The IUPAC name of the compound CH_3-C(OH)=CH- | vedclass

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Question

(A) $1$. Identify the principal functional group: The carboxylic acid group $(-COOH)$ has the highest priority,so the parent chain is named as an alka

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$4-$hydroxypent-$3-$enoic acid

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(A) $1$. Identify the principal functional group: The carboxylic acid group $(-COOH)$ has the highest priority,so the parent chain is named as an alkanoic acid.$2$. Number the chain: Start numbering from the carbon of the $-COOH$ group as $C-1$. The chain is $C_1(OOH)-C_2H_2-C_3H=C_4(OH)-C_5H_3$.$3$. Identify substituents and unsaturation: There is a hydroxyl group $(-OH)$ at $C-4$ and a double bond starting at $C-3$.$4$. Combine the parts: The parent chain has $5$ carbons (pent),the double bond is at position $3$ (pent-$3$-ene),and the $-OH$ group is at position $4$. Combining these,we get $4-$hydroxypent-$3-$enoic acid.

The $IUPAC$ name of the compound $CH_3-C(OH)=CH-CH_2-COOH$ is:

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