The product (A) of the reaction is: | vedclass

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(B) The reaction proceeds in two steps:$1$. Treatment of cinnamic acid $(Ph-CH=CH-COOH)$ with $SOCl_2$ converts the carboxylic acid group into an acid

Vedclass · Accepted Answer

$Ph-CH=CH-C(=O)-NH-(cyclopropyl)$

Vedclass · Answer

(B) The reaction proceeds in two steps:$1$. Treatment of cinnamic acid $(Ph-CH=CH-COOH)$ with $SOCl_2$ converts the carboxylic acid group into an acid chloride group,forming cinnamoyl chloride $(Ph-CH=CH-COCl)$.$2$. The subsequent reaction of cinnamoyl chloride with cyclopropylamine $(NH_2-cyclopropyl)$ is a nucleophilic acyl substitution reaction. The lone pair on the nitrogen atom of the amine attacks the electrophilic carbonyl carbon of the acid chloride,leading to the formation of an amide.Thus,the final product $(A)$ is $N$-cyclopropylcinnamamide,which is $Ph-CH=CH-CONH-(cyclopropyl)$.

Column-$I$ (Formula)	Column-$II$ (General Name) and Column-$III$ ($IUPAC$ Name)
$(A)$ $HCOOH$	$(i)$ Succinic acid,$(x)$ Methanoic acid
$(B)$ $CH_3CH_2CH_2COOH$	$(ii)$ Oxalic acid,$(y)$ Ethanedioic acid
$(C)$ $(COOH)_2$	$(iii)$ Formic acid,$(m)$ Butanedioic acid
$(D)$ $(CH_2)_2(COOH)_2$	$(iv)$ Butyric acid,$(n)$ Butanoic acid

Column-$I$ (Reactant)	Column-$II$ (Product)
$A$. $RCH_2OH \xrightarrow[(ii) H_3O^{+}]{(i) NaOH, KMnO_4} ?$	$i$. $C_6H_5COOH + C_2H_5OH$
$B$. $CH_3(CH_2)_8CH_2OH \xrightarrow{Jones reagent} ?$	$ii$. $RCOOH$
$C$. $RCONH_2 \xrightarrow{H_3O^{+}, \Delta} ?$	$iii$. $CH_3(CH_2)_8COOH$
$D$. $C_6H_5COOEt \xrightarrow{H_3O^{+}} ?$	$iv$. $RCOOH + NH_3$

The product $(A)$ of the reaction is:

Similar Questions

Assertion : The $pK_a$ of acetic acid is lower than that of phenol.
Reason : Phenoxide ion is more resonance stabilised.

The above conversion can be done by:

The product is:

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