Question

Which one of the following carbocation is most stable?

• A. $\left(\mathrm{CH}3\right)3\mathrm{C}\stackrel{+}{\mathrm{C}}\mathrm{H}$
• B. $\left(\mathrm{CH}3\right)3\stackrel{+}{\mathrm{C}}$
• C. ${\mathrm{CH}}_3{\mathrm{CH}}_2\mathrm{C}\stackrel{+}{{\mathrm{H}}_2}$
• D. ${\mathrm{CH}}_3\mathrm{C}\stackrel{+}{\mathrm{H}}{\mathrm{CH}}_2{\mathrm{CH}}_3$

Answer 1

The correct option is B

$\left(\mathrm{CH}3\right)3\stackrel{+}{\mathrm{C}}$

Explanation of the correct options:

The correct answers are options B and C:

Carbocation:

• A molecule in which a carbon atom has a positive charge and three bonds. We can say that they are carbon cations.
• Carbocation is defined as any even-electron cation that possesses a significant positive charge on the carbon atom.
• $\left(\mathrm{CH}3\right)3\stackrel{+}{\mathrm{C}}$ is most stable as it is tertiary carbocation.

Determination of stability of carbocation:

• A positively charged species like carbocation is electron-poor species, and thus any other species (electron-rich species) that can donate electron density to the positively charged carbon atom, will help to stabilize it. It means a carbocation will be stabilized by an electron donating group(electron rich species).
• Alkyl groups such as methyl, ethyl are electron donating groups, and thus stabilize nearby carbocations by donation electron.
• This means that more substituted carbocations ($3°$) are more stable i.e., a tert-butyl carbocation ($3°$), is more stable (due to presence of three methyl groups) than an isopropyl carbocation(due to presence of two methyl groups).
• Primary carbocations (due to presence of one methyl group) are highly unstable. Methyl carbocations are even less stable (due to absence of donation group).

Therefore, the correct order of stability of a given carbocation is $\left(\mathrm{CH}3\right)3\stackrel{+}{\mathrm{C}}$ ($3°$ Carbocation) $>$${\mathrm{CH}}_3\mathrm{C}\stackrel{+}{\mathrm{H}}{\mathrm{CH}}_2{\mathrm{CH}}_3$ ($2°$ Carbocation) $>$${\mathrm{CH}}_3{\mathrm{CH}}_2\mathrm{C}\stackrel{+}{{\mathrm{H}}_2}$($1°$ Carbocation).

Note: $\left(\mathrm{CH}3\right)3\mathrm{C}\stackrel{+}{\mathrm{C}}\mathrm{H}$ is also a $1°$ Carbocation which is the least stable.

Therefore, Option B is correct.