Match List I with List II
| List I | List II |
| A. \(HCl-ZnCl_2\) | I. Reducing agent |
| B. \(LiAlH_4\) | II. Grignard reagent |
| C. \(PCC\) | III. Oxidizing agent |
| D. Ethyl magnesium bromide | IV. Lucas reagent |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → A-IV, B-I, C-III, D-II
A. \( \ce{HCl-ZnCl2} \) — This matches IV. Lucas reagent. Concentrated HCl with anhydrous \( \ce{ZnCl2} \) tells 1°, 2° and 3° alcohols apart by how fast they give the turbid chloride:
\( \displaystyle \ce{R3C-OH + HCl} \xrightarrow{\ce{ZnCl2}} \ce{R3C-Cl + H2O} \)
B. \( \ce{LiAlH4} \) — This matches I. Reducing agent. It reduces aldehydes, ketones, acids and esters to alcohols:
\( \displaystyle \ce{RCHO} \xrightarrow[\text{(ii) }\ce{H3O+}]{\text{(i) }\ce{LiAlH4}} \ce{RCH2OH} \)
C. PCC — This matches III. Oxidizing agent. Pyridinium chlorochromate oxidises a primary alcohol and stops at the aldehyde:
\( \displaystyle \ce{RCH2OH} \xrightarrow{\text{PCC}} \ce{RCHO} \)
D. Ethyl magnesium bromide — This matches II. Grignard reagent. \( \ce{C2H5MgBr} \) adds to carbonyl compounds; with methanal it gives propan-1-ol:
\( \displaystyle \ce{HCHO + C2H5MgBr -> CH3CH2CH2OMgBr} \xrightarrow{\ce{H3O+}} \ce{CH3CH2CH2OH} \)