Consider the following reactions:Ph-CH_2-CH=C | vedclass

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(D) $P$: Acid-catalyzed hydration of $Ph-CH_2-CH=CH_2$ involves the formation of a more stable carbocation. The initial carbocation $Ph-CH_2-CH^+-CH_3

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$P = Ph-CH(OH)-CH_2-CH_3$,$Q = Ph-CH_2-CH(OH)-CH_2D$,$R = Ph-CH_2-CH(D)-CH_2OH$

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(D) $P$: Acid-catalyzed hydration of $Ph-CH_2-CH=CH_2$ involves the formation of a more stable carbocation. The initial carbocation $Ph-CH_2-CH^+-CH_3$ rearranges to the more stable benzylic carbocation $Ph-CH^+-CH_2-CH_3$,which then reacts with $H_2O$ to give $Ph-CH(OH)-CH_2-CH_3$.$Q$: Oxymercuration-demercuration using $Hg(OAc)_2/H_2O$ followed by $NaBD_4$ results in Markovnikov addition of $H$ and $OD$ (from $D$ in $NaBD_4$) across the double bond without rearrangement. The product is $Ph-CH_2-CH(OH)-CH_2D$.$R$: Hydroboration-oxidation using $BD_3/THF$ followed by $H_2O_2/OH^-$ results in anti-Markovnikov addition of $D$ and $OH$ across the double bond. The boron adds to the less substituted carbon,and subsequent oxidation replaces boron with $OH$. The product is $Ph-CH_2-CH(D)-CH_2OH$.

Consider the following reactions:
$Ph-CH_2-CH=CH_2 \xrightarrow{H^+/H_2O} P$
$Ph-CH_2-CH=CH_2 \xrightarrow[(ii) NaBD_4]{(i) Hg(OAc)_2, H_2O} Q$
$Ph-CH_2-CH=CH_2 \xrightarrow[(ii) H_2O_2/OH^-]{(i) BD_3, THF} R$
Identify the products $P, Q$ and $R$.

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Consider the following reactions:$Ph-CH_2-CH=CH_2 \xrightarrow{H^+/H_2O} P$$Ph-CH_2-CH=CH_2 \xrightarrow[(ii) NaBD_4]{(i) Hg(OAc)_2, H_2O} Q$$Ph-CH_2-CH=CH_2 \xrightarrow[(ii) H_2O_2/OH^-]{(i) BD_3, THF} R$Identify the products $P, Q$ and $R$.