Problem 85
Question
Which of the following reaction does not involve a carbocation as intermediate? (a) \(\mathrm{C}_{6} \mathrm{H}_{6}+\mathrm{Br}_{2} \quad \mathrm{AlBr}_{3} \quad \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{Br}\) (b) \(\mathrm{CH}_{2}=\mathrm{CH}_{2}^{2} \overline{+\mathrm{Br}_{2} \stackrel{3}{\longrightarrow}}{\longrightarrow} \mathrm{BrCH}_{2}-\mathrm{CH}_{2} \mathrm{Br}\) (c) \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{COH}+\mathrm{HBr} \mathrm{H}^{+}\left(\mathrm{CH}_{3}\right)_{3} \mathrm{CBr}+\mathrm{H}_{2} \mathrm{O}\) (d) Both (b) and (c)
Step-by-Step Solution
Verified Answer
Option (b) does not involve a carbocation intermediate.
1Step 1: Identify Reaction Types
First, identify the type of organic reactions described in each option. Option (a) is a Friedel-Crafts Bromination involving a Lewis acid (AlBr₃), option (b) is an electrophilic addition of Br₂ to an alkene, option (c) is a substitution reaction of an alcohol with HBr.
2Step 2: Assess Carbocation Formation
Consider how carbocations are typically formed in reactions. They commonly occur in substitution reactions (like SN1) or rearrangements. In (a), the reaction proceeds through an electrophilic aromatic substitution where a carbocation intermediate (arenium ion) is formed on the benzene ring. In (b), bromine adds to the alkene forming a cyclic bromonium ion, not a carbocation. In (c), the alcohol reacts with HBr, forming a carbocation intermediate after the hydroxyl group is protonated and leaves.
3Step 3: Determine Non-Carbocation Reaction
Identify which reaction does not involve a carbocation intermediate. In (a), the arenium ion is a type of carbocation. In (b), no carbocation is formed since the intermediate is a bromonium ion, not a classic carbocation. Reaction (c), the tertiary carbocation forms as an intermediate during the substitution process.
4Step 4: Conclusion
Given the analyses above, option (b), which involves a bromonium ion rather than a carbocation, is the reaction that does not involve a carbocation intermediate.
Key Concepts
Friedel-Crafts BrominationElectrophilic AdditionSubstitution ReactionsBromonium IonOrganic Reaction Mechanisms
Friedel-Crafts Bromination
Friedel-Crafts Bromination is a type of electrophilic aromatic substitution reaction. This process involves the substitution of a hydrogen atom on an aromatic ring with a bromine atom. To facilitate this, a Lewis acid like aluminum bromide
(AlBr₃)
is necessary as a catalyst.
This catalyst helps in generating the bromonium ion, which is a powerful electrophile. In this reaction, the benzene ring acts as a nucleophile, attacking the electrophilic bromonium ion.
This results in the formation of a carbocation intermediate known as an arenium ion. Eventually, the hydrogen is replaced by the bromine, completing the substitution.
This catalyst helps in generating the bromonium ion, which is a powerful electrophile. In this reaction, the benzene ring acts as a nucleophile, attacking the electrophilic bromonium ion.
This results in the formation of a carbocation intermediate known as an arenium ion. Eventually, the hydrogen is replaced by the bromine, completing the substitution.
- The catalyst not only activates the bromine but also stabilizes the carbocation formed during the intermediate stage.
- Although instead of a direct carbocation, the arenium ion is a stabilized version of it.
- Friedel-Crafts reactions are highly useful in organic synthesis for modifying benzene rings.
Electrophilic Addition
Electrophilic addition reactions are key reactions that occur in alkenes or alkynes.
These reactions typically involve the addition of elements across a carbon–carbon double or triple bond.
In the context of the problem, an alkene reacts with bromine to form a dibromo compound. Unlike Friedel-Crafts Bromination, electrophilic addition to alkenes does not form carbocations but rather forms a temporary bromonium ion.
In the context of the problem, an alkene reacts with bromine to form a dibromo compound. Unlike Friedel-Crafts Bromination, electrophilic addition to alkenes does not form carbocations but rather forms a temporary bromonium ion.
- During this reaction, the double bond of the alkene acts as a nucleophile.
- It attacks the electrophilic bromine to form the bromonium ion.
- This intermediate is a three-membered ring, quite distinct from typical carbocations.
Substitution Reactions
Substitution reactions are fundamental in organic chemistry, involving the replacement of one group in a molecule with another.
One classic example is when an alcohol reacts with hydrogen bromide (HBr) to form an alkyl bromide.
This type of reaction might involve carbocations.
One classic example is when an alcohol reacts with hydrogen bromide (HBr) to form an alkyl bromide.
This type of reaction might involve carbocations.
- Initially, the \(\mathrm{OH}\) group in the alcohol is protonated by \(\mathrm{HBr}\) to form \(\mathrm{H}_2\mathrm{O}\) as a leaving group.
- This departure creates a carbocation intermediate, which can be quite stable, especially within tertiary carbons.
- The bromide ion then acts as a nucleophile, attacking the carbocation forming the alkyl bromide.
Bromonium Ion
A bromonium ion is a crucial intermediate in certain electrophilic addition reactions involving alkenes and bromine.
Unlike most carbocations, a bromonium ion is a positively charged ion, where bromine has formed a three-membered ring with two carbons.
This structure is surprisingly stable due to its cyclic nature.
Unlike most carbocations, a bromonium ion is a positively charged ion, where bromine has formed a three-membered ring with two carbons.
This structure is surprisingly stable due to its cyclic nature.
- The creation of the bromonium ion involves initial electrophilic attack of bromine by the alkene.
- This forms a charge-stabilized intermediate, rather than an unstable carbocation.
- Its stability prevents rearrangements common in typical carbocation intermediates.
Organic Reaction Mechanisms
Organic reaction mechanisms detail each step and transformation during chemical reactions involving organic compounds.
They provide a roadmap for understanding how substrates transform into products.
Insights into these steps can clarify reaction conditions and intermediates, including carbocations and bromonium ions.
They provide a roadmap for understanding how substrates transform into products.
Insights into these steps can clarify reaction conditions and intermediates, including carbocations and bromonium ions.
- A deep grasp of mechanisms allows chemists to predict how changes in conditions can affect the outcome.
- They guide the synthesis of complex organic molecules.
- They also determine how and why specific intermediates form, affect reaction rates, and influence product distribution.
Other exercises in this chapter
Problem 83
Most stable carbonium ion is (a) \(\mathrm{p}-\mathrm{NO}_{2}-\mathrm{C}_{6} \mathrm{H}_{4}-\mathrm{CH}_{2}^{+}\) (b) \(\mathrm{C}_{6} \mathrm{H}_{3} \mathrm{CH
View solution Problem 84
Arrange \(\mathrm{CCl}_{3}\left(\mathrm{CH}_{2}\right)_{3} \mathrm{NH}_{2}\) (a), \(\mathrm{CCl}_{3}\left(\mathrm{CH}_{2}\right)_{2} \mathrm{NH}_{2}\) (b) and \
View solution Problem 90
Consider the following nucleophiles: \(\mathrm{H}_{2} \mathrm{O}, \quad \mathrm{CH}_{3} \mathrm{COO}^{-}, \quad \overline{\mathrm{O} H}, \quad \mathrm{CH}_{3} \
View solution Problem 92
Arrange methyl amine (a), pyridine (b) and guan (c) in decreasing order of basic strength (a) \(\mathrm{C}>\mathrm{A}>\mathrm{B}\) (b) \(\mathrm{A}>\mathrm{B}>\
View solution