Question

Which of the following reaction(s) can be used for the preparation of alkyl halides?
(I) ${\text{C}}{{\text{H}}_{\text{3}}}{\text{C}}{{\text{H}}_{\text{2}}}{\text{OH + HCl }}\xrightarrow{{{\text{anh}}{\text{.ZnC}}{{\text{l}}_{\text{2}}}}}$
(II) ${\text{C}}{{\text{H}}_{\text{3}}}{\text{C}}{{\text{H}}_{\text{2}}}{\text{OH + HCl }} \to$
(III) ${\left( {{\text{C}}{{\text{H}}_{\text{3}}}} \right)_3}{\text{C}}{\text{OH + HCl }} \to$
(IV) ${\left( {{\text{C}}{{\text{H}}_{\text{3}}}} \right)_2}{\text{CH}}{\text{OH + HCl }}\xrightarrow{{{\text{anh}}{\text{.ZnC}}{{\text{l}}_{\text{2}}}}}$

A ) (IV) only
B) (III) and (IV) only
C ) (I), (III) and (IV) only
D) (I) and (II) only

Answer 1

Identify the type of substrate such as primary, secondary or tertiary alcohol. Identify which substrate needs a catalyst.

Complete step by step answer:
To prepare alkyl halide, you react with an alcohol with hydrogen chloride. In this reaction, you replace the chlorine atom hydroxyl group. In this reaction, tertiary alcohols are much more reactive than primary or secondary alcohols.
If you start with tertiary alcohol, you do not need a catalyst. Thus, when you mix a tertiary alcohol with hydrogen chloride, you will get tertiary alcohol as the product.
If you start with primary and secondary alcohol, you need a catalyst. The most common catalyst is anhydrous zinc chloride. Thus, when you mix a primary or secondary with hydrogen chloride, you will not get primary or secondary alcohol as the product.
But, when you mix a primary or secondary with hydrogen chloride in presence of anhydrous zinc chloride catalyst, you will get primary or secondary alcohol as the product.
You can use the following reactions for the preparation of alkyl halides
(I) ${\text{C}}{{\text{H}}_{\text{3}}}{\text{C}}{{\text{H}}_{\text{2}}}{\text{OH + HCl }}\xrightarrow{{{\text{anh}}{\text{.ZnC}}{{\text{l}}_{\text{2}}}}}{\text{C}}{{\text{H}}_{\text{3}}}{\text{C}}{{\text{H}}_{\text{2}}}{\text{Cl + }}{{\text{H}}_2}{\text{O }}$
(III) ${\left( {{\text{C}}{{\text{H}}_{\text{3}}}} \right)_3}{\text{C}}{\text{OH + HCl }} \to {\left( {{\text{C}}{{\text{H}}_{\text{3}}}} \right)_3}{\text{CCl + }}{{\text{H}}_2}{\text{O}}$
(IV) ${\left( {{\text{C}}{{\text{H}}_{\text{3}}}} \right)_2}{\text{CH}}{\text{OH + HCl }}\xrightarrow{{{\text{anh}}{\text{.ZnC}}{{\text{l}}_{\text{2}}}}}{\left( {{\text{C}}{{\text{H}}_{\text{3}}}} \right)_2}{\text{CH}}{\text{Cl + }}{{\text{H}}_2}{\text{O}}$
You cannot use the following reaction for the preparation of alkyl halides.
(II) ${\text{C}}{{\text{H}}_{\text{3}}}{\text{C}}{{\text{H}}_{\text{2}}}{\text{OH + HCl }} \to {\text{no reaction}}$

Hence, the correct option is C ) (I), (III) and (IV) only.

Note: In primary alcohol, the carbon atom bearing hydroxyl group is attached to only one other carbon atom. In secondary alcohol, the carbon atom bearing hydroxyl group is attached to two other carbon atoms. In tertiary alcohol, the carbon atom bearing hydroxyl group is attached to three other carbon atoms.