Write the IUPAC name of the given compound. | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) The structure is $C_6H_5-CH=CH-COOH$. The principal functional group is the carboxylic acid $(-COOH)$,which gets the $C-1$ position. The carbon ch

Vedclass · Accepted Answer

$3$-Phenylprop-$2$-enoic acid

Vedclass · Answer

(D) The structure is $C_6H_5-CH=CH-COOH$. The principal functional group is the carboxylic acid $(-COOH)$,which gets the $C-1$ position. The carbon chain consists of $3$ carbons,so the parent alkane is propane. The double bond starts at $C-2$,so it is a prop-$2$-enoic acid derivative. $A$ phenyl group $(-C_6H_5)$ is attached at the $C-3$ position. Therefore,the $IUPAC$ name is $3$-phenylprop-$2$-enoic acid.

Write the $IUPAC$ name of the given compound.

Similar Questions

What is the $IUPAC$ name of the compound $(CH_3)_3C - CH = CH_2$?

The $IUPAC$ name of $(CH_3)_3C \, CH_2CH_2Cl$ is

Sugar of lead is

The correct $IUPAC$ name of $H_2C=CH-CH(CH_3)-CH_2-C \equiv CH$ is:

The $IUPAC$ name of the following structure is: $[CH_3CH(CH_3)]_2C(CH_2CH_3)C(CH_3)C(CH_2CH_3)_2$

Vedclass Test Series

Exam Paper Generator

Online Exam Module