Match List-I with List-II.
| List-I (Reaction) | List-II (Main organic product) |
|---|---|
| (A) \( \ce{(C6H5CH2)2Cd + CH3COCl} \) | (I) Benzophenone |
| (B) \( \ce{CH3CN + C6H5MgBr} \), followed by hydrolysis | (II) Propiophenone |
| (C) Benzene + propanoyl chloride, in the presence of anhydrous AlCl3 | (III) Acetophenone |
| (D) Benzene + benzoyl chloride, in the presence of anhydrous AlCl3 | (IV) 1-Phenylpropan-2-one |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → A-IV, B-III, C-II, D-I
All four reactions make ketones. In each, find which group comes from which reagent: the acyl group (R–CO–) and the second group joined to the carbonyl carbon.
(A) \( \ce{(C6H5CH2)2Cd + CH3COCl} \) — This matches (IV) 1-Phenylpropan-2-one. A dialkylcadmium converts an acyl chloride into a ketone: the acetyl group comes from the acyl chloride and the benzyl group from the cadmium compound.
\( \displaystyle \ce{(C6H5CH2)2Cd + 2CH3COCl -> 2C6H5CH2COCH3 + CdCl2} \)
(B) \( \ce{CH3CN + C6H5MgBr} \), followed by hydrolysis — This matches (III) Acetophenone. The Grignard reagent adds to the C≡N group; hydrolysis of the addition product gives a ketone with the methyl group of the nitrile and the phenyl group of the Grignard reagent.
\( \displaystyle \ce{CH3C#N + C6H5MgBr} \xrightarrow{\text{ether}} \ce{CH3C(=NMgBr)C6H5} \xrightarrow{\ce{H3O+}} \ce{CH3COC6H5} \)
(C) Benzene + propanoyl chloride, in the presence of anhydrous AlCl3 — This matches (II) Propiophenone. In Friedel-Crafts acylation the acyl group of the acid chloride, here CH3CH2CO–, enters the ring.
\( \displaystyle \ce{C6H6 + CH3CH2COCl} \xrightarrow{\text{anhyd. }\ce{AlCl3}} \ce{C6H5COCH2CH3 + HCl} \)
(D) Benzene + benzoyl chloride, in the presence of anhydrous AlCl3 — This matches (I) Benzophenone. The same reaction with the benzoyl group puts C6H5CO– on the ring, giving a ketone with two phenyl groups.
\( \displaystyle \ce{C6H6 + C6H5COCl} \xrightarrow{\text{anhyd. }\ce{AlCl3}} \ce{C6H5COC6H5 + HCl} \)
NCERT (2026-27 edition):
8.2.3 Preparation of Ketones – “1. From acyl chlorides
Treatment of acyl chlorides with dialkylcadmium, prepared by the reaction of cadmium chloride with Grignard reagent, gives ketones.”
“2. From nitriles
Treating a nitrile with Grignard reagent followed by hydrolysis yields a ketone.”
“3. From benzene or substituted benzenes
When benzene or substituted benzene is treated with acid chloride in the presence of anhydrous aluminium chloride, it affords the corresponding ketone. This reaction is known as Friedel-Crafts acylation reaction.”