Compounds P and R in the following reaction a | vedclass

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(A) $1$. The reaction sequence is as follows:$CH_3CHO \xrightarrow[(ii) H_3O^+]{(i) CH_3MgBr} CH_3-CH(OH)-CH_3 (P)$$CH_3-CH(OH)-CH_3 \xrightarrow[\Del

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Position isomers

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(A) $1$. The reaction sequence is as follows:$CH_3CHO \xrightarrow[(ii) H_3O^+]{(i) CH_3MgBr} CH_3-CH(OH)-CH_3 (P)$$CH_3-CH(OH)-CH_3 \xrightarrow[\Delta]{conc. H_2SO_4} CH_3-CH=CH_2 (Q)$$CH_3-CH=CH_2 \xrightarrow[(ii) H_2O_2/OH^-]{(i) B_2H_6} CH_3-CH_2-CH_2OH (R)$$2$. Compound $P$ is propan$-2-$ol $(CH_3-CH(OH)-CH_3)$ and compound $R$ is propan$-1-$ol $(CH_3-CH_2-CH_2OH)$.$3$. Both compounds have the same molecular formula $(C_3H_8O)$ and the same functional group $(-OH)$,but the position of the hydroxyl group is different (carbon-$2$ in $P$ and carbon-$1$ in $R$).$4$. Therefore,$P$ and $R$ are position isomers.

Compounds $P$ and $R$ in the following reaction are:

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