Statement (A): Carboxylic acid can be obtaine | vedclass

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(D) The reaction of $R-C \equiv N$ with $H_3O^+$ (acidic hydrolysis) typically requires heating to proceed to the carboxylic acid stage. Under mild co

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$(A)$ is not correct but $(B)$ is correct

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(D) The reaction of $R-C \equiv N$ with $H_3O^+$ (acidic hydrolysis) typically requires heating to proceed to the carboxylic acid stage. Under mild conditions,it often stops at the amide stage $(R-CONH_2)$. Thus,statement $(A)$ is generally considered incorrect as it implies a simple,mild reaction.When alkyl cyanide $(R-C \equiv N)$ is hydrolysed in an alkaline aqueous medium,it undergoes hydrolysis to form a carboxylate salt $(R-COO^-Na^+)$ and ammonia $(NH_3)$. Thus,statement $(B)$ is correct.

Sets	Property
$I$. $NCCH_2COOH > FCH_2COOH > H_3CCH_2COOH$	Acidity
$II$. $CH_3CH_2CHO > PhCOCH_3 > PhCHO$	Reactivity
$III$. $H_3COCH_2CH_3 < H_3CCH_2CHO < H_3CCH_2CH_2OH$	Boiling points

Statement $(A)$: Carboxylic acid can be obtained by the reaction of $R-C \equiv N$ with $H_3O^+$ under mild reaction conditions.
Statement $(B)$: Hydrolysis of $R-C \equiv N$ in alkaline aqueous medium gives $R-COONa$ and $NH_3$ as products.
The correct answer is:

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The end product of the above reaction is:

Which of the following sets is in the correct order regarding the property mentioned against them?
Sets Property
$I$. $NCCH_2COOH > FCH_2COOH > H_3CCH_2COOH$ Acidity
$II$. $CH_3CH_2CHO > PhCOCH_3 > PhCHO$ Reactivity
$III$. $H_3COCH_2CH_3 < H_3CCH_2CHO < H_3CCH_2CH_2OH$ Boiling points

Product $(G)$ is

In the reaction sequence,$[X]$ is a ketone:
$[X] \xrightarrow{KMnO_4/OH^{-}/\Delta} HOOC-(CH_2)_3-CH(CH_3)-COOH$
$[X]$ will be:

Choose the best sequence of reactions for the transformation given. Semicolons indicate separate reaction steps to be used in the order shown.

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Statement $(A)$: Carboxylic acid can be obtained by the reaction of $R-C \equiv N$ with $H_3O^+$ under mild reaction conditions.Statement $(B)$: Hydrolysis of $R-C \equiv N$ in alkaline aqueous medium gives $R-COONa$ and $NH_3$ as products.The correct answer is: