Question

Which of the following factors does not favour ${S_N}1$ mechanism?
A. Strong nucleophile
B. Polar solvent
C. Low conc. of nucleophile
D. 3rd halide

Answer 1

Hint: We know that ${S_N}1$ stands for nucleophilic substitution and 1 represents unimolecular rate determining step of this mechanism. The mechanism of ${S_N}1$ generally has two steps. One of the examples of ${S_N}1$ mechanism is the hydrolysis reaction of tert-butyl bromide producing tert-butanol. In inorganic chemistry, ${S_N}1$ reaction is called a dissociative mechanism.

Complete step by step answer: We must remember that the ${S_N}1$ reactions form a carbocation intermediate. The reaction of tertiary alkyl halides under strongly basic (or) acidic conditions to form tertiary alcohols is ${S_N}1$ reaction.
We know that the ${S_N}1$ mechanism goes stepwise. The leaving group leaves, and the formation of carbocation that is attacked by the nucleophile.
In ${S_N}1$ reaction, carbocation stability is the big barrier. The stability of the carbocation increases with increasing carbon substitution (tertiary>secondary>>primary) as well as with resonance. The loss of leaving group of ${S_N}1$ reaction gives carbocation, the reaction rate would be proportional carbocation stability. Since stability of carbon increases as we move from primary to secondary to tertiary. The reaction rate moves from primary (slowest)<${S_N}1$ proceed with weak nucleophiles. Option (C) is correct because weak nucleophiles only favour ${S_N}1$ mechanism and strong nucleophiles do not favour ${S_N}1$ mechanism. Strong nucleophiles favour ${S_N}2$ mechanism.
${S_N}1$ proceed with polar protic solvents like water, carboxylic acids, and alcohols. The polar solvent increases the carbocation stability and they both favor ${S_N}1$ mechanism. Therefore, Option (B) is incorrect.
${S_N}1$ proceeds with strong nucleophiles that have low concentration. So, Option (A) is incorrect.
${S_N}1$ reaction would be more favoured if the leaving group is bonded to tertiary, allylic (or) benzylic carbon, the intermediate carbocation would be more stable. Only tertiary halides undergo ${S_N}1$ mechanism.
So, the correct answer is “Option D”.

Note:
One should not confuse them with polar solvents used in ${S_N}1$ and ${S_N}2$ reactions. In ${S_N}1$ reaction, polar protic solvent is required and in ${S_N}2$ reactions polar aprotic solvent is required. Elimination reactions and carbocation rearrangements are the two common side reactions that happen. ${S_N}1$ follows first order kinetic reaction and is multistep reactions.