Question

In which of the following carbocation, there is maximum stabilization due to the hyperconjugation effect?

• A. Neopentyl
• B. Tert-Butyl
• C. Isopropyl
• D. Ethyl

Answer 1

The correct option is B

Tert-Butyl

Explanation for correct option:

Option (b) Tert-Butyl

• Hyperconjugation refers to the interaction of electrons in p systems (multiple bonds) with nearby s bonds (single H–C bonds) of substituent groups in organic compounds.
• The stability order of various carbocations is given as:

$3°>2°>1°>\mathrm{methyl}$

• When compared to other secondary and primary carbocations, the carbon atom in Tertiary Butyl (${\left({\mathrm{CH}}_3\right)}_3\mathrm{C}$) carbocation is wrapped by three methyl groups, resulting in maximal hyper conjugative structures and maximum stability.

• So, Tertiary Butyl has the maximum number of $9\mathrm{\alpha }-$hydrogens and shows nine hyperconjugation structures and is hence the most stable.

Explanation for incorrect options:

Option (a) Neopentyl

• The Neopentyl group (${\left({\mathrm{CH}}_3\right)}_3{\mathrm{CCH}}_2-$) has only eight hyperconjugation structures.
• So, this option is incorrect.

Option (c) Isopropyl

• Isopropyl ($—\mathrm{CH}{\left({\mathrm{CH}}_3\right)}_2$) is $2°$ carbocation and thus is less stable than Tert-butyl.
• So, this option is not correct.

Option (d) Ethyl

• Ethyl ($-{\mathrm{C}}_2{\mathrm{H}}_5$) is the least stable
• This is due to the lesser number of alpha-Hydrogens present in the Ethyl group.
• So, this option is also incorrect.

Hence, correct option is (b) Tert-butyl.