Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 11: Problem 218

(a) The bromination of benzene is catalyzed by small amounts of iodine. Consider a possible explanation for this catalytic effect. (b) The kinetic expression for the bromination of naphthalene in glacial acetic acid involves a term that is first order in naphthalene and second order in bromine. Consider how two molecules of bromine and one of naphthalene could be involved in the rate-determining step of bromination. Suggest a reason why the kinetic expression simplifies to first order in naphthalenes and first order in bromine in 50 per cent aqueous acetic acid.

Short Answer

Expert verified
(a) Iodine catalyzes the bromination of benzene by generating a bromonium ion (Br+), which electrophilically attacks the benzene ring, increasing the overall reaction rate. (b) In glacial acetic acid, the rate-determining step may involve two bromine molecules and one naphthalene molecule, with one bromine forming an intermediate complex. In a 50% aqueous acetic acid, the formation of bromine hydrate (HOBr) changes the reaction pathway, leading to a simplified kinetic expression of first order in naphthalenes and first order in bromine.

Step by step solution

01

(a) Iodine's catalytic effect in bromination of benzene

Iodine can act as a catalyst by generating a bromonium ion (Br+), which can electrophilically attack the benzene ring. This can be done by forming a complex with bromine, which weakens the bond between the two bromine atoms, allowing one to act as an electrophilic bromine (Br+) and the other as a leaving group (Br-). This process increases the overall reaction rate and facilitates the bromination of benzene.
02

(b) Kinetic expression for bromination of naphthalene in glacial acetic acid

The reaction is first order in naphthalene and second order in bromine, which suggests that the rate-determining step (slowest step in the reaction mechanism) might involve two molecules of bromine and one molecule of naphthalene. A possible mechanism could be that one bromine molecule forms an intermediate complex with naphthalene, and then this complex reacts with another bromine molecule to form the final product.
03

(b) Simplification of the kinetic expression in 50% aqueous acetic acid

In a 50% aqueous solution of acetic acid, the solubility of bromine increases, leading to the formation of bromine hydrate (HOBr). This changes the reaction pathway and involves the reaction between naphthalene and bromine hydrate instead of two bromine molecules. The reaction rate would depend significantly on the concentration of naphthalene and bromine hydrate which leads to a kinetic expression of first order in naphthalenes and first order in bromine.

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

Outline all steps in the laboratory synthesis of the following compounds from benzene and/or toluene, using any needed aliphatic or inorganic reagents. Assume that a pure para isomer can be separated from an ortho, para mixture. (a) \(\mathrm{p}\) -nitrotoluene (d) p-bromobenzoic acid (b) \(\mathrm{p}\) -bromonitrobenzene (e) o-iodobenzoic acid (c) m-bromobenzenesulfonic acid (f) \(1,3,5-\) trinitrobenzene (g) 3,5 -dinitrobenzoic acid

Hydrocarbon B, \(\mathrm{C}_{6} \mathrm{H}_{6}\), gave an nmr spectrum with two signals: \(\delta 6.55\) and \(\delta 3.84\), peak area ratio \(2: 1\). When warmed in pyridine for three hours, B was quantitatively converted into benzene. Mild hydrogenation of \(B\) yielded \(C\), whose spectra showed the following: mass spectrum, mol. wt. 82 ; infrared spectrum, no double bonds; nmr spectrum, one broad peak at \(\delta 2.34\). (a)How many rings are there in \(\mathrm{C}\) ? (b) How many rings are there (probably) in B? How many double bonds in \(\mathrm{B}\) ? (c) Can you suggest a structure for \(\mathrm{B}\) ? for \(\mathrm{C}\) ? (d) In the nmr spectrum of B the downfield signal was a triplet. How must you account for these splittings?

Even though \(1,3,5-\) trinitrobenzene (TNB) has more shattering power (more brisance) and is no more dangerous to handle, \(2,4,6-\) trinitrotoluene (TNT) has always been the high explosive in more general use. Can you suggest a reason (connected with manufacture) for the popularity of TNT? (Benzene and toluene are both readily available materials; for many years benzene was cheaper.)

Is \(\mathrm{BH}_{3}\) a nucleophile or an electrophile? Explain.

How do you account for the fact that benzene in the presence of \(\mathrm{ALCl}_{3}\) reacts: (a) with n-propyl bromide to give isopropylbenzene; (b) with isobutyl bromide to yield tert-butylbenzene; (c) with neopentyl bromide to yield tert-pentylbenzene? (d) By which of the alternative mechanisms for the Friedel-Crafts reaction are these products probably formed?
See all solutions

Recommended explanations on Chemistry Textbooks

Kinetics

Read Explanation

Chemical Reactions

Read Explanation

Nuclear Chemistry

Read Explanation

Physical Chemistry

Read Explanation

Organic Chemistry

Read Explanation

Chemical Analysis

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