Question

An alkyl halide, $RX$ reacts with $KCN$ to give propane nitrile is:
a.) ${{C}_{3}}{{H}_{7}}Br$
b.) ${{C}_{4}}{{H}_{9}}Br$
c.) ${{C}_{2}}{{H}_{5}}Br$
d.) ${{C}_{5}}{{H}_{11}}Br$

Answer 1

Hint: $RX$ reacts with $KCN$ to give alkyl nitrile and potassium halide. Alkyl halide $RX$ is a molecule in which R is alkane and X is a halide group in which there can be any halide chlorine, bromine, fluorine or iodine.

Complete step-by-step answer:
The correct option is A.
The molecule $KCN$ is an ionic compound in which K is positively charged and CN is negatively charged because it has high electronegativity.
Whereas, in the $AgCN$ bond between Ag and CN is a covalent bond. And the bond is quite strong because C and N have a triple bond in this structure.
So, in $KCN$ bond is easy to break between K and CN whereas in$AgCN$bond is quite strong.
EXPLANATION:
In the case of KCN,
KCN\text{ }\to \text{ }{{K}^{+}}\text{ }and\text{ }C{{N}^{\\_}}
$R-X\text{ }\to \text{ }{{R}^{+}}\text{ }and\text{ }{{X}^{-}}$
Then the Product is R-CN and K-X
In the case of AgCN,
$AgCN\text{ }\to \text{ }C\text{ }\equiv \text{ }N\text{ }\left( \text{ }expressing\text{ }triple\text{ }bond \right)$
Ag+ and N gets lone pairs. Electron density on N is higher.
So, N gains partially -ve charge and Ag separate to exhibit +ve charge
$R-X\text{ }\to \text{ }{{R}^{+}}\text{ }and\text{ }{{X}^{-}}$
Then, R attaches to N and will form RNC and AgX.
That’s why, when haloalkanes reacts with KCN it will form alkane nitrile or alkyl cyanide and when haloalkanes reacts with AgCN it gives isocyanide.

Overall reaction:
$\begin{aligned} & RX+KCN\to RCN+KX \\\ & RX+AgCN\to RNC+AgX \\\ \end{aligned}$

Resultant product is propane nitrile therefore the reactant should be propane halide. Propane halide has 3 carbons, 6 hydrogens and 1 alkyl group. In the given option alkyl group is bromine.

Note: Potassium cyanide reacts with alkyl halide to alkyl nitrile and potassium halide because potassium cyanide is an ionic compound it readily loses cyanide group.