Problem 2
Question
C-Cl bond of chlorobenzene in comparison to \(\mathrm{C}-\mathrm{Cl}\) bond in methyl chloride is (a) shorter and weaker (b) shorter and stronger (c) longer and weaker (d) longer and stronger
Step-by-Step Solution
Verified Answer
(b) shorter and stronger
1Step 1: Understand Bond Types
First, identify the types of bonds in question. In chlorobenzene, the C-Cl bond is part of an aromatic system, whereas in methyl chloride, the C-Cl bond is in an alkyl halide.
2Step 2: Analyze Bond Length
The C-Cl bond in aromatic compounds like chlorobenzene typically experiences resonance, which can shorten the bond length compared to alkyl chlorides, like methyl chloride, where no resonance occurs.
3Step 3: Evaluate Bond Strength
Due to the resonance stabilization in aromatic systems like chlorobenzene, which delocalizes electron density, the C-Cl bond in chlorobenzene is stronger compared to methyl chloride despite both being single bonds.
4Step 4: Compare and Conclude
In chlorobenzene, the C-Cl bond is shorter due to resonance and stronger due to the delocalization of electrons in the aromatic ring, compared to the C-Cl bond in methyl chloride which is longer and weaker.
Key Concepts
ChlorobenzeneMethyl ChlorideResonance
Chlorobenzene
Chlorobenzene is a chemical compound where a chlorine atom is bonded to a benzene ring. Benzene is an aromatic hydrocarbon, characterized by the sharing of electrons within a ring structure between carbon atoms. This property of sharing, known as resonance, significantly affects the chemical behavior of chlorobenzene.
In chlorobenzene, the chlorine atom is part of the aromatic system, contributing to its unique chemical bond features. The C-Cl bond in chlorobenzene, due to the resonance within the benzene ring, tends to be shorter in length compared to what you find in non-aromatic compounds.
In chlorobenzene, the chlorine atom is part of the aromatic system, contributing to its unique chemical bond features. The C-Cl bond in chlorobenzene, due to the resonance within the benzene ring, tends to be shorter in length compared to what you find in non-aromatic compounds.
- Resonance allows for the delocalization of electrons.
- This delocalization can strengthen the C-Cl bond, making it less reactive.
Methyl Chloride
Methyl chloride, also known as chloromethane, is a simple organic compound where a chlorine atom is bonded to a methyl group (CH extsubscript{3}). This chemical structure forms an alkyl halide, which behaves quite differently from aromatic compounds like chlorobenzene.
In methyl chloride, the C-Cl bond is characterized by the lack of resonance. Unlike the aromatic system of chlorobenzene, the methyl group cannot provide electron delocalization. This lack of resonance in methyl chloride causes the C-Cl bond to be:
In methyl chloride, the C-Cl bond is characterized by the lack of resonance. Unlike the aromatic system of chlorobenzene, the methyl group cannot provide electron delocalization. This lack of resonance in methyl chloride causes the C-Cl bond to be:
- Longer in comparison to aromatic systems.
- Weaker due to the absence of electron delocalization.
Resonance
Resonance is an essential concept in organic chemistry, explaining how certain molecules can stabilize themselves through the delocalization of electrons. This phenomenon is fundamentally a result of electrons being shared over two or more atoms, creating what is known as resonance structures.
In molecules like chlorobenzene, resonance helps shorten and strengthen the C-Cl bond.
In molecules like chlorobenzene, resonance helps shorten and strengthen the C-Cl bond.
- This is because the resonance allows the bond to partake in the electron cloud exchange within the benzene ring.
- Electrons are not stuck between just two atoms, but are instead spread out over multiple atoms, enhancing stability.
Other exercises in this chapter
Problem 1
C-X bond is strongest in (a) \(\mathrm{CH}_{3} \mathrm{Br}\) (b) \(\mathrm{CH}_{3} \mathrm{Cl}\) (c) \(\mathrm{CH}_{3} \mathrm{I}\) (d) \(\mathrm{CH}_{3} \mathr
View solution Problem 3
\(\mathrm{C}_{3} \mathrm{H}_{\mathrm{s}}+\mathrm{Cl}_{2} \stackrel{\text { Light }}{\longrightarrow} \mathrm{C}_{3} \mathrm{H}_{7} \mathrm{Cl}+\mathrm{HCl}\) is
View solution Problem 4
The starting substance for the preparation of iodoform is any one of the following, except (a) \(\mathrm{CH}_{3} \mathrm{CH}(\mathrm{OH}) \mathrm{CH}_{3}\) (b)
View solution Problem 6
\(\mathrm{Cl}_{2}\) reacts with \(\mathrm{CS}_{2}\) in presence of \(\mathrm{I}_{2}\) to form (a) \(\mathrm{CCl}_{4}\) (b) \(\mathrm{Cl}_{3} \mathrm{C}-\mathrm{
View solution