes/alkynes.)

So what is actually happening? Remember that each H nucleus is a spinning charge that produces a magnetic field (spin). When that nucleus is placed in a magnetic field the spin of the nucleus can align parallel (α) or anti-parallel (β) to the outside magnet (B0). The alignment of each nucleus is independent so a multi atom molecule may have some nuclei parallel and some anti-parallel.

If we’re looking at Ha’s resonance frequency (Rf) we’ll see a single frequency in the NMR (a singlet) if Ha has no neighboring H. If Ha is adjacent to Hb, then Hb’s spin will affect the magnetic field that Ha feels. If Hb is parallel to the magnetic field B0 then Ha will feel a slightly

Ha

C C

Hb

C

Ha

C C

Hb

C

HcHc

Ha and Hc are adjacent and will couple.

Ha and Hc will not couple because Hb is in between

Ha

C C

Hb

C

Hc1

2

3Ha

C C

Hb

C

Hc1

2 3

4

Ha and Hb are 3 bonds apart.

Ha and Hc are 4 bonds apart.


82

stronger overall field, the ΔE will increase and the signal will shift slightly downfield (deshielded). If Hb is anti-parallel to the magnetic field B0 then Ha will feel a slightly weaker overall field, the ΔE will decrease and the signal will shift an equal amount upfield (shielded). This results in two possible resonance frequencies for Ha (a doublet). In the analogy Hb is either ‘home’ or ‘not home’ so we have two possible states, Ha and Hb can talk (home/parallel) or cannot talk (not home/anti-parallel).

In a real sample there are billions of molecules and nearly equal populations of parallel/anti-parallel spins so we see two equally possible cases (a doublet) for the environment of Ha, due to its spin being split by the spin on Hb (coupling to Hb). We would also see a corresponding doubling of resonance frequencies in Hb as its spin is split by the spin on Ha (coupling to Ha). The distance between the split peaks is the coupling constant (J value) and it is the same for both Ha and Hb. Coupling Constants:

• The distance between the split peaks in a signal with spin-spin splitting can be measured in hertz (Hz) and the value called is the coupling constant also known as the J-value and abbreviated as J.

• The coupling constant for a particular pair of H is always the same regardless of

spectrophotometer, it depends on the relationship between Ha and Hb not the original magnet.

• The coupling constant is the same number of Hz for both coupled H (they have the same

value as they’re having the same conversation!).

• The values of J range from 0-18 Hz and depend primarily on the angle between the two H (called the Karplus angle). But in cases of complex splitting J-values also depend on the distance (2, 3 or 4 bonds apart).

• In alkanes with free rotation (straight or branched chains) the bonds spin so the angles

average out and the value of J is approx. 7 Hz, so all alkane C-H can be treated as having the same J value.

Ha

C

Rf of Ha

Ha

C C

Hb

B0 B0

Rf of Ha

Ha

C C

Hb

B0

Rf of Ha

Ha

C C

Hb

B0

Rf of Ha

+

Ha alone Ha with Hb parallel to B0

Ha with Hb anti-parallel to B0

Ha with a statistical population of Hb spins.


83

• The similarity of J in alkanes allows for the use of the n+1 rule. • Chemically equivalent nuclei will not couple with each other (they don’t need to talk as

they’re the same thing.)

• OH and NH generally do not show coupling with their neighbors. They’re involved with hydrogen bonds so show up as broad singlets.

Pascal’s Triangle and the n+1 Rule (Alkane Coupling): In coupling (spin-spin splitting) the final shape of the signal, also called a multiplet, depends on the number of neighboring nuclei and their particular coupling constants. In 1H NMR, the signal for a particular 1H nucleus contains the resonance of that particular 1H in all the molecules in the sample, so each signal (peak) shows the probability distribution of each possible combination of parallel and anti-parallel neighbors (spin state). For organic molecules with basic alkane skeletons we can treat all neighboring H as having the same coupling constant, so the shape of the coupled signal depends only on the number of neighbors. This is the n+1 rule.

The n+1 Rule: A single H resonance frequency will be split into a multiplet with n+1 number of peaks, where n is the number of neighboring H.

• Since each nucleus has two possible states (parallel or anti-parallel spin) the n+1 rule follows Pascal’s triangle both for number of peaks in the multiplet and the relative area of each of these peaks.

• In integration all the peaks in a multiplet are added to give the total signal for that type of

H. And the chemical shift is measured from the center of the multiplet where the resonance frequency of the uncoupled H would be.

• H that have an even number of neighboring H will give a multiplet with an ODD number

of peaks, which gives a single tallest central peak. H that have an odd number of neighboring H will give a multiplet with an EVEN number of peaks, which gives a doubled tallest central peak.

• Sometimes it’s helpful to think of the n+1 rule as a series of coin flips. The number of neighbors equals the number of flips. The total number of heads/tails gives the resonance frequency. The number of possibilities that would give that heads/tails total gives the intensity.

So what kind of shapes do we see? Following the analogy: a singlet H has no neighbors so has no conversations and only one possible resonance frequency. A doublet H has one neighbor who may or may not be home so has two possible states and two resonance frequencies. A triplet H has two neighbors who 1) are both home, 2) are both not home, 3) one is home and the second is not, or 4) one is away but the second is home, this gives four possible states at three resonance frequencies. States 3 and 4 have the same ΔE because they both have the same number of neighbors (one home, one not) so the central resonance frequency will have twice the intensity.


84

Formation of Mulitplets by the n+1 Rule:

Ha

C0 neighborsn+1 = 0 + 1 = 1

a singlet (s)

Ha

C C

B0

1 neighborn+1 = 1 + 1 = 2

a doublet (d)

H

Ha

C CB0

2 neighborsn+1 = 2 + 1 = 3

a triplet (t)

H

H

1:1

1:2:1two neigbhors so neither, one, or both may be parallel.

no neighbors so no coupling

one neighbor so it is either parallel or anti-parallel

Ha

C C

B0

3 neighborsn+1 = 3 + 1 = 4

a quartet (q)

HH

H

1:3:3:1three neigbhors so 0, 1, 2, or 3 may be parallel.

J

J

J

Ha

C CC

B0

4 neighborsn+1 = 3 + 1 = 5

a pentet (p)

HH

H

1:4:6:4:1

four neigbhors so 0, 1, 2, 3, or 4 may be parallel.

JH

Ha

C CCH

HH

H

H

Ha

C CCH

HH

H

HH

Ha

C CC

CH

HH

H

H

Ha

C CC

CH

HH

H

HHH H HH

n=5 a hexet

n=6 a heptet n=7

an octetn=8

a nonet


85

Practice:

1) How many peaks (what kind of multiplet) will be seen in the signal for each of the indicated H?

2) What is the name for each of the coupled peaks (e.g. doublet, triplet).

3) What integration ratios do you expect for all the H in each of the above compounds?

O

Br

OHO

O

OCH3 O

H

OHO OH

OOHCl

Compound A = _______ Compound B = _______ Compound C = _______ Compound D = _______

Compound E = _______ Compound F = _______ Compound G = _______ Compound H = _______

Compound I = _______ Compound J = _______ Compound K = _______ Compound L = _______

a ba

b

a

b

a b

a b

ab

ba a

b

a b a

b

a

ba

b


86

Common Coupling Patterns

01234PPM

1.491.49

3.423.42

Cl

01234PPM

0.960.96

1.611.61

3.543.54

Cl

01234PPM

0.940.94

1.471.47

1.761.76

3.543.54

Cl

01234PPM

1.551.55

3.683.68

1.551.55

Cl

01234PPM

0.930.93

2.012.01

3.303.30

0.930.93

Cl

2H

3H

2H

2H

3H

1H

6H

2H

2H2H

3H

2H

1H

6H


87

Common Coupling Patterns Continued:

01234PPM

0.910.91

1.551.553.483.48 1.551.55

Cl

0123PPM

1.591.59 Cl

1.591.591.591.59

01234PPM

0.950.95 3.243.24

0.950.950.950.95Cl

01234567PPM

4.054.05

ClH5.935.93

H5.225.22

H5.185.18

Cl

H6.336.33

H5.435.43

H5.385.38

01234567PPM

1H

5H

3H

1H 1H 1H

1H1H1H

2H

2H

9H

9H


88

Week 8 – Introduction to Spectroscopy IV: Approaching Spectroscopy Problems

Learning Outcomes – • Calculate degrees of unsaturation. • Identify functional groups and coupling patterns from provided spectra. • Use provided spectral information and logic to predict structure of unknown substances.


K. P. C. Vollhardt, N. E. Schore; Organic Chemistry: Structure and Function Chapters -11.6.

Introduction Spectroscopy problems in many ways are logic puzzles. Once you know the chemical shifts and wavenumbers of particular functional groups there really isn’t much actual chemistry involved in solving the structures. Like all logic puzzles a rational approach helps a great deal in solving the structure. Listed below is a way of approaching spectroscopy problems. It is NOT the only way! Most people who do spectroscopy come up with their own favorite way of doing things so try things out until you come up with something that works for you. A step-by-step approach: 1. Degrees of unsaturation a.k.a. double-bond equivalents. This will give the total number

of rings and pi bonds. It’s not impossible to put together a structure if you don’t have the formula, but it’s orders of magnitude more difficult so the formula will nearly always be provided.

a. Each C gets 2 H plus 1 per end in a saturated molecule, so (2 x #C)+2 = hydrogen in a saturated molecule. Every nitrogen atom requires an extra H (+#N), every halogen atom replaces an H (-#X) and oxygen simply inserts and doesn’t change the hydrogen count.

o Unsat. = DBE = ((2 x #C) +2 - #H + #N - #X) / 2

b. If you have a ring this may be the only clue you have to its presence. c. High degrees of unsaturation in a small molecule are a clue that a benzene ring (4

degrees of unsaturation) is present.

2. IR Spectrum. Shows what functional groups are present. a. Look at each of the 4 sections (X-H, triple bond, double bond and fingerprint)

individually to determine what kinds of bonds are present. b. Label each peak with what functional group it represents. c. OH, alkyne H, and aldehyde H will show up distinctly. d. C=O, and triple bonds show up here and may not show up clearly in the 1H NMR as

they tend to be quaternary carbons.

3. 1H NMR Spectrum. Shows exact connections and functional groups in the molecule. Working methodically so you don’t skip any peaks (e.g. left to right) look at each peak and assign three things.

a. Integration – How many H of that type. i. Look for symmetry. There are no molecules (other than CH4) where 4H, 6H,

or 9H are equivalent unless you have two or more symmetrical CH2 or CH3.


89

ii. If your total number of H does not equal the number of H in the formula you likely have a symmetrical molecule.

b. Chemical Shift – What functional groups are nearby? i. sp3

are to the right (0-5 ppm), sp2 are to the left (5-10 ppm) ii. Plain C-H are right most, then C-H next to a pi bond in the middle of the

range, and then C-H next to an electronegative atom are to the left. iii. OH and NH can hydrogen bond so vary a great deal in chemical shift, but will

show up as broad singlets that never couple to neighboring H. c. Coupling – How many nearest neighbors.

i. N+1 rule for all alkanes. ii. Remember that no single neighboring carbon can have more than 3 attached

H. iii. If you’re not sure what the H is coupled to put a number of neighbors, this will

prevent accidental double writing of atoms or errors in how to split up large numbers of neighbors.

4. Check. Make sure you have all the atoms. Check the pieces you found in the 1H NMR

spectrum and the IR. Make sure you have all the degrees of unsaturation accounted for. a. Missing C, O, and degree of unsaturation – likely a carbonyl, check the IR/13C NMR. b. Missing 2C and 2 degrees of unsaturation – likely an internal alkyne, check the

IR/13C NMR. c. Missing H or extra H – probably a mistake in the initial interpretation of the 1H NMR,

check it again. d. Missing just a C – likely a quaternary carbon check the 13C NMR e. Missing degrees of unsaturation but no atoms – likely a ring.

5. Put the puzzle together. Use the information from step 3 to put pieces together. Many

molecules have 'handles' where you can get started, places where there really is only one CH group that can attach at that particular position. What works for one molecule will not necessarily work for others, but places that can be helpful to start are:

a. Ends vs. Middles: Keep track of what pieces are ends (only 1 missing attachment) and which are middles (more than 1 missing attachment). You can't attach two ends together unless they are the last two groups in the molecule or you won't have anywhere to put the rest of the pieces.

b. Coupling: Look at your coupling and join pieces by how many neighbors are needed to get the correct coupling.

i. Often there will be only 1 CH or CH3 or CH2, where does it attach? ii. Chains of CH2 will tend to have two CH2 with two neighbors (ends) and one or

more CH2 with four neighbors (middles). iii. A 6H doublet is always two CH3 attached to a CH (isopropyl) look for the CH

with many neighbors and figure out how many neighbors are unaccounted for.

iv. No coupling often means no neighbors. c. Electron-withdrawing groups: What CH is next to a pi bond? What CH is next to an

O or N or X? i. Placing an OH, or C=O can limit where else you can put pieces and make

middles into ends.


90

ii. Differentiating between groups adjacent to pi bonds is difficult to do by chemical shift, consider the coupling as well.

iii. Alcohols and ethers need to have two groups attached to the oxygen, make sure you can find both of them. If only one is obvious it may be attached to a quaternary carbon.

d. Symmetry: i. Are there symmetrical CH2 or CH3 (4H, 6H or 9H peaks)? ii. Are all the H in the formula accounted for in the 1H spectrum or are all the

peaks doubled (or tripled)? e. Once you’ve attached a few pieces go back and try the tricks that didn’t work again,

often eliminating a few pieces forces the rest to a particular position.

6. CHECK!!!! a. Make sure everything makes sense. Often something that seems out of place or

strange will make sense in the context of the whole molecule – “Oh it’s downfield because it’s next to two different pi bonds…”

b. If you’re sure of your assignments in step 2-4 DO NOT CHANGE THEM. It’s very common for students to ‘know’ that they have a CH2 with two neighbors attached to an OH and instead they draw it as a CH or adjacent to a CH or C=O just to make something else fit. If you change what you know is right to make it fit, the whole molecule is likely wrong. Start over instead.

c. Every peak will be accounted for in the final molecule. d. Watch out for random hanging –CH2 and 5 bonds to C. e. Everything will fit when you have the correct molecule. If things don’t fit then

something is wrong, go back and check carefully! Tips/Tricks:

• Know the chemical shifts in Tables 10.2 in Vollhardt + Schore and the wavenumbers in table 11.4. Without these doing spectroscopy problems is nearly impossible.

• Many times information about a functional group is available in several places. If you

can’t remember a chemical shift, check the IR for the same group. • Every time we cover a new functional group add it’s particular tricks to your process. Ask

is this an alcohol or a phenol (chemical shift)? Ask is this an ether or an alcohol (IR or 1H NMR)?

• Find a way to indicate neighbors without writing the extra atoms. E.g. If you know

something has two neighbors write -2- rather than CH2 or you’ll often get extra atoms in your structure and not know which are real and which are placeholders.

• Everything will make sense in a correct molecule. If one of your pieces doesn’t fit, start

over.

• If you’re confident that all your molecule pieces are correct, don’t change them just to attach them to a structure, likely the structure you’re putting together is incorrect.


91

• Practice practice practice. Like any logic puzzle the more you do, the easier it is to see the patterns and to recognize various common tricks.

Further Readings and References: Silverstein, Robert M.; Webster, Francis X.; Spectroscopic Identification of Organic Compounds; 6th Ed.; John Wiley & Sons, Inc.; New York; 1998 Vollhardt, Peter; Schore, Neil; Organic Chemistry: Structure and Function; 6th Ed.; W. H. Freeman and Co.; New York; 2011 Pretsch, Ernö; Bühlmann, Phillipe; Badertsher, Martin; Structure Determination of Organic Compounds: Tables of Spectral Data; 4th Ed; Springer-Verlag; Berlin; 2009 Wikipedia.com – IR Spectroscopy; NMR Spectroscopy; Mass Spectrometry and Ultraviolet-Visible Spectroscopy YouTube – Spectroscopy.Mov (or search for Sarah Lievens) for a 50 minute lecture walking through a spectroscopy problem.


92

Practice Spectroscopy:

4 3 2 1 0(ppm)

2H2H 2H

2H

1H

2H

3H

3H

3000 20004000

% T

rans

mitt

ance

100%

0%1500 1000

Compound AC8H17BrO

(cm-1)

4 3 2 1 0(ppm)

1H

3H

6H

2H

2H 2H

3000 20004000

% T

rans

mitt

ance

100%

0%1500 1000

(cm-1)

Compound BC9H16


93

4 3 2 1 0(ppm)

2H 2H

9H

1H

6H

3000 20004000

% T

rans

mitt

ance

100%

0%1500 1000

(cm-1)

Compound CC10H20O2

3 2 1 0(ppm)

2H 2H2H

2H

3H

1H

3000 20004000

% T

rans

mitt

ance

0%1500 1000

(cm-1)

Compound DC8H12O


94

3000 20004000

% T

rans

mitt

ance

100%

0%1500 1000

(cm-1)

4 2 1(ppm)

4H

2H

6H

3

1H 2H

1H1H

Compound EC8H17BrO

4 2 1 0(ppm)

2H

9H

2H2H

3

6H

1H

3000 20004000

% T

rans

mitt

ance

100%

0%1500 1000

(cm-1)

Compound FC10H22O

0


95

3000 20004000

% T

rans

mitt

ance

100%

0%1500 1000

(cm-1)

4 2 1 0(ppm)

3

1H1H

6H

1H

Compound GC10H18O2

4 2 1 0(ppm)

2H

2H

3H

3

6H

1H

1H

3000 20004000

% T

rans

mitt

ance

100%

0%1500 1000

(cm-1)

Compound HC8H16O2


96

3000 20004000

% T

rans

mitt

ance

100%

0%1500 1000

(cm-1)

4 2 1 0(ppm)

2H2H

1H

3H

3

3H

Compound IC6H11ClO

4 2 1 0(ppm)

1H

3

1H

3H6H4H 4H

1H

3000 20004000

% T

rans

mitt

ance

100%

0%1500 1000

(cm-1)

Compound JC10H20O2


97

3000 20004000

% T

rans

mitt

ance

100%

0%1500 1000

(cm-1)

10 9 8 7 6 5 4 3 1 02(ppm)

2H

3H

1H

3H

2H1H

2H

Compound KC7H14O2

3000 20004000

% T

rans

mitt

ance

100%

0%1500 1000

(cm-1)

4 2 1 0(ppm)

6H

2H

9H

3

6H

2H

1H

Compound LC12H26O3


98

Week 9 - Approaching Synthesis Problems Learning Outcomes –

• Compare linear vs. convergent synthesis. • Use known chemical reactions in sequence to form more complex molecules • Predict problems with synthetic sequences due to poor yielding or impossible reactions. • Break down a target molecule into building blocks by retrosynthetic analysis.


K. P. C. Vollhardt, N. E. Schore; Organic Chemistry: Structure and Function Chapters -8.9

Introduction – Synthesis can be challenging to study for, as each problem is unique and tricks that work

for one question may not work for another. It is helpful to think of studying synthesis as analogous to writing a poem. To write a poem we need to have a good vocabulary (know many reactions), a solid understanding of grammar (mechanisms and basic principles), and a certain amount of inspiration. Like poetry some synthesis can be elegant 100 step convergent epics, while others may be short but serviceable limericks. A synthetic chemist is always learning new reactions and looking for ways to make their synthesis simpler and more elegant, however even the best epic poet had to start writing somewhere. With practice, particular patterns will show up as familiar, and while there is no magic process, there are ways to approach a problem that give a high likelihood of finding the necessary connections. Flexibility of thinking is as important knowing the chemistry.

NOTE: Writing a synthesis of a large molecule may take an afternoon or even a few days, but doing the same synthesis in a lab may take years! It is important for a synthetic chemist to be flexible in the approach to a target molecule and no synthesis written on paper is every truly ‘correct’ until it’s been done with actual materials.

Linear vs. Convergent Synthesis – The goal of synthesis is to make as much as possible of the target molecule with minimal waste and expense. To that end synthesis may be linear or convergent. Typically convergent syntheses are preferred as material is lost at each step and convergent synthesis allows portions of the molecule to be built independently, resulting in fewer steps directly between starting material and product and better overall yields.

Most syntheses seen in undergraduate classes will be linear synthesis, with only a few steps, but the principle of convergence becomes more important as syntheses more complex.

A B C D E F

A 5 step linear synthesis. If all steps are 90% yield total yield = (.90 x .90 x .90 x .90 x .90) x 100% = 53%

A B C

D E

F

A convergent synthesis (3 step longest linear sequence). If all steps are 90% yield total yield = (.90 x .90 x .90) x 100% = 73%

A Linear Synthesis

A Convergent Synthesis


99

Applications - Method of Approach A: A Question Based Process (Starting molecule is given)

Many coursework problems give a particular starting material as well as a target compound to limit the possible ways of synthesis and to guide students towards paths that will work with the reagents that have been learned. If a starting point is given it can be productive to ask series of questions about the molecule.

1) Look at the target molecule and starting molecule. Find the carbons that are present in both molecules.

a. It can help to circle or mark the original atoms to show what is new/old.

2) Find the changes between the starting material and the product. a. What atoms need to be added (only present in product)? b. What atoms need to be removed (only present in starting material)?

3) Determine which changes are carbon-carbon bond forming reactions, and which are functional group conversions (no additional carbons, just changes in bond order or heteroatoms).

a. There are relatively few reactions that form carbon-carbon bonds reactions so this is often a good place to start.

i. Where do we need to form the new bond? ii. Is there a limitation in the question of how many carbons we can add at

once? iii. What functional group is present in the starting material or formed in the

product? iv. What reactions do we know that can work with the present/formed

functional group? 1. If no reactions work directly do we need to first make a functional

group to add the carbon-carbon bond too? If so go to step 3b then return to step 3aiii.

b. Functional group conversions are more flexible and can often come before or after the carbon-carbon bond forming reactions. Functional group conversions may be needed to convert the starting material to the functional group that will make the carbon-carbon bond (e.g. forming a carbonyl to add a Grignard reagent too.) or form the final product after the carbon-carbon bond is made (e.g. converting a tertiary alcohol to an alkene after a Grignard reaction).

i. What functional group is present in the starting material and what functional group needs to be formed?

ii. Do we need a particular stereochemistry? (e.g. racemic, inversion, cis or trans)

iii. Do we need a particular regiochemistry? (e.g. Markovnikov or Saytzev?) iv. What reactions do we know? Is there a reaction that does this conversion

directly? 1. If not what functional groups can we make directly from the

starting functional group? Can we make one of those into the desired group?

4) List the reactions that will give the desired conversions from steps 3a and 3b.


100

5) Consider the list of reactions in step 4 and determine the order of reactions. Do some reactions need to be done first or last? If there are multiple ways to do a particular conversion is one shorter, or easier?

a. The functional group needed to make the carbon-carbon bonds must be made before the carbon-carbon bond can be made. (i.e. We need to form an alkyne before we can add carbons to an alkynyl anion.)

b. Friedel-Crafts reactions require an activated ring so must be done before adding deactivating groups.

c. Conversions of the new carbon-carbon bond must be done after it is formed. (e.g. eliminating an OH after adding a Grignard to form an alkene)

d. Is there an incompatible functional group (e.g. an alcohol present during a Grignard reaction) that must be converted or protected first?

e. Keep from making two of the same functional group in the same intermediate unless they are going to undergo the same reactions. (i.e. It is very difficult to pick which alcohol will react with PCC, all primary and secondary alcohols will react).

6) Write out the steps with their intermediate products in the order determined in step 5. 7) CHECK!

a. Look for incompatible functional groups. b. Look for extra/missing carbons. It is very common to ‘lose’ a methyl group when

writing quickly. c. Look for side reactions that could occur with the other functional groups. d. Check your stereochemistry. Do the reactions give the desired regiochemistry

and stereochemistry. Example: Synthesis of 1-cyclohexyl-2-methyl-3-propyl-2-hexene Problem A: Synthesize 1-cyclohexyl-2-methyl-3-propyl-2-hexene from

4-cyclohexyl-2-methyl-1-butanol.

OH


101

Question Based Analysis:

Knowing the reactions we need, we can then decide what reagents we can use in what order to give that particular reaction. For example the oxidizing agent for the first step must be PCC since H2CrO4 (aq) conditions will give the carboxylic acid, however it doesn’t matter for the third step. We can use either the Grignard reagent or the corresponding alkyl lithium to add in steps 2 and 4. Synthesis:

2 Groups of new carbons.

How do we add the carbons? New carbons could be added by addition of an organometallic to a carbonyl. Since there are two groups they can be added in series (or to the ester in 118C).

OH H

O Can we make a carbonyl to add too? The starting material doesn't have a carbonyl, but we can make an aldehyde from the primary alcohol.

How do we make the alkene? By elimination of haloalkanes. We can make the OH a better leaving group (haloalkane) and then eliminate.The product is the most stable alkene, so should form easily by E2 with an unhindered base.

OH1) make R-X

2) Base

Ox. agent

Need to form C=C bond.

H

O1) R-M

2) Ox. agent3) R-M

OH

What's new/different? The alcohol has converted to an alkene and two different propyl groups have been added to the carbon that was the alcohol. The presence of the cyclohexane can be used to mark where the original carbons are.

OH

PCC,CH2Cl2

H

O

1) CH3CH2CH2MgBr2) H3O+

OHNa2Cr2O7,

H2SO4, H2O

O OH1) CH3CH2CH2MgBr

2) H3O+

PBr3 or HBr

Br

CH3CH2ONa,CH3CH2OH


102

Applications – Method of Approach B: Retrosynthetic Analysis (Starting molecule is NOT given)

This type of problem is more typical of a real world problem. Any compound could be the starting material, which opens us up to a greater range of possible strategies and reactions that may be used. The lack of a defined starting point allows for greater freedom in putting the molecule together, which means that different people may come up with very different synthesis. Retrosynthetic analysis is commonly also used with convergent synthesis, where several pieces of a molecule are constructed independently and then put together in the last few steps to form the desired product, but it also works for linear synthesis.

If the problem is for a class often there will be a target for the starting material(s) stated such as: no starting material can have more than X carbons. If the problem is real world the target is usually that the starting material(s) are available and relatively inexpensive.

1) Look at the target molecule. What functional groups are present? How many carbons are present?

2) If the carbon skeleton is more than a few carbons, look for places to add carbon-carbon

bonds. a. A branch at a functional group may come from addition of an organometallic to a

carbonyl. b. Branches adjacent to functional groups may be added through enolates. c. Remember functional groups can be converted: what was a ketone may be

masked as a haloalkane, alcohol, alkene or even an alkane. d. Is there particular stereochemistry?

3) Find a precursor or precursors to the target molecule and repeat step 1-3 with the precursor as the new target until the precursor(s) are small inexpensive molecules.

4) You now have a planned retrosynthesis working from the product back to available

starting materials. a. Typically a retrosynthesis skips steps and concerns itself with major molecular

changes so the full synthesis will need reagents added.

5) Now using functional group conversions to convert each precursor exactly to its target until you’ve reached the final target molecule.

a. Watch out for incompatible groups (e.g. organometallics with alcohols or leaving groups with nucleophiles)

b. Watch out for multiple similar functional groups. If one alkene is reduced by hydrogenation, all alkenes (and alkynes) will likely be hydrogenated.

c. Use protecting groups or functional group conversions to force reaction at only the desired position.

i. Carbonyls can be masked as acetals or thioacetals ii. Alcohols can be masked as ethers iii. Amines can be masked as carboxamides

d. Make sure you consider regioselectivity and stereoselectivity in choosing the appropriate reagents.

e. Remember that fewer steps (with reasonable yields) is almost always better than more steps (with equivalent yields).


103

Example: Synthesis of 4-methoxy-2-methyl-4-propyloctane via Retrosynthetic Analysis Problem A: Synthesize 4-methoxy-2-methyl-4-propyloctane (compound A) from haloalkanes containing 4 or fewer carbons.

Compound A has twelve carbons in its main alkane skeleton, the methoxy group is easily added as a substituent by functional group conversions so is not included in the count. To get a twelve carbon skeleton from from four carbon or smaller units we’ll have to add carbons, and will need at least two carbon-carbon bond forming reactions. Compound A has a branched position at C-4, so C-4 will be a good place to start adding carbons. We can picture adding the four carbons of a 2-methylpropyl group to a 4-octanone. The four carbon fragment (2-methylpropyl anion) can be imagined as coming from 1-bromo-2-methylpropane. We can then break the 4-octanone into two pieces of four carbons each by adding a butyl group to butanal. Both the butanal and the butyl anion can come from 1-bromobutane. The order of adding the two anions is somewhat arbitrary on paper; in reality we’d likely try both options and use whichever gave better overall yields. Retrosynthesis:

Once we’ve found a retrosynthesis we can add in the functional group conversions to complete the total synthesis. 1-bromobutane undergoes SN2 replacement with hydroxide and the resulting 1-butanol is oxidized to butanal by PCC. The Grignard reagent is made from 1-bromobutane and butanal is added followed by aqueous workup to give 4-octanol. 4-Octanol can be oxidized to 4-octanone and then added to the 2-methylpropyl lithium reagent followed by aqueous work up to give 2-methyl-4-propyl-4-octanol. Full deprotonation of the alcohol by the strong base sodium hydride will give the alkoxide nucloephile , which can undergo Williamson ether synthesis with iodomethane to give the target ether.

OCH3O+ O

H

+ Br

Br


104

Synthesis:

Applications – Method of Approach C: A Combined Process (Starting molecule is given)

It is possible to use retrosynthetic analysis when the starting molecule is given. This can allow us to work straight backwards (product ! starting material) as in method B. Or work straight forwards (starting material ! product) when the connection between the starting material and product is seen. Or most often both ends can be worked towards the middle. If the connection between the starting material and product is not clear, find the direct precursor(s) to the product and the direct product(s) of the starting material given different reactions that are known. Are there any connections between those substances? If you can’t see any connections, try going out a step further with the most likely product/precursor.

BrNaOH,

DMFOH

PCC

CH2Cl2

Br1) Mg, ether

H

O

OH

Br 1) Li (2 eq)

THF, -78C

3) H3O+

H2SO4, H2ONa2Cr2O7

O

3) H3O+OH

1) NaH, ether2) CH3I

OCH3Synthesis of 4-methoxy-2-methyl-4-propyloctane

6 steps longest linear sequence

Starting Material Productconnection after several steps

A

B

C

X

Y

Z


105

Tricks and Tips: • Know your reactions and reagents. It’s very difficult to put together a synthesis if you don’t

know what functional groups convert to what groups.

• Carbon-carbon bonds are often the key to the synthesis. Once you figure out how to add/lose carbons the rest of the synthesis will often fall into place.

• Remember that your reagents will interact with the whole molecule, not just the functional

group you want them to react with. This can cause side reactions or stop the reaction entirely.

• It is only necessary to write the intermediate products and reagents. There is no need to

write the mechanism for each reaction. Intermediate products are generally those that are stable enough to purify and identify.

• Any answer that works is acceptable, though some are more efficient than others. You are

not limited to reactions learned in this course. If it works it works, however, if it doesn’t work, it doesn’t work so remember the caveats given in lecture.

• It is possible to approach a problem backwards (product ! starting material) or forwards

(starting material ! product) or both ends towards the middle. • In synthesis questions designed for students there will be a functional answer that uses the

reactions you have learned. Real world problems may or may not have a solution, but anything you see in this course will.

• Coursework problems are typically between 3-5 steps (points where the product can be

isolated), though they may be as long as 7 or 8 steps. If you find yourself a dozen steps into a process and not at the answer yet, go back and start over as you’re likely going in an unproductive direction.

• Shorter synthesis are generally preferred if the product can be obtained in good yield. There

are cases where several steps are preferable to one, but these will be cases we talk about specifically.

• When in doubt: think about what functional groups have been discussed in class recently?

It’s likely that they will occur somewhere in the synthesis. • When in doubt: look at the rest of an exam. Often a mechanism or reaction question can

remind you of the key step to a synthesis, it may not be the same reaction, but a similar one, or the opposite regio- or stereoselectivity, etc.


106

Practice –

OHA)

O OHH3COB)

Br

OH

C)

H

O OH

D)

OH

Br

E)

OH

Br

H3CH2CO

OHF)

OH

G)

Br OCH2CH3H)

OHHO BrI)


107

Practice Continued:

J)

Br Cl

ClOH

K)

OH

OL)

BrI

M)

CNN)


108

Appendix A – Other Useful Things:

Functional Groups and Structures of Organic Chemistry

OH

O

Br

H

O

O

OH

O

O

O

NH2

O

Cl

O

NH2

H C C CH

H

H H

H HCH

HH

H C C CH

H

H

HCH

HH

H C C CH

HCH

HH

CC C C

CC

H

HH

H

HH

H C CH

H

H

HBr

H C C OH

H

H

HCH

HCH

HH

H C C NH

H

H

H HH

H C CH

H

OH

H C C OH

H

H

HH

H C C CH

H

O H

HH

H C C OH

H

OH

H C C OH

H

OCH

HCH

HH

H C C NH

H

OH

H

H C CH

H

OCl

FunctionalGroup Name

Bond-LineStructure

CondensedStructure

Kekule Structure

CH3CH2CH2CH3

CH3CHCHCH3

CH3CCCH3

C6H6

CH3CH2Br

CH3CH2OH

CH3CH2OCH2CH3

CH3CH2NH2

CH3CHO

CH3COCH3

CH3COOH

CH3COOCH2CH3

CH3CONH2

CH3COCl

alkane

alkene

alkyne

aromatic/arene

haloalkane

alcohol

ether

amine (primary)

aldehyde

ketone

carboxylic acid

ester

amide (primary)

acyl halide

C C

C C

C C

CC C C

CC

R Br

R O

R NH2

R CO

H

R OH

R C R'

O

CO

R C OR'

O

CO

CO

Cl

General Structure

R'

R OH

R NH2

R

IUPAC Suffix

-ane

-ene

-yne

-ene

-ane

-ol

-ane

-amine

-al

-one

-oic acid

-oate

-amide

-oyl halide

nitrile R C N CH

CH

H NCH3CN CN -nitrile


109

Useful Common Names and Abbreviations Multiple bonds in Hydrocarbons

Positions at or adjacent to a pi bond often have special names these can be used to refer to the position of a charge (e.g. an allyl cation) or a particular substituent (e.g. a propargyl alcohol). These positions often have resonance or particular reactivity due to the presence of the pi bond.

Carbonyls Small chain carbonyls have some particular common names. It is common to make an

alcohol an ester by adding ‘an acetyl group’ rather than adding ‘an ethanoyl group’.

a vinyl position is at the sp2 carbon Clvinyl chloride

an allyl position is adjacent to the sp2 carbonallyl alcohol

OH

a propargyl position is adjacent to the sp carbonpropargyl amine

NH2

a phenyl position is in a benzene ringMgBr

phenyl magnesium bromide

a benzyl position is adjacent to a benzene ringCH2Br

benzyl bromide

H OH

O

H H

OO

H Xa formyl group formaldehyde formic acid

OH

O

H

OO

Xan acetyl group acetaldehyde acetic acid


110

Common Solvents

Many solvents are referred to by common names or by abbreviations. A selection of common solvents along with their commonly used names is shown.

Short Abbreviations

Many common functional groups are abbreviated to two or three letter codes. These codes are then put into formula in the place of actual atoms.

e.g. MeCN = CH3CN, Me stands for methyl or CH3-.

O CH3O OO

O

OO

O

HO OH CH2Cl2

ethyl acetate(EtOAc)

methyl tert-butyl ether(MTBE)

tetrahydrofuran(THF)

ether

1,2-dioxane ethylene glycol dichloromethanemethylene chloride

(DCM)acetone

CH3CN

acetonitrile(MeCN)

O

H N CH3

CH3

SO

H3C CH3

N,N-dimethylformamide(DMF)

dimethylsulfoxide(DMSO)

OPN

NN

CH3H3C

CH3

CH3H3C

H3C

hexamethylphosphoramide(HMPA)

Me-

Abbreviation Stands for

Et-

Pr-

iPr-

Bu-

t-Bu

s-Bu

Ac-

Bn-

Ph-

Tf-

Ts-

Ms-

Methyl (CH3-)

Ethyl (CH3CH2-)

Propyl (CH3CH2CH2-)

Isopropyl ((CH3)2CH-)

Abbreviation Stands for

Butyl (CH3CH2CH2CH2-)

tert-Butyl ((CH3)3C-)

sec-Butyl (CH3CH2CH(CH3)-)

Acetyl (CH3CO-)

Benzyl (C6H5CH2-)

Phenyl (C6H5-)

Triflyl (CF3SO3-)

Tosyl (C6H6SO3-)

Mesyl (CH3SO3-)

Tr- Trityl ((C6H5)3C-)

Bz- Benzoyl (C6H5CO-)


111

Greek labels

In older common names distance from a functional group was given by Greek letters, α, β, γ, δ, ε etc. This type of referent is still used in mechanisms and discussions of the relative positions of functional groups as close proximity affects the interactions between substituents and thus their reactions. (e.g. β-keto-acids can decarboxylate easily, but regular carboxylic acids will not).

OH

O

α

β

γ

δ

ε

O

an α,β−unsaturated ketone

OH

OH2N

γ−aminobutyric acidNH

O

a β−lactamγ

α

β

β

β

α

α

OH

OO

a β−ketoacid


112

Appendix B - Practice Problems Key Week 1 –

1) Constitutional Isomers:

OCH3

OHOH

O

OH

OHOCH3

HOOH

OHOH

HO

OH

O

H

O

HOOH

OHOH

HO

OH

O

H

O OO OCH3

OCH3 OOCH3

ClCl

Cl

ClCl

Cl

ClCl

Cl

Cl

ClCl

ClCl

C4H10O

C4H9Cl

C4H10

C4H8

C4H7Cl

Cl Cl

O O

C4H8O


113

C7H16

O OCH3

OCH3

HOOH

OH

HOOHOH

OH

OH

OCH3 O

OCH3

C5H12

C6H14

BrBr

Br

BrBr Br

Br

BrC5H11Br

C5H12O

C5H10


114

2) A. Molecular formula A: C11H20O, B: C16H23ClO2, C: C9H16O3, D: C13H24O2, E: C12H24, F: C15H33N, G: C10H16ClNO, H: C13H24O2, I: C15H32O2, J: C12H23Br, K: C13H24O3, L: C10H15N

B. Bondline to condensed formula

C8H18

A) CH3CH2CHCHCOCH2CH(CH2CH3)2

B) CH3CH2CH2CH(CH2CH3)CH2CHClCOOCH2(C6H5)

C) CH3(CH2)4COCH(CH3)COOH

D) (CH3)3C(CH2)3COO(CH2)3CHCH2

E) (CH3)2CHCHCHCH2CH2CH(CH2CH3)2

F) CH3CH2CH(CH3)CH(NHCH3)(CH2)5CH(CH3)2

G) CH3CH2CHClCCCH2CH2CONHCH2CH3

H) CH3CH2CH2CHC(CH3)CH2C(CH3)2CHOHCH2CHO

I) CH3(CH2)3CH(CH(CH3)CH2CH3)CHOHCH2CHOHCH2CH2CH3

J) CH3CH2CH2CH(CH2CH3)CBrCHCH(CH3)CH2CH3

K) CH3CHC(CH3)(CH2)3CH(OCH2CH3)CH2CH2COOH

L) (CH3CH2)2CHCCCH2CH2CN


115

3) A. Condensed to bondline structures; B. Add lone pairs.

4) Resonance

Br

O

O

O

O

H

O

NH

Cl Br

Br

O

F

FF

O

OH

O

CN

H2NHN OH

O O

OO

O O

OH

A)

B)

C)

D)

E)

F)

G)

H)

I)

J)

K)

L)

O O

H Br H Br BrH

O O

OH OH

A)

B)

C)

D)

E)

F)

G)

equivalentmajor

all equivalent

major

all equivalent

all equvalent

major

major


116

O O O O

O

NH

O

NHH)

I)

J)

OH

O H

OH

O H

OH

O H

O

H

O

H

O

H

O

O

O

O

O

O

O O O

NH2 NH2 NH2 NH2

Cl O OHClCl OH

NH2 NH2 NH2 NH2

NH

NH

NH N

HNH

K)

L)

M)

N)

O)

P)

Q)

R)

S)

T)

major

major

major

minorequivalent equivalent

major

major

all equivalent

major

all equivalent

major

all equivalent

major

major


117

5) Hybrids

Week 2 – 1) Hybridization

O

NH

OO

H O

O

OH

O H

NH2 Cl OH

OH Br

O OHO

NH

NH2




Compound Q Compound R Compound S Compound T

δ+

δ+

δ+ δ+

δ−δ−

δ−δ−

δ−δ−

δ+δ+

δ+

δ−

δ−

δ−

δ+

δ−

δ−

δ−

δ− δ−

δ−

δ+

δ+δ−

δ−

δ− δ+

δ+

δ+

δ+

δ+

δ+

δ+δ+

δ+

δ−δ−

δ+

δ− δ−δ−

δ− δ−

δ−Compound P

δ−δ−

δ−

δ+δ+δ+δ+

δ+

δ−

δ−δ−

δ−

NH2

O

Cl OH

OH

OCN

NH2

H OH

ON

N

OO

HO

Br




sp3

sp3sp3

sp3

sp3 sp3

sp3

sp3sp3

sp3

sp3

sp3sp3

sp3

sp3sp3 sp3

sp3

sp3

sp3 sp3 sp3

sp3

sp3sp3

sp3

sp3

sp3

sp3 sp3sp3

sp3 sp3sp3

sp3 sp3sp2

sp2sp2

sp2

sp2

sp2

sp2

sp2

sp2

sp2

sp2

sp2

sp2

sp2

sp2

sp2

sp2sp2

sp2

sp2sp2

sp2

sp2

sp2

sp2

sp2

sp2

sp2

sp2

sp2sp2

sp2

sp2

spsp

spsp

spsp

unhyb.

unhyb.

unhyb.unhyb.

all H are unhybridized s orbitals. Halogens bond through unhybridized p orbitals.


118

3) Building Models sp3 1. 109.5o, tetrahedral 4. Electron repulsion occurs between electrons in the aligned C-H bonds. This is

torsional/eclipsing strain. 5.

11. In gauche rotamers the larger groups are closer to each other and the electrons in bonds/lone pairs/electron clouds repel each other when the groups approach each other. This is steric hindrance.

12.

13. The eclipsed where the larger groups pass each other is higher in energy as it has the

normal eclipsing strain, but the groups are also much larger so get closer together which increases repulsion as they pass (steric hindrance).

sp2 Hybridization

1. 120o, trigonal planar 2. Ethene and other sp2 hybridized groups cannot rotate because rotation will involve

breaking the pi bond.

H

H HC

H

HHH

H HC

H

HHH

H HC

H

HH

0o 60o 120o 180o 240o 300o 360o

H

H HC

H

HHH

H

HC

H

HHH

H

HC

H

HHH

H

HC

H

HH

Energy

staggeredeclipsed staggered staggeredstaggered eclipsedeclipsed

CH3

H HC

CH3

HHH

H3C HC

CH3

HHH

H CH3C

CH3

HH

0o 60o 120o 180o 240o 300o 360o

CH3

H HC

CH3

HHH3C

H

HC

CH3

HHH

H3C

HC

CH3

HHH

H

CH3C

CH3

HH

Energy

gaucheeclipsed gauche antianti eclipsedeclipsed


119

3,4.

6. The trans isomer is more stable because the cis isomer has increased steric hindrance

by having the two CH3 closer together. 7. The two isomers will not interconvert, because it would require breaking the pi bond to

rotate from one to the other. 8. The two isomers will not have the same physical properties, they have different shapes

and bonds so will be different compounds.

sp Hybridization 1. 180o, linear 2. Ethyne acts more like ethene because it also has a pi bond so cannot rotate. 3. There are no rotamers or isomers of butyne since there is only one group attached to

each CC bond and the molecule is linear and so symmetrical.

Week 3 –

1) Compound to Name Compound A – 3-bromo-6-isopropylnonane or 3-bromo-6-(1-methylethyl)nonane Compound B – 5-ethyl-1-fluoro-2,6-dimethylheptane Compound C – 2-fluoro-1,3-diisopropylcyclopentane or

2-fluoro-1,3-bis-(1-methylethyl)cyclopentane Compound D – 2-bromo-4,4-dimethylhexane Compound E – 3-tert-butyl-1,6-dichloro-1,5-diethylcyclohexane or

1,6-dichloro-3-(1,1-dimethylethyl)-1,5-diethylcyclohexane Compound F – 6-sec-butyl-3,7-diethyl-2-methyldecane or

3,7-diethyl-2-methyl-6-(1-methylpropyl)decane Compound G – 6-bromo-1,1,1-trifluoro-4,5-dimethylheptane Compound H – 3-cyclohexyl-4-ethyl-6-methylnonane Compound I – 3-chloro-2,2,4,4-tetramethylpentane Compound J – cis-1-cyclopentyl-3-isobutylcyclohexane or

cis-1-cyclopentyl-3-(2-methylpropyl)cyclohexane Compound K – 1-bromo-6-sec-butyl-10-chloro-5-isopropyldecane or

1-bromo-10-chloro-5-(1-methylethyl)-6-(1-methylpropyl)decane Compound L – trans-1-bromo-2-ethylcyclopentane Compound M – 1,3-diethyl-2-iodo-5-methylcycloheptane

Compound N – 6-cyclopropyl-3-fluoro-5-isopropylnonane or 6-cyclopropyl-3-fluoro-5-(1-methylethyl)nonane Compound O – 2-bromo-4-methylpentane Compound P – 2,3-diethyl-1,1-dimethylcyclohexane Compound Q – 3-chloro-3,4-diethylhexane Compound R – cis-1-ethyl-3-methylcyclohexane

H3CCH3

H

HH3C

HH

CH3

cis isomertrans isomer

H3C CH3 butyne


120

Compound S – 2-bromo-6-ethyl-3-methylnonane Compound T – 2,3-dichloro-4-propylheptane Compound U – 5,5,6,6-tetrabromo-2-methyloctane Compound V – trans-1-sec-butyl-3-isopropylcyclopentane or trans-1-(1-methylethyl)-3-(1-methylpropyl)cyclopentane Compound W – 2,3-dibromo-5,6,8-triethyl-11-methyldodecane Compound X – 1,3-dibromo-1,3-dimethylcyclobutane Compound Y – cis-1,4-diisobutylcyclodecane or cis-1,4-bis-(2-methylpropyl)cyclodecane

Compound Z – 5-bromo-4,5,6-triethylnonane 2) Name to Compound

F F BrCl

Cl

F3C

Br

Cl

Cl F

Cl

F

F F

Cl

Cl

Cl

Cl

Br

Compound AA Compound BB Compound CC Compound DD

Compound EE Compound FF Compound GG Compound HH

Compound II Compound JJ Compound KK Compound LL

Compound MM Compound NN Compound OO Compound PP

Compound QQ Compound RR


121

Week 4 – 1) Circle the optically active molecules and write the mirror plane for meso substances.

2) Assign R vs. S

(S)(S)

(R)(R)

Br

Br

(R)(R)(R)(R)

OH

OH

(S)(S) (R)(R)

(S)(S)(R)(R)

(R)(R)O

(S)(S)(S)(S)(R)(R)

(S)(S)

(S)(S)(R)(R)

(R)(R)

(S)(S)(S)(S)

(R)(R)(R)(R)

(Z)

(S)(S)(R)(R)

(R)(R)(R)(R)

(R)(R)

(S)(S)

(S)(S)

(R)(R) (S)(S)

(R)(R) (S)(S)(R)(R)

(R)(R)(S)(S)

O

Cl

OH

OH

Br

F

OHCl Br

Cl

Br

CN

Cl

H3CO

(E)

(S)(S) (R)(R) (S)(S)

CN CN

(S)(S)(R)(R)

OCH3

OCH3

(R)(R)(R)(R)

(E)

H(R)(R) (S)(S)

(S)(S)(R)(R) Br

(E)(R)(R)

(R)(R)

(S)(S)(S)(S)(R)(R)

(S)(S)(R)(R)

(Z)

(R)(R)(Z)

(S)(S)(R)(R)

(S)(S)

(S)(S)

(R)(R) (R)(R)Cl

H

(S)(S)H

(S)(S)CH3

HO CH2CH3

COOH (S)(S)F

Br

(S)(S) H HSH

CHOCH3

HOOCH

BrH2CH3

Br (S)(S)

(R)(R)

COOHH

H3CO (R)(R)

(S)(S)CH3

H OHCH3

H3C BrH

OH

Br OCH3

OH

OH

CN

Br

SCH3

CNNC

O

O

O

OH

O

O

BrOH

NC O O

O H

(S)(S) (R)(R)

(R)(R)OHBr

H (S)(S)CH2CH3CF3

COOH(R)(R) (R)(R)Cl

(R)(R)(R)(R)

OH





Compound U Compound V Compound W Compound X Compound Y

(S)

(S)


122

3) Enantiomers, diastereomers, identical, meso, constitutional isomers.

4) IUPAC nomenclature.

Compound A – (3S,4S)-4-bromo-3-ethyl-1,1,1-trifluorohexane Compound B – (5R,6S)-5-isopropyl-2,2,6-trimethyloctane Compound C – (4R)-1,4-dichlorohexane Compound D – (1S,3R)-1-fluoro-3-isopropyl-1-methylcyclohexane Compound E – (1S,2R)-1-ethyl-2-methylcyclopentane

(S)(S)(R)(R)

(S)(S)(R)(R)

(R)(R)(R)(R)

(R)(R)(R)(R)

(S)(S) (R)(R)

(R)(R)(S)(S)

(R)(R)(R)(R)

H (S)(S)BrCH3

(R)(R)CH2CH3HOH

H3C(R)(R)FH

(S)(S)HH3COCH3

Br (R)(R)HCH2CH3

(S)(S)CNHCH3

H (S)(S)CNCH2CH3

(R)(R)CNHCH2CH3

H (S)(S)OHCH3

(R)(R)BrHCH3

(R)(R)

(R)(R)

(S)(S)

(S)(S)

(S)(S)

(R)(R)

(R)(R)

(S)(S) (S)(S) (S)(S)

(S)(S) (R)(R)

(S)(S) (R)(R)

Cl

OCH3

Br

Cl

Cl

OH

OH

Cl

Br

Cl

Br

NC Br

Br

HO

(S)(S)

(R)(R)

(S)(S) (R)(R)

OH OH

(S)(S)(S)(S)

(S)(S)(S)(S)

Br (R)(R)HCH3

(S)(S)HHOCH2CH3

(R)(R) (S)(S)

Cl

OCH3

HOCN Brvs.

vs.

vs.

vs. E

E

E

E

F (S)(S)CH3

H

(S)(S)OCH3H3CH

(R)(R)

(S)(S) (S)(S)

(S)(S)

HO (S)(S) (S)(S) OH

F

F

vs.

vs. vs.

vs.

D

D D

D(R)(R)(S)(S)

(R)(R)(R)(R)

H (S)(S)OHCH3

(R)(R)CH3BrH

(R)(R) (S)(S)

(R)(R)Cl

OH

Cl

vs.

vs.

vs.

vs.

I

I

I

I

(R)(R)(S)(S)

(S)(S)(R)(R)

NC (S)(S)CH2CH3

H

(R)(R)HH3CH2CCN

(R)(R) (S)(S)

Cl

Br

Cl

Br

OHOH

OH

vs.

vs.

vs.

vs.

M

M

M

M

(S)(S)(R)(R)

NC (S)(S)HCH2CH3

(R)(R)BrHCH3

(S)(S)

(S)(S)

(S)(S)(R)(R)(R)(R)

Br

Br

Br

Br

vs.

vs.

vs.

vs.

C

C

C

C


123

Compound F – (4R)-4-bromo-2,2,5-trimethylhexane Compound G – (1R,3R)-1-(1,1-dimethylethyl)-3-ethylcyclohexane Compound H – (2R,3S)-3-bromo-2-chloropentane

Building Models

1) CH4 – a: yes, b: yes, c: 6, d: 4, 3-fold, e: no, 2) CH3Cl – a: yes, b: yes, c: 3, d: 3-fold; e: 0. f: no 3) CH2Cl2 – a: yes, b: yes, c: 2, d: no 4) CH2BrCl – a: yes, b: yes, c: 1, d: no 5) CHBrClF – a: no, b: 0, c: yes 6) C6H12 –

a. i- 120°, ii- eclipsed, iv- No, because the bond angles are not close to tetrahedral and eclipsed structures are high in energy

c. i- 109.50, ii- staggered, iv- Yes. The bonds are at an ideal angle and staggered structures are low in energy.

7) C6H10BrCl – a: cis, b: equatorial, c: axial 8) C6H10BrCl – a: no, yes, b: no, yes, c: yes, d: enantiomers 9) C6H10BrCl – a: trans, b: no, yes, c: no, yes, d: yes, e: enantiomers, f: diastereomers 10) C6H10Br2 – a: trans, b: no, yes c: no, yes, d: yes, e. enantiomers 11) C6H10Br2 – a: cis, b: yes, yes c: yes, yes, d: no, e. meso, f: diastereomers 12) C6H12Br2 – b: no 13) C6H12Br2 – b: yes

H

CH H

H

H

CHH

H

1) CH4

Cl

CH H

H

Cl

CHH

H

2) CH3Cl

Cl

CCl H

H

Cl

CClH

H

3) CH2Cl2

Cl

CBr H

H

Cl

CBrH

H

4) CH2BrCl

Cl

C(R)(R)

Br FH

Cl

C(S)(S)

BrFH

5) CHBrClF

C C

H HC

C CC

H

H

H

H

H H

H H

H H

C

H

C

C C C

CH

H

HH

H

H

H

H

H

H

H

6a) planar 6c) chair

BrCl

(S)(S)

(R)(R) (S)(S)

(R)(R)

Br Br

Cl Cl

7, 8) C6H10BrCl

BrCl

(S)(S)

(S)(S) (R)(R)

(R)(R)

Br BrCl Cl

9) C6H10BrCl

BrBr

(S)(S)

(S)(S) (R)(R)

(R)(R)

Br BrBr Br

10) C6H10Br2

BrBr

(S)(S)

(R)(R) (S)(S)

(R)(R)

Br Br

Br Br

11) C6H10BrCl

(S)(S)

(R)(R)

Br

Br

HH

(S)(S)(R)(R)

Br

Br

12) C6H12Br2

(S)(S)(S)(S)

H

Br

Br

H

(S)(S)(S)(S)

Br

Br

13) C6H12Br2


124

Week 5 – 1) Compound A –

This molecule contains an aldehyde as evidenced by the C=O and the aldehyde C-H. The presence of the aldehyde C-H ensures that the C=O is an aldehyde and not a ketone or ester. The broad OH/NH peak could be either an alcohol or a secondary amine, but the C-O single bond in the fingerprint region indicates an alcohol. The molecule most likely contains AN ALCOHOL and AN ALDEHYDE. Compound B -

This molecule contains an alkyne as shown by the sp C-H and the triple bond. The triple bond could be an alkyne or nitrile, but the presence of the alkyne C-H ensures it’s a terminal alkyne. There is also a carbonyl, there is no acid –OH, and no aldehyde C-H, so it’s likely a ketone or ester. There is no obvious C-O in the fingerprint region to indicate an ester, but it’s possible that the peak is masked or weak. The molecule most likely contains A TERMINAL ALKYNE and A KETONE, but may instead have an ester. Compound C – This molecule contains a primary amine as shown by the doubled –NH2 peak. It also contains a C=C and sp2 hybridized C-H indicating the presence of an alkene. None of the peaks are particularly ambiguous for other functional groups, however, the presence of sp2 C-H does help confirm the C=C is not a shifted

3000 2000

% T

rans

mitt

ance

100%

0%1500 1000

(cm-1)

Compound A

OH or NHsp3 C-H

aldehydeC-H C=O

fingerprint region

C-O

3000 20004000

% T

rans

mitt

ance

100%

0%1500 1000

(cm-1)

Compound Bfingerprint region

spC-H

sp3

C-H

CC orCN C=O


125

C=O or fingerprint peak. The molecule most likely contains AN ALKENE and A PRIMARY AMINE. Compound D –

The very broad OH peak in conjunction with a C=O peak strongly indicates a carboxylic acid rather than an alcohol or amine. There is also a C=C bond that is distinct from the fingerprint and carbonyl. No sp2 C-H are visible so they may be overwhelmed by the carboxylic acid peak or there may not be any alkene C-H. Additionally there is a triple bond, but without a formula or an alkyne C-H peak the determination between a nitrile and an internal alkyne peak isn’t obvious. This molecule most likely contains A CARBOXYLIC ACID, AN ALKENE, and either an INTERNAL ALKYNE or A NITRILE. 2)

3000 20004000

% T

rans

mitt

ance

100%

0%1500 1000

(cm-1)

Compound C

-NH2

sp2 C-H C=Csp3

C-H

fingerprint region

3000 20004000

% T

rans

mitt

ance

100%

0%1500 1000

(cm-1)

Compound D

fingerprint region

-OH(acid) sp3

C-H

CC orCN

C=OC=C C-O

Compound E - H

OO

H

OO

or Compound F -

Compound G -

OH OH N or

OO

O

or


126

Week 6 – Compound A = 8 Compound B = 8 Compound C = 7 Compound D = 5

Compound E = 8 Compound F = 5 Compound G = 4 Compound H = 9

Compound I = 9 Compound J = 5 Compound K = 2 Compound L = 6

Week 7 – 1) Coupling Compound A: a = 7, b = 4 Compound B: a = 5, b = 5 Compound C: a = 5, b = 3 Compound D: a = 1, b = 6 Compound E: a = 6, b = 2 Compound F: a = 7, b = 2

Compound G: a = 4, b = 2 Compound H: a = 3, b = 5 Compound I: a = 4, b = 2 Compound J: a = 6, b = 1 Compound K: a = 3, b = 4 Compound L: a = 5, b = 1

2) Integration

10 9 8 7 6 5 4 3 1 02(ppm)

alkene

HO

Haldehyde

alcoholOH

α to amine

NH

α to ketone

OH

arene/aromatic

H

ether OH

alkaneH

O

Br

OHO

O

OCH3 O

H

OHO OH

OOHCl


Compound E Compound F(appears as 6:1:1:3:1)

Compound G(appears as 3:1:1:1)

Compound H

Compound I Compound J Compound K(appears as 3:2)

Compound L

6

6

41 2 2 22

3

6

9

1

2

2

22

3

3

1

3

3

12

2

21

13

H

HH

H

2

22

2

1

1

1

1

1

1

213 3

122

9

6

22HH

6

32

22

22

31

31

3

3

2

2 2

2

2

22

3

3

3

3

44

3131

231

31 31


127

Week 8 –

OBr O

O

O

Cl

OH

O

O

O

OH

OH

Br

H

OOCH3

O

OH

OH

O

OO

Compound AChemical Formula: C8H17BrO

Compound HChemical Formula: C8H16O2

Compound BChemical Formula: C9H16

Compound IChemical Formula: C6H11ClO

Compound CChemical Formula: C10H20O2

Compound JChemical Formula: C10H20O2

Compound D Chemical Formula: C8H12O

Compound EChemical Formula: C8H17BrO

Compound KChemical Formula: C7H14O2

Compound FChemical Formula: C10H22O

Compound LChemical Formula: C12H26O3

Compound GChemical Formula: C10H18O2


128

Week 9 –

OH

Pd, H2

Br2, hvBr CH3COONa

DMF

O

O

NaOH,H2O

A)

O OHH3CO

CH3ONaCH3OH

PCC, CH2Cl2

or Jones Reagent

1) CH3CH2CH2CH2Li2) H3O+

H3COOH O

H3CO

B)

Br

OH

NaOH,H2O

PCC, CH2Cl2


OH H

O

C)

H

O OH

1) CH3CH2Li2) H3O+

PCC, CH2Cl2

or Jones Reagent


OH OH

D)


129

OH

Br

PCC,CH2Cl2

1) (CH3)2CHCH2MgBr

2) H3O+

HBr or PBr3

H

OBr

E)

OH

Br

H3CH2CO

OHF)

NaH,DMSO

1) CH3Li2) H3O+

O CH3CH2OH

H2SO4

H3CH2CO

OH

PCC, CH2Cl2

or Jones Reagent

H3CH2CO

OOH

Cl2 or Br2hv 1) CH3Li

2) H3O+

1) CH3COONa, DMF

2) NaOH, H2O

PCC, CH2Cl2

or Jones Reagent

Br OH O

G)

Br OCH2CH3

1) NaH, DMSO

2) CH3CH2Br

O

CH3COONa,DMF

OOH

NaOH, H2O

H)


130

OHHO Br

ONaH,

DMSO1) CH3CH2Li, ether2) H3O+

I)

J)Br Cl

PCl5

or SOCl2

CH3COONa,DMF

NaOH, H2OO OH

O

Cl

OH

NaOH,H2O

1) CH3CH2CH2CH2Li2) H3O+

PCCCH2Cl2

PCC, CH2Cl2or Jones Reagent

OH

H

O 1) CH3CH2MgBr

2) H3O+

OH

O

K)

OH

O

Pd, H2

F2,or Cl2, hv FNaOH

H2OOH

Jones Reagent

L)


131

BrI

(CH3)3COK(CH3)3COH

H2SO4, H2O HO

HI

M)

KCN, DMFHBr Br

CNN)

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