Question

Which one of the following is the correct statement?

• A. Acetone undergoes reaction in presence of \( \mathrm{Ba(OH)_2} \) on heating to form 4-Methylpent-3-en-2-one.
• B. Acetone reacts with \( \mathrm{NH_2NH_2/KOH} \) to form Butane.
• C. Acetophenone cannot be prepared from Benzoyl chloride and Dimethyl cadmium.
• D. Acetophenone does not undergo iodoform test.

Hint:

Aldol condensation of acetone first forms diacetone alcohol, which upon dehydration gives mesityl oxide.

Answer 1

The Correct Option is A

Understanding the Concept: This question is based on aldol condensation reactions of carbonyl compounds and characteristic reactions of ketones.
Step 1: Understanding the reaction of acetone with base. Acetone contains \(\alpha\)-hydrogen atoms and therefore undergoes aldol condensation in the presence of a dilute base such as \( \mathrm{Ba(OH)_2} \). Initially, two molecules of acetone combine to form diacetone alcohol: \[ 2CH_3COCH_3 \xrightarrow{\mathrm{Ba(OH)_2}} CH_3COCH_2C(OH)(CH_3)_2 \] The product formed is 4-hydroxy-4-methylpentan-2-one.
Step 2: Dehydration on heating. On heating, the aldol product loses one molecule of water and forms an \(\alpha,\beta\)-unsaturated ketone called mesityl oxide. \[ CH_3COCH_2C(OH)(CH_3)_2 \xrightarrow{\Delta} CH_3COCH=C(CH_3)_2 \] The IUPAC name of this compound is: \[ \text{4-Methylpent-3-en-2-one} \] Thus, statement (1) is correct.
Step 3: Checking the other options. Acetone with \( \mathrm{NH_2NH_2/KOH} \) undergoes Wolff–Kishner reduction and forms propane, not butane. Acetophenone \emph{can} be prepared from benzoyl chloride using dimethyl cadmium. Acetophenone contains the \( \mathrm{CH_3CO-} \) group and therefore gives a positive iodoform test. Hence, all remaining statements are incorrect.