Search any question & find its solution
Question:
Answered & Verified by Expert
The ' $\mathrm{X}$ ' in the following conversion is
$$
\mathrm{H}_3 \mathrm{CCH}_2 \mathrm{COOH} \xrightarrow{\mathrm{x}} \mathrm{CH}_3 \mathrm{CHBrCOOH}
$$
Options:
$$
\mathrm{H}_3 \mathrm{CCH}_2 \mathrm{COOH} \xrightarrow{\mathrm{x}} \mathrm{CH}_3 \mathrm{CHBrCOOH}
$$
Solution:
1469 Upvotes
Verified Answer
The correct answer is:
(i) $\mathrm{Br}_2 / \mathrm{P}$ red, (ii) $\mathrm{H}_2 \mathrm{O}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.