Problem 114 - A Complete Resource Book in Chemistry for JEE Main

Question

Consider the following carbanions: (1) \(\mathrm{CH}_{3}-\stackrel{\Theta}{\mathrm{C}} \mathrm{H}_{2}\) (2) \(\mathrm{CH}_{3}=\mathrm{C} \mathrm{H}\) (3) \(\mathrm{CH} \equiv \stackrel{\theta}{\mathrm{C}}\) Correct order of stability of these carbanions in decreasing order is (a) \(1>2>3\) (b) \(2>1>3\) (c) \(3>2>1\) (d) \(3>1>2\)

Step-by-Step Solution

Answer

The correct order of stability is (c) \(3 > 2 > 1\).

Step 1: Understanding Carbanions

A carbanion is a negatively charged carbon atom with a lone pair of electrons. The stability of a carbanion is influenced by the hybridization of the carbon atom carrying the negative charge, as well as the inductive and resonance effects of surrounding groups.

Step 2: Determine Hybridization

Identify the hybridization of the carbon atom in each carbanion:1. \(\mathrm{CH}_{3}-\stackrel{\Theta}{\mathrm{C}} \mathrm{H}_{2}\): The carbon is \(sp^3\) hybridized.2. \(\mathrm{CH}_{3}=\mathrm{C} \mathrm{H}\): The carbon is \(sp^2\) hybridized.3. \(\mathrm{CH} \equiv \stackrel{\theta}{\mathrm{C}}\): The carbon is \(sp\) hybridized.

Step 3: Evaluate Stability Based on Hybridization

The stability of a carbanion increases with increasing \s\text{character}\ due to higher electronegativity in more \s\text{character}\. \(sp\) hybridized (50% s-character) is more stable than \(sp^2\) (33% s-character), which is more stable than \(sp^3\) (25% s-character).

Step 4: Arranging the Carbanions

Based on hybridization effects and their s-character, the correct order of stability in decreasing order is: \(3 > 2 > 1\).

Key Concepts

Hybridization

Hybridization is an essential concept in understanding molecular geometry and the stability of compounds like carbanions. It refers to the mixing of atomic orbitals to form new hybrid orbitals, which are used in bonding. When carbon forms stable carbanions, it engages different hybridization states: sp, sp2, and sp3. These states tell us about the shape and properties of the molecule. In simple terms, hybridization alters the electron cloud around a carbon, impacting how it holds onto its electrons. This is critical in determining a carbanion's stability.

s-Character

The term s-character refers to the proportion of an s orbital present in the hybrid orbital of a carbon atom. The more s-character an orbital has, the closer electrons can be to the nucleus, enhancing electronegativity and stability. For example, an sp hybridized carbanion, which has a 50% s-character, is more stable because the electrons are held more tightly. Contrastingly, an sp3 hybridization has only a 25% s-character, making those electrons less tightly bound and contributing to lower stability.

Electronegativity Effect

Electronegativity is a measure of an atom's ability to attract electrons within a bond. In carbanions, electronegativity plays a crucial role in determining stability. The higher the s-character, the higher the electronegativity of the carbon. A negative charge on an sp hybridized carbon is more stable due to its higher electronegativity. The electrons are held closer to the nucleus, decreasing the overall energy of the molecule and increasing stability.

sp, sp2, sp3 Hybridization

Each type of hybridization describes a different bonding state for the carbon atom.

• sp Hybridization: Involves a mix of 50% s-character and 50% p-character. The electrons are closest to the nucleus, offering the greatest stability for a carbanion.
• sp2 Hybridization: Combines 33% s-character with 67% p-character. This setup is less stable than sp but more stable than sp3.
• sp3 Hybridization: Provides a 25% s-character and results in the least stability for carbanions, as the electrons are more loosely held.

Hybridization not only dictates molecular geometry but also plays a central role in the net stability of ions.

Inductive and Resonance Effects

The stability of a carbanion is influenced by both inductive and resonance effects: - Inductive Effect: This involves the transmission of charge through a chain of atoms in a molecule by electrostatic induction. It can stabilize or destabilize a carbanion depending on the surrounding atoms or groups in the molecule, which can either donate or withdraw electron density.
- Resonance Effect: This is the delocalization of electrons across multiple atoms, allowing a carbanion to distribute its electron density and achieve greater stability through resonance forms. In compounds where resonance is possible, it greatly enhances stability since the negative charge can be spread across several atoms, reducing localized charge and lowering energy.