Question

Which of the following alkyl halides is hydrolysed by \[{S_{{N^1}}}\] mechanism?
A. \[C{H_3}Cl\]
B. \[C{H_3}C{H_2}Cl\]
C. \[C{H_3}C{H_2}C{H_2}Cl\]
D. \[{\left( {C{H_3}} \right)_3}CCl\]

Answer 1

Hint: Try to prove that \[{S_{{N^1}}}\] is a nucleophilic substitution reaction and this reaction proceeds through the formation of carbocation. The alkyl halide which forms the most stable carbocation is hydrolysed by \[{S_{{N^1}}}\] mechanism. Now, by using this you can easily find the correct option from the given options.

Complete step by step solution:
It is known to you that \[{S_{{N^1}}}\] is substitution nucleophilic unimolecular because only one molecule is involved in determining the rate of the reaction. Such types of reactions are generally carried out in polar protic solvents such as water, alcohol, acetic acid, etc.
The \[{S_{{N^1}}}\] reaction proceeds through the formation of carbocation. The carbocations have the unique property of rearrangement. Carbocations rearrange if they become more stable as a result of rearrangement. Therefore, the products formed are different than expected products.
The order of reactivity depends upon the stability of carbocation in the first step. Greater the stability of the carbocation, greater will be its ease of formation from alkyl halide and hence faster will be the rate of reaction.
As we know that 3 degree carbocation is most stable, therefore, the tert-alkyl halides will undergo \[{S_{{N^1}}}\] reaction very fast. The order of reactivity will be tertiary alkyl halide> secondary alkyl halide> primary alkyl halide> methyl halide.
Therefore, from above we can conclude that option D is the correct option to the given question.

Note: It should be remembered to you that when a reaction forms a carbocation intermediate, always check for the possibility of a carbocation rearrangement.
Also, you should remember in case of optically active alkyl halide, the product obtained during \[{S_{{N^1}}}\] reaction is a racemic mixture.