Chapter 7, Problem 7.59P

Pick the reactant or solvent in each part that gives the faster ${\text{SN}}_{\text{1}}$ reaction.

a. reaction of ${\text{H}}_2\text{O}$ with ${\left({\text{CH}}_{\text{3}}\right)}_3\text{CCl}$ or ${\left({\text{CH}}_{\text{3}}\right)}_3\text{Cl}$

b. reaction of ${\text{CH}}_{\text{3}}\text{OH}$ with ${\left({\text{CH}}_{\text{3}}\right)}_3\text{CBr}$ or ${\left({\text{CH}}_{\text{3}}\right)}_2{\text{CHCH}}_2\text{Br}$

c. reaction of ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{CH}\left(\text{I}\right){\text{CH}}_{\text{3}}$ with ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{OH}$ in water or DMSO

Interpretation Introduction

(a)

Interpretation: The reactant that gives faster ${\text{S}}_{\text{N}}\text{1}$ reaction is to be identified.

Concept introduction: Nucleophilic substitution reaction takes place by two mechanisms, ${\text{S}}_{\text{N}}\text{1}$ and ${\text{S}}_{\text{N}}\text{2}$. In ${\text{S}}_{\text{N}}\text{1}$ mechanism, formation of carbocation takes place by removal of halide and then nucleophile attack on that carbocation. However in ${\text{S}}_{\text{N}}\text{2}$ mechanism, removal of halide and attack of nucleophile takes place simultaneously. The presence of good leaving group increases the rate of both ${\text{S}}_{\text{N}}\text{1}$ and ${\text{S}}_{\text{N}}\text{2}$ reactions.

Answer to Problem 7.59P

The faster ${\text{S}}_{\text{N}}1$ reaction will take place with ${\left({\text{CH}}_{\text{3}}\right)}_3\text{CI}$ than ${\left({\text{CH}}_{\text{3}}\right)}_3\text{CCl}$.

Explanation of Solution

The rate of ${\text{S}}_{\text{N}}1$ reaction increases in the presence of good leaving group. Out of ${\text{I}}^{-}$ and ${\text{Cl}}^{-}$, ${\text{I}}^{-}$ is a good leaving group because its conjugate acid has low $p{K}_a$ value as compared to the $p{K}_a$ value of conjugate acid of ${\text{Cl}}^{-}$.

Therefore, the faster ${\text{S}}_{\text{N}}1$ reaction will take place with ${\left({\text{CH}}_{\text{3}}\right)}_3\text{CI}$ than ${\left({\text{CH}}_{\text{3}}\right)}_3\text{CCl}$.

Conclusion

The faster ${\text{S}}_{\text{N}}1$ reaction will take place with ${\left({\text{CH}}_{\text{3}}\right)}_3\text{CI}$ than ${\left({\text{CH}}_{\text{3}}\right)}_3\text{CCl}$.

Interpretation Introduction

(b)

Interpretation: The reactant that gives faster ${\text{S}}_{\text{N}}\text{1}$ reaction is to be identified.

Concept introduction: Nucleophilic substitution reaction takes place by two mechanisms, ${\text{S}}_{\text{N}}\text{1}$ and ${\text{S}}_{\text{N}}\text{2}$. In ${\text{S}}_{\text{N}}\text{1}$ mechanism, formation of carbocation takes place by removal of halide and then nucleophile attack on that carbocation. However in ${\text{S}}_{\text{N}}\text{2}$ mechanism, removal of halide and attack of nucleophile takes place simultaneously. The presence of good leaving group increases the rate of both ${\text{S}}_{\text{N}}\text{1}$ and ${\text{S}}_{\text{N}}\text{2}$ reactions.

Answer to Problem 7.59P

The faster ${\text{S}}_{\text{N}}1$ reaction will take place with ${\left({\text{CH}}_{\text{3}}\right)}_{\text{3}}\text{CBr}$ than ${\left({\text{CH}}_{\text{3}}\right)}_2{\text{CHCH}}_{\text{2}}\text{Br}$.

Explanation of Solution

The tertiary halide undergoes nucleophilic substitution by ${\text{S}}_{\text{N}}1$ mechanism in which carbocation is formed. The removal of halide leads to the formation of carbocation. Bromine atom is bonded to tertiary carbon atom in ${\left({\text{CH}}_{\text{3}}\right)}_{\text{3}}\text{CBr}$ whereas in ${\left({\text{CH}}_{\text{3}}\right)}_2{\text{CHCH}}_{\text{2}}\text{Br}$, it is bonded to primary carbon atom. The order of increasing reactivity of alkyl halides toward ${\text{S}}_{\text{N}}\text{1}$ reaction is $1°<2°<3°$. Therefore, the faster ${\text{S}}_{\text{N}}1$ reaction will take place with ${\left({\text{CH}}_{\text{3}}\right)}_{\text{3}}\text{CBr}$ than ${\left({\text{CH}}_{\text{3}}\right)}_2{\text{CHCH}}_{\text{2}}\text{Br}$.

Conclusion

The faster ${\text{S}}_{\text{N}}1$ reaction will take place with ${\left({\text{CH}}_{\text{3}}\right)}_{\text{3}}\text{CBr}$ than ${\left({\text{CH}}_{\text{3}}\right)}_2{\text{CHCH}}_{\text{2}}\text{Br}$.

Interpretation Introduction

(c)

Interpretation: The solvent that gives faster ${\text{S}}_{\text{N}}\text{1}$ reaction is to be identified.

Concept introduction: Nucleophilic substitution reaction takes place by two mechanisms, ${\text{S}}_{\text{N}}\text{1}$ and ${\text{S}}_{\text{N}}\text{2}$. In ${\text{S}}_{\text{N}}\text{1}$ mechanism, formation of carbocation takes place by removal of halide and then nucleophile attack on that carbocation. However in ${\text{S}}_{\text{N}}\text{2}$ mechanism, removal of halide and attack of nucleophile takes place simultaneously. The presence of good leaving group increases the rate of both ${\text{S}}_{\text{N}}\text{1}$ and ${\text{S}}_{\text{N}}\text{2}$ reactions.

Answer to Problem 7.59P

The faster ${\text{S}}_{\text{N}}1$ reaction will take place in water than DMSO.

Explanation of Solution

The polar protic solvent favors ${\text{S}}_{\text{N}}\text{1}$ reaction whereas the polar aprotic solvent favors reaction through ${\text{S}}_{\text{N}}\text{2}$ mechanism. Out of water and DMSO, water is a polar protic solvent because hydrogen atom is bonded to electronegative $\text{O}$ atom. Whereas, DMSO is a polar aprotic solvent. Therefore, the reaction of ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{CH}\left(\text{I}\right){\text{CH}}_3$ with ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{OH}$ in water gives faster ${\text{S}}_{\text{N}}\text{1}$ reaction.

Conclusion

The faster ${\text{S}}_{\text{N}}1$ reaction will take place in water than DMSO.