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) \(\mathrm{CH}_{3}=\stackrel{\ominus}{\mathrm{C}} \mathrm{H}\) (3) \(\mathrm{CH} \equiv \stackrel{\ominus}{\mathrm{C}}\) 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 is (c) \(3>2>1\).

Step 1: Recognize the Structure of Carbanions

A carbanion is an anion with a negatively charged carbon atom. The structures given are: \( \mathrm{CH}_{3}-\overset{\Theta}{\mathrm{C}}\mathrm{H}_2 \) for carbanion (1), \( \mathrm{CH}_{3}=\overset{\ominus}{\mathrm{C}}\mathrm{H} \) for carbanion (2), and \( \mathrm{CH}\equiv \overset{\ominus}{\mathrm{C}} \) for carbanion (3). In these molecules, carbon carries one negative charge.

Step 2: Evaluate Stability through Hybridization

Carbanion stability is greatly influenced by the hybridization of the negatively charged carbon: The order of stability is \( \text{s} \) gives more stability than \( \text{sp}^2 \) which gives more stability than \( \text{sp}^3 \). Therefore, carbanion 3, with triple bonds and an \( \text{sp} \) hybridization, is more stable than carbanion 2, which has \( \text{sp}^2 \), and carbanion 1, which has \( \text{sp}^3 \).

Step 3: Consider Inductive and Resonance Effects

Check for any inductive effects or resonance stabilization. In the provided structures, resonance effect mostly applies for carbanion 2 due to conjugation, slightly increasing its stability but not surpassing the stability provided by the \( \text{sp} \) hybridization of carbanion 3.

Step 4: Conclusion by Hierarchical Stability

Based on hybridization stability and inductive effects, the most stable carbanion is 3 (sp hybridized), followed by 2 (sp2 hybridized), and finally 1 (sp3 hybridized). Therefore, the correct order of decreasing stability is \( 3 > 2 > 1 \).

Key Concepts

Hybridization

Hybridization refers to the way atomic orbitals mix to form new hybrid orbitals, which can hold the valence electrons of atoms involved in bonding. This concept is essential in understanding the stability of carbanions, molecules where carbon atoms bear a negative charge. The hybridization state affects how tightly the negative charge is held by carbon.

Carbanions are heavily influenced by the type of hybridization the negatively charged carbon atom has. Specifically, the greater the percentage of s-character in the hybrid orbital, the more stable the carbanion. Here's why:

• sp hybridized orbitals have 50% s-character, leading to a stronger hold on the negative charge because s orbitals are closer to the nucleus.
• sp2 hybridized orbitals have 33% s-character, providing less stability than sp orbitals.
• sp3 hybridized orbitals have only 25% s-character, offering the least stability of the three.

For example, in the provided exercise, carbanion 3, which has an sp hybridized carbon, is more stable than both carbanion 2 (sp2) and carbanion 1 (sp3). This hybridization concept establishes a hierarchy where sp > sp2 > sp3 in terms of stability.

Inductive Effects

Inductive effects involve the transmission of charge through a chain of atoms in a molecule by electrostatic induction. They can either stabilize or destabilize a carbanion. When considering carbanions, electronegative atoms can pull electron density away from the negative charge, thereby stabilizing it.

For the carbanions you're examining, the inductive effect might not be as pronounced since they are primarily hydrocarbons. However, it's essential to recognize how nearby electronegative atoms could impact stability through electron withdrawal. Some key points include:

• Electronegative substituents close to a carbanion can help stabilize the negative charge.
• Alkyl groups, on the other hand, can donate electron density, which typically destabilizes a carbanion due to increased electron repulsion.

While inductive effects don't play a significant role in the exercise you've been given, they can be crucial in complex or more substituted carbanion systems.

Resonance Stabilization

Resonance stabilization is a key concept in organic chemistry that describes the delocalization of electrons across multiple atoms in a molecule, creating structures that contribute to overall stability. This effect can significantly enhance the stability of a carbanion.
For instance, in carbanion 2 from your exercise, resonance stabilization can occur due to conjugation, which means that the lone pair of electrons on the negatively charged carbon can delocalize through pi bonds. This delocalization allows the electron density to spread over multiple atoms, which stabilizes the carbanion.

• Carbanion 2 gains this slight additional stability because its negative charge can resonate with adjacent double bonds.
• Resonance structures are only possible if the carbanion is adjacent to a system capable of resonance, like pi bonds or an aromatic ring.

While carbanion 2 benefits from resonance, it still does not surpass the stability given by the sp hybridization of carbanion 3. Nonetheless, understanding resonance helps explain why carbanion 2 is more stable than carbanion 1, despite both having sp2 and sp3 hybridization, respectively.