Chapter 8: Problem 34
\text { What two different alkyl halides yield }\left(\mathrm{CH}_{3}\right)_{2} \mathrm{C}=\mathrm{CH}_{2} \text { as the only product of dehydrohalogenation? }
Short Answer
Expert verified
The alkyl halides are 2-bromo-2-methylpropane and 1-bromo-2-methylpropane.
Step by step solution
01
Understanding Dehydrohalogenation
Dehydrohalogenation is a chemical reaction that involves the elimination of a hydrogen halide (like HCl or HBr) from an alkyl halide to form an alkene. For the alkene \( ext{(CH}_3) ext{C} = ext{CH}_2\), there are two potential starting alkyl halides that upon elimination of HX will form this alkene.
02
Identifying Alkyl Halides
To obtain \((\text{CH}_3)_2\text{C} = \text{CH}_2\), we need to lose an HX from alkyl halides that have the structure which results in double bond formation. Possible alkyl halides include 2-bromo-2-methylpropane (tert-butyl bromide) and 1-bromo-2-methylpropane (isobutyl bromide).
03
Structure of the Alkyl Halides
2-bromo-2-methylpropane has the structure \((\text{C}_3\text{H}_7)\text{CBr}\), where elimination of HBr from the quaternary carbon produces the desired alkene immediately. 1-bromo-2-methylpropane has the structure \((\text{CH}_3)_2\text{CHCH}_2\text{Br}\), leading to the alkene by formation of a more substituted product after elimination of HBr.
04
Verifying Dehydrohalogenation
When 2-bromo-2-methylpropane undergoes dehydrohalogenation, it directly forms \((\text{CH}_3)_2\text{C}=\text{CH}_2\). For 1-bromo-2-methylpropane, elimination happens by abstraction of hydrogen from the terminal carbon having a Br atom, followed by rearrangement during the reaction to give the same alkene.
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 91Ó°ÊÓ!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Alkyl Halides
Alkyl halides are organic compounds where a halogen atom (such as chlorine, bromine, or iodine) is bonded to an alkyl group.
- The alkyl group is typically a carbon chain, making the compound a derivative of an alkane.
- Alkyl halides serve as a primary substrate in various chemical reactions, often leading to the formation of more complex compounds.
- Primary Alkyl Halides: The halogen is attached to a carbon atom that is bonded to one other carbon.
- Secondary Alkyl Halides: The halogen is attached to a carbon that is bonded to two other carbons.
- Tertiary Alkyl Halides: The halogen is attached to a carbon bonded to three other carbons.
Alkene Formation
Alkene formation involves the creation of a carbon-carbon double bond within an organic molecule. This occurs via elimination reactions, where elements are removed from a precursor molecule. In our context, the precursor is an alkyl halide.
- The process typically replaces a carbon-halogen bond with a carbon-carbon double bond.
- The result is an unsaturated compound known as an alkene.
- When the hydrogens and the halide are removed, a double bond is formed between the two carbons.
- This leads to the creation of the desired alkene, which in this case is \((\text{CH}_3)_2\text{C} = \text{CH}_2\).
Elimination Reaction
Elimination reactions are a class of reactions where two substituents are removed from a molecule, leading to the formation of a double bond. They are a key step in converting alkyl halides into alkenes. The primary goal here is transforming the saturated alkyl halide into an unsaturated alkene.There are two main mechanisms for elimination reactions:
- E1 Mechanism: Involves two steps; formation of a carbocation intermediate followed by the elimination of a proton to form a double bond.
- E2 Mechanism: A one-step process; simultaneous removal of a proton and a halide ion, directly forming the double bond.