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Chapter 24: Problem 125

When a methyl radical is formed from \(\mathrm{CH}_{3} \mathrm{Cl}\), select the correct statement: (1) bond angle of \(109^{\circ} 28\) ' is retained (2) number of sigma bonds is three (3) carbon undergoes geometric change from tetrahedral to planar (4) hybridization changes \(\mathrm{sp}^{3}\) to \(\mathrm{sp}^{2}\) (a) 2,3 and 4 (b) 1,3 and 4 (c) 2 and 4 (d) 3 and 4

Short Answer

Expert verified
The correct answer is (a) 2, 3, and 4.

Step by step solution

01

Identify the changes in hybridization

When the methyl radical is formed from methane chloride (\(\mathrm{CH}_{3}\mathrm{Cl}\)), a chlorine atom and its bond with carbon are removed, resulting in the loss of one hydrogen atom and its associated bonding electron. This means the central carbon in the methyl radical changes from \(\mathrm{sp}^3\) (tetrahedral) hybridization to \(\mathrm{sp}^2\) (planar) hybridization because it only has three sigma bonds left.
02

Determine the sigma bonds

In the methyl radical, there are three hydrogen atoms each forming a single sigma bond with the central carbon atom. Thus, there are a total of three sigma bonds in the methyl radical.
03

Assess the structure's geometry

Due to the change in hybridization to \(\mathrm{sp}^2\), the geometry of the carbon atom alters from tetrahedral to planar. This planar structure implies a bond angle of approximately 120掳 instead of the tetrahedral bond angle of \(109^{\circ}28'\).
04

Select the correct statement

Given these considerations, the correct statements are 2, 3, and 4: three sigma bonds remain, the structure transitions from tetrahedral to planar, and the hybridization changes from \(\mathrm{sp}^3\) to \(\mathrm{sp}^2\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Hybridization
Hybridization is a fundamental concept in organic chemistry that helps us understand how atomic orbitals mix to form new hybrid orbitals suitable for bonding. When dealing with carbon, as in the case of a methyl radical formed from \(\text{CH}_3\text{Cl}\), hybridization plays an important role.Let's explore it:
  • Tetrahedral Configuration: In \(\text{CH}_3\text{Cl}\), carbon is initially \(\text{sp}^3\) hybridized. This means that one s orbital and three p orbitals mix to create four equivalent \(\text{sp}^3\) orbitals. These orbitals form a tetrahedral shape with bond angles of approximately \(109^\circ 28'\).
  • Change to \(\text{sp}^2\) Hybridization: Upon the formation of the methyl radical, a chlorine atom and its associated electron pair are removed. This causes carbon to shift to \(\text{sp}^2\) hybridization. Here, one s orbital combines with two p orbitals, forming three hybrid orbitals that position themselves in a planar, trigonal arrangement.
  • Significance in Reactions: Understanding changes in hybridization is crucial for predicting how molecules will react. \(\text{sp}^2\) hybridization indicates a higher reactivity due to the presence of an unhybridized p orbital, which can participate in forming \(\pi\)-bonds or in radical reactions.
Each shift in hybridization not only affects the molecule's geometry but also its reactivity and how it will interact with other molecules.
Geometric Structure
The geometric structure of molecules is determined largely by their hybridization. In our case of the methyl radical derived from \(\text{CH}_3\text{Cl}\), the change to \(\text{sp}^2\) hybridization significantly alters its geometry.Here's how:
  • From Tetrahedral to Planar: Initially, the \(\text{CH}_3\text{Cl}\) molecule's carbon atom has a tetrahedral geometry due to its \(\text{sp}^3\) hybridization. This gives rise to a three-dimensional shape with bond angles of \(109^\circ 28'\).
  • Effect of Hybridization Change: When a methyl radical is formed, the carbon atom switches to \(\text{sp}^2\) hybridization, resulting in a planar structure. This means the molecule lies flat, with bond angles around each hydrogen-carbon bond being approximately \(120^\circ\).
  • Planar Structures: This planar configuration is often found in compounds where stability and specific reactive properties, such as resonance, are required.
Understanding these geometric transformations helps in visualizing molecules and predicting their chemical behavior in reactions and interactions with other substances.
Sigma Bonds
Sigma bonds (\(\sigma\)-bonds) are the strongest type of covalent bond and form the foundation of molecular stability in organic compounds. Let's dive into how these bonds play a role in the methyl radical derived from \(\text{CH}_3\text{Cl}\).Here's what you should know:
  • Nature of Sigma Bonds: Sigma bonds are a result of the head-on overlap of atomic orbitals. This overlap can occur between s-s, s-p, or p-p orbitals, leading to formation of a strong bond along the internuclear axis.
  • Three Sigma Bonds in a Methyl Radical: Upon the conversion from \(\text{CH}_3\text{Cl}\), there are still three hydrogen atoms bound to the carbon, each with a single sigma bond. These are the only \(\sigma\)-bonds present in the methyl radical structure, signifying direct overlap of orbitals.
  • Properties and Role: Sigma bonds are responsible for the basic framework and stability of a molecule, permitting rotations in the molecular structure. They are extremely crucial in determining the physical and chemical properties of the compound.
Grasping the concept of sigma bonds and their implications on molecular stability is essential for understanding chemical structures and reactivity.

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Most popular questions from this chapter

Amongst the following the most basic compound is (a) benzylamine (b) aniline (c) acetanilide (d) p-nitroaniline

Cc1ccc(N)cc1 Cc1ccc(N)cc1 I II III IV (a) IV \(>\鈥� # Arrange the following in decreasing order of basicity: Nc1cccc(N)c1 N#Cc1ccc(N)cc1 Cc1ccc(N)cc1 I II III IV (a) IV \)>\mathrm{I}>\mathrm{III}>\mathrm{II}\( (b) \)\mathrm{IV}>\mathrm{I}>\mathrm{II}>\mathrm{III}\( (c) IV \)>\mathrm{III}>\mathrm{I}>\mathrm{II}\( (d) \)\mathrm{I}>\mathrm{II}>\mathrm{III}>\mathrm{IV}$

Consider the following compounds: (1) \(\mathrm{C}_{6} \mathrm{H}_{5}-\mathrm{NH}_{2}\) (2) \(\mathrm{C}_{6} \mathrm{H}_{3}-\mathrm{NH}-\mathrm{C}_{6} \mathrm{H}_{5}\) (3) \(\left(\mathrm{C}_{6} \mathrm{H}_{5}\right)_{3} \mathrm{~N}\) (4) \(\mathrm{CH}_{3}-\mathrm{CH}_{2}-\mathrm{NH}_{2}\) Arrange these compounds in decreasing order of their basicity. (a) \(1>2>3>4\) (b) \(2>3>1>4\) (c) \(3>2>1>4\) (d) \(4>1>2>3\)

The major product obtained on treatment of \(\mathrm{CH}_{3}\) \(\mathrm{CH}_{2} \mathrm{CH}(\mathrm{F}) \mathrm{CH}_{3}\) with \(\mathrm{CH}_{3} \mathrm{O}^{-/ \mathrm{CH}_{3} \mathrm{OH}}\) is (a) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}\left(\mathrm{OCH}_{3}\right) \mathrm{CH}_{3}\) (b) \(\mathrm{CH}_{3} \mathrm{CH}=\mathrm{CHCH}_{3}\) (c) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}=\mathrm{CH}_{2}\) (d) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{OCH}_{3}\)

Which of the following statements are correct. (1) \(\overline{\mathrm{N}} \mathrm{H}_{2}\) is better nucleophile than \(\mathrm{NH}_{3}\) but latter \(\left(\mathrm{NH}_{3}\right)\) is better nucleophile than \(\mathrm{NH}_{4}^{+}\) (2) \(\mathrm{C}_{6} \mathrm{H}_{3} \mathrm{O}^{-}\)is better nucleophile than CC(=O)O (3) \(\mathrm{OH}^{-}\)is better nucleophile than \(\mathrm{SH}^{-}\)and \(\mathrm{H}_{2} \mathrm{O}\), but \(\mathrm{H}_{2} \mathrm{O}\) is better nucleophile than \(\mathrm{H}_{3} \mathrm{O}^{+}\) (4) \(\mathrm{ClO}^{-1}\) is weaker nucleophile than \(\mathrm{ClO}_{4}^{-}\) (a) 1,2 and 3 (b) 1,3 and 4 (c) 2,3 and 4 (d) \(1,2,3\) and 4
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