Out of the following given statements, the incorrect statements are;…

Out of the following given statements, the incorrect statements are;

(A) 2, 2, 6-Trimethylcyclohexanone forms a cyanohydrin in good yield, but the cyclohexanone doesn't.
(B) There are two amino groups in semicarbazide and both can be involved in the formation of semicarbazones.
(C) Mononitration of benzaldehyde will form m-nitrobenzaldehyde.
(D) Mononitration of acetophenone will form m-nitroacetophenone.

Choose the correct answer from the options given below:

Answer & explanation

Correct answer: option 4

The correct answer is Option (4) → (A) and (B) only

Note: None of the given options was correct in the NTA question paper. To provide additional question practice to the students, the question has been slightly modified so that the question has one clear answer.

Each statement rests on a different property of the carbonyl group. Nucleophilic addition to C=O is slowed down by bulky groups around the carbonyl carbon. A nitrogen nucleophile adds only if its lone pair is free, not tied up by resonance. And in electrophilic substitution, a C=O group attached to the ring withdraws electrons, so it deactivates the ring and directs meta.

(A) 2, 2, 6-Trimethylcyclohexanone forms a cyanohydrin in good yield, but the cyclohexanone doesn’t. — Incorrect. The reverse is true. Cyclohexanone forms a cyanohydrin in good yield, but in 2,2,6-trimethylcyclohexanone the three methyl groups next to C=O hinder the attack of \( \ce{CN^{-}} \).
(B) There are two amino groups in semicarbazide and both can be involved in the formation of semicarbazones. — Incorrect. In \( \ce{H2N-NH-CO-NH2} \) the –\( \ce{NH2} \) next to C=O has its lone pair delocalised into the carbonyl group (amide resonance). Only the free –\( \ce{NH2} \) of the hydrazine end attacks: \( \ce{R2C=O + H2N-NH-CO-NH2 -> R2C=N-NH-CO-NH2 + H2O} \).
(C) Mononitration of benzaldehyde will form m-nitrobenzaldehyde. — Correct. The –CHO group is a deactivating, meta-directing group: \( \ce{C6H5CHO} \xrightarrow[\text{273--283 K}]{\text{conc. }\ce{HNO3}\text{ / conc. }\ce{H2SO4}} m\text{-}\ce{O2NC6H4CHO} \).
(D) Mononitration of acetophenone will form m-nitroacetophenone. — Correct. The acetyl group, like –CHO, is a deactivating, meta-directing group, so m-nitroacetophenone is the main product: \( \ce{C6H5COCH3} \xrightarrow{\text{conc. }\ce{HNO3}\text{ / conc. }\ce{H2SO4}} m\text{-}\ce{O2NC6H4COCH3} \).

Out of the following given statements, the incorrect statements are;

Choose the correct answer from the options given below:

Answer & explanation

Correct answer: option 4

The correct answer is Option (4) → (A) and (B) only

(A) 2, 2, 6-Trimethylcyclohexanone forms a cyanohydrin in good yield, but the cyclohexanone doesn’t. — Incorrect. The reverse is true. Cyclohexanone forms a cyanohydrin in good yield, but in 2,2,6-trimethylcyclohexanone the three methyl groups next to C=O hinder the attack of \( \ce{CN^{-}} \).
(B) There are two amino groups in semicarbazide and both can be involved in the formation of semicarbazones. — Incorrect. In \( \ce{H2N-NH-CO-NH2} \) the –\( \ce{NH2} \) next to C=O has its lone pair delocalised into the carbonyl group (amide resonance). Only the free –\( \ce{NH2} \) of the hydrazine end attacks: \( \ce{R2C=O + H2N-NH-CO-NH2 -> R2C=N-NH-CO-NH2 + H2O} \).
(C) Mononitration of benzaldehyde will form m-nitrobenzaldehyde. — Correct. The –CHO group is a deactivating, meta-directing group: \( \ce{C6H5CHO} \xrightarrow[\text{273--283 K}]{\text{conc. }\ce{HNO3}\text{ / conc. }\ce{H2SO4}} m\text{-}\ce{O2NC6H4CHO} \).
(D) Mononitration of acetophenone will form m-nitroacetophenone. — Correct. The acetyl group, like –CHO, is a deactivating, meta-directing group, so m-nitroacetophenone is the main product: \( \ce{C6H5COCH3} \xrightarrow{\text{conc. }\ce{HNO3}\text{ / conc. }\ce{H2SO4}} m\text{-}\ce{O2NC6H4COCH3} \).