Chapter 22, Problem 58PP

Interpretation Introduction

Interpretation: The missing reagents have to be filled from the knowledge of the conversion of starting materials into the products.

Concept Introduction:

Sodium cyanoborohydride is a strong reducing agent than sodium borohydride.  It reduces the carbonyl group into amine group in a rapid way.  So, it is called as reductive amination reactions.  Aldehyde or ketone group is reacted with ammonia in the presence of sodium cyanoborohydride as a reducing agent and a proton source in the reaction medium to produce primary amines.

Aldehyde or ketone group is reacted with primary amine in the presence of sodium cyanoborohydride as a reducing agent and a proton source in the reaction medium to produce secondary amines.

Aldehyde or ketone group is reacted with secondary amine in the presence of sodium cyanoborohydride as a reducing agent and a proton source in the reaction medium to produce tertiary amines.

The reagents such as methyl iodide and silver oxide are given and used as the reaction conditions in the given reactions.  This reaction is called Hofmann elimination.  The amines undergo exhaustive alkylation in the presence of excess methyl iodide producing quaternary ammonium salt which is a excellent leaving group.  After treatment with strong base, aqueous silver oxide and heat, quaternary ammonium salt gives alkenes.  Alkene formation involves in E2 elimination.  Through E2 elimination, the major product is less substituted alkene with trace amount of more substituted alkene.

Alkenes on treatment with ozone followed by DMS give aldehydes.

To find: The missing reagents in the given set of reactions

Find the reagents involved in the reaction (a)