Chapter 16: Problem 47
Complete the following reactions. (a) \(\mathrm{CH}_{3} \mathrm{CH}=\mathrm{CHCH}_{3}+\mathrm{HBr}\) (b) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}=\mathrm{CHCH}_{3}+\mathrm{Br}_{2}\)
Short Answer
Expert verified
The products of reactions are (a) \(\mathrm{CH}_{3}\mathrm{CHBrCH}_{2} \mathrm{CH}_{3}\) and (b) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CHBrCHBrCH}_{3}\)
Step by step solution
01
Identify theType of Reaction
These reactions are electrophilic additions, where an electrophile (such as a halogen) adds to a double bond, giving a single bond. Markovnikov's Rule helps predict how the structure will change.
02
Solve for Reaction (a)
The reaction is: \(\mathrm{CH}_{3} \mathrm{CH}=\mathrm{CHCH}_{3}+\mathrm{HBr}\) According to Markovnikov's Rule, the hydrogen (H) from HBr will attach to the carbon which already has more hydrogen atoms attached to it. The bromine (Br) attaches to the other carbon. Thus, the product becomes: \(\mathrm{CH}_{3}\mathrm{CHBrCH}_{2} \mathrm{CH}_{3}\).
03
Solve for Reaction (b)
The reaction is: \(\mathrm{CH}_{3} \mathrm{CH}_{2}\mathrm{CH}=\mathrm{CHCH}_{3}+\mathrm{Br}_{2}\). Here bromine (Br2) acts as the electrophile. Each carbon involved in the double bond will add a Br atom. The product is \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CHBrCHBrCH}_{3}\).
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.
Markovnikov's Rule
Markovnikov's Rule is a handy guideline in organic chemistry that helps predict the outcome of an addition reaction involving alkenes. When an alkene reacts with a hydrogen halide (like HBr), this rule states that the hydrogen atom will bond with the carbon that already has more hydrogen atoms attached. In other words, it will go to the more hydrogen "rich" carbon. This leaves the halogen (like Br) to attach to the carbon with fewer hydrogens. This rule is useful because it helps chemists anticipate the major product formed during the reaction, and it generally holds true for many electrophilic addition reactions.
Understanding this concept is crucial when predicting the structures of products in electrophilic additions. For example, in the reaction of 1-butene with HBr, application of Markovnikov's Rule tells us that the bromine atom will attach to the secondary carbon atom, resulting in the formation of 2-bromobutane. This rule applies because the hydrogen atom prefers to add to the carbon that can support extra hydrogens, thus stabilizing the carbocation intermediate formed during the reaction. By following this principle, chemists can reliably predict how alkenes will react in various addition reactions.
Understanding this concept is crucial when predicting the structures of products in electrophilic additions. For example, in the reaction of 1-butene with HBr, application of Markovnikov's Rule tells us that the bromine atom will attach to the secondary carbon atom, resulting in the formation of 2-bromobutane. This rule applies because the hydrogen atom prefers to add to the carbon that can support extra hydrogens, thus stabilizing the carbocation intermediate formed during the reaction. By following this principle, chemists can reliably predict how alkenes will react in various addition reactions.
Alkenes
Alkenes are hydrocarbons that contain carbon-carbon double bonds, making them unsaturated. This feature lends them unique chemical reactivity compared to their saturated cousins, alkanes. The double bond present in alkenes consists of one sigma bond and one pi bond. This pi bond is particularly reactive, conferring electrophilic characteristics to alkenes.
Alkenes are often represented by the general formula \ \(C_{n}H_{2n} \). The presence of the double bond in alkenes allows them to undergo addition reactions where the pi bond breaks, and new single bonds form. This is the characteristic reaction type for alkenes. They can participate in various reactions, forming a wide range of products. For instance, when an alkene reacts with a halogen like Br2, the double bond breaks, allowing each carbon involved to bond with a bromine atom, as observed in the addition reactions discussed.
Alkenes are often represented by the general formula \ \(C_{n}H_{2n} \). The presence of the double bond in alkenes allows them to undergo addition reactions where the pi bond breaks, and new single bonds form. This is the characteristic reaction type for alkenes. They can participate in various reactions, forming a wide range of products. For instance, when an alkene reacts with a halogen like Br2, the double bond breaks, allowing each carbon involved to bond with a bromine atom, as observed in the addition reactions discussed.
- Alkenes are characterized by their double bonds.
- They are more reactive than alkanes due to these double bonds.
- Common reactions include hydrogenation, halogenation, and hydrohalogenation.
Reaction Mechanisms
In organic chemistry, understanding reaction mechanisms is key to predicting how and why chemical reactions occur. A reaction mechanism provides a detailed, step-by-step account of the changes in molecular structure and electron redistribution that happen during a chemical reaction.
For electrophilic addition reactions like those involving alkenes and HBr, the mechanism typically involves the following steps:
Understanding these mechanisms is crucial for chemists. They allow the prediction and rationalization of the products of a reaction, as well as offering insights into creating new synthetic pathways for desired compounds. A clear grasp of reaction mechanisms enhances a chemist’s ability to manipulate and control chemical reactions effectively.
For electrophilic addition reactions like those involving alkenes and HBr, the mechanism typically involves the following steps:
- First, the pi electrons of the alkene attack the electrophilic part of the reagent, such as H in HBr, forming a carbocation intermediate.
- Next, the negatively charged bromine ion attacks the positively charged carbocation, resulting in the formation of a stable alkyl bromide.
Understanding these mechanisms is crucial for chemists. They allow the prediction and rationalization of the products of a reaction, as well as offering insights into creating new synthetic pathways for desired compounds. A clear grasp of reaction mechanisms enhances a chemist’s ability to manipulate and control chemical reactions effectively.