Chapter 9.2, Problem 9.7P

Interpretation Introduction

Interpretation:

The carbocation that will be formed upon loss of leaving group will be resonance-stabilized or not has to be identified.

Concept Introduction:

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 major reason for ${\text{S}}_{\text{N}}\text{1}$ is the stability of the carbocation formed.  The tertiary carbocation is more stable than the secondary followed by primary.  Apart from this, the another way by which the carbocation can be stabilized was through resonance.  If the formed carbocation is resonance stabilized, then that carbocation will be formed more readily than the others.  In simple words, if the leaving group is in a benzylic or allylic position, the carbocation formed from this will be resonance stabilized when the leaving group leaves.

In case if a double bond is near the leaving group, then the carbocation that will form has to be drawn and it has to be checked for resonance stabilization.