How many distinct alkenes are obtained from 3 | vedclass

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(B) The reaction of $3-$bromo$-3-$methylhexane with alcoholic $KOH$ proceeds via an $E2$ elimination mechanism.$3-$bromo$-3-$methylhexane has the stru

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$3$

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(B) The reaction of $3-$bromo$-3-$methylhexane with alcoholic $KOH$ proceeds via an $E2$ elimination mechanism.$3-$bromo$-3-$methylhexane has the structure $CH_3-CH_2-C(Br)(CH_3)-CH_2-CH_2-CH_3$.The $\beta$-carbons are the carbons adjacent to the carbon bearing the bromine atom.There are three distinct types of $\beta$-hydrogens available for elimination:$1$. Elimination from the $C2$ position ($CH_2$ group) gives $3-$methylhex$-2-$ene.$2$. Elimination from the $C4$ position ($CH_2$ group) gives $3-$methylhex$-3-$ene.$3$. Elimination from the methyl group attached to $C3$ gives $2-$ethylpent$-1-$ene.Thus,$3$ distinct alkenes are obtained.

How many distinct alkenes are obtained from $3-$bromo$-3-$methylhexane upon treatment with alcoholic $KOH$?

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