Chapter 9.6, Problem 9.35P

Interpretation Introduction

Interpretation:

The reaction will proceed via ${\text{S}}_{\text{N}}\text{2}$ or ${\text{S}}_{\text{N}}\text{1}$ pathway or neither has to be determined considering the reagents and conditions given.

Concept Introduction:

Electrophile:

The compound that is been attacked by nucleophile is known as electrophile.  Usually electrophile is referred as substrates.  Valency of carbon atom is four.  Therefore, carbon atom has four bonds.  Hence, apart from the bond that is with the leaving group, the carbon atom contains three other bonds also.

Out of the three bonds that are present apart from the leaving group, if one is an alkyl group, then the substrate is known as “primary”.  If there are two alkyl groups, then the substrate is known as “secondary”.  If there are three alkyl groups, then the substrate is known as “tertiary”.  This is represented as shown below,

In ${\text{S}}_{\text{N}}\text{2}$ reaction, the alkyl groups are more crowded.  Hence, the nucleophile cannot attack the electrophilic center easily if there are three alkyl groups.  Therefore, for ${\text{S}}_{\text{N}}\text{2}$ reaction, the primary substrate is more susceptible for nucleophilic attack, while the tertiary substrate is unreactive.

In ${\text{S}}_{\text{N}}\text{1}$ reaction, the first step is the loss of leaving group to form a carbocation.  The second step is the attack of nucleophile on carbocation.  In this case, the stability of the carbocation formed is the major issue.  As we know that the alkyl groups are electron donating groups, the more the alkyl groups substituted, the more the carbocation formed will be stabilized.  Hence, tertiary is more stable than the secondary, and the primary is the worst as it contains only one alkyl group.

In simple words, we can say that, if the substrate is a primary or secondary one, then the reaction will proceed through ${\text{S}}_{\text{N}}\text{2}$ mechanism.  The final product has inversion of configuration.  If the substrate is tertiary means, then the reaction will proceed through ${\text{S}}_{\text{N}}\text{1}$ mechanism and the final product is racemic.

The strength of the nucleophile also determines through which reaction the given compound undergoes.  The ${\text{S}}_{\text{N}}\text{2}$ process depends on the strength of nucleophile as the rate depends on it.  Generally, a strong nucleophile will speed up the rate at which the ${\text{S}}_{\text{N}}\text{2}$ reaction occurs, and a weak nucleophile slows the rate of ${\text{S}}_{\text{N}}\text{2}$ reaction.

Nucleophile:

The rate of ${\text{S}}_{\text{N}}\text{1}$ reaction does not depend upon the nucleophile.  This is because, the rate of ${\text{S}}_{\text{N}}\text{1}$ reaction is purely dependent on the concentration of substrate and not on concentration of nucleophile.  Hence, for ${\text{S}}_{\text{N}}\text{1}$ reaction, the strength of nucleophile is an irrelevant one.

In simple words it can be said that,

• Strong nucleophile favors ${\text{S}}_{\text{N}}\text{2}$ reaction.
• Weak nucleophile does not favor ${\text{S}}_{\text{N}}\text{2}$ reaction.

The first important thing that has to be identified was whether the given nucleophile is a strong or weak one.  The strength of the nucleophile depends on many factors and they are charge, polarizability etc.

Presence or absence of a negative charge determines the strength of nucleophile.  If a negative charge is present means, the nucleophile will be strong.  More than this polarizability is an important factor that determines the strength of nucleophile.  The size of the atom is directly related to polarizability.  If the size of the atom is more means, then polarizability will be more resulting in strong nucleophile and vice-versa.

Some of the strong nucleophiles are,

    

Some of the weak nucleophiles are,

    

Leaving Group:

The leaving group is the important one that also affects the mechanism by which the reaction proceeds.  The ${\text{S}}_{\text{N}}\text{1}$ and ${\text{S}}_{\text{N}}\text{2}$ mechanism are very much sensitive to the nature of leaving group.  If there is no leaving group in the substrate, then both the mechanism cannot operate.  In comparison, ${\text{S}}_{\text{N}}\text{1}$ mechanism is more sensitive than ${\text{S}}_{\text{N}}\text{2}$ mechanism, because the formation of carbocation by the loss of leaving group is the rate determining step in ${\text{S}}_{\text{N}}\text{1}$ mechanism.

The stability of the carbocation formed is more sensitive which means that the stability of the leaving groups is much more important.  Generally, the leaving groups are conjugate base of strong acids.  If the base is not stabilized, then it is a bad leaving group.  In few cases the bad leaving group (hydroxide) can be converted into good leaving group (protonated form) treating with strong acids.

Solvent:

Aprotic solvents are the solvents that do not contain a proton that is present on an electronegative atom.  This does not mean that there should not be any hydrogen atoms present in the compound.  Even though the compound contain hydrogen atom, none of them should be bonded to an electronegative atom.  Polar aprotic solvents are not very good in forming solvent shells around the negative charge.  In simple words, we can say that, if the solvent is polar aprotic, then the reaction will proceed through ${\text{S}}_{\text{N}}\text{2}$ mechanism.