Skip to main content
Back to ALCOHOLS & PHENOLS & ETHERS

JEE CHEMISTRY MCQ: ALCOHOLS & PHENOLS & ETHERS

1 Medium
1 of 1
Analyzing +4 | -1
C4H8O2(Ester A)+CH3MgBr(2 parts)→H3O+C4H10OAlcohol(B)\underset{\text{(Ester A)}}{\mathrm{C}_4 \mathrm{H}_8 \mathrm{O}_2}+\underset{\text{(2 parts)}}{\mathrm{CH}_3 \mathrm{MgBr}} \xrightarrow{\mathrm{H}_3 \mathrm{O}^{+}} \underset{\text{Alcohol(B)}}{\mathrm{C}_4 \mathrm{H}_{10} \mathrm{O}}

Alcohol B\text{B} reacts faster with Lucas reagent. Hence A\text{A} and B\text{B} are
View answer and explanation
Correct option: A
CH3−COO−C2H5,(CH3)3COH\mathrm{CH}_3-\mathrm{COO}-\mathrm{C}_2 \mathrm{H}_5,\left(\mathrm{CH}_3\right)_3 \mathrm{COH}
Option A is correct because the ester CH3−COO−C2H5\mathrm{CH}_3-\mathrm{COO}-\mathrm{C}_2 \mathrm{H}_5 reacts with excess Grignard reagent to form the tertiary alcohol (CH3)3COH(\mathrm{CH}_3)_3 \mathrm{COH}, which reacts fastest with Lucas reagent.

More Questions from ALCOHOLS & PHENOLS & ETHERS

Practice All Chapter MCQs →
Type at least 2 characters or jump directly:
🔓

Save your learning progress

Sign in with a free account to save your practice answers and quiz results, then return to them when you revise.