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Question: (a) How will you convert methyl bromide to dimethylamine? (b) How will you prepare propanamine from ...

(a) How will you convert methyl bromide to dimethylamine? (b) How will you prepare propanamine from methyl chloride using suitable reactions?

Explanation

Solution

In (a) and (b), both alkyl halides are converted into amine compounds, both are aliphatic alkyl halides. Using appropriate alkyl amine, ethanol, and sodium hydroxide they can be converted into respective amines. From this method, secondary amines can be prepared.

Complete answer:
(a)- We have to convert methyl bromide to dimethylamine. The methyl bromide is a chain of one carbon atom and a bromine atom is also present. The formula of methyl bromide is given below:
CH3−BrC{{H}_{3}}-Br
Dimethylamine is an amine in which there are two carbon atoms and between them, an amine group is present. The formula is CH3−NH−CH3C{{H}_{3}}-NH-C{{H}_{3}}.
We can convert methyl bromide to dimethylamine by treating the methyl bromide with methylamine and then treating it with sodium hydroxide. The reaction is given below:
CH3Br+NH2CH3→CH3NH2CH3+Br−C{{H}_{3}}Br+N{{H}_{2}}C{{H}_{3}}\to C{{H}_{3}}N{{H}_{2}}CH_{3}^{+}B{{r}^{-}}
CH3NH2CH3+Br−+NaOH→CH3NHCH3+NaBr+H2OC{{H}_{3}}N{{H}_{2}}CH_{3}^{+}B{{r}^{-}}+NaOH\to C{{H}_{3}}NHC{{H}_{3}}+NaBr+{{H}_{2}}O
(b)- We have to convert methyl chloride to propanamine, but the propanamine formed will be N-methylethylamine. Methyl chloride is a chain of one carbon atom and a chlorine atom is also present. The formula of methyl chloride is given below:
CH3−ClC{{H}_{3}}-Cl
N-methylethylamine is an amine in which there are three carbon atoms and between them, an amine group is present. The formula is CH3−NH−CH2−CH3C{{H}_{3}}-NH-C{{H}_{2}}-C{{H}_{3}}
We can convert methyl chloride to N-methylethylamine by treating the methyl chloride with ethylamine and then treating it with sodium hydroxide. The reaction is given below:
CH3Cl+NH2CH2CH3→CH3CH2NH2CH3+Cl−C{{H}_{3}}Cl+N{{H}_{2}}C{{H}_{2}}C{{H}_{3}}\to C{{H}_{3}}C{{H}_{2}}N{{H}_{2}}CH_{3}^{+}C{{l}^{-}}
CH3CH2NH2CH3+Cl−+NaOH→CH3NHCH2CH3+NaCl+H2OC{{H}_{3}}C{{H}_{2}}N{{H}_{2}}CH_{3}^{+}C{{l}^{-}}+NaOH\to C{{H}_{3}}NHC{{H}_{2}}C{{H}_{3}}+NaCl+{{H}_{2}}O

Note:
It is difficult to make primary amines from this method. If you want to make tertiary mines, then take secondary amines in the reactant, ethanol, and sodium hydroxide. And it must be noted that the solvent used is ethanol.