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(D) The given molecule is $3$-iodo-$3$-methyloctane. In an $E_2$ elimination reaction,the base abstracts a proton from the $\beta$-carbon atoms adjace

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$5$

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(D) The given molecule is $3$-iodo-$3$-methyloctane. In an $E_2$ elimination reaction,the base abstracts a proton from the $\beta$-carbon atoms adjacent to the carbon bearing the iodine atom. The $\beta$-carbons are: $1$. The $CH_2$ group of the ethyl chain ($C$-$2$ position). $2$. The $CH_2$ group of the pentyl chain ($C$-$4$ position). $3$. The $CH_3$ group attached to the $C$-$3$ position. Elimination from the ethyl group gives $3$-methyl-oct-$2$-ene (which can exist as $E$ and $Z$ isomers). Elimination from the pentyl group gives $3$-methyl-oct-$3$-ene (which can exist as $E$ and $Z$ isomers). Elimination from the methyl group gives $2$-ethyl-hept-$1$-ene. Total products = $2$ (from ethyl) + $2$ (from pentyl) + $1$ (from methyl) = $5$ products.

How many different products are formed from the $E_2$ elimination reaction of the following alkyl iodide?

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