Provide the IUPAC name of the given compound. | vedclass

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(B) $1$. Identify the principal functional group: The compound is a cyclic alkene. The double bond gets priority in numbering.$2$. Numbering the ring:

Vedclass · Accepted Answer

$5-$ethyl$-1-$methylcyclohex$-1-$ene

Vedclass · Answer

(B) $1$. Identify the principal functional group: The compound is a cyclic alkene. The double bond gets priority in numbering.$2$. Numbering the ring: Start numbering from the double bond such that the substituents get the lowest possible locants. If we start at the carbon with the double bond and move towards the ethyl group,we get $1-$ethyl$-5-$methylcyclohex$-1-$ene. However,alphabetical order is followed for naming substituents.$3$. Alphabetical order: Ethyl $(E)$ comes before Methyl $(M)$.$4$. Correct numbering: To give the lowest locants to substituents while prioritizing the double bond,we number the double bond carbons as $1$ and $2$. If we assign $1$ to the carbon with the ethyl group and $2$ to the adjacent double bond carbon,the methyl group is at position $5$. This gives $1-$ethyl$-5-$methylcyclohex$-1-$ene.$5$. Comparing with options: The correct name is $1-$ethyl$-5-$methylcyclohex$-1-$ene. Since this specific option is not listed,we re-evaluate the structure. The structure shows the double bond between two carbons. If we number the double bond as $1$ and $2$,and the ethyl group is at $1$ and methyl at $5$,the name is $1-$ethyl$-5-$methylcyclohex$-1-$ene. Looking at the provided options,option $B$ is $5-$ethyl$-1-$methylcyclohex$-1-$ene,which is the same structure if numbered differently. However,based on standard $IUPAC$ rules,$1-$ethyl$-5-$methylcyclohex$-1-$ene is preferred.

Provide the $IUPAC$ name of the given compound.

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