Search any question & find its solution
Question:
Answered & Verified by Expert
Arrange the carbanions, $\left(\mathrm{CH}_3\right)_3 \overline{\mathrm{C}}, \overline{\mathrm{C}} \mathrm{Cl}_3,\left(\mathrm{CH}_3\right)_2 \overline{\mathrm{C}} \mathrm{H}, \mathrm{C}_6 \mathrm{H}_5 \overline{\mathrm{C}} \mathrm{H}_2$, in order of their decreasing stability :
Options:
Solution:
2901 Upvotes
Verified Answer
The correct answer is:
$\overline{\mathrm{C}} \mathrm{Cl}_3>\mathrm{C}_8 \mathrm{H}_5 \overline{\mathrm{C}} \mathrm{H}_2>\left(\mathrm{CH}_3\right)_2 \overline{\mathrm{C}} \mathrm{H}>\left(\mathrm{CH}_3\right)_3 \overline{\mathrm{C}}$
$\overline{\mathrm{C}} \mathrm{Cl}_3>\mathrm{C}_8 \mathrm{H}_5 \overline{\mathrm{C}} \mathrm{H}_2>\left(\mathrm{CH}_3\right)_2 \overline{\mathrm{C}} \mathrm{H}>\left(\mathrm{CH}_3\right)_3 \overline{\mathrm{C}}$
$2^{\circ}$ carbanion is more stable than $3^{\circ}$ and $\mathrm{Cl}$ is $-1$ effect group.
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.