In the given pairs of alcohols,in which pair | vedclass

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(D) The reaction of alcohols with $HBr$ proceeds via an $S_N1$ mechanism,where the rate-determining step is the formation of a carbocation intermediat

Vedclass · Accepted Answer

$CH_3-CH(OH)-CH_2-CH_3$ and $(CH_3)_2C(OH)-CH_2-CH_3$

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(D) The reaction of alcohols with $HBr$ proceeds via an $S_N1$ mechanism,where the rate-determining step is the formation of a carbocation intermediate. The stability of the carbocation determines the reactivity of the alcohol; more stable carbocations are formed more easily,making the parent alcohol more reactive.$(a)$ $1-$methylcyclopentanol forms a tertiary $(3^{\circ})$ carbocation,while cyclohexanol forms a secondary $(2^{\circ})$ carbocation. Thus,the first is more reactive.$(b)$ $1-$ethylcyclopentanol forms a tertiary $(3^{\circ})$ carbocation,while $1-$cyclopentylethanol forms a secondary $(2^{\circ})$ carbocation. Thus,the first is more reactive.$(c)$ $CH_3-CH(OH)-CH_2-CH_3$ (butan$-2-$ol) forms a secondary $(2^{\circ})$ carbocation,while $CH_3-CH_2-CH(CH_3)-CH_2OH$ ($2$-methylbutan$-1-$ol) forms a primary $(1^{\circ})$ carbocation. Thus,the first is more reactive.$(d)$ $CH_3-CH(OH)-CH_2-CH_3$ (butan$-2-$ol) forms a secondary $(2^{\circ})$ carbocation,while $(CH_3)_2C(OH)-CH_2-CH_3$ ($3$-methylpentan$-3-$ol) forms a tertiary $(3^{\circ})$ carbocation. Since the tertiary carbocation is more stable than the secondary one,the second alcohol is more reactive than the first.

In the given pairs of alcohols,in which pair is the second alcohol more reactive than the first towards hydrogen bromide $(HBr)$?

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