NEET 2022 Chemistry Reactions of Carbonyl Compounds MCQ Question

Chapter: Aldehydes, Ketones & Carboxylic Acids

Class: 12 | Topic: Aldol Condensation | Difficulty: 🟡 Moderate

✅ Ans: C — 3

Both acetone and²⁻pentanone contain $\alpha$-hydrogens, so in dilute NaOH they can undergo self-aldol and cross-aldol condensation. On heating, the $\beta$-hydroxy ketones lose $H_2O$ to give $\alpha,\beta$-unsaturated ketones. (Neet Physics Classes)

The important possible products include:

$$CH_3COCH=C(CH_3)CH_2CH_2CH_3$$ $$CH_3CH_2CH_2C(CH_3)=CHCOCH_3$$ $$CH_3COCH=C(CH_3)CH_3$$

and other products depending on which $\alpha$-carbon forms the enolate. (Neet Physics Classes)

❌ Why Option 3 is NOT formed?

Option 3 has the connectivity corresponding to:

$$CH_3-C(CH_3)=C(CH_2CH_3)-CO-CH_3$$

For this structure, the carbonyl group is retained from acetone, while both a $CH_3$ and an $CH_2CH_3$ group appear on the adjacent alkene carbon.

That carbon skeleton cannot arise from a valid enolate–carbonyl combination of acetone and²⁻pentanone.

Therefore:

$$\boxed{\text{Option 3 is NOT formed}}$$

❌ Why other options are formed?

• Option 1 → possible cross-aldol product.
• Option 2 → possible condensation product from the two ketones.
• Option 4 → possible product from self-aldol condensation of acetone:

$$2CH_3COCH_3 \rightarrow CH_3COCH=C(CH_3)_2+H_2O$$

📌 NCERT: Aldol condensation occurs in aldehydes/ketones having at least one $\alpha$-hydrogen in the presence of dilute base, followed by dehydration on heating.

🧠 NEET Trick: Draw the enolate first, then check which carbon retains the carbonyl. If the required carbon skeleton cannot come from an enolate attacking a carbonyl carbon, that product is impossible.