Consider the following reaction sequence: CH_ | vedclass

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(D) Step $1$: Reaction of $CH_3CHO$ with $CH_3MgBr$ followed by hydrolysis gives $(A)$,which is propan$-2-$ol $(CH_3CH(OH)CH_3)$.Step $2$: Dehydration

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Position isomers

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(D) Step $1$: Reaction of $CH_3CHO$ with $CH_3MgBr$ followed by hydrolysis gives $(A)$,which is propan$-2-$ol $(CH_3CH(OH)CH_3)$.Step $2$: Dehydration of propan$-2-$ol with $H_2SO_4$ and heat gives $(B)$,which is propene $(CH_3CH=CH_2)$.Step $3$: Hydroboration-oxidation of propene $(CH_3CH=CH_2)$ gives $(C)$,which is propan$-1-$ol $(CH_3CH_2CH_2OH)$.Step $4$: $(A)$ is propan$-2-$ol and $(C)$ is propan$-1-$ol. These are position isomers because the position of the $-OH$ group differs.

Consider the following reaction sequence: $CH_3CHO$ $\xrightarrow[(ii) H_2O/H^+]{(i) CH_3MgBr} (A)$ $\xrightarrow{H_2SO_4, \Delta} (B)$ $\xrightarrow[(ii) H_2O_2/OH^-]{(i) B_2H_6} (C)$. $(A)$ and $(C)$ are

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