Question

Heterolysis of propane gives

• A. Methyl and ethyl free radical
• B. Methylium and ethyl free radical
• C. Methyl anion and ethylium cation
• D. Methylium and ethylium cation

Answer 1

The correct option is C

Methyl anion and ethylium cation

The explanation for the correct answer:-

Option (C): Methyl anion and ethylium cation

Heterolysis of propane:

• A covalent bond breaks in such a way that one product gets both of the shared electrons is known as heterolytic cleavage.
• The two electrons in the bond are divided equally between the products is known as homolytic cleavage,
• In propane, the electron density on the second carbon atom is greater than the electron density on terminal methyl groups.
• This is because of the electron-releasing (+I) effect of two methyl groups.
• The positive charge on ethylene carbon atoms is stabilized through the electron-donating (+I ) effect of the adjacent methyl groups.
• The heterolytic cleavage of propane could also have formed methyl cations and ethylium anion.

$\underset{\mathrm{Propane}}{{\mathrm{CH}}_3-{\mathrm{CH}}_2-{\mathrm{CH}}_3\to }\underset{\mathrm{Methylium}\mathrm{cation}}{{\mathrm{CH}}_3^{-}}+\underset{\mathrm{Ethylium}\mathrm{anion}}{{\mathrm{CH}}_3{\mathrm{CH}}_2^{+}}$

Therefore, Option C is correct methylium anion and ethylium cation.