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Why is delocalization not possible in A, B and C options? Also why is +/- charge on benzene ring not delocalized?

Why is delocalization not possible in A, B and C options? Also why is +/- charge on benzene ring not delocalized?

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Grade:11

1 Answers

Vikas TU
14149 Points
4 years ago
Dear stuent 
The presence of alternating  π  and  σ  bonds in a molecule such as benzene is known as a conjugated system, or conjugated  π  bonds. Conjugated systems can extend across the entire molecule, as in benzene, or they can comprise only part of a molecule. A conjugated system always starts and ends with a  π  bond (i.e. an  sp2  or an  sp -hybridized atom), or sometimes with a charge. The atoms that form part of a conjugated system in the examples below are shown in blue, and the ones that do not are shown in red. Most of the times it is  sp3  hybridized atoms that break a conjugated system. Practically every time there are  π  bonds in a molecule, especially if they form part of a conjugated system, there is a possibility for having resonance structures, that is, several valid Lewis formulas for the same compound. What resonance forms show is that there is electron delocalization, and sometimes charge delocalization. All the examples we have seen so far show that electrons move around and are not static, that is, they are delocalized. Charge delocalization is a stabilizing force because it spreads energy over a larger area rather than keeping it confined to a small area. Since electrons are charges, the presence of delocalized electrons brings extra stability to a system compared to a similar system where electrons are localized. The stabilizing effect of charge and electron delocalization is known as resonance energy.
Since conjugation brings up electron delocalization, it follows that the more extensive the conjugated system, the more stable the molecule (i.e. the lower its potential energy). If there are positive or negative charges, they also spread out as a result of resonance.

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