Compound 'A' (molecular formula $\left.\mathrm{C}_{3} \mathrm{H}_{8} \mathrm{O}ight)$ is treated with…

Compound 'A' (molecular formula $\left.\mathrm{C}_{3} \mathrm{H}_{8} \mathrm{O}ight)$ is treated with acidified potassium dichromate to form a product ' B' (molecular formula $\left.\mathrm{C}_{3} \mathrm{H}_{6} \mathrm{O}ight)$. 'B' forms a shining silver mirror on warming with ammonical silver nitrate. 'B' when treated with an aqueous solution of $\mathrm{H}_{2} \mathrm{NCONHNH}_{2} . \mathrm{HCl}$ and sodium acetate gives a product ' $\mathrm{C}$ '. Identify the structure of ' $\mathrm{C}^{\prime}$
  1. $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}=\mathrm{NNHCONH}_{2}$
  4. $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}=\mathrm{NCONHNH}_{2}$

Solution

$\mathrm{C}_{3} \mathrm{H}_{8} \mathrm{O} \stackrel{\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7} / \mathrm{H}^{+}}{\longrightarrow} \mathrm{C}_{3} \mathrm{H}_{6} \mathrm{O} \stackrel{\mathrm{H}_{2} \mathrm{NCONHNH}_{2}}{\longrightarrow} \mathrm{C}$
$\quad$$\quad$ A $\quad$$\quad$ $\quad$$\quad$ B $(-\mathrm{CHO})$
Since B reduces Tollen's reagent, it indicates that it has an -CHO group, so it must be $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CHO}$. Hence
$\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{OH} ightarrow \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CHO} \stackrel{\mathrm{H}_{2} \mathrm{NNHCONH}_{2}}{\longrightarrow} \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}=\mathrm{NNHCONH}_{2}$
[A] $\quad$$\quad$$\quad$$\quad$$\quad$$\quad$ [B] $\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$[C] ,

Asked in: JEE-TOPICTESTS-CHEMISTRY

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