Aldehydes Ketones Question 228
Question: Compound ‘A’ (molecular formula $ C_3H_8O $ ) is treated with acidified potassium dichromate to form a product ‘B’ (molecular formula $ C_3H_6O $ ). ‘B’ forms a shining silver mirror on warming with ammonical silver nitrate. ‘B’ when treated with an aqueous solution of $ H_2NCONHNH_2.HCl $ and sodium acetate gives a product ‘C’. Identify the structure of ‘C’-
Options:
A) $ CH_3CH_2CH=NNHCONH_2 $
B) $ CH_3-\underset{CH_3}{\mathop{\underset{|}{\mathop{C}}}}=NNHCONH_2 $
C) $ CH_3-\underset{CH_3}{\mathop{\underset{|}{\mathop{C}}}}=NCONHNH_2 $
D) $ CH_3CH_2CH=NCONHNH_2 $
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Answer:
Correct Answer: A
Solution:
[a] $ \underset{A}{\mathop{C_3H_8O\xrightarrow{K_2Cr_2O_7/{H^{+}}}C_3H_6O}} $
$ \underset{B(-CHO)}{\mathop{\xrightarrow{H_2NCONHNH_2}C}} $ Since B reduces Tollen’s reagent, it indicates that it has an $ -CHO $ group, so it must be $ CH_3CH_2CHO $ . Hence $ \underset{[A]}{\mathop{CH_3CH_2CH_2OH}}\to \underset{[B]}{\mathop{CH_3CH_2CHO}}\xrightarrow{H_2NHCONH_2} $
$ \underset{[C]}{\mathop{CH_3CH_2CH}}=NNHCONH_2 $
BETA
