Question

Draw the products of the following reactions:

a. diethyl heptanedioate: (1) sodium ethoxide; (2) HCl

b. pentanoic acid + PBr3 + Br2, followed by hydrolysis

c. acetone + LDA/THF: (1) slow addition of ethyl acetate; (2) HCl

d. diethyl 2-ethylhexanedioate: (1) sodium ethoxide; (2) HCl

e. diethyl malonate: (1) sodium ethoxide; (2) isobutyl bromide; (3) HCl, H2O + ∆

f. acetophenone + LDA/THF: (1) slow addition of diethyl carbonate; (2) HCl

Short answer

a)

b)

c)

d)

e)

f)

Step-by-step solution

Step-by-Step Solution Step 1: Reaction of diethyl heptanedioate in the presence of sodium ethoxide and HCl (a)

In this reaction, the ethoxide ion abstracts the acidic hydrogen adjacent to a carbonyl group. In the next step, the negatively charged ion attacks the carbonyl group, and ethoxide is removed as the leaving group. The reaction is carried out as follows;

Reaction (a)

The reaction of pentanoic acid + PBr3 + Br2, followed by hydrolysis (b)

Pentanoic acid reacts with phosphorus tribromide and bromide giving alpha bromoalkonyl bromide and $\mathrm{\alpha }$-Bromo carboxylic acid without any further hydrolysis. The reaction is as follows;

Reaction (b)

The reaction of acetone + LDA/THF: (1) slow addition of ethyl acetate; (2) HCl (c)

In this reaction in the first step, LDA abstracts the acidic hydrogen of acetone which further attacks the ethyl acetate so that${}^{\text{-}}\text{OEt}$ is removed as a leaving group so that we get the desired product. The reaction is carried out as follows;

Product (c)

The reaction of diethyl 2-ethylhexanedioate in the presence of sodium ethoxide and HCl (d)

Firstly, the acidic hydrogen is abstracted by the ethoxide ion and in the next step negatively charged carbon attacks the carbonyl carbon and ethoxide is removed as a leaving group.

The product of the reaction is as follows;

Product (d)

Reaction of diethyl malonate: (1) sodium ethoxide; (2) isobutyl bromide; (3) HCl, H2O + ∆ (e)

In the first step of the reaction, acidic hydrogen diethyl malonate is abstracted by the ethoxide ion, and negatively charged carbon of diethyl malonate attacks the isobutyl bromide, and on further hydrolysis gives the desired product. The product of the reaction is as follows;

Product (e)

Reaction of acetophenone + LDA/THF: (1) slow addition of diethyl carbonate; (2) HCl (f)

First of all, LDA is a base and it removes the acidic hydrogen of acetophenone. In the next step, the negatively charged carbon attacks the diethyl carbonate and ethoxide group is removed as a leaving group and we get the desired product as follows;

Product (f)