Question

Compounds A and B, having the same molecular formula \( C_4H_8O \), react separately with \( CH_3MgBr \), followed by reaction with dil. HCl to form compounds X and Y respectively. Compound Y undergoes acidic dehydration in the presence of Conc. \( H_2SO_4 \) much more readily than X. Compound Y also reacts with Lucas reagent, much more readily than X, with the appearance of turbidity. Identify X and Y.

• A. X = Pentan - 1 - ol Y = Pentan - 2 - ol
• B. X = Butan - 2 - ol Y = 2-Methylpropan - 2 - ol
• C. X = Pentan - 1 - ol Y = 2 - Methylpentan - 2 - ol
• D. X = Pentan - 2 - ol Y = 2 - Methylbutan - 2 - ol

Hint:

In reactions with Lucas reagent, tertiary alcohols are the fastest to react, followed by secondary alcohols, and primary alcohols react the slowest. This helps in distinguishing alcohol types based on reactivity.

Answer 1

The Correct Option is D

Given that compounds A and B have the same molecular formula, both will likely be alcohols that react with CH\(_3\)MgBr to form alkylated products. Since compound Y undergoes acidic dehydration more readily, this suggests it has a structure that allows for easier elimination of water. When reacting with Lucas reagent, compound Y also reacts more readily, indicating it is a more reactive alcohol. 

The Lucas test typically distinguishes between primary, secondary, and tertiary alcohols, with tertiary alcohols reacting most rapidly. Thus, Y must be a more substituted alcohol, which fits the structure of 2-Methylbutan-2-ol, a secondary alcohol. On the other hand, X is a primary alcohol, pentan-2-ol, which reacts less readily under the same conditions. 

Thus, the correct identification for X is Pentan-2-ol and for Y is 2-Methylbutan-2-ol.