Compound A (molecular formula C3H8O) is treated with acidified potassium dichromate to form a product B (molecular formula C3H6O). B forms a shining silver mirror on warming with ammoniacal silver nitrate. B when treated with an aqueous solution of H2 NCONHNH2. HCl and sodium acetate gives a product C. Identify the structure of C.

Question

Compound A (molecular formula C3H8O) is treated with acidified potassium dichromate to form a product B (molecular formula C3H6O). B forms a shining silver mirror on warming with ammoniacal silver nitrate. B when treated with an aqueous solution of H2 NCONHNH2. HCl and sodium acetate gives a product C. Identify the structure of C. (A) CH3CH2CH = NNHCONH2 (B)  (C)  (D) CH3CH2CHNCONHNH2

Tardigrade Admin · Accepted Answer

Reaction of B indicates that B is an aldehyde thus B should be C2H5CHO or CH3CH2CHO and therefore C should be CH3CH2CH = NNHCONH2.

$C H_{3} C H_{2} C H = NN H CON H_{2}$

$C H_{3} C H_{2} C H NCON H N H_{2}$

Reaction of B indicates that B is an aldehyde thus B should be $C_{2} H_{5} C H O$ or $C H_{3} C H_{2} C H O$ and therefore C should be $C H_{3} C H_{2} C H = NN H CON H_{2} .$

CH3​CH2​CH=NNHCONH2​

CH3​CH2​CHNCONHNH2​

Reaction of B indicates that B is an aldehyde thus B should be C2​H5​CHO or CH3​CH2​CHO and therefore C should be CH3​CH2​CH=NNHCONH2​.

$C H_{3} C H_{2} C H = NN H CON H_{2}$

$C H_{3} C H_{2} C H NCON H N H_{2}$

Reaction of B indicates that B is an aldehyde thus B should be $C_{2} H_{5} C H O$ or $C H_{3} C H_{2} C H O$ and therefore C should be $C H_{3} C H_{2} C H = NN H CON H_{2} .$