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: Understanding the Question:
The question asks us to trace a multi-step organic synthesis starting from a half-ester of glutaric acid (methyl 4-carboxybutanoate) and identify the final product R.

Step 2: Key Formula or Approach:
1. Thionyl chloride ($\text{SOCl}_2$) converts carboxylic acids to acid chlorides.
2. Organocadmium reagents ($\text{R}_2\text{Cd}$) are mild nucleophiles that react selectively with acid chlorides to yield ketones, leaving esters untouched.
3. Grignard reagents (R'MgX) react with ketones (using 1 equivalent) to yield tertiary alcohols and with esters (using 2 equivalents) to yield tertiary alcohols containing two identical R' groups.

Step 3: Detailed Explanation:

• The starting material is a half-ester, methyl 4-carboxybutanoate:
\[ \text{MeOOC-CH}_2\text{-CH}_2\text{-CH}_2\text{-COOH} \]

• Step 1: Reaction with thionyl chloride ($\text{SOCl}_2$) targets the carboxylic acid group selectively, converting it to an acid chloride to produce compound P:
\[ \text{MeOOC-CH}_2\text{-CH}_2\text{-CH}_2\text{-COCl} \]

• Step 2: Reaction of P with dimethyl cadmium ($\text{Me}_2\text{Cd}$) selectively converts the highly reactive acid chloride group into a methyl ketone group, while the less reactive ester remains completely unreacted, forming Q:
\[ \text{MeOOC-CH}_2\text{-CH}_2\text{-CH}_2\text{-CO-CH}_3 \]

• Step 3: Compound Q is treated with three equivalents of ethyl magnesium bromide (EtMgBr):
itemize

• One equivalent of EtMgBr attacks the more electrophilic ketone group ($-\text{CO-CH}_3$) to form, after workup, a tertiary alcohol:
\[ -\text{C(OH)(Me)(Et)} \]

• Two equivalents of EtMgBr are required to react with the ester group ($-\text{COOMe}$). The first equivalent replaces the methoxy group to form an ethyl ketone intermediate, and the second equivalent immediately attacks this ketone to form a tertiary alcohol containing two ethyl groups:
\[ -\text{C(OH)(Et)}_2 \]

Water workup protonates both alkoxide intermediates to yield the diol product R:
\[ \text{Et}_2\text{C(OH)-CH}_2\text{-CH}_2\text{-CH}_2\text{-C(OH)(Me)(Et)} \]
itemize

Step 4: Final Answer:
Therefore, the final product R is represented by the diol structure shown in Option A.