Problem 39
Question
Anti-Markovnikov's addition of HBr is not observed in (a) propene (b) but-1-ene (c) but-2-ene (d) pent-3-ene
Step-by-Step Solution
Verified Answer
Anti-Markovnikov's addition of HBr is not observed in but-2-ene.
1Step 1: Understand Anti-Markovnikov's Rule
Anti-Markovnikov's addition involves the addition of a hydrogen atom to the less substituted carbon atom of a double bond and bromine to the more substituted carbon. This means that in anti-Markovnikov addition, the bromine atom will attach to the more substituted carbon.
2Step 2: Analyze Each Option
For anti-Markovnikov addition to occur, the substrate must be a terminal alkene (with at least one hydrogen attached to the carbon of the double bond at one end), and presence of peroxides is necessary. Let’s analyze each option for their suitability:
- (a) Propene: Has terminal alkene suitable for anti-Markovnikov.
- (b) But-1-ene: Also has terminal alkene.
- (c) But-2-ene: Has an internal double bond (not suitable for anti-Markovnikov addition).
- (d) Pent-3-ene: Has an internal double bond (also not suitable).
3Step 3: Identify the Correct Answer
Anti-Markovnikov addition requires terminal alkenes. Both but-2-ene and pent-3-ene have internal double bonds, meaning they do not meet the requirement for anti-Markovnikov addition. Since but-2-ene is the first option listed among those unsuitable, it is the one where anti-Markovnikov’s addition is not observed in this context.
Key Concepts
Terminal AlkenesInternal Double BondsMarkovnikov's Rule
Terminal Alkenes
Terminal alkenes, also known as alpha-olefins, are organic compounds where the carbon-carbon double bond is located at the end of the carbon chain. This positioning makes them more reactive in certain chemical reactions, like the anti-Markovnikov addition.
In terminal alkenes, one of the carbon atoms involved in the double bond is attached to only one other carbon atom. This structural feature affects how chemical reactions proceed.
In terminal alkenes, one of the carbon atoms involved in the double bond is attached to only one other carbon atom. This structural feature affects how chemical reactions proceed.
- Terminal alkenes are often preferred in anti-Markovnikov reactions because the hydrogen atom of the added reagent will bind to the less substituted carbon.
- This reactivity is due in part to steric factors, where there is less crowding around the terminal carbon, allowing better interaction with the reactants.
Internal Double Bonds
Internal double bonds are carbon-carbon double bonds that are not located on the terminal end of the carbon chain but rather within the chain itself. These bonds are characteristic of substances like but-2-ene and pent-3-ene. Their position affects the chemical behavior:
- Internal double bonds typically have carbon atoms that are each attached to two or more other carbon atoms.
- This arrangement makes them less suitable for anti-Markovnikov addition because the site available for adding a hydrogen atom is equally substituted, reducing selectivity.
- The added steric hindrance around internal double bonds means there is less available space for reactants to approach and bond.
Markovnikov's Rule
Markovnikov's rule is a principle in organic chemistry used to predict the products of the addition of proton acids to alkenes. It states that, in the addition of a halogen and a hydrogen to an alkene, the hydrogen atom attaches to the carbon with the most hydrogen atoms already present (the less substituted carbon), while the halogen attaches to the more substituted carbon.
- This rule applies in standard addition reactions of H-X compounds (such as HBr) without the presence of peroxides.
- It explains why certain products are favored based on the stability of carbocation intermediates formed during the reaction.
- Interestingly, anti-Markovnikov addition is the exception to this rule, typically facilitated by the presence of peroxides, which reverses the typical regioselectivity due to radical mechanisms.
Other exercises in this chapter
Problem 37
Acidic hydrogen is present in (a) ethyne (b) ethene (c) benzene (d) ethane
View solution Problem 38
Which is not a product when O=C1C(O)CCC(O)C1O is treated with \(\mathrm{HIO}_{4} ?\) (a) \(\mathrm{HCOOH}\) (b) \(\mathrm{H}_{2} \mathrm{O}\) (c) \(\mathrm{HCHO
View solution Problem 40
Which is an incorrect statement regarding oxymercuration-demercuration? (a) here rearrangement takes place (b) in the second step demercuration occurs that is,
View solution Problem 41
Baeyer's reagent is (a) acidified permanganate solution (b) alkaline permanganate solution (c) neutral permanganate solution (d) aqueous bromine solution
View solution