Compound $A$ (molecular formula $\mathrm{C}_3 \mathrm{H}_8 \mathrm{O}$) is treated with acidified potassium…

Compound $A$ (molecular formula $\mathrm{C}_3 \mathrm{H}_8 \mathrm{O}$) is treated with acidified potassium dichromate to form a product $B$ (molecular formula $\mathrm{C}_3 \mathrm{H}_6 \mathrm{O}$). $B$ forms a shining silver mirror on warming with ammoniacal silver nitrate. $B$ when treated with an aqueous solution of $\mathrm{H}_2 \mathrm{NCONHNH}_2 . \mathrm{HCl}$ and sodium acetate gives a product $C$. Identify the structure of $C$.
  1. $\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}=\mathrm{NNHCONH}_2$


  2. $\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}=\mathrm{NCONHNH}_2$

Solution


Reaction of $B$ indicates that $B$ is an aldehyde thus $B$ should be $\mathrm{C}_2 \mathrm{H}_5 \mathrm{CHO}$ or $\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CHO}$ and therefore $C$ should be $\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}=\mathrm{NNHCONH}_2$.

Asked in: NEET 2022 (Phase 2)

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