The final product A formed in the following m | vedclass

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(A) Step $1$: Hydration of styrene $(C_6H_5CH=CH_2)$ with $H_2O, H^+$ follows Markovnikov's rule to yield $1$-phenylethanol $(C_6H_5CH(OH)CH_3)$.Step

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Ethylbenzene

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(A) Step $1$: Hydration of styrene $(C_6H_5CH=CH_2)$ with $H_2O, H^+$ follows Markovnikov's rule to yield $1$-phenylethanol $(C_6H_5CH(OH)CH_3)$.Step $2$: Oxidation of $1$-phenylethanol with $CrO_3$ yields acetophenone $(C_6H_5COCH_3)$.Step $3$: Wolff-Kishner reduction of acetophenone using $H_2N-NH_2, KOH$ and heating reduces the carbonyl group to a methylene group,yielding ethylbenzene $(C_6H_5CH_2CH_3)$ as the final product $A$.

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	$IV$. $Zn-Hg / HCl$

The final product $A$ formed in the following multistep reaction sequence is:

Similar Questions

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 $IV$. $Zn-Hg / HCl$

Silver Mirror test is given by which one of the following compounds?

An alkene $A$ $(C_4H_8)$ exhibits cis/trans isomerism. $A$ on ozonolysis gives $B$,which when reacted with $NaOH$ followed by hydroxylamine gives $C$. What are $B$ and $C$?

The product of the reaction between ammonia and formaldehyde is

The reaction $C_6H_5CHO + CH_3CHO \to C_6H_5CH=CHCHO + H_2O$ is known as:

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