VerifiedHint: We will solve each reaction step by step and see what will happen. As we know, alc. KOH is a reducing agent and hence it will reduce the haloalkane. Also when chlorobenzene is treated with hydrolysis phenol is formed.
Complete Answer:
(A) When n- butyl chloride reacts with alcoholic KOH, the product formed is butene. This reaction is known as hydrohalogenation.
$ C{H_3} - C{H_2} - C{H_2} - C{H_2} - Cl\xrightarrow[{Dehydrogenation}]{{KOH\left( {alc} \right)}}C{H_3} - C{H_2} - C{H_2} = C{H_2} + {H_2}O $
(B) The reaction of bromobenzene with Mg in the presence of dry ether, the product of this reaction is Phenylmagnesium bromide
$ Ph - Br + Mg\xrightarrow[{}]{{dry\,\,ether}}Ph - MgBr $
(C) The hydrolysis of chlorobenzene is not possible under normal conditions. In order to subject chlorobenzene for hydrolysis, we need to heat chlorobenzene in an aqueous medium with sodium hydroxide solution at temperature 623K and a pressure of 300 atm to form phenol.
$ Ph - Cl\xrightarrow[{{H^ + }}]{{NaOH,623K,300atm}}Ph - OH $
(D) The reaction of ethyl chloride with aqueous KOH, the product formed is ethanol
$ C{H_3} - C{H_2} - Cl\xrightarrow[{\left( {hydrolysis} \right)}]{{KOH\left( {aq} \right)}}C{H_3} - C{H_2} - OH + KCl $
(E) The reaction of methyl bromide with sodium in the presence of dry ether yields ethane. This reaction is also known as the wurtz reaction.
$ 2C{H_3} - Br + 2Na\xrightarrow[{\left( {wurtz\,\,\,reaction} \right)}]{{Dry\,\,ether}}C{H_3} - C{H_3} + 2NaBr $
(F) The reaction of methyl chloride with KCN, the product formed is methyl cyanide. This reaction is a substitution reaction.
$ C{H_3} - Cl + KCN\xrightarrow[{Nucleophilic\,\,substitution}]{{}}C{H_3} - CN + KCl $
Note: The reactions above are some named reactions known as wurtz reactions which are used to produce symmetrical alkanes. Also remember aqueous KOH is used when the reactant is subjected to hydrolysis while alcoholic KOH is used as an armstrong base. Such as in the above reaction we convert the alkyl halide to alkene using alcoholic KOH and when we use aqueous KOH with alkyl halide, the product formed will be alcohols with functional group -OH.
