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Chapter 1: Problem 72

The most stable free radical among the following is (a) C=CC(C)C (b) c1ccccc1 (c) \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{C}\) (d) \(\mathrm{CH}_{2}=\mathrm{CH}-\mathrm{CH}_{2}-\mathrm{CH}_{2}\).

Short Answer

Expert verified
(a) C=CC(C)C (b) c1ccccc1 (a benzene ring) (c) (CH3)3C (tert-butyl radical) (d) CH2=CH-CH2-CH2 (allyl radical) Answer: (a) C=CC(C)C

Step by step solution

01

Assess the stability of the free radicals in options (a) and (d)

The free radicals in options (a) and (d) are allylic radicals. Allylic radicals have a resonance structure where the unpaired electron is delocalized between two adjacent carbon atoms. This electron delocalization stabilizes the radical. For (a): The stability of the free radical in option (a) is also influenced by the electron-donating inductive effect of the alkyl groups (CH2=CHC(CH3)2). The two methyl groups attached to the carbon holding the unpaired electron will donate electron density, further stabilizing the radical. For (d): The free radical in option (d) is the allyl radical. Compared to option (a), there are no additional electron-donating groups to stabilize the radical further.
02

Assess the stability of the free radicals in options (b) and (c)

For (b): Option (b) is a benzene ring, which is an aromatic compound. Free radicals are highly reactive and destroy the highly stable aromaticity of the benzene ring. As a result, free radicals are not easily formed and are very unstable in an aromatic compound like benzene. For (c): Option (c) is a tert-butyl radical, with the central carbon atom holding the unpaired electron and attached to three methyl groups. It is a highly stabilized structure due to the electron-donating inductive effect of the three methyl groups. However, there is no resonance structure contributing to the stability, unlike allylic radicals.
03

Compare and determine the most stable free radical

Now that we have assessed the stability of the free radicals in each option, let's compare them: (a) Allylic radical stabilized by electron delocalization and electron-donating alkyl groups (b) Unstable benzene ring due to destruction of aromaticity (c) Tert-butyl radical stabilized by electron-donating methyl groups, but no resonance stabilization (d) Allylic radical stabilized by electron delocalization, with fewer electron-donating alkyl groups than option (a) From this comparison, option (a) – the allylic radical CH2=CHC(CH3)2 – appears to be the most stable free radical due to both resonance stabilization and electron-donating inductive effects of the attached alkyl groups.

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Most popular questions from this chapter

Configuration of a chiral molecule can be changed only if (a) the compound is rotated out of the plane of paper. (b) the confirmation is changed by rotation. (c) the Fischer projection is changed to the Newman projection. (d) the bond to the chiral carbon is broken.

ClCCI ClC(Cl)(Cl)Cl The \(\mathrm{Cl}^{\text { }} \mathrm{Cl}\) bond angle in \(1,1,2,2\) -tetrachloroethene and tetrachloromethane respectively are about (a) \(90^{\circ}\) and \(109.5^{\circ}\) (b) \(109.5^{\circ}\) and \(120^{\circ}\) (c) \(120^{\circ}\) and \(109.5^{\circ}\) (d) \(120^{\circ}\) and \(120^{\circ}\)

The IUPAC name of the compound given is (a) \(2,2-\) diethyl propane \(-1,3-\) dialdehyde (b) 3,3 - diethyl pentane- 1,5 -dial (c) \(2,2-\) diethyl propane \(-1,3-\) dial (d) 3,3 -dimethyl keto pentane

IUPAC name of the compound (a) Trans-3,4-dimethylhex-3-ene (b) Cis- 3,4 -dimethylhex-3-ene (c) Cis- 3,4 -dimethylhex-4-ene (d) Cis-2,3-diethylbut-2-ene

Which of the following are correct statements? (a) \(\mathrm{BF}_{3}\) and are: \(\mathrm{CCl}_{2}\) electrophiles. (b) Dimethylsulphoxide and dimethyl formamide (DMF) are aprotic solvents. (c) In butadiene \(\mathrm{H}_{2} \mathrm{C}=\mathrm{CH}-\mathrm{CH}=\mathrm{CH}_{2}\) all the three \(\mathrm{C}-\mathrm{C}\) bond lengths are equal. (d) \(\mathrm{CH}_{2}=\mathrm{CH}-\mathrm{Cl}\) has low reactivity because chlorine is attached to an \(\mathrm{sp}^{2}\) carbon.
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