ACID-BASE CHEMISTRY A. Bronsted-Lowry Acid-Base Theory Acid – H+ (proton) donor Base – H+ acceptor Acid + Base ↔ Conj Base + Conj Acid (must be able to identify acids versus bases) H2SO4 + H2O ↔ HSO4

- + H3O+ HA + H2O ↔ A- + H3O+ Not all acids donate protons equally well. The position of the equilibrium is evaluated by seeing how well an “acid” donates the proton to H2O. The acidity constant, Ka = [A-][H3O+]/[HA] Strong acids favor donation due to stability of resultant anion (conjugate base).

HA + H2O ↔ A- + H3O+ Equilibrium shifts to the right! Lots of product formation – Large Ka Value. Ka values range from 1015 to 10-60. Can use logarithmic relationship to express acid strengths. pKa = -log Ka (inverse relationship!) So – Strong Acids – Large Ka – Small pKa Also – there is an inverse relationship between acids and bases and their conjugate bases and acids. H2SO4 (strong acid), HSO4

- (weak conjugate base) Compare and determine which is the stronger acid?

HF pKa = 3.2 HNO3 pKa = -1.3 Compare and determine which has the stronger conjugate base:

CH3CO2H pKa = 4.8 HCN pKa = 9.1 Could also solve for [H3O+] of weak acids (and thus pH but we will not need to do that this session.

Stability drives this process – the reaction will always proceed to produce the most stable acid/base combination.

SA + SB ↔ WA + WB Why does the equilibrium move to the more stable acid and base? Strong acids WANT to donate a proton because they form stable anions (a weak conjugate bases) that are lower in energy. Weak acids do NOT donate protons because the resultant anion would be unstable… In which direction will the following reactions be favored? a.

CH3CO2H + NaOH ↔ CH3CO2Na + H2O pKa = 4.75 pKa = 15.7 b. HF + H2O ↔ F- + H3O+ pKa = 3.2 pKa = -1.7 c. As an organic chemist, it is helpful to know acid/base strength so you know whether a reaction you are considering is even worth pursuing in the lab. Is sodium hydroxide a strong enough base to do what we need to deprotonate acetylene?

H-C≡C-H + NaOH ↔ NaC≡C-H + H2O pKa = 25 pKa = 15.7

Nope… Epic fail… waste of time… L

d. What about sodium bicarbonate? Is this a better choice?

H-C≡C-H + NaHCO3 ↔ NaC≡C-H + H2CO3

pKa = 25 pKa = 6.3 Nope…

Can we use sodium bicarbonate to deprotonate a carboxylic acid like acetic acid, CH3CO2H? e. CH3CO2H + NaHCO3 ↔ CH3CO2Na + H2CO3

pKa = 4.75 pKa = 6.3

Yes! A success! This reaction works! J

So – What factors make a proton acidic? What factors stabilize an anion? 1. Electronegativity – the more EN an atom is, the tighter the e- are held to the positively charged nucleus (same discussion seen in Ch 17, alcohols) Explain the following trend:

It’s all about the anion stability:

2. Resonance – An anion is “delocalized” through the π system. The splitting up and spreading out of an anion across two or more atoms stabilizes the anion, making the anion more stable and easier to form (i.e. the proton is more acidic). For Carboxylic Acids:

Which is more acidic?

3. Inductive Effect – movement of electron density thru bonds in response to differences in electronegativities, etc.

• EWG – electron-withdrawing groups pull electron density towards themselves (like electronegative atoms)

• EDG – electron-donating groups have an excess of electron density and push it out to other areas of molecule (such as alkyl groups)

EWG’s stabilize anions by pulling electron density away from one specific atom, spreading out electron density over many atoms. CH3CO2H pKa = 4.75

O

HNH

O

HO

O

H

pKa = 19

pKa = 9.5 pKa = 4-5

O

NH

O

O

O

R OH

O

R O

O

R O

O

OH OH

O

OR

Cl-CH2CO2H pKa = 2.81 Additional EWG’s further increase acidity even more: Cl2CHCO2H pKa = 1.29 Cl3CO2H pKa = 0.70

Consider benzoic acid and para-nitrobenzoic acid:

If benzoic acid has a pKa of 4.2 and para-nitrobenzoic acid has a pKa of 3.47, does the presence of the nitro group make the carboxylic acid proton more or less acidic? MORE… So, is a nitro group an EWG or an EDG? EWG! Proximity – the closer the EWG, the stronger the effect:

The closer the electron-withdrawing chlorine atom is to the anion that will need to be stabilized, the more stability that anion will have. What effect would an EDG have on pKa? EDG destabilize anions (adding electron density to an already electron-rich system creates an unstable system) therefore causing protons to be less acidic Consider the following: Benzoic Acid, C6H5CO2H, shown below on the left, has a pKa of 4.2 and para-methoxybenzoic acid, shown on the right, has a pKa of 4.5:

CO2H CO2H

O2N

OH

O

OH

O

OH

O

OH

O

Cl

Cl

ClpKa = 4.8 pKa = 4.5 pKa =4.0 pKa = 2.9

O

OCl

δ+δ+

O

OCl

δ+ O

OCl

δ+Cl Cl

Cl

Which acid is the stronger acid? Based on pKa values, Benzoic acid is the stronger acid (lower pKa = more acidic proton = more stable anion). The compound on the right, para-methoxybenzoic acid, is the weaker acid. The less acidic proton = less stable anion… The methoxy ether group must be destabilizing the anion making the ether an EDG. 4. Solvation – Use of a polar solvent creates a solvent cage and stabilizes an anion

Solvent effect – smaller solvents = more solvation = more stable anion = more acidic 5. Size of the atom – (for atoms in the same column of the periodic table) - the larger the atom, the less e- repulsions Acid: HF HCl HBr HI pKa: 3.45 -7 -9 -9.5 Consider phenol, shown on left, and thiophenol, shown on right:

Which is more acidic? The anion on sulfur is more stable, due to the atom’s larger size and ability to stabilize the extra electron density of an anion. The pKa of phenol is 10 and the pKa of thiophenol is 6, exactly as predicted.

CO2H CO2H

H3COpKa = 4.2 pKa = 4.5

R O

O

H OH

HOH

HO

HHO

H

HOH

HO

H

HOH

HOH

OH SH

and finally… 6. Hybridization – the larger the amount of s character, the shorter, rounder the orbitals. Shorter orbitals stabilize an anion by holding the e- closer to the nucleus. Alkanes sp3 C-H pKa = 60 Alkenes sp2 C-H pKa = 45 Alkynes sp C-H pKa = 25 B. Lewis Acids and Bases Lewis Acids – electron pair acceptor

• Requires presence of an empty orbital) Lewis Bases – electron pair donor

• Requires presence of a lone pair Commonly found in “mechanisms” where we show how electrons are moving during the course of a reaction process. Curved arrows show the direction of the movement of the electron pair – always starting from the Lewis Base and moving towards the Lewis Acid, ALWAYS NEGATIVE TO POSITIVE! Ex. H+ + H2O → BF3 + CH3OH →

OH

+ HO

H H

+ HH

O

HO

H

H H

+

+ Br

Br

Chapter 3/4: Basics of Alkanes and Cycloalkanes Alkanes are considered to be the simplest family of organic compounds because they contain only sp3C-sp3C and sp3C-H bonds. Any other bonding patterns (structural features), like the alcohols (OH) shown below, will be used to classify organic compounds and are called “functional groups”. Thus, alkanes do not have any functional groups. Functional Group: An atom or group of atoms that have a characteristic chemical behavior, where regardless of size/shape of molecule, the basic reactivity is the same.

Functional groups are made of bonds BESIDES sp3C-sp3C and sp3C-H. You need to be able to identify functional groups, on sight, by identifying significant bonds in each type. 1. Alkenes – Contain C=C

2. Alkynes – Contain CΞC

3. Aromatic rings (or arenes) – typically seen in six-membered rings, with or without groups attached

4. Alcohols – contain a hydroxyl group (OH) attached to a sp3C

HO

OH

simple alcohol

steroidal type alcohol

OH

5. Phenol – contains a hydroxyl group (OH) attached to sp2C of an aromatic ring.

6. Alkyl halides - contain sp3C-X (where X = F, Cl, Br, I)

7. Aryl Halides – contains sp2-X (where X = F, Cl, Br, I), attached to the sp2C of an aromatic ring

8. Amines – Contain –N-R and –N-H bonds. The R groups are all carbon groups and do not have to be the same type of carbon group but cannot be C=O attached to N.

9. Ethers – Contain a C-O-C sequence, but neither C can be C=O

10. Ketones – contain C=O, which is surrounded by carbon groups on both sides

11. Aldehydes – contains C=O which has at least one H attached

12. Carboxylic Acids – contains C=O with an O-H attached

OH

BrCl

F

NH2 NH

N

OO

O O O

H

O

H

O

H H

O

13. Esters - contains C=O with one –OR group attached

14. Amide – contains C=O, with one N group attached (-NH2, -NHR or -NR2).

15. Acid Anhydride – contains two C=O and C-O-C

16. Acid Halide – contains C=O and sp2C-X (usually X = Cl, Br)

17. Nitrile – contains CΞN [Note there is no hydrogen on that nitrogen atom, no N-H bond]

18. Nitro – contains N-O/N=O

Find the “different types of bonds” and name the functional groups in each:

HO

O

OH

O

OCH3

O

O

O

NH2 NH

N

O O O

O

O

O

O O O

Br

O

Cl

O

NC

N

NO

O

NO

O

Alkanes – very little reactivity (thus boring!), difficult to control what reactions that do occur (thus troublesome!): a. Combustion:

b. Halogenation:

Physical Properties: While we are discussing alkanes, these principals apply to all organic compounds Melting Point/Boiling Point – Always remember than molecules do not exist as lone entities; recall that quantity affectionately called a mole = 6.02 x 1023 molecules. The melting point or boiling point of a compound is affected by those attractive forces found between the molecules (aka: Intermolecular Forces). a. London or Dispersion Forces: fleeting attractions caused by shifts in e- clouds (always in alkanes and wherever globs of branches of carbons/hydrogens are found)

OH

H3CO

O

H3CO H

O O

Alkane CO2 + H2OheatO2

CH4 CH3X + HXlight

Cl2

b. Dipole-Dipole Forces: permanent attractions caused by electrostatic attractions of partial positive charges

c. Hydrogen bonds: permanent attraction between an acidic proton attached to an electronegative atom (N, O, F) and an electronegative atom in another molecule (N, O, F). Alcohols commonly form hydrogen bonds (as do amines, carboxylic acids and amides):

Hydrogen bonds are the strongest of the intermolecular forces due to the small size of the hydrogen atom (very close and tight intermolecular force) and thus the small size of the hydrogen bond length. General Trends: 1. As size of molecules increases, the electron cloud increases, and with more surface area, the amount of Van Der Waal’s Forces increases, thus the MP/BP increases. Which has the higher MP/BP? CH3CH3 CH3CH2CH2CH2CH3

2. For those molecules with the same molecular formula, as they become more branched, they become more spherical and rounder. With less surface area, there are less Van Der Waal’s Forces and a lower MP/BP. Which has the higher MP/BP?

O O O O

O

H O

H O

H

O

H

O

H

O

H

CH3 CH2 CH2 CH2 CH3 CH3 C CH3

CH3

CH3

Alkanes – referred to as “saturated” hydrocarbons Saturated – maximum number of H’s per C Hydrocarbon – hydrogen and carbon atoms only All carbons are sp3 hybridized General Formula: CnH2n+2 n = 1 CH4 methane n = 2 C2H6 CH3CH3 ethane n = 3 C3H8 CH3CH2CH3 propane n = 4 C4H10 CH3CH2CH2CH3 butane or CH3CH(CH3)2 isobutane Butane – straight chain or “normal” alkane Also known as “n-butane” Isobutane – branched alkane Isomers: Compounds with the same molecular formula (same number and kinds of atoms) that differ in the way the atoms are arranged. Constitutional Isomers: Compounds with the same molecular formula but different connectivity a. different C skeleton b. different functional groups c. different positions of functional groups Remember that the connectivity must be DIFFERENT. Are these the same or different?

OHO

OHOH

NOMENCLATURE: the naming of organic compounds IUPAC – International Union of Pure and Applied Chemists Alkanes are named in three parts: a. Prefix: Branches or pieces attached to main chain b. Parent: main carbon chain of molecule c. Suffix: determined by dominant functional group and defines the family that the molecule belongs to

Prefixes – Parent – Suffix SUFFIX for alkanes – “-ane” PARENT: the longest chain in the alkane molecule Together: CH4 1 carbon meth- methane CH3CH3 2 carbons eth- ethane CH3CH2CH3 3 carbons prop- propane CH3CH2CH2CH3 4 carbons but- butane 5 carbons pent- pentane 6 carbons hex- hexane 7 carbons hept- heptane 8 carbons oct- octane 9 carbons non- nonane 10 carbons dec- decane PREFIX: Branches or “substituents” that are attached to the main chain. The most common prefix in an organic molecule is called an alkyl group (generic name) and is a branch made of just carbon and hydrogen atoms, all singly bonded together. Alkyl groups are similar to alkanes but are a partial structure, the fragment that forms when you remove a H from an alkane. All alkyl groups end in “yl”

Straight chain or “normal” alkanes can be converted to normal alkyl groups (“n-alkyl” groups) by the removal of a H atom from the end carbon. Thus – propane will form n-propyl, butane will form n-butyl, etc. More often then not, the “n-“ is left off. IF you see “propyl”, “butyl”, “pentyl”, then you are looking at the straight chain alkyl group:

Pentyl, hexyl, etc continue as straight chains… Propane can form two types of alkyl groups because the hydrogen atoms on the central carbon are different from (not symmetrical to) the hydrogen atoms on the end carbons: So: Normal version (H removed from end C):

HCH

H HHCH H

methane methyl group

HCH

H CHCH

H C

ethane ethyl group

HH

H

HH

C C C H

H

H

H

H

H

H

H

propane

C C C

H

H

H

H

H

H

H

propyl group

HCH

H CHCH

H C

butane butyl group

HC

H

H

HC

HC

H

HC

H

HH

H

H

H

C C C H

H

H

H

H

H

H

H

propane

C C C

H

H

H

H

H

H

H

propyl group

Other possibility (H removed from middle C):

From Dictionary.Com: “Iso-“

“a term used to indicate an isomer of an organic compound, especially a branched isomer of a compound that normally consists of a straight chain. The isomer is characterized by a Y-shaped branch at the end of the chain that consists of two "prongs". Each prong consists of one carbon atom. “

The butyls come in different shapes and sizes because butane can be straight (n-butane) or branched (isobutane): From butane:

From isobutane:

HCH

H CHCH

H C

propane isopropyl group

HC

H

HC

HH

H

HH

H

HCH

H CHCH

H C

butane butyl group

HC

H

H

HC

HC

H

HC

H

HH

H

H

H

HCH

H CHCH

H C

butane sec-butyl group

HC

H

HC

HC

H

HC

H

HH

H

HH

H

HC

HH C

isobutane isobutyl group

CC

H

HH

H

HH H

HC

HH C

CC

H

H

H

HH H

What’s that “sec-“ and “tert-“ all about?

Classification of carbon atoms- based on number of carbon groups attached Primary – 1º

Secondary (sec-) – 2º Tertiary (tert-) – 3º Quaternary – 4º Let’s draw in the groups as zig-zag structures:

HC

HH C

isobutane tert-butyl group

CC

H

HH

H

HH H

HC

HH C

CC

HH

H

HH H

H C C C C

H

H

H

H

H

HH

HH C C C H

CH H H

H

H

H

H

sec-butyl tert-butyl

HCH

H R

HCR

H R

HCR

R R

RC

RR R

Cl

Cl

Also going to add in some halides, for the nomenclature problems. Warning: Spelling Counts!

F- Fluoro Cl- Chloro Br- Bromo I- Iodo

Pull all of the pieces together now – General Rules for IUPAC Nomenclature: Acyclic Molecules: 1. Find the longest possible chain (going around corners if necessary)

If there are two chains of the same length, the parent is defined as the one with the most branches 2. Number the chain starting from the end closest to the first branch. If the first branches are equidistant, utilize the second branches, etc.

methyl

ethyl

propyl

isopropyl

butyl

sec-butyl

isobutyl tert-butyl

CH3 CH2 CH CH CH3

CH2 CH3

CH3

3. Identify and assign position numbers to each substituent. Find the parent and number:

3-methyl 4-propyl

3-ethyl 4-methyl 7-methyl

Can combine by using extra prefixes like di-, tri-, etc.

3-methyl 4-methyl 7-methyl can be written as 3,4,7-trimethyl Sometimes you may find two substituents on the same carbon atom – still need two position numbers.

4. Write the name as a single word, Prefixes–Parent-Suffix, using hyphens between the substituents and commas between the numbers. Write the full name:

4,4-dimethylheptane

4-methyl4-methyl

4,4-dimethyl

Write the full name:

3,4,7-trimethylnonane

If there is more than one kind of substituent, list them alphabetically (but do not alphabetize using the “prefix” of a prefix, i.e. di-, tri-, sec-, tert, etc.). NOTE: “ISO” is not a prefix. Write the full name:

3-ethyl-4, 7-dimethylnonane

5. Nomenclature of a Complex branch: Branches that do not qualify as one of the simple names that you’ve learned. Don’t let the name “Complex” branch throw you off…. They aren’t that complex!

After finding the complex branch, begin by numbering the carbon of attachment as #1 and find the longest chain FROM THAT POINT.

Find any substituents on this “miniature parent” and name them. This group has a “propyl” parent and an “ethyl” on position #1:

Write the name of the complex branch and set entire complex branch in brackets or parenthesis: [1-ethylpropyl] Write the full name:

1 32

Perhaps you recall that the branch shown below is a sec-butyl group. How would you name it if you couldn’t remember the difference between sec-butyl and isobutyl? Name this sec-butyl group as a complex branch:

Name? 5-[1-methylpropyl]nonane Name the complex branch:

[1,1-dimethylpropyl] Now write the full name. Note that when complex branches are alphabetized with other prefixes, they are alphabetized based on their first letter, regardless of the use of di-, tri-, etc. So – which comes first?

[1,1-dimethylpropyl] or ethyl

Full name: 5-[1,1-dimethylpropyl]-4-ethyldecane Name the complex branch:

[1-bromopropyl]

Cycloalkanes: Repeating units of –CH2- in a ring General Formula – CnH2n

1. Determine which is the parent – the ring or any chain attached?

5-[1-ethylpropyl]decane

Br

If the ring is larger than any chain attached, the ring is the parent. If a chain exists that is longer in length than the number of carbons in the ring, the chain is the parent and the ring becomes a “cycloalkyl” group (generically).

4-cyclopropyloctane When the ring contains the most carbons, the ring is the parent and they are named as “cycloalkanes” based on the number of carbon atoms in the ring.

1-ethylcyclohexane The name can be simplified to ethylcyclohexane, since there is only one group and it HAS to be on position #1. 2. With more than one substituent, the ring is numbered so as to arrive at the lowest possible sum of the substituent position numbers. IF more than one possibility exists, THEN alphabetize to prioritize. Where is position #1?

1-bromo-2-methylcyclopentane Which is correct and why?

Name? 1-bromo-4-chloro-2-ethylcyclopentane Number the following:

CH3

Br

Br

Cl

Br

Cl

12

4 4

21

Final Topic: Cis-Trans Isomers - Stereoisomers Stereoisomers: compounds with the same molecular formula and same connectivity but a different 3-dimensional arrangement in space. Geometric Isomers: Same molecular formula, same connectivity but different 3-dimensional arrangement due to rigidity Acyclic – flexible Cyclic – rigid Acyclic chain molecules are flexible and twist and turn. Cyclic molecules are more rigid and have limited motions. Cyclic molecules, generally speaking, cannot fully rotate their C-C bonds. This results in the cyclic molecule having a “plane” (a top-side and a bottom-side, so to speak). Trans – Substituents on opposites sides of molecule

Cis – Substituents on same side of molecule.

Draw: Trans – 1- bromo-2-methylcyclobutane Cis – 1,4-dimethylcyclooctane

Br

F

CH3

F

Br

CH3

CH3

BrF

HR

RH R

R

or

RR

HH R

R

R

R