Q18 / 160 · TS EAMCET 2024 (11 May Shift 1) · Chemistry
Consider the following carbocations
I. $\quad \mathrm{C}_6 \mathrm{H}_5 \stackrel{+}{\mathrm{C}} \mathrm{H}_2$
II. $\mathrm{CH}_2=\stackrel{+}{\mathrm{C}} \mathrm{H}$
III. \(\mathrm{CH}_3-\underset{| \atop \large \mathrm{CH}_3}{\stackrel{+}{\mathrm{C}}}-\mathrm{H}\)
IV. $\mathrm{CH}_3-\stackrel{+}{\mathrm{C}} \mathrm{H}_2$
V. $\quad \mathrm{HC} \equiv \stackrel{+}{\mathrm{C}}$
Arrange the above carbocations in the order of decreasing stability
- Correct answerA$\mathrm{I}\gt\mathrm{III}\gt\mathrm{IV}\gt\mathrm{II}\gt\mathrm{V}$
- BV $\gt$ II $\gt$ IV $\gt$ III $\gt$ I
- C$\mathrm{V}\gt\mathrm{II}\gt\mathrm{III}\gt\mathrm{I}\gt\mathrm{IV}$
- DII $\gt$ III $\gt$ IV $\gt$ V $\gt$ I
Solution
Benzylic carbocation $\left(\mathrm{C}_6 \mathrm{H}_5-\stackrel{+}{\mathrm{C}} \mathrm{H}_2\right)$ is more stable due to $(+\mathrm{M})$ effect of $\mathrm{C}_6 \mathrm{H}_5$ then $3^{\circ}$ is more stable due to three methyle group then $1^{\circ}$ carbocation.
In $\mathrm{CH}_2=\stackrel{+}{\mathrm{C}} \mathrm{H}$ and $\mathrm{HC} \stackrel{\oplus}{\mathrm{C}}$, due to more S character in alkyne its stability is lower then alkene. So order will ue $\mathrm{I}\gt\mathrm{III}\gt\mathrm{IV}\gt\mathrm{II}\gt\mathrm{V}$
Hence, the correct answer is option: 'A'.