JEE Main Chemistry: General Organic Chemistry High-Yield Practice Questions, Formulas & PYQ Breakdown (2026-27)

Correct Answer: (c)

Explanation: Stability of carbanions A carbanion is a nucleophile, which stability and reactivity determined by several factors. The inductive effect, electronegative atoms adjacent to the charge will stabilise the charge. Hybridisation of the charge-bearing atom The greater the $s$-character of the charge bearing atom, the more stable the anion. The extent of conjugation of the anion Resonance effects can stabilise the anion. This is especially true, when the anion is stabilised as a result of aromaticity. $ \left(\mathrm{C}_6 \mathrm{H}_5\right)_2 \mathrm{CH}^{-}>\mathrm{CH}_2=\mathrm{CH}-\mathrm{CH}_2^{-}>\mathrm{HC} \equiv \mathrm{C}^{-}> $ $ \begin{array}{lll} s p^2 \text { more stable } & \text { Stable due } & \text { More } \quad \mathrm{H}_2 \mathrm{C}=\mathrm{CH}^{-} \\ \text {due to } & \text { to conjugation } & s \text {-character } \\ \text { aromaticity } & \text { of charge } & s p \text {-hybridised } \end{array} $ Therefore, IV $>$ II $>$ I $>$ III Hence, correct option is (c).

Correct Answer: (b)

Explanation: The delocalisation of $\sigma$ electrons of $\mathrm{C}-\mathrm{H}$ bond of an alkyl group linked directly to an atom of unsaturated system or to an atom having unshared $p$-orbital (as in benzene) is observed in hyperconjugation effect.  

Correct Answer: (d)

Explanation: An alkyl group exerts electron releasing inductive effect ( +I ) which stabilizes positive charge on atom bonded to it. As we move from conjugate acid of ammonia $\left(\mathrm{NH}_4^{+}\right)$to that of tertiary amine $\left(\mathrm{R}_3 \stackrel{+}{\mathrm{N}}-\mathrm{H}\right)$, the number of alkyl groups $(\mathrm{R})$ bonded to nitrogen goes on increasing steadily. This results in increasing stabilization of the positive charge. Order of stabilization: $ \stackrel{+}{\mathrm{N}} \mathrm{H}_4

Correct Answer: (c)

Explanation: ---

Correct Answer: (a)

Explanation: The key to radical stability is how well the unpaired electron can be delocalized. Here’s a quick comparison: Methyl radical, $\dot{\mathrm{C}}\mathrm{H}_3$ (B) – No resonance, only 3 hyperconjugative C–H bonds. Vinyl radical, $\mathrm{CH}_2=\dot{\mathrm{C}}\mathrm{H}$ (D) – Radical on an sp² carbon, very little stabilization. Tertiary radical, $(\mathrm{CH}_3)_3\dot{\mathrm{C}}$ (C) – Hyperconjugation from 9 C–H bonds, moderately stabilized. Diphenylmethyl (benzhydryl) radical, $(\mathrm{C}_6\mathrm{H}_5)_2\dot{\mathrm{C}}\mathrm{H}$ (A) – Unpaired electron is delocalized over two aromatic rings (resonance), giving the greatest stabilization. Because resonance stabilization (especially over two phenyl rings) outweighs hyperconjugation, the most stable radical here is option A.