Chemistry 14D Lecture 1 Fall 2012 Final Exam Part A Solutions Page 1 Statistics: High score, average and low score will be posted on the course web site after exam grading is complete. A note about exam keys: The answers presented here may be significantly longer than expected from a student taking the exam. An exam key serves not only to reveal what was expected, but to instruct you as well. To see the final course grade cutoffs, consult the grading scale on the Chem 14D course web page.

1.

H

OFNaBH4

CH3OH

OHF

2.

H

OF

H BH3

OF H OCH3 OHF

3. (a) Molecule A reacts slower than molecule B because it lacks the electron-withdrawing inductive effect of the fluorine atom, which enhances the +C=O.

(b) Molecule C reacts slower than molecule B because it has more steric hindrance at the C=O. In addition, a

ketone has a smaller +C=O due to the electron-donating methyl group.

4. Reduction. There is an increase in the number of bonds between carbon and atoms (H) less electronegative than carbon.

5. Cl

ONaOH

H2OO

O

6.

Cl

O

OH

Cl

OHO

O

O

H OH O

O

7. (a) Molecule E reacts slower than molecule D because -OCH3 is a poorer leaving group than -Cl. In addition, the

resonance lost upon nucleophilic attack at an ester is more than the resonance stabilization lost upon nucleophilic attack at an acid chloride.

(b) Molecule F reacts slower than molecule D because -N(CH3)2 is a poorer leaving group than -Cl. In addition,

the resonance lost upon nucleophilic attack at an amide is more than the resonance stabilization lost upon nucleophilic attack at an acid chloride.

8. O

O

OCH3(CH2)16

O

(CH2)16CH3

O

H2SO4

CH3OH

O

(CH2)16CH3

CH3(CH2)16COOCH3 + HO

HO

OH

Formed in greatest amount (3 moles) Formed in lesser amount (1 mole)

Chemistry 14D Lecture 1 Fall 2012 Final Exam Part A Solutions Page 2 9.

Ph O CH3

OO

Ph OH

O

H3C OH

O

H OH2

Ph O CH3

OHO

OH2

Ph O CH3

OOHHO

H OH2

Ph O CH3

OOHHO

H OH2

Ph O CH3

OHOHHO

Ph OH

OH OH2

+

Initial protonation occurs at the PhC=O because the conjugate acid formed is more stable than the conjugate acid formed from CH3C=O protonation. Draw resonance contributors to verify.

10. (a) N(CH3)2

OO

CH3O

1. LiAlH4

2. H2ON(CH3)2

HO

LiAlH4 reduces every C=O to an alcohol, except for an amide, which is reduced to an amine.

(b) H OCH3

O O

1. C CH (excess)

2. H3O+

C

HO

CC

H

CHO

CH

C

H

(c) Br1. Mg, ether2. CO23. H3O+

OH

O

(d) Ph OPh

ONaOCH3

CH3OHPh OCH3

O

(e)

OK C N

cat. H2SO4

C

OH

N

11. Organometallic, nucleophile, Grignard reagent, polar bond, strong base. Other answers are possible. 12. Keq > 1. An amide has greater resonance stabilization than a carboxylic acid. An equilibrium favors the more

stable side. 13. Glycine's OH group must be converted into a pyrophosphate group before coupling with the next amino acid

because OH is a poor leaving group.

14. (a) Fischer esterification: OH

OH2SO4

CH3OHOCH3

O

Chemistry 14D Lecture 1 Fall 2012 Final Exam Part A Solutions Page 3

(b) Grignard reaction: CH3

O1. CH3CH2MgBr

2. H3O+CH3

CH2CH3HO

(c) Chichibabin reaction:

N

1. NaNH2, NH3 (l)

2. H2ON NH2

(d) Claisen condensation: OCH2CH3

O1. NaOCH2CH3

2. H3O+ OCH2CH3

OO

or

OCH2CH3

OOH

Other examples are possible in each case.

15. Major product: Ph H

O

+NaOH

CH3OH

Ph

O Ph

O

Ph

Mechanism:

Ph

O

H OH

Ph

O

Ph H

O

Ph

O

OPh H OCH3

Ph

O

OHPh

H OH

Ph

O

Ph

Initial deprotonation occurs at the PhCH2 because this is the most acidic proton (its removal results in the most stable enolate). NaOH does not convert all of the ketone into the enolate, so we must decided which carbonyl the enolates attacks. An aldehyde has less steric hindrance and greater +C=O than a ketone, so the ketone enolate attacks the aldehyde.

16. LDA = Lithium diisopropyl amide

17. Ph

OO 1. LDA2. H3O+

3. PhCH2BrPh

OO

or

Ph

OOH

18. The "does not form" product is not formed because the enolate formed by LDA protonated by H3O+. This gives

the starting diketone (or its enol). These do not react with PhCH2Br.

19. Problems 8, 9, 10(e), and 14(a).