block 1 _ introduction to organic chemistry

8/12/2019 Block 1 _ Introduction to Organic Chemistry

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Block 1 Introduction to Organic chemistry

This topic forms the basis for which the three blocks later on organics (includingspectroscopy) will be built upon, thus it is absolutely essential that studentsfamiliarize and consolidate upon their knowledge of the main concepts describedbelow. This set of notes will include a mixture of content from your course guide,lecture slides as well as drawing relevant content from the textbookFundamentals of Organic Chemistry 6 th edition by McMurry and Simanek. Theaim of this is to provide a whole picture and inspire interest for those passionateabout chemistry. However, vital ideas of which you will definitely need to reviseand drill in firmly into your heads will be explicit pointed out.

Organic chemicals are Molecules

Organic chemicals are predominantly molecules; exceptions include compounds

such as the carboxylate ion or ethoxide ion, which do contain ionic characters.Thus we begin our tutorial by exploring the important features to bear in mindwhen describing a particular molecule.

Features of a molecule:

Molecules are formed by covalent bonds, which themselves are the sharing ofelectrons. The aim of this alliance between individual atoms in achieving themolecular state is to gain an outer shell (full valence). When we consider howmany bonds an atom can make, we need to firstly consider how many outerelectrons it already has, and subsequently deduce how many further electrons it

will need.

Lets consider the reaction below:

The nitrogen atom in ammonia needs three more electrons as it already has fiveto start with and thus it forms an alliance with 3 hydrogen atoms via singlebonds, which means that each atom is donating one electron to the bond. Doublebonds are thus the investment of two electrons each from each atom, Triplebonds: three electrons from each. *Why cant a quadruple bond form?


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Notice however, N also has a lone pair in its valence; this simple means a pair ofelectrons not involved in a particular bond. This lone pair will have vital roles inthe organic chemistry reactions to come.

The combination of NH3 and H+ is a unique one. This is an example of aformation of a new covalent bond, but whats special about it? Well, if you lookclosely, NH3 is providing both electrons to the new bond while H+ has nothing todonate. This type of bond is called a dative bond, in which one atom providesboth electrons to the newly formed covalent bond. This type of relationship willalso be quite prevalent in reactions to come.

Thus in summary, when inspecting a particular molecule we should be observantof features such as:

Covalent bond; and whether is it a single, double or a triple bond. Lone pairs

Presence of dative bonds

New concept:There are molecules that have an unpaired electron in their outer shells andthese are called free radicals. As you can imagine, due to the instability of thelone electron, these molecules will be extremely reactive. They are presented inthe manner below , and play a vital role in the Free Radical Substitution reactiondiscussed in the next tutorial.

This is an example of the formation of chlorine free radicals.

Covalent bondsSo if covalent bonds make up molecules, what are some features of a bond weneed to be aware of?

Bond energy and bond length Bond polarity Bond geometry

Bond energy and length are inversely proportional. These terms and conceptspertinent will not be a part of CHEM110.

Bond polarityBond polarity is the result of the difference in electronegativity between theatoms involved in the bond. Electronegativity is a feature possessed by all atomsto different extents, and describes the ability of the bonded atom to pull thebonded electrons towards itself. Thus you can imagine, the higher the

electronegativity of an atom, the more potent it is at pulling the electrons


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towards itself and thus making its surrounding region slightly more negativethan the neutral state we expect it to be.

The rule of electronegativity is based on atomic structure and presents itself inthe way below:

Thus atoms, N, O and F exhibit the highest amounts of electronegativity and willshow delta ( -) in any bond that it is involved in. *Whats the exception?

Why bond polarity?

The appreciation of polarity is crucial for the discussion of two further concepts:boiling point and solubility.

Boiling point:The determinant of variation in boiling points between different organicmolecules is the type of intermolecular force present. Note although the type ofintermolecular force present is determined by the type of covalent bond present;the term intermolecular is specific in describing the forces between moleculesand not within. It is essential that you acknowledge that in boiling a substance,we are not interrupting covalent bonds (intra-molecular bonds), you arebreaking down the intermolecular forces.

Important learning point:A molecule with polar bonds does not necessarily have to be a polar molecule,why?


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Polarity of molecule Type of intermolecular forcesNon-polar:1.No polar bonds present2. Symmetrical molecule

Temporary dipole forces

Polar Permanent dipole forcesPresence of Hydrogen bonds Hydrogen bonds

When we consider molecules with purely temporary dipole-dipole forcespresent, the bigger the molecular mass the higher the boiling point. Thus butanewill have a higher boiling point than ethane. *How do you think the boiling pointof pentane will compare to 2-methylbutane?

However if we consider two molecules with the same molecular mass, butdifferent types of intermolecular forces present, the rank of the strength ofintermolecular forces will follow that:

H-bonds > Permanent dipole-dipole > Temporary dipole dipole

The reasoning of this is quite intuitive. The absence of a permanently polarmolecule will lead to a weakened intermolecular attraction as electro-staticitybetween the (-) and (+) dipoles between neighboring molecules will be short-lived. Permanent dipole-dipole forces will provide a more concreteintermolecular adhesion. Finally, H-bonds which is strictly defined as the:

The presence of Hydrogen bonded to one of the following atoms: N, O, F.

Will provide the most intensive intermolecular force. One can consider H-bondas an extreme type of permanent dipole-dipole as from the table above, you cansee that N, O and F are the most electronegative atoms in the periodic table andthus such molecules will establish a very prominent permanent dipole. *Despiteso, how do you think the presence of an ionic bond will compare to H2O (H-bond) in terms the boiling point?

Specific examples are discussed in the table on the following page.

Solubility:

The concept of solubility is really founded upon this fundamental concept that:

Like dissolves Like.

Solvents that contain similar intermolecular forces as the solute will readilyaccept that solute and vice versa. This information is encrypted in the tablebelow. *Why do some of these molecules, despite having compatible bonds towater, only dissolve with carbon chains up to three or seven?


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Compound type Relativepolarity

Reasoning Watersolubility

Hydrocarbons (CH) 1 Nonpolar molecule andthus only exhibittemporary dipoleforces

Insoluble

Halides (CX) 2 Permanent dipolepresent but quite weak

Ethers (COC) 3 Permanent dipoles butstronger than oneabove due to presenceof oxygen atom

Esters/aldehydes/ketones(C=O)

4 Permanent dipole forcepresent, however a C=Obond exhibits a

stronger bond polaritythan C-O from ethers.This is due to the factthat there are moreelectrons between thetwo atoms and thusallows distortion of amore rich electron fieldby the oxygen atom.

Up to threecarbon in chain

Amides (COHN) 5 Permanent dipole + H-bond

Up to sevencarbons inchainAmines (NH) 6 H-bonds present

Alcohols (OH) 7 H-bond present also,but stronger than theone above due to factthat oxygen is moreelectronegative thannitrogen.

Carboxylic acids (COOH) 8 H-bond involvingoxygen + permanentdipole (C=O)

Carboxylate and aminesalts (COO-)/(NH3+) 8 These are on the top ofthe food chain, due thepresence of complete (-)/(+) charges on theoxygen and nitrogencompared to moleculesabove which onlyexhibit delta (-)/(+) s.


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Bond geometry:

Hybridizationnotation

Carbon configuration Shape Bondangle

Sp3 Tetrahedral 109.5

Sp2 Trigonalplanar

120

sp linear 180

*Can you think of another bonding configuration of carbon that will also providethe linear shape?

Key learning point:

This table is very crucial, and remembering the information contained withinwill be key to attaining some easy marks. The background and derivation ofhybridization notations will not be examined and thus the majority ofinformation below will not be examinable but merely provide the overall picturefor those with more interest. Important things are highlighted and need to belearned as normally.

The theory of hybridization:

Lets begin by looking at the electron configuration of c arbon: 1s2 2s2 3px1 3py13pz0. Immediately we realize that there is no way carbon can form four bonds ifit only has two single electrons available in its electron configuration. Thus theachievement of four equivalent bonds in methane and so on will rely on thereasoning provided by the hybridization theory.

1. Electron promotion


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In this step, the atom promotes an electron from the 2s orbital to the empty2py orbital, and now we have four single electrons occupying four individualorbitals.

2. Hybridization

i)sp3The atom raises the energy level of the 2s orbital and lowers the energy level ofthe three 2p orbitals so that they are all at the same energy state. They atom hasHYBRIDIZED one s orbital and three p orbitals and this is why it is given the termsp3. Consequently, the four bonds this carbon atom will make with the usage ofthese four orbitals will be equivalent in energy and allow the formation of atetrahedral shape.

Key learning point:The formation of covalent bonds has always been known to students as thesharing of electrons; this concept is modified in this course as the overlapping oforbitals. Thus in this context, after the process of hybridization, the orbitals withthe electrons that they possess will go on to OVERLAP with another orbital. Inmethane, this would be hydrogen. Through overlap, the newly formed bond willcontain an electron from each atom, thus the idea of sharing remains the same.When hybridized orbitals overlap, the orientation under which it happens can bein two ways:

1. If the orbitals overlap head to head they will form a sigma bond2. if the orbitals overlap side to side they will form a pi bond.

In this scenario of a sp3 carbon, the carbon will have four sigma bonds. This isbecause the inter-nuclear axis between carbon and the four hydrogens is thepreferred site of orbital overlap and thus overlap will always be head to headwhere possible.

ii)sp2

In this case, the atom will hybridize one s orbital and 2 p orbitals, leavinganother p orbital un-hybridized. The hybridized orbitals will undergo overlap asabove, forming sigma bonds. At this instant, the remaining p orbital will beforced form another type of bond, that does not take place along the internuclearaxis as it has already been taken by an existing sigma bond. It can only form aside-to-side orbital overlap up and below the pre-existing sigma bond; a pi bond.

Key learning point:A sigma bond + a pi bond = a double bond.

iii)

In this case, the atom will hybridize one s orbital and one p orbital, leaving two p

orbitals un-hybridized. The hybridized orbitals will form sigma bonds and the p


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orbitals will form two pi bonds, one above and below the pre-existing sigmabond, and the other in front of and behind the sigma bond.

Key learning point:

One sigma bond + two pi bonds = a triple bond

Questions:1. why is the carbon in =C= also considered sp?2. why cant a carbon atom have a quadriple bond?

Nomenclature, DBE and skeletal formula:

In terms of the different functional groups that this course will cover, we willtake a slightly different approach in discussing each functional group specificallyand in detail when we get to it instead of introducing them all at the same time.Students should definitely revise the organics part of their high-schoolcurriculum in preparation for block one, because even though it wont includespecific reactions, it will include nomenclature and formulae presentation.

In terms of nomenclature, a quick debrief of the principles that you should stick

to will be:1. Identify longest carbon chain.2. Identify side groups.3. Identify the side group with the highest priority, and number carbons

from the side which this group is closes to.4. Attach correct prefixes and suffixes according to side groups present.

DBE:DBE stands for double bond equivalents. One can think of it literally as the totalnumber of double bonds and structures equivalent in character to double bondswhich can be considered to be a triple bond or a ring. We will explore thisconcept in more detail below.

The concept of DBE is really to deduce the level and pattern of saturation.Whether it is the presence of a double bond or a triple bond or a ring, it willreduce the number of hydrogens present and thus alter the molecular formulasignificantly, thus they are collectively considered to be DBE. We can find out theDBE in two ways:

By inspection:

This method of investigating DBE requires that we count the number of doublebonds, triple bonds and rings in the displayed/skeletal formula. Note:


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1 double bond = 1 DBE1 ring = 1 DBE1 triple bond = 2DBE

Key learning point: we can only find out DBE by inspection if the molecule ispresented to us in its displayed or skeletal formula (maybe also structural),because only these presentations will specifically show the positions andnumbers of the structures relevant.

By Calculation:

DBE = (2n4+n3-n1+2)

The subscripted numbers imply the number of atoms, which can form the statednumber of maximum bonds. Thus n4 represents all the numbers of carbonatoms, n3 represent the number of nitrogen atoms and n1 represent the totalnumbers of hydrogen atoms and halogen atoms.

DBE provides a good tool for getting a quick impression of the type of functionalgroups that might be present, and it will be very handy in the topic ofspectroscopy.

Skeletal formula:

This is a crucial component of your assessment in CHEM110, it requires you to

extend your knowledge of the variety of formulae used in organic chemistry.Organic molecules in reality are big and complex, therefore this form of formulaprovides a more compact and dense way of illustrating a molecule.

The process of translation from any other formular to a skeletal formula followsthat:

1. Any joint or end of a line represents the presence of a carbon atom.2. Any carbon atoms or hydrogen atoms directly joined to a carbon are not

shown.

Handy note: this seems like a short set of rules, but it is really hard not to makemistakes, thus plentiful of exercises will be really crucial.


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Isomerism* Insert tree of isomerism here


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Constitutional isomerism

A fancy term of structural isomerism that you learned in highschool. It is definedto be molecules with the same molecular formula but different structuralformula or more specifically different atom-to-atom bonding sequence.

key learning point:this means that consititutional isomers can be functionally the same or different,i.e. C3H6O can be an aldehyde or a ketone.

Stereoisomerism

This encompasses all categories of isomerism in which the isomers have thesame molecular and structural formula but have different spatial arrangements.

Configurational isomerismTo interconvert between the isomers of this type, one has to physically break abond and reattach it at another position.

Geometric isomerism

These isomers arise due to restricted rotation around a bond. The most commonexample that students are aware of is alkenes, but the structure not to be

overlooked is rings that also restrict complete bond rotation.

Students are expected to build upon their previous interactions with the Cis-Trans system and work with the E/Z system. This system provides a morecomprehensive way of dealing with this class of isomers as it allows the presenceof four different substituents on both carbons involved in the double bond. Thesame applies to rings.

The principles of assigning priorities (discussed on next page) are as follows:1. Consider the two groups on the same carbon, and prioritize them as

either Hi or Lo.2. If both His are on the same side, it will be considered Z, if they are on the

opposite side, it will be considered E.

Key learning point: the two groups on the same carbon involved in the doublebond must be different.

Handy note: E is for Deefferent.


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Enantiomers:

These are isomers, which are mirror images that arise from a carbonsimultaneously joined to four different groups. This carbon is known as a chiralcarbon or a stereocentre, and the two isomers might seem similar but they areactually different as mirror images are non-superimposable.

One enantiomer will rotate plane-polarized light to the left, this is known aslaevorotatory, the other will rotate plane-polarized light to the right and this isknown as dextrorotatory. A mixture containing exactly 50:50 of each is known asa racemic; it will not rotate plane-polarized light. Distinguishing enantiomers isan important process especially in the drug industry as one can be much morepotent than another.

One is expected to work with the R/S system when considering a stereocentre;the principles are as follows:

1. Prioritize the four groups. ( Discussed below)2. Have the lowest group turning directly away from you and thus the

remaining three should point at you like a steering wheel.3. If the priorities rank down in a clockwise manner, it is considered R, if it is

anti-clockwise, it will be considered S.

Handy note: R for turning to the Right i.e. clockwise.

Assigning priorities:1. The higher the atomic number of the immediate atom bonded to the C, the

higher the priority.2. If the atomic number of the immediate atoms discussed above is the

same, then one should keep inspecting downstream until a cleardifference can be established.

3. Atoms bonded by a double bond or a triple bond can be considered asbeing bonded to an equivalent number of single bonds.

Conformational isomerism:

Isomers arise due the rotation around single bonds.

Diastereomers

Stereoisomers, which are, not mirror images. This is a perplexing topic and oneshouldnt spend too much time reading upon this. However one should be awareof its definition and what that implies in context: is covers geometric and

conformational isomers in theory.


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block 1 _ introduction to organic chemistry

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