How will you convert methyl bromide to dimethylamine?

Hint: We know that here we have 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:
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: $ C{{H}_{3}}-Br. $ The steps for conversion are,
React it with excess $ N{{H}_{2}}C{{H}_{3}}, $ ethanol and $ NaOH $
To make propyl amine from methyl chloride, unless you are talking about $ N-methylethylamine. $ If we are, it's the same process as in (a), except that the reactant is $ {{C}_{2}}{{H}_{5}}N{{H}_{2}}. $
Dimethylamine is an amine in which there are two carbon atoms and between them, an amine group is present. The formula is; $ C{{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. Amines can be synthesized from haloalkanes in nucleophilic substitution reactions. The product of this substitution is an ammonium salt. An amine is generated by reacting the salt with alkali.
 $ C{{H}_{3}}B+N{{H}_{2}}C{{H}_{3}}\to C{{H}^{3}}N{{H}_{2}}CH_{3}^{+}B{{r}^{-}}\text{ } $
Which on further reaction with sodium hydroxide gives;
 $ C{{H}_{3}}N{{H}_{2}}C{{H}_{3}}B{{r}^{-}}+NaOH\to C{{H}_{3}}NHC{{H}_{3}}+NaBr+{{H}_{2}}O $ .

Note:
Note that 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.