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Chapter 14: Problem 49

Predict the major product(s), if any, for the reaction of \(\mathrm{HBr}\) in THF (solvent) with each of the following: (a) styrene (b) \((Z)-2\)-phenyl-2-hexene (c) 2 -methyl-1-phenyl-1-propene (d) 3-phenyl-1-propene (e) naphthalene

Short Answer

Expert verified
The main products are: (a) 1-bromo-1-phenylethane, (b) 2-bromo-2-phenylhexane, (c) 1-bromo-2-methyl-1-phenylpropane, (d) 3-phenyl-1-bromopropane, (e) no reaction.

Step by step solution

01

- Understanding the Reaction Mechanism

The reaction involves an addition of \(\b{HBr}\) to an unsaturated hydrocarbon in THF solvent. In the case of each compound listed, determine whether the double bond is present for addition and consider Markovnikov's rule, which states that \(\b{H}\) will attach to the carbon with more hydrogen atoms (less substituted) and \(\b{Br}\) will attach to the carbon with fewer hydrogen atoms (more substituted).
02

- Reacting with Styrene

Styrene contains an ethylene group attached to a benzene ring. Applying Markovnikov's rule, the \(\b{H}\) from HBr will add to the \(\b{CH_2}\) carbon of styrene (the carbon in the double bond that has more hydrogen atoms), and the \(\b{Br}\) will add to the carbon that is directly attached to the benzene ring. The product is 1-bromo-1-phenylethane.
03

- Reacting with (Z)-2-Phenyl-2-Hexene

(Z)-2-Phenyl-2-Hexene has a central double bond between the second and third carbon atoms. According to Markovnikov's rule, \(\b{H}\) will add to the carbon that has more hydrogen atoms (the third carbon), and \(\b{Br}\) will add to the second carbon which is also bonded to a phenyl group. This creates (2-bromo-2-phenylhexane).
04

- Reacting with 2-Methyl-1-Phenyl-1-Propene

2-Methyl-1-Phenyl-1-Propene has a double bond between the first carbon and the second carbon in the propene group. Applying Markovnikov's rule, the \(\b{H}\) will add to the carbon with more hydrogen atoms, and \(\b{Br}\) will add to the carbon bonded to the phenyl and methyl groups. The product is 1-bromo-2-methyl-1-phenylpropane.
05

- Reacting with 3-Phenyl-1-Propene

3-Phenyl-1-Propene has a double bond between the first and second carbon atoms. According to Markovnikov's rule, the \(\b{H}\) will add to the carbon with two hydrogen atoms (first carbon), and \(\b{Br}\) will add to the second carbon, yielding 3-phenyl-1-bromopropane.
06

- Reacting with Naphthalene

Naphthalene has a stable conjugated ring system with no isolated double bond, making it resistant to simple addition reactions. Therefore, no reaction will occur with \(\b{HBr}\).

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Key Concepts

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

Markovnikov's rule
Markovnikov's rule is a key principle in organic chemistry that helps predict the outcomes of addition reactions. When a hydrogen halide like \(\b{HBr}\) adds to an unsaturated hydrocarbon, the hydrogen (\b{H}) attaches to the carbon with more hydrogen atoms, while the bromine (\b{Br}) attaches to the carbon with fewer hydrogen atoms. This creates a more stable, more substituted carbocation intermediate, which ultimately determines the major product of the reaction.
Addition reaction
An addition reaction is a fundamental type of reaction where atoms are added to a double or triple bond in an unsaturated hydrocarbon. In the case of \(\b{HBr}\) adding to various hydrocarbons:
  • Styrene
  • (Z)-2-phenyl-2-hexene
  • 2-methyl-1-phenyl-1-propene
  • 3-phenyl-1-propene

In each of these cases, the double bond breaks, allowing new atoms to form bonds with the carbon atoms. Markovnikov's rule guides us to predict which carbon the hydrogen and bromine atoms will attach to.
Styrene
Styrene is an unsaturated hydrocarbon consisting of an ethylene group (\(\b{CH_2=CH}\)) attached to a benzene ring. When \(\b{HBr}\) adds to styrene, Markovnikov's rule predicts that the hydrogen (\b{H}) will attach to the \(\b{CH_2}\) carbon, and the bromine (\b{Br}) will attach to the carbon directly attached to the benzene ring. The product of this reaction is 1-bromo-1-phenylethane. This makes styrene an important compound in the study of addition reactions.
HBr addition mechanism
The mechanism of \(\b{HBr}\) addition to an unsaturated hydrocarbon involves the following steps:
  • Initiation: The double bond in the hydrocarbon reacts with \(\b{HBr}\).
  • Intermediate formation: A carbocation (positively charged carbon atom) intermediate is formed, guided by Markovnikov's rule.
  • Product formation: The \(\b{Br}\) anion attaches to the carbocation, resulting in the final product.

This sequence ensures the most stable and predictable placement of the hydrogen and bromine atoms in the product.
Unsaturated hydrocarbons
Unsaturated hydrocarbons contain at least one double or triple bond between carbon atoms. This bond makes them reactive sites for addition reactions. Examples include:
  • Styrene: A single double bond attached to a benzene ring.
  • (Z)-2-phenyl-2-hexene: A double bond in a hexene chain with a phenyl group attached.
  • 2-methyl-1-phenyl-1-propene: A double bond in a propene group with methyl and phenyl substituents.

These compounds readily undergo addition reactions, such as with \(\b{HBr}\), making them crucial in organic reaction mechanisms.

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Most popular questions from this chapter

The nitronium ion \(\left({ }^{+} \mathrm{NO}_{2}\right)\) is known and can react with simple benzenes to give nitrobenzenes. Write an arrow formalism mechanism for this reaction.

Design an aromatic heterocyclic compound that (a) contains one boron atom. (b) contains one boron and one oxygen atom in a sixmembered ring. (c) contains one oxygen and one nitrogen atom in a five-membered ring. (d) contains two nitrogen atoms in a five-membered ring. (e) contains three boron and three nitrogen atoms in a sixmembered ring.

Predict the major product(s), if any, for the bromination \(\left(\mathrm{Br}_{2}, \mathrm{CH}_{2} \mathrm{Cl}_{2}\right)\) of the following compounds: (a) styrene (b) 1-ethyl-3-nitrobenzene (c) \((Z)-2\)-phenyl-2-butene (d) 3 -phenyl-1-propene (e) 3 -phenylheptane

One early problem with the Kekulé structure was that only one 1,2 -disubstituted isomer of benzene could be found for any given substituent. That is, there is only one 1,2 -dimethylbenzene (o-xylene). Explain why this observation would have been hard for Kekulé to explain.

Select the reaction "Benzylic oxidation." Observe the reaction several times. The animation uses oxygen \(\left(\mathrm{O}_{2}\right)\) as the initiator of the reaction, which also occurs in nature. One of the reasons for the common practice of packing organic foods over nitrogen \(\left(\mathrm{N}_{2}\right)\) is to avoid the reaction of \(\mathrm{O}_{2}\) with benzylic (and allylic) positions in natural products present in food. Draw each step of the reaction between \(\mathrm{O}_{2}\) and 2 -phenylpropane. Include a termination reaction. Label your steps as initiation, propagation, or termination.
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