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

Question

the fo (1) \(\mathrm{CH}_{3}-\stackrel{\Theta}{\mathrm{C}} \mathrm{H}_{2}\) (2) \(\mathrm{CH}_{3}=\stackrel{\Theta}{\mathrm{C}} \mathrm{H}\) (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 of stability is (c) 3>2>1, due to increasing s-character of hybrid orbitals.

Step 1: Understand the Problem

We need to determine the order of stability of different carbanions: \(\mathrm{CH}_{3}-\stackrel{\Theta}{\mathrm{C}}\mathrm{H}_{2}\), \(\mathrm{CH}_{3}=\stackrel{\Theta}{\mathrm{C}}\mathrm{H}\), and \(\mathrm{CH}\equiv\stackrel{\ominus}{\mathrm{C}}\). Carbanion stability depends largely on the hybridization of the carbon bearing the negative charge, and the availability of s-character that stabilizes the negative charge.

Step 2: Analyze Hybridization of Each Carbanion

1. The first carbanion, \(\mathrm{CH}_{3}-\stackrel{\Theta}{\mathrm{C}}\mathrm{H}_{2}\), has the carbanion center as \(sp^3\) hybridized. 2. The second carbanion, \(\mathrm{CH}_{3}=\stackrel{\Theta}{\mathrm{C}}\mathrm{H}\), has an \(sp^2\) hybridized carbanion center.3. The third carbanion, \(\mathrm{CH}\equiv\stackrel{\ominus}{\mathrm{C}}\), has the carbanion center as \(sp\) hybridized.

Step 3: Discuss Hybridization Influence on Stability

Carbanions with more s-character (\(sp\) has 50% s-character, \(sp^2\) has 33%, and \(sp^3\) has 25%) hold the negative charge closer to the nucleus for better stabilization. Consequently, the stability order is usually \(sp > sp^2 > sp^3\).

Step 4: Determine Correct Stability Order

Given the s-character influence: \(\mathrm{CH}\equiv\stackrel{\ominus}{\mathrm{C}}\) (sp hybridized) > \(\mathrm{CH}_{3}=\stackrel{\Theta}{\mathrm{C}}\mathrm{H}\) (sp^2 hybridized) > \(\mathrm{CH}_{3}-\stackrel{\Theta}{\mathrm{C}}\mathrm{H}_{2}\) (sp^3 hybridized). Thus, option (c) \(3>2>1\) is the correct order of stability.

Key Concepts

Hybridization

In organic chemistry, hybridization helps explain the shape and stability of molecular structures by combining atomic orbitals into new hybrid orbitals. These hybrids are typically sp, sp², or sp³, depending on the types of atomic orbitals involved.

For carbanions, hybridization affects stability. This is because hybridization determines the degree to which the negative charge (extra electron) can be delocalized or stabilized by the nucleus.

• sp hybridization: An sp hybridized carbon has 50% s-character and 50% p-character. This means electrons are closer to the nucleus, allowing better stabilization of any negative charge.
• sp² hybridization: Carbons with sp² hybridization have 33% s-character. They hold electrons somewhat closer to the nucleus than sp³ but not as close as sp.
• sp³ hybridization: This type features 25% s-character, meaning the electrons are held further from the nucleus compared to sp or sp², reducing stability of a negative charge.

These hybridizations play a crucial role in determining the order of stability for the carbanions in the given exercise, where more s-character means greater stability.

s-character

The s-character of an orbital refers to the proportion of spherical 's' orbital mixed into the hybrid orbitals formed during hybridization. This can significantly affect electron behavior and, consequently, molecular stability.

s-character contributes to carbanion stability as follows:

• Higher s-character: Orbitals with more s-character allow the negative charge in a carbanion to occupy a space closer to the positively charged nucleus, leading to increased attraction and stabilization. In an sp hybridized orbital, s-character is notably higher (50%), stabilizing negative charges well.
• Intermediate s-character: sp² hybridized orbitals average 33% s-character, balancing between stability and reactivity.
• Lower s-character: With an sp³ hybridized orbital, the s-character is the least (25%), leading to lower stability of negative charges because they are held farther from the nucleus.

Understanding the s-character helps predict why some carbanions are more stable than others, making it an essential concept in determining stability in organic molecules.

Organic Chemistry

Organic chemistry is the branch dedicated to the study of carbon-containing compounds, including their structure, properties, composition, reactions, and preparation. Carbanions are a key topic in organic chemistry, as they are intermediates in many organic reactions.

A carbanion is essentially a negatively charged carbon atom, which plays a pivotal role in many mechanisms and reactions.

• Reactivity: The negative charge on a carbanion makes it reactive. It can act as a nucleophile, attacking positive centers in other molecules.
• Stability and Hybridization: In organic applications, understanding how hybridization influences the s-character of a carbanion explains its stability. In practice, this knowledge helps chemists predict likely reaction pathways and products.
• Mechanisms: Carbanions are integral to various mechanisms, such as nucleophilic substitutions and additions. They can be generated via deprotonation of hydrocarbons or by reduction reactions.

Organic chemistry, with its focus on the hybridization and stabilization of carbanions, fuels innovations in designing new compounds and reactions essential for pharmaceuticals, materials science, and more.