NEET 2020 Chemistry Question 112 Solution

Carbocations possess a sp2-hybridized planar central carbon with an unhybridized, empty p-orbital perpendicular to the molecular plane.

Hyperconjugation involves the overlap of $$\sigma$$-electrons of adjacent $$\text{C}-\text{H}$$ bonds of the alkyl group with the empty $$p$$-orbital of the carbocationic carbon (also called no-bond resonance). This delocalizes the positive charge across multiple hydrogen atoms, dispersing charge and stabilizing the cation.

- Tertiary butyl carbocation, $$(\text{CH}_3)_3\text{C}^+$$, has three methyl groups directly attached to the carbocationic carbon, giving a total of $$3 \times 3 = 9\text{ }\alpha\text{-hydrogens}$$, yielding 9 hyperconjugative canonical structures. - Secondary butyl carbocation, $$\text{CH}_3-\text{CH}^+-\text{CH}_2-\text{CH}_3$$, has one methyl ($$3\text{ }\alpha\text{-H}$$) and one methylene ($$2\text{ }\alpha\text{-H}$$), totaling only $$5\text{ }\alpha\text{-hydrogens}$$, yielding 5 hyperconjugative structures.

- Methyl groups exhibit $$+I$$ effect, NOT $$-I$$ effect (Option 1 is false). - Methyl groups lack lone pairs or $$\pi$$-bonds and CANNOT exhibit resonance ($$+R$$ or $$-R$$) effects (Options 2 and 3 are chemically invalid). - Thus, Hyperconjugation uniquely and correctly accounts for the superior stability of tert-butyl carbocation.