Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Which of the following molecules has all the effects: inductive, mesomeric and Baker Nathan effect? (a) \(\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{Cl}\) (b) \(\mathrm{CH}_{3}-\mathrm{CH}=\mathrm{CH}_{2}\) (c) \(\mathrm{CH}_{2}=\mathrm{CH}-\mathrm{CH}=\mathrm{CH}_{2}\) (d) \(\mathrm{CH}_{3}-\mathrm{CH}=\mathrm{CH}-\mathrm{C}-\mathrm{CH}_{3}\)

Short Answer

Expert verified
The correct answer is (c) \( \mathrm{CH}_{2}=\mathrm{CH}-\mathrm{CH}=\mathrm{CH}_{2} \).

Step by step solution

01

Understanding Effects

Let's understand the terms. The inductive effect is the electron-donating or withdrawing effect exerted through sigma bonds. The mesomeric effect exerts an electron-withdrawing or donating effect through pi bonds from conjugated pi systems. The Baker-Nathan effect relates to the stabilization of conjugated systems.
02

Analyze Option (a)

For \ \( \mathrm{C}_{2} \mathrm{H}_{3} \mathrm{Cl} \ \), Chlorine exhibits an inductive effect by withdrawing electrons through sigma bonds, but it lacks conjugated pi systems needed for mesomeric and Baker-Nathan effects.
03

Analyze Option (b)

The molecule \ \( \mathrm{CH}_{3}-\mathrm{CH}=\mathrm{CH}_{2} \ \) has neither extended pi systems nor does it show significant effects required for mesomeric or the Baker-Nathan effect.
04

Analyze Option (c)

Option \ \( \mathrm{CH}_{2}=\mathrm{CH}-\mathrm{CH}=\mathrm{CH}_{2} \ \) involves full conjugation, allowing for inducement, mesomeric, and Baker-Nathan effects due to its symmetric, fully conjugated double-bonded chain.
05

Analyze Option (d)

The molecule \ \( \mathrm{CH}_{3}-\mathrm{CH}=\mathrm{CH}-\mathrm{C}-\mathrm{CH}_{3} \ \) is lacking some elements of conjugation necessary for the Baker-Nathan effect and has mostly sigma bonds which don't support mesomerics.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Mesomeric Effect
The mesomeric effect is a concept that explains how electrons are distributed in a molecule when pi bonds are present. When we say a molecule has a mesomeric effect, we're talking about how its structure allows electrons to move through a series of overlapping p orbitals. This is because electrons in pi bonds can be shared across adjacent atoms that have p orbitals in close proximity.
This movement of electrons results in two scenarios: when it is positive, the molecule donates electrons; when negative, the molecule withdraws electrons. This electron sharing across pi bonds helps stabilize or destabilize the entire molecule, changing its chemical properties.
  • Positive Mesomeric Effect (+M): An electron-donating group increases electron density over another atom or group. Examples include groups like -OH or -NH₂, which push electrons towards the conjugated system.
  • Negative Mesomeric Effect (-M): An electron-withdrawing group decreases electron density, which polarizes pi electrons, pulling them away. Typical groups include -NO₂ or -CN.
The mesomeric effect is especially significant in molecules with extensive conjugated systems, since these electrons play a crucial role in the chemistry of the molecule, affecting reactivity and stability.
Baker Nathan Effect
The Baker-Nathan effect explores the stabilization that is provided in conjugated systems via hyperconjugation. It's a subtle factor but can profoundly impact the behavior of organic molecules. This effect is typically observed in molecules where conjugated pi bonds are present alongside sigma bonds.
Within this context, the sigma bonds neighboring the pi bonds can help to reinforce the stabilization. This is often described as "no bond resonance," where electrons from the sigma bond flow towards the neighboring pi system, stabilizing the entire system.
For this effect to be significant, a molecule needs a contiguous chain of conjugated bonds. Through hyperconjugation, it enhances stability similar to traditional resonance. It contributes to various molecular properties including lower energy, enhanced stability, and potentially increased reactivity.
It's vital for chemistry students to consider the Baker-Nathan effect when analyzing systems with multiple bonds, as it often guides the understanding of molecular stability and reactivity, complementing both the inductive and mesomeric effects.
Conjugated Systems
Conjugated systems are special structures in chemistry where a sequence of alternating single and multiple bonds exist. These are the backbone of many organic compounds. A molecule with conjugated pi bonds can delocalize its electron density across all the connected pi orbitals through overlapping, leading to a lower overall energy state.
A classic example of a conjugated system is a molecule like 1,3-butadiene, where four carbon atoms are connected by alternating single and double bonds. This setup allows the pi electrons to move freely along the sequence, resulting in extensive delocalization of electrons.
Some key features of conjugated systems include:
  • Delocalization of electrons, which results in stability and typically lowers the molecule's energy.
  • Introduction of significant color in certain compounds due to electronic transitions that occur with certain wavelengths of light.
  • Influence on the chemical properties, contributing to increased reactivity and making them more susceptible to some reactions like electrophilic addition.
Understanding conjugated systems is crucial as they form the basis for understanding more complex chemical phenomena, influencing both physical and chemical properties.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The most stable free radical among the following is (1) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{2}\) (2) \(\mathrm{CH}_{3}-\mathrm{CH}-\mathrm{CH}_{3}\) (3) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CHCH}_{3}\) (4) \(\mathrm{CH}_{3} \mathrm{CH}_{2}\) (a) \(3>1>4>2\) (b) \(1>3>2>4\) (c) \(3>1>2>4\) (d) \(3>2>1>4\)

\(\left(\mathrm{H}_{3} \mathrm{C}\right)_{2} \mathrm{C}=\mathrm{CHCH}_{3}+\mathrm{NOBr} \longrightarrow \mathrm{A}\) The structure of the product (a)is given as (a) \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{C}(\mathrm{Br})-\mathrm{CH}(\mathrm{NO}) \mathrm{CH}_{3}\) (b) \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{C}(\mathrm{NO})-\mathrm{CH}(\mathrm{Br}) \mathrm{CH}_{3}\) (c) \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CH}-\mathrm{C}(\mathrm{NO})(\mathrm{Br}) \mathrm{CH}_{3}\) (d)

Phenol is more reactive than benzene towards eletrophillic substitution due to (a) strong mesomeric effect (b) hyperconjugative effect (c) Inductive effect only (d) hydrogen bonding

Which of the following statements are correct? 1\. SN mechanism is most commonly given by tertiary alkyl halides. 2\. \(\mathrm{SN}^{1}\) mechanism proceeds through the formation of a carbocation. 3\. SN \(^{2}\) mechanism involves retention of configuration. 4\. \(\mathrm{SN}^{2}\) mechanism proceeds through the formation of a transition state. (a) 1,2 and 4 (b) 1,3 and 4 (c) 2,3 and 4 (d) \(1,2,3\) and 4

Amongst the following, the compound that can most readily get sulphonated is (a) benzene (b) toluene (c) nitrobenzene (d) chlorobenzene

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free