Ncert Exercises & Solutions

Compounds A and C in the following reaction are:

$\small{\mathrm{CH}_3 \mathrm{CHO} \xrightarrow[{(ii) H_{2}O}]{(i)CH_{3}MgBr}(\mathrm{A}) \stackrel{\mathrm{H}_2 \mathrm{SO}_4}{\longrightarrow}(\mathrm{B}) \xrightarrow[]{Hydroboration \ oxidation}(C)}$

1. Identical

2. Positional isomers

3. Functional isomers

4. Optical isomers

Hint: A and C both are alcohols

The reactions are as follows:

${CH}_{3} CHO \to_{(ii) H_{2} O}^{(i) {CH}_{3} MgBr} CH — \underset{|}{C}$
${CH}_{3} H — OH \to_{Dehydration}^{H_{2} {SO}_{4}, ∆} {CH}_{3} — \underset{|}{C}$
$H = {CH}_{2}$
$\to_{(Hydroboration Oxidation)}^{(i) {BH}_{3} (ii) H_{2} O_{2} / {OH}^{⊝}} {CH}_{3} — {CH}_{2} {CH}_{2} OH$

So structure A and C are the positional isomers because parent chain is same in both the compound but position of -OH group is different.

${CH}_{3} \underset{|}{C} H — OH and {CH}_{3} — {CH}_{2} - {CH}_{2} - OH$
${CH}_{3}$

Hence, option (2) is correct.

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