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An organic compound with molecular formula ${{\text{C}}_{\text{3}}}{{\text{H}}_{\text{6}}}{\text{O}}$ did not give a silver mirror with Tollen’s reagent, but gave an oxime and hydroxylamine, it may be:
A.${\text{C}}{{\text{H}}_{\text{3}}}{\text{COC}}{{\text{H}}_{\text{3}}}$
B. ${{\text{C}}_2}{{\text{H}}_5}{\text{CHO}}$
C.${\text{C}}{{\text{H}}_2} = {\text{CH}} - {\text{C}}{{\text{H}}_{\text{2}}} - {\text{OH}}$
D. ${\text{C}}{{\text{H}}_3} - {\text{O}} - {\text{CH}} = {\text{C}}{{\text{H}}_2}$

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Last updated date: 24th Jul 2024
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Answer
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Hint: The chemical compound having a ketone functional group gives the oxime and hydroxylamine with Tollen’s reagent whereas the chemical compound having an aldehyde functional group gives the silver mirror.

Complete step-by-step answer:Tollen’s reagent is used to determine the presence of the carbonyl group in chemical compounds.
The carbonyl group contains aldehyde and ketone. The aldehyde gives a silver mirror with Tollen’s reagent and ketone forms the oxime and hydroxylamine with Tollen’s reagent.
The formation of silver mirror is shown as follows:
${\text{RCHO}}\, + \,\mathop {{{\left[ {{\text{Ag(N}}{{\text{H}}_{\text{3}}}{{\text{)}}_{\text{2}}}} \right]}^ + }\, + \,{{\text{H}}_2}{\text{O}}}\limits_{{\text{Tollen reagent}}} \,\, \to \mathop {{\text{2Ag}}}\limits_{{\text{silver mirror}}} \,{\text{ + }}\,\,{\text{4N}}{{\text{H}}_{\text{3}}}\, + \,{\text{RCOOH}}\, + \,2\,{{\text{H}}^ + }$
The formation of oxime is shown as follows:
${\text{ROR}}\, + \,\mathop {{{\left[ {{\text{Ag(N}}{{\text{H}}_{\text{3}}}{{\text{)}}_{\text{2}}}} \right]}^ + }}\limits_{{\text{Tollen reagent}}} \, \to \mathop {{\text{C}}{{\text{H}}_{\text{3}}}{\text{C = N(OH)C}}{{\text{H}}_{\text{3}}}\,}\limits_{{\text{oxime}}} $
So, the organic compound having a molecular formula ${{\text{C}}_{\text{3}}}{{\text{H}}_{\text{6}}}{\text{O}}$is a ketone.
The organic compound ${\text{C}}{{\text{H}}_{\text{3}}}{\text{COC}}{{\text{H}}_{\text{3}}}$is a ketone so, option (A) is correct.
The formation of oxime is shown as follows:
${\text{C}}{{\text{H}}_{\text{3}}}{\text{COC}}{{\text{H}}_{\text{3}}}\, + \,\mathop {{{\left[ {{\text{Ag(N}}{{\text{H}}_{\text{3}}}{{\text{)}}_{\text{2}}}} \right]}^ + }}\limits_{{\text{Tollen reagent}}} \, \to \mathop {{\text{C}}{{\text{H}}_{\text{3}}}{\text{C = N(OH)C}}{{\text{H}}_{\text{3}}}\,}\limits_{{\text{oxime}}} $
The organic compound ${{\text{C}}_2}{{\text{H}}_5}{\text{CHO}}$ is an aldehyde. Aldehyde gives silver mirror with Tollen’s reagent but the organic compound having molecular formula ${{\text{C}}_{\text{3}}}{{\text{H}}_{\text{6}}}{\text{O}}$is not giving silver mirror so, option (B) is incorrect.
The organic compound${\text{C}}{{\text{H}}_2} = {\text{CH}} - {\text{C}}{{\text{H}}_{\text{2}}} - {\text{OH}}$is an alcohol. Alcohol does not give a positive test with Tollen’s reagent, so option (C) is incorrect.
The organic compound${\text{C}}{{\text{H}}_3} - {\text{O}} - {\text{CH}} = {\text{C}}{{\text{H}}_2}$is an ether. Ether does not give a positive test with Tollen’s reagent, so option (D) is incorrect.

Therefore, option (A) ${\text{C}}{{\text{H}}_{\text{3}}}{\text{COC}}{{\text{H}}_{\text{3}}}$is correct.

Note:Tollen’s reagent is a solution of silver nitrate, sodium hydroxide, and ammonia which forms silver oxide and sodium nitrate. When aldehyde reacts with Tollen’s reagent, the elemental silver precipitates which get deposited on the wall of the test tube. The precipitate is known as a silver mirror. Tollen’s reagent gives a positive test for alpha-hydroxy ketones.