Question

: Which of the following statements is not true about carbanions?
(A) Stability of carbanions increases with an increase in $s-$character of orbital
(B) Electron withdrawing groups like $-NO_2​, -CN, -C​=O$ increase the stability of carbanions.
(C) Order of stability of carbanions is $3^\circ > 2^\circ > 1\circ$
(D) The negatively charged carbon is $sp_3$ hybridized and tetrahedral.

Answer 1

Hint: A carbanion is an organic ion that contains a negative charge on the carbon atom. There are three types of carbanion - $1^\circ$, and $3^\circ$. The general formula for $1^\circ$ carbanion is $RCH_2^-$. The general formula for $2^\circ$ carbanion is $R_2CH^-$, and the general formula for $3^\circ$ carbanion is $R_3C:^-$. The stability of a carbanion is inversely proportional to the electron density on the central ion. The greater the electron density, the lesser the stability of the carbanion.

Complete step by step answer:
The $s-$ character in an orbital is defined as the percentage of contribution of $s-$ orbital in a compound. The stability of a carbanion is directly proportional to the $s-$ character of the orbitals present in it. The greater the $s-$ character, the greater the stability of the carbanion.
Electron-withdrawing groups are the groups that have a negative inductive effect $(-I\ effect)$. Groups with $-I$ effect pull or withdraw the electrons towards themselves. This reduces the electron density on the carbon atom by pulling the electrons towards themselves. As the electron density decreases, the stability of the carbanion increases. Therefore, electron-withdrawing groups like $-NO_2, -CN, -C=O$ increase the stability of carbanions.

As the degree of carbanions increases from $1$ to $3$, the number of methyl groups attached to the central carbon anion also increases. $1\circ$ carbanion contains one methyl $(CH_3)$ group and two hydrogen atoms attached to the negative carbon ion. $2\circ$ carbanion contains two methyl $(CH_3)$ groups and one hydrogen atom attached to the negative carbon ion, and $3\circ$ carbanion contains three methyl $(CH_3)$ groups attached to the negative carbon ion.

Now, the methyl group is an electron-donating group with a positive inductive $(+I)$ effect. It means that a methyl group donates an electron to a carbon ion. As the number of methyl groups increases, the electron density on the carbon anion increases. Due to an increase in the electron density on the carbon ion from $1\circ$ to $3\circ$, the stability of the carbanion decreases in the same order.

The negatively charged carbon in a carbanion is surrounded by single bonds. So, it is $sp^3$ hybridized. An $sp^3$ hybridized orbital has a bond angle of $109.5\circ$ and hence, forms a tetrahedral shape.

Therefore, it can be concluded that the only incorrect statement among the given options is that $3^\circ > 2^\circ > 1\circ$.

So, the correct answer is Option C .

Note: For electron-donating groups, the direction of the movement of electrons is towards the negative carbon ion, which increases the electron density. For electron-withdrawing groups, the direction of the movement of electrons is away from the negative carbon ion, which decreases the electron density.