When (CH₃)₃CBr reacts with CH₃ONa in methanol,two products are formed: a major one is an alkene and minor one is an ether.Identify the two products and by what mechanism(s)they are formed?

Question

Quizplus · Accepted Answer

The reaction between CH₃CBr and CH₃ONa in methanol is a nucleophilic substitution reaction. The mechanism for this reaction is the SN2 mechanism, where the nucleophile (CH₃ONa) attacks the carbon atom attached to the leaving group (CBr) causing the leaving group to leave and the formation of a new bond between the nucleophile and the carbon atom. The major product formed is an alkene which is formed by the elimination of the leaving group (CBr) and a hydrogen atom on the neighboring carbon atom. This is a β-elimination reaction and follows the E2 mechanism, where the leaving group and the hydrogen atom are eliminated simultaneously from the molecule to form an alkene. The minor product formed is an ether which is formed by the attack of the nucleophile (CH₃ONa) on the carbon atom attached to the leaving group (CBr) followed by a second nucleophilic substitution reaction with methanol (CH₃OMe) present in the reaction medium. The mechanism for this reaction is SN2 followed by SN1, where the first nucleophilic substitution reaction results in the formation of an intermediate carbocation which then reacts with methanol to form the ether.

When (CH3)3CBr Reacts with CH3ONa in Methanol,two Products Are Formed

When (CH₃)₃CBr Reacts with CH₃ONa in Methanol,two Products Are Formed