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Question
What is the correct order of reactivity of alcohols in the following reaction?
\[\ce{R-OH + HCl ->[ZnCl2] R-Cl + H2O}\]
Options
1° > 2° > 3°
1° < 2° > 3°
3° > 2° > 1°
3° > 1° > 2°
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Solution
3° > 2° > 1°
Explanation:
The given reaction is nucleophilic substitution reaction in which –OH group is replaced by –Cl. Tertiary alcohols, when react with \[\ce{HCl}\] in presence of \[\ce{ZnCl2}\], form tertiary carbocation.
This intermediate 3° carbocation is more stable than 2° carbocation as well as 1° carbocation. The higher the stability of intermediate, the higher will be the reactivity of reactant molecule.
So, the order of reactivity of alcohols in the given reaction is 3° > 2° > 1°.
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