The final product A,formed in the following r | vedclass

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(D) Step $1$: Hydroboration-oxidation of styrene $(Ph-CH=CH_2)$ gives anti-Markovnikov addition of water,resulting in $Ph-CH_2-CH_2-OH$.Step $2$: Reac

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$Ph-CH_2-CH_2-CH_2-OH$

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(D) Step $1$: Hydroboration-oxidation of styrene $(Ph-CH=CH_2)$ gives anti-Markovnikov addition of water,resulting in $Ph-CH_2-CH_2-OH$.Step $2$: Reaction with $HBr$ converts the alcohol to an alkyl bromide: $Ph-CH_2-CH_2-OH + HBr ightarrow Ph-CH_2-CH_2-Br + H_2O$.Step $3$: Formation of Grignard reagent and reaction with formaldehyde $(HCHO)$ followed by hydrolysis:$Ph-CH_2-CH_2-Br + Mg \xrightarrow{	ext{ether}} Ph-CH_2-CH_2-MgBr$$Ph-CH_2-CH_2-MgBr + HCHO ightarrow Ph-CH_2-CH_2-CH_2-OMgBr$$Ph-CH_2-CH_2-CH_2-OMgBr \xrightarrow{H_3O^{+}} Ph-CH_2-CH_2-CH_2-OH$.Thus,the final product $A$ is $Ph-CH_2-CH_2-CH_2-OH$.

The final product $A$,formed in the following reaction sequence is:
$Ph-CH=CH_2$ $\xrightarrow[\text{(iii) } HBr, \text{ (iv) } Mg, \text{ ether, then } HCHO / H_3O^{+}]{\text{(i) } BH_3, \text{ (ii) } H_2O_2, OH^{-}} A$

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The final product $A$,formed in the following reaction sequence is:$Ph-CH=CH_2$ $\xrightarrow[\text{(iii) } HBr, \text{ (iv) } Mg, \text{ ether, then } HCHO / H_3O^{+}]{\text{(i) } BH_3, \text{ (ii) } H_2O_2, OH^{-}} A$