Question

CH3COOAg + Br2 in presence of CCl4 gives?

Answer 1

It's Borodine Hunsdiecker Reaction. Reaction is in the attachment:

Answer 2

Methyl bromide is the product formed in the reaction of silver acetate with bromine in presence of carbon tetrachloride.

• This is an example of the Hunsdiecker reaction, where silver salts of carboxylic acids react with halogens to form alkyl halides.
• The reaction involves decarboxylation, where one carbon atom is lost in the form of CO₂. So the product formed has one carbon less than the precursor carboxylic acid.
• The further reaction involves halogenation, combining the alkyl and halogen's free radicals, forming the alkyl halide.
• Silver acetate loses one carbon, forming methyl free radical on decarboxylation. This reacts with bromine free radical forming methyl bromide.
• The reactions are depicted in the attachment.