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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 $C$.
Options:
- A $$
\text { } \mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}=\mathrm{NNHCONH}_2
$$ - B
- C
- D $$
\text { } \mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}=\mathrm{NCONHNH}_2
$$
Solution:
1994 Upvotes
Verified Answer
The correct answer is:
$$
\text { } \mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}=\mathrm{NNHCONH}_2
$$
\text { } \mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}=\mathrm{NNHCONH}_2
$$
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 $\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}=\mathrm{NNHCONH}_2$.
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