Match the LIST-I with LIST-II.
| LIST-I Reaction of Alkyl Halides | LIST-II Major Product |
| A. CH3CH2Br + KOH (aq.) → | I. CH2=CH2 |
| B. CH3CH2Br + KOH (hot alc.) → | II. CH3CH2CN |
| C. CH3CH2Br + AgCN → | III. CH3CH2OH |
| D. CH3CH2Br + KCN → | IV. CH3CH2NC |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → A-III, B-I, C-IV, D-II
An alkyl halide can undergo nucleophilic substitution (the halogen is replaced) or elimination (HX is removed to give an alkene). The reagent and the solvent decide which one happens. With cyanides, the nature of the M–CN bond decides which atom attacks: ionic KCN releases \( \ce{CN^{-}} \), which attacks through carbon, while AgCN is mainly covalent, so only the nitrogen lone pair is free to attack.
(A) \( \ce{CH3CH2Br} \) + KOH (aq.) → — This matches (III) CH3CH2OH. In water, \( \ce{OH^{-}} \) acts as a nucleophile and substitutes Br.
\( \displaystyle \ce{CH3CH2Br + KOH(aq) -> CH3CH2OH + KBr} \)
(B) \( \ce{CH3CH2Br} \) + KOH (hot alc.) → — This matches (I) CH2=CH2. In hot ethanol, the ethoxide/hydroxide acts as a base and removes H and Br from adjacent carbons (dehydrohalogenation).
\( \displaystyle \ce{CH3CH2Br} \xrightarrow{\text{alc. }\ce{KOH},\ \Delta} \ce{CH2=CH2 + KBr + H2O} \)
(C) \( \ce{CH3CH2Br} \) + AgCN → — This matches (IV) CH3CH2NC. The covalent Ag–C bond leaves only the nitrogen free to attack, giving the isocyanide.
\( \displaystyle \ce{CH3CH2Br + AgCN -> CH3CH2NC + AgBr} \)
(D) \( \ce{CH3CH2Br} \) + KCN → — This matches (II) CH3CH2CN. The free \( \ce{CN^{-}} \) ion attacks through carbon, forming the stronger C–C bond and giving the nitrile.
\( \displaystyle \ce{CH3CH2Br + KCN -> CH3CH2CN + KBr} \)
Shortcut: (A)–(III) appears in Options (2) and (3), and (C)–(IV) then leaves only Option (2).