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Toluene when refluxed with \(\mathrm{Br}_{2}\) in the presence of light mainly gives (a) p-bromotoluene (b) benzyl bromide (c) o-bromotoluene (d) mixture of o- and p-bromotoluene

Short Answer

Expert verified
The main product is benzyl bromide (b).

Step by step solution

01

Understand the Reaction Conditions

Toluene is being refluxed with bromine \( \mathrm{Br}_2 \) in the presence of light. These conditions indicate a free radical substitution reaction rather than an electrophilic aromatic substitution. Light and heat promote the formation of free radicals, especially in alkyl side chains.
02

Identify the Reactivity of Toluene

Toluene, when subjected to bromination under these conditions, will not undergo substitution on the benzene ring, but rather on the side chain. The methyl group of toluene is susceptible to free radical attack, targeting the benzylic hydrogen atoms.
03

Determine the Product of the Reaction

The free radical reaction primarily occurs at the benzylic position of toluene. Bromine replaces a hydrogen atom from the methyl group, converting it into a benzyl bromide. This reaction does not result in substitution directly on the benzene ring.

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Key Concepts

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

Bromination Reaction
Bromination is a chemical reaction involving the addition or substitution of bromine atoms in a compound. When bromination occurs in the presence of heat or light, as with toluene, it typically involves free radicals, leading to a free radical substitution. This kind of reaction begins with the homolytic cleavage of the bromine molecule (\( \mathrm{Br}_2 \)) to form bromine radicals.
  • Initiation: Light energy breaks the bromine molecule into two radicals, \( 2\,\mathrm{Br}^\bullet \).
  • Propagation: A bromine radical abstracts a hydrogen atom from the toluene’s methyl group, generating a tolyl radical.
  • Termination: The radicals combine to form stable products, such as benzyl bromide.
This mechanism results in the substitution of a hydrogen atom in the benzylic position, rather than electrophilic substitution on the aromatic ring.
Benzylic Position
The benzylic position refers to the hydrogen atoms in the methyl group attached to a benzene ring, like in toluene. This position is notably reactive due to the stability of the radicals formed during reactions. When the tolyl radical forms in the bromination reaction, it is more stable because of resonance stabilization, where the unpaired electron is delocalized into the aromatic system. This delocalization stabilizes the radical, making the benzylic position a preferred site for radical reactions. Understanding the benzylic position's reactivity helps explain why, in the presence of bromine and light, substitution happens at the side chain and not directly on the aromatic ring.
Reactivity of Toluene
Toluene is a common organic compound, noted for its aromatic benzene ring and a methyl side chain. The methyl group in toluene is a primary radical stabilization site, making it more reactive compared to the benzene ring during certain conditions. In free radical substitution, toluene's methyl group undergoes bromination at the benzylic position.
  • Under radical bromination conditions, bromine selectively adds to the side chain.
  • This selectivity arises because the aromatic ring’s electron-rich nature generally favors different reaction conditions, like electrophilic aromatic substitution, not supported by free radical pathways.
  • Consequently, such conditions do not alter the aromatic structure but instead result in benzyl bromide formation.
Thus, in the presence of light and bromine, the benzylic position is always the target for reactive intermediates, showcasing toluene's unique reactivity profile.

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Most popular questions from this chapter

Match the following: List I (Reactants) 1\. \(\mathrm{CH}_{3}\left(\mathrm{CH}_{2}\right)_{3} \mathrm{OH} \frac{\mathrm{NaBr}, \mathrm{H}_{2} \mathrm{SO}_{4}, \Delta}{\longrightarrow}\) 3\. \(\mathrm{CH}_{3} \mathrm{CH}(\mathrm{OH})\left(\mathrm{CH}_{2}\right)_{2} \mathrm{CH}_{3} \stackrel{\mathrm{PBr}_{3}}{\longrightarrow}\) 4\. \(\mathrm{Me}_{2} \mathrm{CHCH}_{2} \mathrm{OH} \stackrel{\mathrm{sOC}_{3}}{\longrightarrow}\) List II (Alkyl halides) A. \(\mathrm{CH}_{3} \mathrm{CHBr}\left(\mathrm{CH}_{2}\right)_{2} \mathrm{CH}_{3}\) B. \(\mathrm{Me}_{2} \mathrm{CHCH}_{2} \mathrm{Cl}\) C. \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{CCl}\) D. \(\mathrm{CH}_{3}\left(\mathrm{CH}_{2}\right)_{3} \mathrm{Br}\) The correct matching is \(\begin{array}{rrrrr}1 & 2 & 3 & 4 \\ \text { (a) } \mathrm{C} & \mathrm{D} & \mathrm{B} & \mathrm{A}\end{array}\) (b) \(\mathrm{C} \quad \mathrm{D} \quad \mathrm{A} \quad \mathrm{B}\) (c) \(\mathrm{D} \quad \mathrm{C} \quad \mathrm{A} \quad \mathrm{B}\) (d) D \(\mathrm{C} \quad \mathrm{B} \quad \mathrm{A}\)

The number of \(\mathrm{H}\) - atoms that can participate when CCC(C)(Br)CC is subjected to elimination is

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\(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{C}-\mathrm{Cl} \stackrel{\text { Acl. } \mathrm{KOH}}{\Delta} \rightarrow \mathrm{A} \frac{\mathrm{SO}_{2} \mathrm{Cl}_{2}}{475 \mathrm{~K}}-\mathrm{B}\) The compound \(\mathrm{B}\) in the above sequence is (a) 1,2 -dichloro-2-methyl propene (b) 1 -chloro-2-methyl propene (c) 1,2 -dichloro-2-methyl propene (d) 3 -chloro-2-methyl propene

Which of the following compounds is not chiral? [2004] (a) 2 -chloropentane (b) 1 -chloropentane (c) 3-chloro-2-methylpentane (d) 1-chloro-2-methylpentane

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