Match List I with List II
| LIST I | LIST II |
| A. \(Benzophenone \longrightarrow Diphenylmethane\) | I. \(LiAlH_4\) |
| B. \(Benzaldehyde \longrightarrow 1-Phenylethanol\) | II. \(DIBAL-H\) |
| C. \(Cyclohexanone \longrightarrow Cyclohexanol\) | III. \(NH_2-NH_2, KOH\) |
| D. \(Methylbenzoate \longrightarrow Benzaldehyde\) | IV. \(CH_3MgBr, H_2O\) |
Choose the correct answers from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → A-III, B-IV, C-I, D-II
Each reagent does one kind of change to a carbonyl compound: Wolff–Kishner reduction turns C=O into \( \ce{CH2} \); a Grignard reagent adds a carbon and gives an alcohol; LiAlH4 reduces C=O to CH–OH; DIBAL-H reduces an ester only as far as the aldehyde.
(A) Benzophenone → Diphenylmethane — This matches (III) NH2–NH2, KOH. The C=O group becomes \( \ce{CH2} \): the hydrazone formed with hydrazine is heated with KOH in ethylene glycol (Wolff–Kishner reduction).
\( \ce{(C6H5)2C=O} \xrightarrow[\ce{KOH},\ \Delta]{\ce{NH2NH2}} \ce{(C6H5)2CH2 + N2 + H2O} \)
(B) Benzaldehyde → 1-Phenylethanol — This matches (IV) CH3MgBr, H2O. The Grignard reagent adds a methyl group to the C=O carbon; hydrolysis gives the secondary alcohol.
\( \ce{C6H5CHO} \xrightarrow[\text{(ii) }\ce{H2O}]{\text{(i) }\ce{CH3MgBr}} \ce{C6H5CH(OH)CH3} \)
(C) Cyclohexanone → Cyclohexanol — This matches (I) LiAlH4. Lithium aluminium hydride reduces the ketone to a secondary alcohol.
\( \ce{C6H10O} \xrightarrow[\text{(ii) }\ce{H2O}]{\text{(i) }\ce{LiAlH4}} \ce{C6H11OH} \)
(D) Methylbenzoate → Benzaldehyde — This matches (II) DIBAL-H. Diisobutylaluminium hydride reduces an ester to the aldehyde and stops there.
\( \ce{C6H5COOCH3} \xrightarrow[\text{(ii) }\ce{H2O}]{\text{(i) DIBAL-H}} \ce{C6H5CHO + CH3OH} \)