Question

In the above reaction,the product will be:

• A. $C{H}_3-\underset{COC{H}_3}{\underset{|}{C}H-}C{H}_2-Cl$
• B. $C{H}_3-\underset{Cl}{\underset{|}{C}}H-C{H}_2-\underset{O}{\underset{||}{C}}-C{H}_3$
• C. $C{H}_3-\underset{Cl}{\underset{|}{C}}H-\underset{O}{\underset{||}{C}}-C{H}_2-C{H}_3$
• D. $C{H}_3-\underset{Cl}{\underset{|}{\stackrel{C{H}_3}{\stackrel{|}{C}}}}-C{H}_2-\underset{O}{\underset{||}{C}}-C{H}_{2}C{H}_3$

Answer 1

The correct option is B $C{H}_3-\underset{Cl}{\underset{|}{C}}H-C{H}_2-\underset{O}{\underset{||}{C}}-C{H}_3$

First step of the reaction is Hoffmann elimination. Tertiary amine is methylated with methyl iodide to form quaternary ammonium iodide.

Quaternary ammonium iodide is heated with $AgOH$ to form propene and tertiary amine.

Next step is electrophilic addition reaction. In presence of anhydrous aluminium chloride, acetyl chloride is added across $C=C$ double bond of propene,