Consider the following reactions:(a) ; (CH_3) | vedclass

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(A) The Saytzeff rule states that in elimination reactions,the more substituted alkene (the one with more alkyl groups attached to the double-bonded c

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$(c)$ only

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(A) The Saytzeff rule states that in elimination reactions,the more substituted alkene (the one with more alkyl groups attached to the double-bonded carbons) is the major product.$(a)$ $(CH_3)_3CCH(OH)CH_3$ $\xrightarrow{H^+} (CH_3)_3CCH^+(CH_3)$ $\xrightarrow{1,2-methyl shift} (CH_3)_2C^+(CH_2CH_3)CH_3$ $\xrightarrow{-H^+} (CH_3)_2C=C(CH_3)_2$. This is a highly substituted alkene,which is the Saytzeff product.$(b)$ $(CH_3)_2CHCH(Br)CH_3 \xrightarrow{alc. KOH} (CH_3)_2C=CHCH_3$. This is the more substituted alkene,which is the Saytzeff product.$(c)$ $(CH_3)_2CHCH(Br)CH_3 \xrightarrow{(CH_3)_3CO^-K^+} (CH_3)_2CHCH=CH_2$. The bulky base $(CH_3)_3CO^-$ abstracts the less hindered proton,leading to the less substituted alkene,which is the Hofmann product.$(d)$ $(CH_3)_2C(OH)CH_2CHO \xrightarrow{\Delta} (CH_3)_2C=CHCHO$. This is the more substituted alkene (conjugated with the carbonyl group),which is the Saytzeff product.Thus,only reaction $(c)$ does not produce the Saytzeff product as the major product.

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