Achiral amine a having the r configuration undergoes hofmann elimination to form an alkene b as the major product. b is oxidatively cleaved with ozone, followed by ch3sch3, to form ch2═o and ch3ch2ch2cho. what are the structures of a and b? indicate stereochemistry where appropriate.

Hoffman Product is always a less substituted alkene. In given scenario Structure A (Given Below) was taken as a starting chiral Amine. For the sake of elimination the amine was taken tertiary so that on methylation it becomes a good leaving group. This chiral amina (A) when treated with Methyl Iodide gives a quarternary amine which on treatment with Silver oxide yields less substituted Alkene (B) as shown Below.

Alkene B on Ozonolysis give two aldehydes i.e. Formaldehyde and Butyraldehyde.

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