Question

Identify compound A from following reaction.
$$\text{A} + \text{C}_2\text{H}_5\text{MgBr} \xrightarrow{\text{dry ether}} \text{B} \xrightarrow{\text{H}_3\text{O}^+} \text{3-methylpentan-3-ol}$$

• A. Butanal
• B. Propanone
• C. Propanal
• D. Butanone

Hint:

Count the total carbons to save time! The product 3-methylpentan-3-ol contains exactly 6 carbons (pentane = 5, plus 1 from methyl). The Grignard reagent provides 2 carbons (ethyl). Therefore, starting compound A must contain exactly $6 - 2 = 4$ carbons. Since it must be a ketone, Butanone is the only logical answer!

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
We are given a Grignard reaction sequence where an unknown carbonyl compound A reacts with ethylmagnesium bromide ($\text{C}_2\text{H}_5\text{MgBr}$) in dry ether, followed by acid hydrolysis to yield a tertiary alcohol, 3-methylpentan-3-ol. We need to find the structural identity of A.

Step 2: Key Formula or Approach:
The reaction of a Grignard reagent (R'-MgX) with a carbonyl compound followed by hydrolysis yields an alcohol:
Formaldehyde gives a $1^\circ$ alcohol.
Higher aldehydes give a $2^\circ$ alcohol.
Ketones give a $3^\circ$ alcohol.
Since the product 3-methylpentan-3-ol is a tertiary ($3^\circ$) alcohol, compound A must be a ketone.

Step 3: Detailed Explanation:
Let's reverse-engineer the structure of the product, 3-methylpentan-3-ol:
$$\begin{array}{ccccccc} & & \text{CH}_3 & & \\ & & | & & \\ \text{CH}_3\text{-CH}_2 & - & \text{C} & - & \text{CH}_2\text{-CH}_3 \\ & & | & & \\ & & \text{OH} & & \end{array}$$ The Grignard reagent used is ethylmagnesium bromide, which introduces one ethyl group ($-\text{CH}_2\text{-CH}_3$) to the central carbon.
If we disconnect this ethyl group from the tertiary carbon atom of 3-methylpentan-3-ol, the remaining molecular fragment attached to the oxygen atom is:
$$\text{CH}_3\text{-CH}_2\text{-CO-CH}_3$$ Converting this remnant fragment back into its parent carbonyl configuration gives a 4-carbon chain ketone. Let's trace it out:
Longest chain has 4 carbons $\rightarrow$ butane.
Containing a ketone group at position 2 $\rightarrow$

butanone (ethyl methyl ketone).
Let's verify forward: Butanone ($\text{CH}_3\text{-CH}_2\text{-CO-CH}_3$) $+$ Ethylmagnesium bromide ($\text{C}_2\text{H}_5\text{MgBr}$) creates an adduct that hydrolyzes exactly into 3-methylpentan-3-ol. This perfectly matches option (D).

Step 4: Final Answer:
Compound A is Butanone, which corresponds to option (D).