Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 3: Problem 121

Give, in the form of simple equations, the method of generation of electrophiles in the following aromatic electrophilic substitutions. (i) Sulphonation (ii) Friedel-Crafts alkylation (iii) Friedel-Crafts acylation

Short Answer

Expert verified
Question: Explain the generation of electrophiles in sulphonation, Friedel-Crafts alkylation, and Friedel-Crafts acylation of aromatic rings. Answer: In sulphonation, the electrophile SO3 is generated by the reaction of concentrated sulfuric acid (H2SO4) with the aromatic compound. In Friedel-Crafts alkylation, the carbocation electrophile (R+) is created by reacting an alkyl halide (RX) with a Lewis acid catalyst, such as aluminum chloride (AlCl3). In Friedel-Crafts acylation, the acylium ion electrophile (RCO+) is formed by reacting an acid chloride (RCOCl) with a Lewis acid catalyst, typically aluminum chloride (AlCl3).

Step by step solution

01

(i) Sulphonation of Aromatic Rings

Sulphonation is the process of introducing a sulphonic acid group (-SO3H) to an aromatic ring. The most common electrophile used in sulphonation is the sulphur trioxide (SO3) molecule. It is generated by the reaction of concentrated sulfuric acid (H2SO4) with the aromatic compound. The equation for the generation of the electrophile is: H2SO4 -> H+ + HSO4- HSO4- + SO3 -> H2SO4SO3 (Electrophile: SO3) The aromatic compound then reacts with the generated SO3 electrophile, leading to the sulphonation.
02

(ii) Friedel-Crafts Alkylation of Aromatic Rings

Friedel-Crafts alkylation is a reaction in which an alkyl group is added to an aromatic ring. This involves an electrophile in the form of a carbocation (R+). The carbocation is usually generated by reacting an alkyl halide (RX) with a Lewis acid catalyst, mostly aluminum chloride (AlCl3). The equation for the generation of the electrophile is: AlCl3 + R-Cl -> R+ + AlCl4- (Electrophile: R+) The aromatic compound reacts with the generated R+ electrophile, leading to alkylation.
03

(iii) Friedel-Crafts Acylation of Aromatic Rings

Friedel-Crafts acylation is a reaction in which an acyl group (-COR) is added to an aromatic ring. The electrophile in this case is an acylium ion (RCO+). The acylium ion is generated by reacting an acid chloride (RCOCl) with a Lewis acid catalyst, mostly aluminum chloride (AlCl3). The equation for the generation of the electrophile is: AlCl3 + RCO-Cl -> RCO+ + AlCl4- (Electrophile: RCO+) The aromatic compound then reacts with the generated RCO+ electrophile, leading to acylation.

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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

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

Statement 1 On heating with con. \(\mathrm{H}_{2} \mathrm{SO}_{4}\), benzene gives benzene sulphonic acid which when heated with superheated steam under pressure gives back benzene.

$$ \mathrm{HCHO}+\mathrm{HCl} \stackrel{\mathrm{ZnCl}_{2}}{\longrightarrow} \mathrm{P} $$ The organic compound P formed in the above reaction is

The compound (A) when treated with methanol and few drops of conc.H \(_{2} \mathrm{SO}_{4}\) gives the smell of oil of winter green. The compound (A) is (a) phenol (b) benzene (c) \(\mathrm{p}\) -cresol (d) salicylic acid

The reaction of 4 -hydroxybenzene sulphonic acid with excess bromine water gives (a) 3 -Bromo-4-hydroxybenzenesulphonic acid (b) 3,5 -Dibromo-4-hydroxybenzenesulphonic acid (c) \(2,4,6\) -Tribromophenol (d) 4-Bromobenzenesulphonic acid

Column I \(\quad\) Column II (a) \(\mathrm{C}_{6} \mathrm{H}_{6}+\mathrm{CH}_{3}-\mathrm{CH}=\mathrm{CH}_{2} \stackrel{\mathrm{H}_{2} \mathrm{SO}_{4}}{\longrightarrow}\) (p) Schotten-Baumann reaction (b) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{OH}+\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COCl} \stackrel{\mathrm{NaOH}}{\longrightarrow}\) (q) Friedel-Crafts reaction (c) \(\mathrm{C}_{6} \mathrm{H}_{6}+\stackrel{\mathrm{AlCl}_{3}}{\longrightarrow}\) (r) Reimer-Tiemann reaction (d) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{OH} \frac{{ }_{5}^{(\mathrm{i}) \mathrm{CCl}_{4} / \mathrm{NaOH}}}{{ }_{(\text {iii }) \mathrm{H}_{3} \mathrm{O}^{+}}}{\longrightarrow}\) (s) Isopropyl benzene (t) Electrophilic substitution
See all solutions

Recommended explanations on Chemistry Textbooks

Kinetics

Read Explanation

Chemistry Branches

Read Explanation

Inorganic Chemistry

Read Explanation

Chemical Reactions

Read Explanation

Nuclear Chemistry

Read Explanation

The Earths Atmosphere

Read Explanation
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