Question

To obtain 3-methylbutan-2-ol from acetaldehyde, the Grignard's reagent used is

• A. $\mathrm{CH_3MgBr}$
• B. $\mathrm{CH_3CH_2CH_2MgBr}$
• C. $\mathrm{CH_3CH_2MgBr}$
• D. $\mathrm{(CH_3)_2CHMgBr}$

Hint:

To find the components of a Grignard synthesis instantly, mentally cut the alcohol at the carbon bearing the $\mathrm{-OH}$ group ($\mathrm{C-OH}$ bond). One side represents your aldehyde/ketone, and the other side reveals the exact alkyl structure of the Grignard reagent!

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
We need to determine which Grignard reagent must react with acetaldehyde ($\mathrm{CH_3CHO}$) to synthesize the secondary alcohol 3-methylbutan-2-ol.

Step 2: Key Formula or Approach:
The reaction of an aldehyde with a Grignard reagent ($\mathrm{R-MgX}$) followed by acid hydrolysis yields a secondary alcohol: $$\mathrm{R'CHO + R-MgX \xrightarrow{ether} R'-CH(OMgX)-R \xrightarrow{H_3O^+} R'-CH(OH)-R}$$ By breaking down the structure of the target alcohol, we can identify the carbonyl-derived part and the nucleophilic alkyl group derived from the Grignard reagent.

Step 3: Detailed Explanation:
Let's look closely at the structural formula of the product, 3-methylbutan-2-ol: $$\mathrm{\overset{4}{C}H_3-\overset{3}{C}H(CH_3)-\overset{2}{C}H(OH)-\overset{1}{C}H_3}$$ The starting material provided is acetaldehyde, which contains a two-carbon fragment: $\mathrm{\overset{2}{C}H_3-\overset{1}{C}HO}$.
In the product molecule, this acetaldehyde component corresponds perfectly to the $\mathrm{-CH(OH)-CH_3}$ portion (carbons 1 and 2).
The remaining fragment attached to carbon 2 is the isopropyl group: $\mathrm{-CH(CH_3)_2}$.
Therefore, the nucleophilic alkyl group provided by the Grignard reagent must be an isopropyl group. The matching Grignard reagent is isopropyl magnesium bromide, written structurally as $\mathrm{(CH_3)_2CHMgBr}$.

Step 4: Final Answer:
The required Grignard reagent is $\mathrm{(CH_3)_2CHMgBr}$, which corresponds to option (D).