How is methyl bromide preferentially converted into methyl isocyanide? Why is methyl isocyanide obtained? Give reason.

Hint: Here we have to give the reaction methyl bromide to methyl isocyanide. This reaction depends on the reagent used for this reaction. It can be cyanide or isocyanide depending on reagent and type of reaction or bonding between carbon and nitrogen.

Complete step by step answer:
So, the methyl isocyanide was first prepared by A. Gautier. And the common method for preparing methyl isocyanides is the dehydration of N-methylformamide.

The chemical formula of methyl bromide is \[C{{H}_{3}}Br\]. Also the chemical formula of methyl cyanide is \[C{{H}_{3}}CN\] and methyl isocyanide is \[C{{H}_{3}}NC\] . In \[C{{H}_{3}}NC\]covalent bond is made by nitrogen with a methyl group.

Below given is the reaction of making methyl isocyanide from methyl bromide:
\[C{{H}_{3}}Br\xrightarrow{AgCN}C{{H}_{3}}N\equiv C\]

The reaction of making methyl cyanide from methyl bromide:
\[C{{H}_{3}}Br\xrightarrow{KCN}C{{H}_{3}}C\equiv N\]

So, the reason for obtaining methyl isocyanide from the reaction \[AgCN\] is that \[AgCN\] is largely covalent and available negative charge is present at nitrogen. In \[AgCN\] the bond between Ag and C is covalent so nucleophilic attack occurs through nitrogen carrying lone pairs of electrons.

\[KCN\] is predominantly ionic, that is it gives ions in solution. \[KCN\] is ionic and dissociates to give cyanide which has the negative charge concentrated on carbon. Thus, reaction with haloalkanes yields cyanides. And both C and N are in position to give electrons but the C-C bond is more stable than C-N bond so cyanide is formed.

Note: Nucleophilic substitution at haloalkanes gives isocyanates as a result. Cyanide ion is considered an ambident nucleophile meaning that the negative charge can be associated with either carbon or nitrogen.