Question

What are ${\text{S}}{{\text{N}}_1}$, ${\text{S}}{{\text{N}}_2}$, ${{\text{E}}_1}$, ${{\text{E}}_2}$ reactions . How to identify which reaction is occurring in any given reaction?

Answer 1

To answer this question we should know what primary, secondary or tertiary carbon is and what is protic and aprotic solvent and how do they affect the mechanism of the reaction. First, we will see the reactant and then solvent and then decide the type of mechanism that should be followed by the reaction.

Complete answer:
The full name of ${\text{S}}{{\text{N}}_1}$reaction is unimolecular nucleophilic substitution reaction. In ${{\text{S}}_{\text{N}}}1$mechanism, the reaction takes place in two steps. In first step carbocation forms. In the second step the nucleophile attacks on the carbocation. The carbocation is the intermediate of ${{\text{S}}_{\text{N}}}1$types.
The full name of ${\text{S}}{{\text{N}}_2}$reaction is a bimolecular nucleophilic substitution reaction. In ${\text{S}}{{\text{N}}_2}$reaction, a nucleophile substitutes another nucleophile. The whole reaction takes place in one step.
As the nucleophile removes from the reactant another nucleophile attacks from the opposite side. The formed structure is known as a transition state in which both nucleophiles remain bound with weak bonds. Due to the attack from the opposite side the stereo of the product changes. This is known as Walden inversion.
${{\text{E}}_1}$ is known as a unimolecular elimination mechanism. In the first step the leaving group leaves, forming a carbocation. In the second step, base abstract the proton from the alkyl halide forming a negative charge. The negative charge shifts to the next carbon forming a carbon-carbon double bond.
${{\text{E}}_{\text{2}}}$ is known as a bimolecular elimination mechanism. In this mechanism a base abstract the proton from the alkyl halide forming a negative charge. The negative charge shifts to the next carbon forming a carbon-carbon double bond and simultaneously the leaving group leaves.
The order of decreasing reactivity of alkyl halide towards the ${\text{S}}{{\text{N}}_1}$ reaction is as follows:
${3^ \circ } > \,{2^ \circ }\, > \,{1^ \circ }$
The order of decreasing reactivity of alkyl halide towards the ${\text{S}}{{\text{N}}_2}$ reaction is as follows:
${1^ \circ } > \,{2^ \circ }\, > \,{3^ \circ }$
The primary and secondary alkyl halide goes through ${\text{S}}{{\text{N}}_{\text{2}}}$ or ${{\text{E}}_{\text{2}}}$ mechanism. The tertiary alkyl halide ${\text{S}}{{\text{N}}_{\text{1}}}$ or ${{\text{E}}_1}$ mechanism.
Polar protic solvent favour the ${{\text{E}}_{\text{2}}}$ mechanism and polar aprotic solvent favours the ${\text{S}}{{\text{N}}_{\text{2}}}$.
Weak nucleophile favour ${{\text{E}}_1}$ whereas the strong nucleophile favour the ${\text{S}}{{\text{N}}_{\text{1}}}$.

Note: In primary and secondary alkyl halide if the reaction proceed through ${\text{S}}{{\text{N}}_{\text{1}}}$ it cannot form highly stable tertiary carbocation so, the reaction will proceed through ${\text{S}}{{\text{N}}_{\text{2}}}$ or ${{\text{E}}_{\text{2}}}$. The stability order of carbocation is tertiary > secondary > primary. The solvent which can form hydrogen bonds or we can say have hydroxyl groups is known as polar protic solvent. The solvent that cannot form hydrogen bonds is known as a polar aprotic solvent.