Question

Consider the following compound.

The compound reacts with $\text{SOCl}_2$ to produce P. P reacts with dimethyl cadmium to produce Q. Q upon reaction with ethyl magnesium bromide (3 equiv) followed by water workup produces R. The structure of R is:

• A. Structure with one tertiary alcohol containing two ethyl groups and another containing one methyl and one ethyl group.
• B. Structure with one secondary alcohol and one tertiary alcohol.
• C. Structure with two tertiary alcohols containing only methyl groups.
• D. Structure with two tertiary alcohols containing only ethyl groups.

Hint:

Organocadmium and organocopper (Gilman) reagents are less nucleophilic than Grignard reagents and will not react with esters, allowing the selective conversion of acid chlorides to ketones.

Answer 1

The Correct Option is A

Step 1: Read the starting material.
We start from a half ester of glutaric acid, that is a five carbon chain with a free $-\text{COOH}$ at one end and a $-\text{CO}_2\text{Me}$ ester at the other end. We will follow each reagent step by step and watch only the two end groups.

Step 2: React with thionyl chloride.
$\text{SOCl}_2$ attacks the free carboxylic acid and turns it into an acid chloride ($-\text{COCl}$). The ester end is untouched. So intermediate P has one $-\text{COCl}$ and one $-\text{CO}_2\text{Me}$.

Step 3: React with dimethyl cadmium.
Organocadmium reagents are gentle. They react with the reactive acid chloride but leave the less reactive ester alone. So $\text{Me}_2\text{Cd}$ converts the $-\text{COCl}$ into a methyl ketone ($-\text{COCH}_3$), while the ester stays as it is. That gives intermediate Q with one ketone and one ester.

Step 4: Add ethyl magnesium bromide.
Grignard reagents are strong and attack both the ketone and the ester. With the ketone, one ethyl group adds to give a tertiary alcohol that carries one methyl and one ethyl group.

Step 5: Handle the ester with Grignard.
An ester reacts with two equivalents of the Grignard reagent, so two ethyl groups add to that carbon. After water workup this end becomes a tertiary alcohol carrying two ethyl groups. This is why three equivalents of EtMgBr are used in total.

Step 6: Write product R.
The molecule now has two tertiary alcohols: one bearing one methyl plus one ethyl, the other bearing two ethyl groups.

\[ \boxed{\text{One tertiary alcohol (methyl + ethyl) and one tertiary alcohol (two ethyls)}} \]