Question

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 ammoniacal silver nitrate. B when treated with an aqueous solution of $N{H_2}CONHN{H_2}$, $HCl$ and sodium acetate gives a product C. Identify the structure of C.
A.${\text{C}}{{\text{H}}_{\text{3}}}{\text{C}}{{\text{H}}_{\text{2}}}{\text{CH = NNHCON}}{{\text{H}}_{\text{2}}}$
B.${\text{C}}{{\text{H}}_{\text{3}}}{\text{C}}\left( {{\text{C}}{{\text{H}}_{\text{3}}}} \right){\text{ = NNHCON}}{{\text{H}}_{\text{2}}}$
C.${\text{C}}{{\text{H}}_{\text{3}}}{\text{C}}\left( {{\text{C}}{{\text{H}}_{\text{3}}}} \right){\text{ = NCONHN}}{{\text{H}}_{\text{2}}}$
D.${\text{C}}{{\text{H}}_{\text{3}}}{\text{C}}{{\text{H}}_{\text{2}}}{\text{CH = NCONHN}}{{\text{H}}_{\text{2}}}$

Answer 1

Hint:Ammoniacal silver nitrate is the Tollen’s reagent and is used for the identification of aldehydes in a given organic sample through the Tollen’s test. To answer this question, you must recall the reactions shown by different reagents in organic chemistry.

Complete answer:
It is given to us that a certain compound A with molecular formula ${C_3}{H_8}O$ is treated with acidified potassium dichromate to form a new compound B with molecular formula ${C_3}{H_6}O$. We are further given that B on heating with ammoniacal silver nitrate solution forms a silver mirror on the walls of the test tube. Later the compound B is made to react with $N{H_2}CONHN{H_2}$ in presence of hydrochloric acid and sodium acetate to form C.
We know that Tollen’s test is given by aldehydes. Thus the formation of silver mirror on reaction with ammoniacal silver nitrate signifies that compound B is an aldehyde. An aldehyde with the molecular formula ${C_3}{H_6}O$ is propanal.
We are given that the compound B, i.e., propanal is produced by oxidation of a certain compound A with molecular formula ${C_3}{H_8}O$ with acidified potassium dichromate. Thus we can conclude that A is a terminal alcohol or primary alcohol.
$A:{\text{ }}C{H_3}C{H_2}C{H_2}OH$
$B:{\text{ }}C{H_3}C{H_2}CHO$
When a carbonyl compound is treated with $N{H_2}CONHN{H_2}$ in presence of hydrochloric acid and sodium acetate, the oxygen atom is replaced by $N{H_2}CONHN{H_2}$.
Thus, the compound C is ${\text{C}}{{\text{H}}_{\text{3}}}{\text{C}}{{\text{H}}_{\text{2}}}{\text{CH = NNHCON}}{{\text{H}}_{\text{2}}}$

The correct answer is A.
Note:
Nitrogen is a highly electronegative atom and thus, $ - N{H_2}$ group is an electron withdrawing group. More the number of electron withdrawing groups, more is the reactivity. Thus, the electrophile $N{H_2}CONHN{H_2}$ attacks from that side which has more number of$ - N{H_2}$ groups.