Q.Which of the following reactions is/are correct? (Two or more options may be correct.)

Reaction (i) — correct

CH3Cl + 2 NH3 → CH3NH2 + NH4Cl. Ammonia acts as a nucleophile (ammonolysis of an alkyl halide); the second molecule of NH3 neutralises the HCl formed. Correct.

Reaction (ii) — incorrect

(CH3)3C–Cl (a tertiary halide) with AQUEOUS KOH undergoes mainly nucleophilic substitution (SN1) to give tert-butyl alcohol, (CH3)3C–OH. Elimination to 2-methylprop-1-ene is favoured by ALCOHOLIC KOH (a stronger, less solvated base). As written (aqueous KOH → alkene), the reaction is not correct.

Reaction (iii) — correct

Cyclohexyl chloride + alcoholic KOH → cyclohexene + KCl + H2O. Alcoholic KOH promotes dehydrohalogenation (E2 elimination). Correct.

Reaction (iv) — correct …

Method: Evaluating a Set of Named Reactions Independently by Reagent Selectivity

Core Concept

When several unrelated reactions are presented together for verification, each must be checked on its own against the reagent-dependent outcome it should give - common traps include confusing aqueous vs. alcoholic base (substitution vs. elimination) and forgetting that aliphatic diazonium salts are far too unstable to isolate, unlike their aromatic counterparts.

Steps

1. Treat every numbered reaction as a separate, independent claim - do not assume the whole set shares one theme.
2. For a reaction with NH3 attacking an alkyl halide, identify it as nucleophilic substitution (ammonolysis); check the drawn amine and ammonium-salt by-product against the stoichiometry.
3. For a reaction with a tertiary alkyl halide and a hydroxide/alkoxide base, check EXACTLY which solvent system is specified: aqueous KOH favours substitution (SN1, giving the alcohol), while alcoholic KOH favours elimination (E1/E2, giving the alkene) - verify the stated product actually matches the stated solvent.
4. For a reaction with a secondary cyclic halide and alcoholic KOH, identify it as dehydrohalogenation (elimination) to the corresponding cycloalkene, and verify.
5. For a reaction with a primary ALIPHATIC amine and HNO2 (from NaNO2/HCl) at low temperature, recall that the resulting aliphatic diazonium salt is far too unstable to be isolated even near 0 degrees C - it decomposes immediately, releasing N2 and forming a carbocation trapped by water to give the alcohol; verify this matches the stated product (this is in sharp contrast to aromatic diazonium salts, which ARE stable at 273-278 K). …