CH_3-CH_2-C(Ph)(OH)-CH_3 can not be prepared | vedclass

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Question

(A) The target molecule is $2	ext{-phenylbutan-2-ol}$,which is a tertiary $(3^o)$ alcohol. Option $A$: $HCHO + PhCH(CH_3)CH_2MgX$ gives a primary $(1

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$HCHO + PhCH(CH_3)CH_2MgX$

Vedclass · Answer

(A) The target molecule is $2	ext{-phenylbutan-2-ol}$,which is a tertiary $(3^o)$ alcohol. Option $A$: $HCHO + PhCH(CH_3)CH_2MgX$ gives a primary $(1^o)$ alcohol,$PhCH(CH_3)CH_2CH_2OH$. Option $B$: $PhCOCH_2CH_3 + CH_3MgX$ gives $CH_3-CH_2-C(Ph)(OH)-CH_3$. Option $C$: $PhCOCH_3 + CH_3CH_2MgX$ gives $CH_3-CH_2-C(Ph)(OH)-CH_3$. Option $D$: $CH_3CH_2COCH_3 + PhMgX$ gives $CH_3-CH_2-C(Ph)(OH)-CH_3$. Thus,option $A$ cannot prepare the target molecule.

$CH_3-CH_2-C(Ph)(OH)-CH_3$ can not be prepared by:

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