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

Chlorination of toluene in presence of light and heat followed by treatment with aqueous NaOH gives (a) o-cresol (b) p-cresol (c) 2,4 -dihydroxytoluene (d) benzoic acid

Short Answer

Expert verified
The product is (b) p-cresol.

Step by step solution

01

Identify the reaction conditions

The problem involves two main reactions: chlorination in the presence of light and heat, followed by treatment with aqueous NaOH. The first step, chlorination, is a free radical substitution reaction that is accelerated by light.
02

Understand the effect of chlorination

In the presence of light, toluene undergoes chlorination primarily at the methyl position due to the formation of a stable benzylic radical. Therefore, the product after chlorination will be benzyl chloride.
03

Analyze the reaction with aqueous NaOH

Benzyl chloride can react with aqueous NaOH through a nucleophilic substitution reaction. The chlorine atom in benzyl chloride is replaced by an OH group, converting it into benzyl alcohol, which is the same as o/p-cresol after hydroxylation at the benzylic position.
04

Determine the possible products

Under the conditions given, the major products are ortho and para cresol due to the formation of benzyl chloride followed by hydrolysis. However, the position of substitution is typically indicated upon conditions favoring para substitution.

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.

Chlorination
Chlorination is a vital organic reaction that involves the addition of chlorine atoms to an organic molecule. In the context of this exercise, chlorination primarily refers to the addition of chlorine to the methyl group of toluene. This process is initiated by exposing the reactants to light and heat, which helps generate chlorine radicals. These radicals are highly reactive and attack the hydrogen atoms on the methyl group of toluene.
  • Chlorine radicals are formed in the presence of light and heat.
  • Toluene, due to its methyl group, becomes susceptible to radical attack.
  • The end product after this reaction is the chlorinated compound benzyl chloride.
Understanding the nature of chlorination and the influence of light and heat helps in predicting the structural changes in the organic compound. The stability of the benzylic radical due to resonance makes the methyl group the favored site for chlorination.
Free Radical Substitution
Free radical substitution is a type of chemical reaction where a free radical replaces an atom in a molecule. It's a chain reaction that occurs in three steps: initiation, propagation, and termination.
  • Initiation: In this step, chlorine molecules (Cl_2) dissociate into radicals under the influence of light.
  • Propagation: These radicals react with toluene, abstracting hydrogen from the methyl group, forming a benzylic radical.
  • Termination: The radicals combine to form stable molecules, ending the chain reaction.
Free radical substitution reactions are significant in organic chemistry because they help in the functionalization of alkanes to more reactive intermediates, like benzyl chloride in this case. It's important to control the reaction conditions to direct the substitution effectively.
Nucleophilic Substitution
Nucleophilic substitution is a crucial mechanism in organic chemistry. It involves the replacement of a leaving group (like a chlorine atom) by a nucleophile (such as the hydroxide ion). In this case, the reaction of benzyl chloride with aqueous NaOH exemplifies this process.
  • The chlorine atom in benzyl chloride acts as a leaving group.
  • The hydroxide ion from NaOH attacks the carbon atom bonded to the chlorine, forming an intermediate complex.
  • This results in the formation of a new product, benzyl alcohol (or o/p-cresol).
Understanding nucleophilic substitution is vital because it highlights the reactivity of carbon-halogen bonds and the facile nature of such transformations in organic synthesis. It also underscores how different conditions and reagents influence the pathway and outcome of a reaction.
Toluene Reactions
Toluene is an aromatic hydrocarbon that can undergo various reactions due to its stable benzene ring and reactive methyl group. In this context, the reaction sequence of chlorination followed by hydrolysis is central to understanding toluene's behavior.
  • The methyl group on toluene can be chlorinated to give benzyl chloride.
  • Subsequent treatment with aqueous NaOH leads to the formation of o/p-cresol.
  • This demonstrates the interplay between radical and nucleophilic substitution mechanisms starting from toluene.
The reactions involving toluene underscore its versatility in organic synthesis, allowing chemists to produce a variety of compounds depending on the conditions employed and subsequent transformations. This sequence leads mainly to ortho and para products due to steric and electronic factors in the benzene ring.

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

Which one of the following is most reactive towards nucleophillic substitution reaction? (a) \(\mathrm{C}_{6} \mathrm{H}_{3} \mathrm{Cl}\) (b) \(\mathrm{CH}_{2}=\mathrm{CH}-\mathrm{Cl}\) (c) \(\mathrm{ClCH}_{2}-\mathrm{CH}=\mathrm{CH}_{2}\) (d) \(\mathrm{CH}_{3} \mathrm{CH}=\mathrm{CH}-\mathrm{Cl}\)

n-propyl bromide on treatment with ethanolic potassium hydroxide produces (a) propane (b) propene (c) propyne (d) propanol

In the following chemical reactions: \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{NH}_{2}+\mathrm{CHCl}_{3}+3 \mathrm{KOH} \longrightarrow(\mathrm{A})+(\mathrm{B})+3 \mathrm{H}_{2} \mathrm{O}\), the compounds (A) and (B) are respectively [2007] (a) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CONH}_{2}\) and \(3 \mathrm{KCl}\) (b) \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NC}\) and \(\mathrm{K}_{2} \mathrm{CO}_{3}\) (c) \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NC}\) and \(3 \mathrm{KCl}\) (d) \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{CN}\) and \(3 \mathrm{KCl}\)

\(\mathrm{R}-\mathrm{OH}+\mathrm{HX} \longrightarrow \mathrm{R}-\mathrm{X}+\mathrm{H}_{2} \mathrm{O}\) In the above reaction, the reactivity of different alcohols is (a) tertiary > secondary > primary (b) tertiary \(>\) secondary \(<\) primary (c) tertiary \(<\) secondary \(>\) primary (d) secondary < primary \(<\) tertiary

The reaction of the formation of bromobenzyl chloride with \(\mathrm{NaCN}\) in ethanol leads to (a) 4-bromo-2-cyanobenzyl chloride (b) 4 -cyanobenzyl cyanide (c) 4-cyanobenzyl chloride (d) 4-bromobenzyl cyanide

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