Question

What happens when (Write chemical equations only): (i) \textit{n}-Butyl chloride reacts with alcoholic KOH?
(ii) Bromobenzene reacts with magnesium in the presence of dry ether?
(iii) Ethyl chloride reacts with aqueous KOH?
(iv) Methyl bromide reacts with sodium in the presence of dry ether?
(v) Ethyl bromide reacts with KCN (alc.)?

Hint:

Remember: Alcoholic KOH → elimination (alkenes), Aqueous KOH → substitution (alcohols), Dry ether + Mg → Grignard reagent, Wurtz reaction → higher alkane, and KCN → alkyl nitrile.

Answer 1

(i) \textit{n-Butyl chloride + alcoholic KOH:}
Dehydrohalogenation occurs, producing 1-butene.
\[ CH_3CH_2CH_2CH_2Cl + KOH(alc.) \; \xrightarrow{\Delta} \; CH_3CH_2CH=CH_2 + KCl + H_2O \] (ii) Bromobenzene + Mg in dry ether:
Grignard reagent is formed.
\[ C_6H_5Br + Mg \; \xrightarrow{\text{dry ether}} \; C_6H_5MgBr \] (iii) Ethyl chloride + aqueous KOH:
Hydrolysis takes place, giving ethyl alcohol.
\[ C_2H_5Cl + KOH(aq) \; \longrightarrow \; C_2H_5OH + KCl \] (iv) Methyl bromide + Na in dry ether:
Wurtz reaction occurs, forming ethane.
\[ 2CH_3Br + 2Na \; \xrightarrow{\text{dry ether}} \; C_2H_6 + 2NaBr \] (v) Ethyl bromide + KCN (alc.):
Nucleophilic substitution occurs, forming propionitrile.
\[ C_2H_5Br + KCN(alc.) \; \longrightarrow \; C_2H_5CN + KBr \] Conclusion:
These reactions show elimination (E2), substitution (SN1/SN2), Grignard reagent formation, and carbon–carbon bond formation, highlighting important pathways in organic synthesis.