Question

On treatment of propanone with dilute \(Ba(OH)_2\), the product formed is

• A. aldol
• B. phorone
• C. propionaldehyde
• D. 4-hydroxy-4-methyl-2-pentanone

Hint:

Acetone under dilute base gives diacetone alcohol (aldol addition product). Strong base/heating gives dehydration products like mesityl oxide or phorone.

Answer 1

The Correct Option is D

Step 1: Identify reaction type.
Propanone (acetone) in presence of dilute base undergoes aldol addition (self-condensation).
Step 2: Product of aldol addition of acetone.
Two acetone molecules combine to form diacetone alcohol:
\[ (CH_3)_2CO + (CH_3)_2CO \xrightarrow{dil.\ Ba(OH)_2} (CH_3)_2C(OH)CH_2COCH_3 \] Step 3: IUPAC name.
Diacetone alcohol is:
\[ \text{4-hydroxy-4-methyl-2-pentanone} \] Final Answer: \[ \boxed{\text{4-hydroxy-4-methyl-2-pentanone}} \]