Which one of the following compounds will be | vedclass

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(C) The dehydration of alcohols involves the formation of a carbocation intermediate. The stability of the carbocation determines the ease of dehydrat

Vedclass · Accepted Answer

$4-$hydroxypentan$-2-$one

Vedclass · Answer

(C) The dehydration of alcohols involves the formation of a carbocation intermediate. The stability of the carbocation determines the ease of dehydration. The more stable the carbocation,the easier the dehydration. In the given compounds,the hydroxyl group is located at different positions relative to the carbonyl group. For option $C$ ($4$-hydroxypentan-$2$-one),the carbocation formed after the loss of the $-OH$ group is a secondary carbocation that is relatively more stable due to its position,allowing for easier formation of the conjugated enone system. Comparing the stability of the intermediate carbocations: $(b)$ and $(d)$ form secondary carbocations adjacent to the carbonyl group (destabilized by the electron-withdrawing effect of the carbonyl group). $(a)$ forms a secondary carbocation adjacent to the carbonyl group. $(c)$ forms a secondary carbocation at the $4$-position,which is further from the electron-withdrawing carbonyl group,making it more stable than the others. Therefore,the compound in option $C$ undergoes dehydration most readily.

List-$I$ Reaction	List-$II$ Reagents
$A$. Hoffmann Degradation	$I$. Conc. $KOH, \Delta$
$B$. Clemmensen reduction	$II$. $CHCl_3, NaOH / H_3O^{+}$
$C$. Cannizzaro reaction	$III$. $Br_2, NaOH$
$D$. Reimer-Tiemann reaction	$IV$. $Zn-Hg / HCl$

Which one of the following compounds will be most readily dehydrated?

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Match List-$I$ with List-$II$.

List-$I$ Reaction List-$II$ Reagents

$A$. Hoffmann Degradation $I$. Conc. $KOH, \Delta$

$B$. Clemmensen reduction $II$. $CHCl_3, NaOH / H_3O^{+}$

$C$. Cannizzaro reaction $III$. $Br_2, NaOH$

$D$. Reimer-Tiemann reaction $IV$. $Zn-Hg / HCl$

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