Question

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.

Short answer

${{\text{(CH}}_3\text{)}}_3\text{CBr}$ will react faster.

Step-by-step solution

Inductive effect

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.

Carbocation

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

Stability of carbocation

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.

S CHARACTER

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.

Comparison of given structures

• Both compounds are tertiary.
• ${{\text{(CH}}_3\text{)}}_3\text{CBr}$has more hyperconjugation effect than ${\text{H}}_2{\text{C=CHC(CH}}_3{\text{)}}_2\text{Br}$.
• role="math" localid="1651500395854" ${{\text{(CH}}_3\text{)}}_3\text{CBr}$has carbon with role="math" localid="1651500433317" ${\mathrm{sp}}^3$hybridisation while ${\text{H}}_2{\text{C=CHC(CH}}_3{\text{)}}_2\text{Br}$alkene has hybridisation, which shows a negative inductive effect, making carbocation less stable${{\text{(CH}}_3\text{)}}_3\text{CBr}$.

• So,${{\text{(CH}}_3\text{)}}_3\text{CBr}$ will react faster.