Question:
[P] on treatment with $\mathrm{Br}_{2} / \mathrm{FeBr}_{3}$ in $\mathrm{CCl}_{4}$ produced a single isomer $\mathrm{C}_{8} \mathrm{H}_{7} \mathrm{O}_{2} \mathrm{Br}$ while heating $[\mathrm{P}]$ with sodalime gave toluene. The compound [P] is :
Correct Option:
Solution:
A certain orbital has no angular nodes and two radial nodes. The orbital is:
(1) $2 \mathrm{~s}$
(2) $3 \mathrm{~s}$
(3) $3 \mathrm{p}$
(4) $2 \mathrm{p}$
Click here to get exam-ready with eSaral
For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.