Chapter 19: Problem 55
Which of the following reactions will yield 2,2 -dibromopropane? (1) \(\mathrm{CH}_{2}=\mathrm{CHBr}+\mathrm{HBr}\) (2) \(\mathrm{CH}_{3} \mathrm{C}=\mathrm{CCH}_{3}\) (3) \(\mathrm{CH}_{3} \mathrm{C} \equiv \mathrm{CH}+2 \mathrm{HBr}\) (4) \(\mathrm{CH}_{3} \mathrm{CH}=\mathrm{CHBr}+\mathrm{HBr}\)
Short Answer
Step by step solution
- Identify the structure of 2,2-dibromopropane
- Analyze Reaction (1)
- Analyze Reaction (2)
- Analyze Reaction (3)
- Analyze Reaction (4)
- Conclusion
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
organic chemistry reactions
One of the primary focuses in organic chemistry is how different reactants interact and transform, leading to the formation of desired products. These reactions can be classified into several types, such as addition, substitution, elimination, and rearrangement reactions.
In our case, the synthesis of 2,2-dibromopropane involves an addition reaction—a type of reaction where atoms are added to a molecule, typically across a double or triple bond.
addition reactions
In an addition reaction, a reactant is added across the carbon-carbon multiple bond, breaking the bond, and forming new single bonds. Addition reactions can be further divided into different mechanisms, such as electrophilic addition and nucleophilic addition.
For instance, when we look at Reaction (3) from our exercise, it involves the addition of hydrogen bromide (HBr) across the triple bond of \(\text{CH}_3 \text{C} \equiv \text{CH}\). This addition converts the triple bond to a single bond, resulting in the formation of 2,2-dibromopropane (\text{CH}_3\text{CBr}_2\text{CH}_3).
Key points about addition reactions:
- Involves breaking multiple bonds to add new atoms.
- Commonly occurs with alkenes and alkynes.
- Can lead to the formation of saturated products from unsaturated reactants.
alkyne halogenation
When an alkyne undergoes halogenation, halogen atoms add across the triple bond, converting it first to a double bond and then to a single bond, if excess halogen is provided. This results in the formation of di- or tetra-substituted products.
For example, in Reaction (3), the alkyne \(\text{CH}_3 \text{C} \equiv \text{CH}\) reacts with two molecules of HBr. The first molecule of HBr adds across the triple bond, forming a bromo-alkene (\text{CH}_3 \text{CBr}=\text{CH}_2). The second molecule of HBr then adds across the double bond, yielding 2,2-dibromopropane (\text{CH}_3 \text{CBr}_2 \text{CH}_3).
Important aspects of alkyne halogenation:
- Stepwise addition of halogens.
- Formation of intermediate alkenes for reactions involving excess halogen.
reaction mechanisms
A reaction mechanism details the step-by-step process by which reactants transform into products. This involves the breaking and forming of bonds, the movement of electrons, and the formation of intermediates.
For Reaction (3), the mechanism is as follows:
- The triple bond in \(\text{CH}_3 \text{C} \equiv \text{CH}\) reacts with the first molecule of HBr. The electrons in the triple bond are attracted to the hydrogen atom of HBr, leading to the formation of the bromo-alkene intermediate (\text{CH}_3 \text{CBr}=\text{CH}_2).
- The second molecule of HBr then reacts with the bromo-alkene. The electrons in the double bond again target the hydrogen atom of HBr, forming the final product \(\text{CH}_3 \text{CBr}_2 \text{CH}_3\).