Q. Compound (A) (molecular formula $C_{3}H_{8}O$) is treated with acidified potassium dichromate to form a product B (molecular formula $C_{3}H_{6}O$).’B’ forms a shining silver mirror on warming with ammonical silver nitrate. ’B’ when treated with an aqueous solution of $H_{2}NCONHNH_{2}.HCl$ and sodium acetate gives a product ‘C’. Identify the structure of ‘C’.
A) $CH_{3}CH_{2}CH = NNHCONH_{2}$
B) $\left( CH_{3} \right)_{2}C = NNHCONH_{2}$
C) $\left( CH_{3} \right)_{2}C = NCONHNH_{2}$
D) $CH_{3}CH_{2}CH = NCONHNH_{2}$
Explanation
(a)
$CH_{3}CH_{2}CH_{2}OH\underset{H_{2}So_{4}}{\overset{\ \ K_{2}{Cr}_{2}O_{7}\ \ \ \ }{\rightarrow}}CH_{3}CH_{2}CHO$
(B )
propanol propanal
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \overset{H_{2}NCONHNH_{2}}{\rightarrow}CH_{3}CH_{2}CH = NNHCONH_{2}$
(C)
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.