Question

Ethanal \(\xrightarrow{i. (CH_3)_2CHMgBr,\ ii.\ H_2O}\) X;
Propanone \(\xrightarrow{i.\ C_2H_5MgBr,\ ii.\ H_2O}\) Y.
Consider statements: I. Ease of dehydration \(Y>X\); II. Acidic character \(X>Y\); III. Reactivity towards Lucas reagent \(X>Y\). I. Ease of dehydration is Y > X
II. Acidic character is X > Y
III. Reactivity towards Lucas reagent is X > Y

• A. I, II only
• B. I, III only
• C. II, III only
• D. I, II, III

Hint:

Lucas reagent test follows the order: \[ 3^\circ > 2^\circ > 1^\circ \] because tertiary carbocations form most easily.

Answer 1

The Correct Option is A

Step 1: Recall what Grignard reagents do.
A Grignard reagent adds to an aldehyde or ketone and, after water workup, gives an alcohol. The number of carbons on the carbonyl decides the alcohol type.
Step 2: Find X.
Ethanal (an aldehyde) reacts with isopropyl magnesium bromide to give a secondary alcohol. So $X$ is a secondary alcohol.
Step 3: Find Y.
Propanone (a ketone) reacts with ethyl magnesium bromide to give a tertiary alcohol. So $Y$ is a tertiary alcohol.
Step 4: Test statement I (dehydration).
Tertiary alcohols lose water most easily, so dehydration ease is $Y>X$. Statement I is correct.
Step 5: Test statement II (acidity).
A secondary alcohol is a little more acidic than a tertiary one because fewer electron pushing alkyl groups, so acidity is $X>Y$. Statement II is correct.
Step 6: Test statement III (Lucas).
The Lucas reagent reacts faster with tertiary alcohols, so it should be $Y>X$, making statement III wrong. So correct are I and II only. \[ \boxed{I,\ II\ only} \]