The major product of the above reaction is:

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(C) The reaction is a Hofmann elimination of a quaternary ammonium salt. In the given Newman projection,the leaving group is $-NMe_3^+$. For the elimi

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$1,2-$diphenylpropene

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(C) The reaction is a Hofmann elimination of a quaternary ammonium salt. In the given Newman projection,the leaving group is $-NMe_3^+$. For the elimination to occur,the base $(EtO^-)$ must abstract a $\beta$-hydrogen that is anti-periplanar to the leaving group. Looking at the Newman projection,the $-NMe_3^+$ group is on the front carbon. The back carbon has a $Ph$ group,a $H$ atom,and a $Ph$ group. Wait,re-examining the structure: The front carbon has $-NMe_3^+$,$-Me$,and $-Ph$. The back carbon has $-H$,$-H$,and $-Ph$. Actually,the structure shows the front carbon with $-NMe_3^+$,$-Me$,and $-Ph$. The back carbon has $-H$,$-H$,and $-Ph$. For $E2$ elimination,we need an anti-periplanar $H$. The $H$ on the back carbon is anti to the $-NMe_3^+$ group. Removing the $H$ and the $-NMe_3^+$ group results in the formation of a double bond between the two carbons. The resulting product is $Ph(Me)C=C(H)Ph$,which is $1,2$-diphenylpropene.

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