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Which of the following compound(s) has/have delocalized \(\pi\) - electrons? (a) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{NHCH}=\mathrm{CH}-\mathrm{CH}_{3}\) (b) \(\mathrm{CH}_{3}-\mathrm{C}_{-}\left(\mathrm{CH}_{3}\right)-\mathrm{CH}_{2}-\mathrm{CH}=\mathrm{CH}_{2}\) (c) \(\mathrm{CH}_{3} \mathrm{CH}=\mathrm{CH}-\mathrm{CH}_{2}-\mathrm{OCH}_{3}\) (d)C1=CCC=C1

Short Answer

Expert verified
Only compound (d) has delocalized \( \pi \) - electrons.

Step by step solution

01

Understand Delocalization

Delocalized \( \pi \)-electrons refer to electrons in \( \pi \) bonds that can be shared across multiple atoms rather than being confined between two atoms. This usually happens in conjugated systems where double bonds are separated by a single bond, allowing for overlapping \( \pi \) orbitals.
02

Analyze Compound (a)

Compound (a) is \( \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{NHCH}=\mathrm{CH}-\mathrm{CH}_{3} \). In this molecule, the double bond \( \mathrm{CH}=\mathrm{CH} \) is isolated with no possibility of conjugation with other lone pairs or double bonds. Therefore, it does not have delocalized \( \pi \)-electrons.
03

Analyze Compound (b)

Compound (b) is \( \mathrm{CH}_{3}-\mathrm{C}(\mathrm{CH}_{3})-\mathrm{CH}_{2}-\mathrm{CH}=\mathrm{CH}_{2} \). Again, the double bond here is isolated with no adjacent \( \pi \)-bonds or available lone pairs to participate in delocalization. Thus, this molecule lacks delocalized \( \pi \)-electrons.
04

Analyze Compound (c)

Compound (c) is \( \mathrm{CH}_{3} \mathrm{CH}=\mathrm{CH}-\mathrm{CH}_{2}-\mathrm{OCH}_{3} \). Here, the \( \mathrm{OCH}_{3} \) group could potentially engage lone pairs in delocalization with the \( \pi \)-bond. However, this structure typically doesn't support efficient delocalization as the required adjacent \( \pi \)-bonds are absent.
05

Analyze Compound (d)

For compound (d), the structure \( \mathrm{C}_{5}\mathrm{H}_{6} \) appears. Since the bonding structure is like cyclopentadiene (cyclic without specified hydrogens), it forms a conjugated system – a cyclic molecule with alternating double bonds which is highly supportive of delocalized \( \pi \)-electrons.
06

Conclusion: Identify Compounds with Delocalization

After analyzing the compounds, it's clear that only compound (d) supports delocalized \( \pi \)-electrons due to its alternate double-bond structure in a cyclic formation.

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Key Concepts

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

Conjugated Systems
Conjugated systems are fascinating elements of organic chemistry, often featured in molecules with alternating single and double bonds. The essence of a conjugated system is its potential to enable delocalization of electrons, particularly the \( \pi \)-electrons. This occurs when \( \pi \)-bonds, or sometimes lone pairs, are in sequence separated by single bonds, allowing \( \pi \)-orbitals to interact across adjacent atoms. The result is a lower energy state for the molecule, which can lead to unique chemical properties.
Conjugated systems are not only found in aromatic compounds but also in dienes and polyenes. They can involve the entire structure when cyclic and planar, promoting aromaticity. The delocalized electrons contribute to properties such as increased stability and the potential for exciting features like color. In some molecules, these systems are vital for their function, where the ability to transfer or absorb energy is crucial, such as in chromophores and photosynthetic organisms.
Pi Bonds
Pi bonds are the cornerstone of understanding organic molecules' reactivity and structure. Formed by the side-to-side overlap of \( p \)-orbitals, \( \pi \)-bonds play a crucial role in double and triple bonds. Unlike \( \sigma \)-bonds, formed by direct overlap and allowing free rotation, \( \pi \)-bonds restrict rotation, introducing a rigidity that is pivotal in defining the geometry of compounds.
This rigidity explains why conjugated systems can facilitate electron delocalization. With \( \pi \)-orbitals capable of extending over multiple atoms efficiently in conjugated settings, their interaction leads to a shared electron cloud above and below the bonding axis. This delocalization is especially important in understanding compounds like benzene, where \( \pi \)-bonds collectively foster increased molecule stability. Examining a molecule's \( \pi \)-bonds and their interactions can provide insights into its reactivity and possible participation in resonance.
Organic Chemistry
Organic chemistry is the study of carbon-containing compounds and their diverse properties and reactions. It explores how carbon's versatile bonding allows the formation of stable structures from simple hydrocarbons to complex polysaccharides and proteins. It's within this field that concepts like conjugated systems and \( \pi \)-bonds come to life.
In organic chemistry, delocalized electrons can influence a molecule's chemical behavior, impacting how it interacts with other substances. This is crucial in fields ranging from pharmaceuticals to materials science, where controlled electron flow can lead to breakthroughs in drug design or new materials. By understanding the principles governing carbon's interaction with other elements, including hydrogen, oxygen, nitrogen, and others, scientists can predict reaction outcomes and create novel compounds!
Aromatic Compounds
Aromatic compounds are a special class of cyclic hydrocarbons that owe their unique properties to the phenomenon of electron delocalization. Characterized by one or more planar rings of atoms connected by alternating single and double bonds, these compounds follow Huckel's rule: they contain an odd number of \( \pi \)-electron pairs (expressed as \( 4n + 2 \) \( \pi \)-electrons, where \( n \) is an integer).
This delocalization within aromatic compounds gives them enhanced stability over non-aromatic structures – a concept known as aromaticity. The nuances of aromatic compounds manifest in their lower reactivity compared to alkenes and the ability to participate in unique reactions such as electrophilic aromatic substitution.
Understanding aromatic compounds involves appreciating how delocalized \( \pi \)-bonds influence not just their chemical properties, but also their physical attributes, like their tendency to absorb light, influencing color in dyes and pigments.

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Most popular questions from this chapter

In which case, the intermediate involved is incorrect? (a) Pinacol - pinacolone rearrangement - carbocation (b) Hofmann's bromamide reaction \(-\) Nitrene (c) Aldol condensation - carbocation (d) Anti Markonikoff's addition of HBr to propene Free radical.

Arrange the following compounds in the order of increasing tendency to undergo electrophillic substitution: (1) O=[N+]([O-])c1ccccc1 (2) c1ccccc1(3) Oc1ccccc1 (4) Cc1ccccc1 (5) (a) \(5<1<2<4<3\) (b) \(3>2>1>4>5\) (c) \(1>2>5>3>4\) (d) \(5<1<4<2<3\)

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

Match the following: List I List II (Reagents) (Nature) 1\. \((\mathrm{CN})\) (i) Neutral nucleophile 2\. (ONO) (ii) electrophile 3\. \(\mathrm{NH}_{3}\) (iii) Ambident nucleophile 4\. \(\mathrm{AlCl}_{3}\) (iv) both El and Nu The correct matching is: 1 \(2 \quad 3\) 4 (a) (i) (iii) (ii) (iv) (b) (iii) (iii) (i) (ii) (c) (iv) (iii) (iii) (i) (d) (ii) (iv) (i) (iii)

In the following groups: \(-\mathrm{OAc}(\mathrm{I})\) \- OMe (II) \(-\mathrm{OSO}_{2} \mathrm{Me}(\mathrm{III}), \quad-\mathrm{OSO}_{2} \mathrm{CF}_{3}(\mathrm{IV})\) the order of the leaving group ability is (a) \(\mathrm{I}>\mathrm{II}>\mathrm{III}>\mathrm{IV}\) (b) IV > III > I > II (c) \(\mathrm{III}>\mathrm{II}>\mathrm{I}>\mathrm{IV}\) (d) \(\mathrm{II}>\mathrm{III}>\mathrm{IV}>\mathrm{I}\).

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