Two alkenes,X (91% yield) and Y (9% yield),ar | vedclass

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(C) The reaction is a Hoffmann elimination (also known as the Cope elimination if an amine oxide is used,but here it is a quaternary ammonium hydroxid

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methylenecyclopentane and $1-$methylcyclopentene

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(C) The reaction is a Hoffmann elimination (also known as the Cope elimination if an amine oxide is used,but here it is a quaternary ammonium hydroxide). In Hoffmann elimination,the less substituted alkene (the one with fewer alkyl groups on the double-bonded carbons) is the major product because the bulky leaving group (trimethylamine) prefers to abstract a proton from the less sterically hindered $\beta$-carbon. $1$. The starting material is $1$-methylcyclopentyltrimethylammonium hydroxide. $2$. The $\beta$-hydrogens are available on the ring carbons adjacent to the quaternary nitrogen. $3$. Removing a proton from the $CH_2$ group adjacent to the quaternary center leads to the formation of methylenecyclopentane (the less substituted,Hoffmann product). $4$. Removing a proton from the $CH$ group (the tertiary carbon) leads to the formation of $1$-methylcyclopentene (the more substituted,Zaitsev product). Since Hoffmann elimination favors the less substituted alkene as the major product,$X$ $(91\%)$ is methylenecyclopentane and $Y$ $(9\%)$ is $1$-methylcyclopentene. Therefore,the correct option is $(C)$.

Two alkenes,$X$ ($91\%$ yield) and $Y$ ($9\%$ yield),are formed when the following compound is heated:
The structures of $X$ and $Y$,respectively,are:

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Two alkenes,$X$ ($91\%$ yield) and $Y$ ($9\%$ yield),are formed when the following compound is heated:The structures of $X$ and $Y$,respectively,are: