Question

In which case formation of butane nitrile is possible?
(A) \[{{\rm{C}}_{\rm{3}}}{{\rm{H}}_{\rm{7}}}{\rm{Br}} + {\rm{KCN}}\]
(B) \[{{\rm{C}}_4}{{\rm{H}}_9}{\rm{Br}} + {\rm{KCN}}\]
(C) \[{{\rm{C}}_{\rm{3}}}{{\rm{H}}_{\rm{7}}}{\rm{OH}} + {\rm{KCN}}\]
(D) \[{{\rm{C}}_4}{{\rm{H}}_9}{\rm{OH}} + {\rm{KCN}}\]

Answer 1

Hint: Butane nitrile is a compound that belongs to the group of cyanides. The functional group present in butane nitrile is cyanide. The chemical symbol of butane nitrile is \[{{\rm{C}}_{\rm{3}}}{{\rm{H}}_{\rm{7}}}{\rm{CN}}\] . The reaction of KCN with alkyl halides increases the carbon chain by one atom.

Complete Step by Step Answer:
Let’s discuss the options one by one:

Option B says the reaction of bromobutane with potassium cyanide gives butanenitrile. We know that a halogen is replaced by the cyanide group. Therefore, the reaction of bromobutane with potassium cyanide gives pentane nitrile.
\[{\rm{C}}{{\rm{H}}_{\rm{3}}} - {\rm{C}}{{\rm{H}}_{\rm{3}}} - {\rm{C}}{{\rm{H}}_{\rm{3}}} - {\rm{C}}{{\rm{H}}_{\rm{3}}} - {\rm{Br}} + {\rm{K}} - {\rm{CN}} \to \mathop {{\rm{C}}{{\rm{H}}_{\rm{3}}} - {\rm{C}}{{\rm{H}}_{\rm{3}}} - {\rm{C}}{{\rm{H}}_{\rm{3}}} - {\rm{C}}{{\rm{H}}_{\rm{3}}} - {\rm{CN}}}\limits_{{\rm{Pentane}}\,\,{\rm{nitrile}}} + {\rm{KBr}}\]
Therefore, option B is wrong.

Option C says the reaction of propanol with potassium cyanide gives butanenitrile. But, this reaction will not occur because potassium cyanide does not undergo a reaction with the alcohol group.
\[{{\rm{C}}_{\rm{3}}}{{\rm{H}}_{\rm{7}}}{\rm{OH}} + {\rm{KCN}} \to {\rm{No}}\,{\rm{reaction}}\]
Therefore, option B is wrong.

Option D says the reaction of butanol with potassium hydroxide gives butanenitrile. But, we know, this reaction will not happen.
\[{{\rm{C}}_4}{{\rm{H}}_9}{\rm{OH}} + {\rm{KCN}} \to {\rm{No}}\,{\rm{reaction}}\]
Therefore, option D is wrong.

Option A says the reaction of bromopropane with potassium cyanide gives butane nitrile. This is true because the cyanide ion replaces the halogen from the alkyl halide and gives cyanide. The reaction is,
\[{{\rm{C}}_{\rm{3}}}{{\rm{H}}_{\rm{7}}}{\rm{Br}} + {\rm{KCN}} \to \mathop {{{\rm{H}}_{\rm{3}}}{\rm{C}} - {\rm{C}}{{\rm{H}}_{\rm{2}}} - {\rm{C}}{{\rm{H}}_{\rm{2}}} - {\rm{CN}}}\limits_{{\rm{Butanenitrile}}} + {\rm{KBr}}\]
Therefore, the reaction given in option A gives butane nitrile.

Note: It is to be noted that cyanide is an ambidentate ligand. An ambidentate ligand is a ligand that has two donor sites. For example, in the cyanide functional group, the nucleophilic attack happens through both carbon and nitrogen atoms. One other ambidentate ligand is the nitro group (\[{\rm{N}}{{\rm{O}}_{\rm{2}}}\] ).