The starting material is propene (CH_3-CH=CH_ | vedclass

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(C) $1$. The starting material is propene $(CH_3-CH=CH_2)$.$2$. Acid-catalyzed hydration $(H_3O^+)$ of propene follows Markovnikov's rule,where the hy

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Position isomer

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(C) $1$. The starting material is propene $(CH_3-CH=CH_2)$.$2$. Acid-catalyzed hydration $(H_3O^+)$ of propene follows Markovnikov's rule,where the hydroxyl group attaches to the more substituted carbon,yielding propan$-2-$ol $(CH_3-CH(OH)-CH_3)$ as product $A$.$3$. Hydroboration-oxidation $((i) BH_3/THF, (ii) H_2O_2/OH^-)$ of propene follows anti-Markovnikov's rule,where the hydroxyl group attaches to the less substituted carbon,yielding propan$-1-$ol $(CH_3-CH_2-CH_2OH)$ as product $B$.$4$. Comparing $A$ (propan$-2-$ol) and $B$ (propan$-1-$ol),both are alcohols with the same molecular formula $(C_3H_8O)$,but the position of the functional group $(-OH)$ differs. Therefore,they are position isomers.

Similar Questions

Identify the pairs of compounds that represent chain isomerism among the following:
$I. CH_3-CH_2-CH_2-CH_2-OH$ (Butan$-1-$ol)
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$IV. (CH_3)_2CH-CH_2-OH$ ($2$-Methylpropan$-1-$ol)

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