Optically active 3-methylpent-1-en-3-ol loses | vedclass

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Question

(A) The compound $3-$methylpent$-1-$en$-3-$ol is an allylic alcohol. In the presence of an acid,it undergoes protonation of the $-OH$ group followed b

Vedclass · Accepted Answer

$CH_3-CH_2-C(CH_3)=CH-CH_3$

Vedclass · Answer

(A) The compound $3-$methylpent$-1-$en$-3-$ol is an allylic alcohol. In the presence of an acid,it undergoes protonation of the $-OH$ group followed by the loss of water to form a resonance-stabilized carbocation: $CH_3-CH_2-C^+(CH_3)-CH=CH_2 \leftrightarrow CH_3-CH_2-C(CH_3)=CH-CH_2^+$.This carbocation can lose a proton to form a more stable,conjugated diene or a substituted alkene. Specifically,the loss of optical activity occurs because the chiral center at $C-3$ is destroyed upon the formation of the planar carbocation intermediate. The major product formed is $3-$methylpenta$-1,3-$diene or a related stable alkene isomer depending on the specific rearrangement,but among the given options,the most stable alkene formed via rearrangement is $3-$methylpent$-2-$ene $(CH_3-CH_2-C(CH_3)=CH-CH_3)$.

Optically active $3-$methylpent$-1-$en$-3-$ol loses its optical activity after standing in water containing a few drops of acid,mainly due to the formation of:

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