Assertion: Acetamide has a more polar C=O gro | vedclass

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Question

(A) In acetamide $(CH_{3}CONH_{2})$,the nitrogen atom has a lone pair that participates in resonance with the $C=O$ group. However,the $-NH_{2}$ group

Vedclass · Accepted Answer

If both Assertion and Reason are correct and the Reason is a correct explanation of the Assertion.

Vedclass · Answer

(A) In acetamide $(CH_{3}CONH_{2})$,the nitrogen atom has a lone pair that participates in resonance with the $C=O$ group. However,the $-NH_{2}$ group is a stronger electron-donating group by resonance compared to the $-OC_{2}H_{5}$ group in ethyl acetoacetate $(CH_{3}COCH_{2}COOC_{2}H_{5})$.Because nitrogen is less electronegative than oxygen,the lone pair on nitrogen is more readily donated into the carbonyl system,increasing the electron density on the oxygen atom of the $C=O$ bond,which makes the $C=O$ bond more polar.Thus,both the Assertion and the Reason are correct,and the Reason is the correct explanation for the Assertion.

List-$I$ (Chemical Reaction)	List-$II$ (Reagent used)
$(a)$ $CH_{3}COOCH_{2}CH_{3} \rightarrow CH_{3}CH_{2}OH$	$(i)$ $CH_{3}MgBr / H_{3}O^{+}$ ($1$ equivalent)
$(b)$ $CH_{3}COOCH_{3} \rightarrow CH_{3}CHO$	$(ii)$ $H_{2}SO_{4} / H_{2}O$
$(c)$ $CH_{3}C \equiv N \rightarrow CH_{3}CHO$	$(iii)$ $DIBAL-H / H_{2}O$
$(d)$ $CH_{3}C \equiv N \rightarrow CH_{3}COCH_{3}$	$(iv)$ $SnCl_{2}, HCl / H_{2}O$

Assertion: Acetamide has a more polar $C=O$ group than ethyl acetoacetate.
Reason: $-\ddot{N}H_{2}$ is more electron-donating than $-\ddot{O}C_{2}H_{5}$.

Similar Questions

Arrange the following compounds in the correct order of their boiling points:
$(I) \, CH_3CH_2CH_2CH_2OH$
$(II) \, CH_3CH_2CH_2CHO$
$(III) \, CH_3CH_2CH_2COOH$

Choose the correct option$(s)$ for the following reaction sequence. Consider $Q$,$R$ and $S$ as major products.

Which of the following best describes the nucleophilic addition step in the acid-catalyzed hydrolysis of acetonitrile $(CH_3CN)$?

$CH_3-C(=O)-OH$ $\xrightarrow{NaH^{14}CO_3} (A) (\text{gas})$ $\xrightarrow[(ii) H_3O^{+}]{(i) PhMgBr} (B)$
$CH_3-S(=O)_2-OH$ $\xrightarrow{NaH^{14}CO_3} (C) (\text{gas})$ $\xrightarrow[(ii) H_3O^{+}]{(i) PhMgBr} (D)$
Product $(B)$ and $(D)$ in the above reaction are:

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Assertion: Acetamide has a more polar $C=O$ group than ethyl acetoacetate.Reason: $-\ddot{N}H_{2}$ is more electron-donating than $-\ddot{O}C_{2}H_{5}$.