YOUR STUDENT NUMBER:

Q2 continued overleaf

Year 2 Seminars, 06524, Semester 2, 2004 Bifunctional Chemistry Test ANSWERS

Q1 (a) Label all of the acidic hydrogen atoms on the molecules A - E shown.

OH

O

O

O

O OO

O

O

O

O OA B C

D E

H H

HH

H

HH

HHH

HH

H H H H H H

HH

H

HH

H

Q1 (b) Considering the most acidic hydrogen(s) of each molecule only, place the

molecules in order of acidity starting with the most acidic. A (CO2H) > E (CH2) > B (CH2) > C > D Q2 Consider the scheme below then answer the questions:

OEt

O OX

2 eq. NaOEt

Br-(CH2)4-Br

i) aq. NaOH r.t.

ii) dilute H+ heat

O

(a) Identify X and give a mechanism for its synthesis.

OEt

O O

X

Br-CH2CH2CH2CH2-Br

HHOEt

O O

HOEt

O O

H

OEt

O O

H

Br

OEt

O O

Br

OEt

O O

resonance stablilized

OEt

EtO

OEt

O O

H

intramolecular reaction faster than intermolecular, especially if making 5-ring

END

(b) Give a mechanism for conversion of X into acetylcyclopentane (methyl cyclopentyl ketone).

Xaq. NaOH r.t.

ONa

O

OCH3

OH

O

OCH3

O

O

H3CO H

C

O

O

CH3

OH

CH3

OC

HCl (aq) heat

tautomerism

(c) What would happen if acetone were allowed to react with 2 eq. NaOEt and 2 eq.

Br(CH2)4Br? Acetone (propanone) has hydrogen atoms considerably less acidic than ethyl acetoacetate (see Q2a). Therefore, in reaction with NaOEt (weak base) there is always ketone present in the equilibrium mixture as well as enolate. An aldol reaction is therefore one possible side reaction. Also the second deprotonation could take place at the same carbon as the first, or on the other side of the C=O. Two moles of dibromide also complicates matters and so dimers/polymers and dialkylated / and or cyclic products can form. Overall a right old mixture would be produced!