Question

Difference between ${{\text{S}}_{\text{N}}}\text{1}$ and ${{\text{S}}_{\text{N}}}2$ reactions.

Answer 1

Hint: ${{\text{S}}_{\text{N}}}\text{1}$and ${{\text{S}}_{\text{N}}}2$ represents unimolecular nucleophilic substitution reaction or bimolecular nucleophilic substitution reaction respectively.
Nucleophilic substitution reaction takes place in the presence of nucleophiles, where a nucleophile replaces the leaving group (could be a weak base) attached with an organic compound.
$\text{R-L}\,\,\text{+}\,\text{N}{{\text{u}}^{\text{-}}}\,\to \,\text{R-Nu}\,\,\text{+}\,{{\text{L}}^{\text{-}}}$
Nucleophilic substitution reactions are the characteristic reactions of alkyl halide, alcohol and ether organic compounds.
A nucleophilic substitution reaction completed in two steps –
In the first step breaking of the bond between the alkyl group and leaving group (halogens in alkyl halide) takes place.
In the second step bond formation in between the nucleophile and alkyl group takes place.

Complete answer:
On the basis of rate of reaction, nature of solvent and leaving group nucleophilic substitution reactions are mainly two types; unimolecular substitution reaction and bimolecular reaction.
Unimolecular substitution reaction$\text{(}{{\text{S}}_{\text{N}}}\text{1)}$-
This reaction follows the first order of kinetics, and the rate of reaction depends on the substrate so the nucleophile does not play in the rate determination of the reaction. This reaction takes place in two steps-
In the first step intermediate carbocation formed, so the rate of reaction depends upon the stability of carbocation. In the second step the nucleophile will attack the carbocation and formation of a new bond takes place.
Polar solvent favors the ${{\text{S}}_{\text{N}}}\text{1}$ reaction because in polar solvent complete solvation of the leaving group takes place. Since intermediate carbocation formation takes place, so branched alkyl group (tertiary alkyl group) gives ${{\text{S}}_{\text{N}}}\text{1}$ reaction. The order toward ${{\text{S}}_{\text{N}}}\text{1}$ is ${{3}^{\circ }}>\,{{2}^{\circ }}>\,{{1}^{\circ }}$ alkyl halide.
Carbocation is planar, so attack of nucleophile takes place from front side. If an optically active compound is used in ${{\text{S}}_{\text{N}}}\text{1}$ reaction then racemization takes place, it means both retention and inversion of configuration takes place.
Hydrolysis of $\text{tert-butylchloride}$ in the presence of water is an example of${{\text{S}}_{\text{N}}}\text{1}$reaction. $\underset{Tert-butylchloride}{\mathop{{{\left( C{{H}_{3}} \right)}_{3}}C-Cl}}\,\xrightarrow[20%{{H}_{2}}O]{80%{{C}_{2}}{{H}_{5}}OH}\underset{Stable\,\,carbocation}{\mathop{{{\left( C{{H}_{3}} \right)}_{3}}{{C}^{+}}}}\,+Cl\xrightarrow{O{{H}^{-}}}\underset{Tert-butylalcohol}{\mathop{{{\left( C{{H}_{3}} \right)}_{3}}C-OH}}\,$
Bimolecular substitution reaction$\text{(}{{\text{S}}_{\text{N}}}\text{2)}$-
This reaction follows the second order of kinetics and rate of reaction depends on the concentration of the substrate. In ${{\text{S}}_{\text{N}}}\text{2}$ reaction is a single step process and there is no intermediate state, this reaction takes place through a transition state. The conversion of reactant into transition state will determine the rate of reaction. In ${{\text{S}}_{\text{N}}}\text{2}$ reaction back attack of nucleophile takes place so the rate of reaction also depends on the steric hindrance. So the primary alkyl group (unhindered) and non-polar protic solvent favors the ${{\text{S}}_{\text{N}}}\text{2}$ reaction. The order of ${{\text{S}}_{\text{N}}}\text{2}$ reaction is ${{1}^{\circ }}>\,{{2}^{\circ }}>\,{{3}^{\circ }}$ alkyl halide. If an optically active substrate is used in ${{\text{S}}_{\text{N}}}\text{2}$ reaction inversion of configuration takes place this inversion is known as Walden inversion.
Hydrolysis of methyl bromide in the presence of $\text{NaOH}$ is the typical example of ${{\text{S}}_{\text{N}}}\text{2}$ reaction.
$\underset{\text{Methyl}\,\text{bromide}}{\mathop{\text{C}{{\text{H}}_{\text{3}}}\text{-Br}}}\,\xrightarrow{\text{aq}\text{.KOH}}\underset{\text{Methyl}\,\text{alcohol}}{\mathop{\text{C}{{\text{H}}_{\text{3}}}\text{-OH}}}\,$

Note :
Weaker bases are a good leaving group, thus favors the ${{\text{S}}_{\text{N}}}\text{1}$ reaction.
Substrate containing carbonyl group on β-carbon does not give ${{\text{S}}_{\text{N}}}\text{1}$ reaction because carbonyl group has very strong negative inductive effect which destabilize the intermediate.
Since ${{\text{S}}_{\text{N}}}\text{2}$ is a single step reaction, and rate of reaction depends on the concentration of substrate and nucleophile hence a strong nucleophile increase the rate of ${{\text{S}}_{\text{N}}}\text{2}$ reaction while weak nucleophile will decrease the rate of chemical reaction.