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Question: Identify the product 'B' in the following sequence of reactions. Propanone $\xrightarrow{Ba(OH)_2}$...

Identify the product 'B' in the following sequence of reactions.

Propanone →Ba(OH)2\xrightarrow{Ba(OH)_2} A →Δ−H2O\xrightarrow{\Delta \atop -H_2O} B

A

4-Hydroxy-4-methylpentan-2-one

B

2-methylpentan-3-one

C

2-methylpent-2-en-4-one

D

4-methylpent-3-en-2-one

Answer

4-methylpent-3-en-2-one

Explanation

Solution

  1. Step 1: Under basic conditions, acetone (propanone) undergoes aldol condensation to form diacetone alcohol:

    2 CH3COCH3→Ba(OH)24-hydroxy-4-methylpentan-2-one (Compound A)2 \, \text{CH}_3COCH_3 \xrightarrow{Ba(OH)_2} \text{4-hydroxy-4-methylpentan-2-one} \, (\text{Compound A})

  2. Step 2: On heating, diacetone alcohol loses water (dehydration) to form mesityl oxide:

    Diacetone alcohol→Δ, −H2O4-methylpent-3-en-2-one (Compound B)\text{Diacetone alcohol} \xrightarrow{\Delta,\ -H_2O} \text{4-methylpent-3-en-2-one} \, (\text{Compound B})