Tertiary alkyl halides,${\mathbf{R}}_3\mathbf{CX}$, undergo spontaneous dissociation to yield a carbocation,${\mathbf{R}}_3{\mathbf{C}}^{+}$, plus halide ion. Which do you think reacts faster,${{\text{(CH}}_3\text{)}}_3{\text{CBrorH}}_2{\text{C=CHC(CH}}_3{\text{)}}_2\text{Br}$? Explain.

When atoms of different electronegativity form a covalent bond, the electron density is more toward a more electronegative atom in the bond. This is known as an inductive effect.

–I-Effect is when the substituent attached is electron-withdrawing. +I-Effect is when the substituent attached is electron-donating. The Halogen group is an electron-withdrawing group, so it shows the –I effect, while the alkyl group attached to the halogen are electron-donating, showing the +I effect.

Carbocations are species in which a carbon atom has a positive charge. They are electron-deficient species.

There are two factors for the stability of carbocations:

• Inductive effect: More the alkyl groups will be an inductive effect. Hence, tertiary carbocation is most stable.
• Hyperconjugation effect: Hyperconjugation refers to the delocalisation of electrons with sigma bonds. More the hyperconjugation will be the stability of carbocation.

Alkenes have hybridized orbitals, due to which the percentage s character will be more than that of alkanes which are hybridized. S-orbital is near the nucleus, so they don’t donate electrons; instead, electrons are easily attracted to the nucleus.

S character is more for alkenes, so they withdraw electrons by – I effect.