Consider the following reaction sequence:CH_3 | vedclass

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(B) $1$. Reaction of $CH_3CHO$ with $CH_3MgBr$ followed by hydrolysis gives $A$,which is $CH_3-CH(OH)-CH_3$ (Propan$-2-$ol).$2$. Dehydration of $A$ wi

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Position isomers

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(B) $1$. Reaction of $CH_3CHO$ with $CH_3MgBr$ followed by hydrolysis gives $A$,which is $CH_3-CH(OH)-CH_3$ (Propan$-2-$ol).$2$. Dehydration of $A$ with $Conc. H_2SO_4$ at high temperature gives $B$,which is $CH_3-CH=CH_2$ (Prop$-1-$ene).$3$. Hydroboration-oxidation of $B$ $(CH_3-CH=CH_2)$ with $(i) B_2H_6$ and $(ii) H_2O_2/OH^-$ follows anti-Markovnikov addition of water to give $C$,which is $CH_3-CH_2-CH_2OH$ (Propan$-1-$ol).$4$. Comparing $A$ $(CH_3-CH(OH)-CH_3)$ and $C$ $(CH_3-CH_2-CH_2OH)$,the position of the $-OH$ group is different (carbon-$2$ vs carbon-$1$). Therefore,they are position isomers.

Consider the following reaction sequence:
$CH_3CHO$ $\xrightarrow[(ii) H_3O^+]{(i) CH_3MgBr} A$ $\xrightarrow[\Delta]{Conc. H_2SO_4} B$ $\xrightarrow[(ii) H_2O, OH^-]{(i) B_2H_6} C$
$A$ and $C$ are:

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Consider the following reaction sequence:$CH_3CHO$ $\xrightarrow[(ii) H_3O^+]{(i) CH_3MgBr} A$ $\xrightarrow[\Delta]{Conc. H_2SO_4} B$ $\xrightarrow[(ii) H_2O, OH^-]{(i) B_2H_6} C$$A$ and $C$ are: