Match the reactions in List-I with their majo | vedclass

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(A) . Hydration of propyne $(CH_3-C \equiv CH)$ in the presence of $Hg^{2+}/H^+$ follows Markovnikov's rule to form an enol,which tautomerizes to acet

Vedclass · Accepted Answer

$A-II, B-IV, C-I, D-III$

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(A) . Hydration of propyne $(CH_3-C \equiv CH)$ in the presence of $Hg^{2+}/H^+$ follows Markovnikov's rule to form an enol,which tautomerizes to acetone $(CH_3-CO-CH_3)$,which is $II$.$B$. Kolbe's electrolysis of sodium acetate $(CH_3COONa)$ yields ethane $(CH_3-CH_3)$,which is $IV$.$C$. Acid-catalyzed hydration of propene $(CH_3-CH=CH_2)$ follows Markovnikov's rule to form propan$-2-$ol $(CH_3-CH(OH)-CH_3)$,which is $I$.$D$. Hydroxylation of propene using cold dilute alkaline $KMnO_4$ (Baeyer's reagent) yields propane$-1,2-$diol $(CH_3-CH(OH)-CH_2OH)$,which is $III$.Therefore,the correct match is $A-II, B-IV, C-I, D-III$.

List-$I$ (Reaction)	List-$II$ (Major product)
$A. CH_3-C \equiv CH \xrightarrow{H_2O, Hg^{2+}/H^+, 333K}$	$I. CH_3-CH(OH)-CH_3$
$B. CH_3COONa \xrightarrow{H_2O, Pt, \text{electrolysis}}$	$II. CH_3-CO-CH_3$
$C. CH_3-CH=CH_2 \xrightarrow{H_2O/H^+}$	$III. CH_3-CH(OH)-CH_2OH$
$D. CH_3-CH=CH_2 \xrightarrow{H_2O, \text{dil. } KMnO_4, 273K}$	$IV. CH_3-CH_3$
	$V. CH_4$

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