Problem 50 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

Give the structures of the major organic products from3-ethyl-2pentene under each of the following reaction conditions. [1996 - 3 Marks] (a) HBr in the presence of peroxide (b) \(\mathrm{Br}_{2} / \mathrm{H}_{2} \mathrm{O}\) (c) \(\mathrm{Hg}(\mathrm{OAc})_{2} / \mathrm{H}_{2} \mathrm{O} ; \mathrm{NaBH}_{4}\)

Step-by-Step Solution

Answer

(a) 1-bromo-3-ethylpentane; (b) 3-bromo-4-ethyl-2-pentanol; (c) 3-ethyl-2-pentanol.

Step 1: Understand Reactant Structure

3-ethyl-2-pentene is an alkene. Its structure can be broken down as follows: the name indicates it's a five-carbon chain (pentene) with a double bond between the second and third carbon, and an ethyl group (CH₂-CH₃) on the third carbon. The structure is: ``` CH₃-CH=C(CH₂CH₃)-CH₂-CH₃ ```

Step 2: Reaction (a) – HBr with Peroxide

In the presence of peroxides, HBr follows the anti-Markovnikov rule, where the Br⁻ attaches to the less substituted carbon of the double bond. The radical chain mechanism leads to the bromine attaching to the first carbon in the alkene: Resulting Structure: CH₃-CH₂-Br-CH-(CH₂CH₃)-CH₂-CH₃

Step 3: Reaction (b) – \(\text{Br}_2/\text{H}_2\text{O}\)

The reaction of Br₂ in water results in halohydrin formation. The Br attaches to the more substituted carbon (following Markovnikov's rule), and the OH group attaches to the less substituted carbon: Resulting Structure: CH₃-CHOH-CHBr-(CH₂CH₃)-CH₂-CH₃

Step 4: Reaction (c) – Oxymercuration-Demercuration

The reaction with Hg(OAc)₂ followed by NaBH₄ in water leads to Markovnikov addition where the OH group attaches to the more substituted carbon without rearrangement: Resulting Structure: CH₃-CH(OAc)-CH-(CH₂CH₃)-CH₂-CH₃ (after NaBH₄, the acetate is replaced by H to give CH₃-CH(OH)-CH(CH₂CH₃)-CH₂-CH₃)

Key Concepts

Alkene Reactions

Alkenes are a type of hydrocarbon that contain at least one carbon-carbon double bond. They are highly reactive due to this double bond, which can participate in various chemical reactions. These reactions typically involve the addition of atoms or groups of atoms across the carbon-carbon double bond.
Some common types of alkene reactions include:

• Hydrogenation, where hydrogen is added across the double bond, converting it to a single bond.
• Halogenation, where halogens like chlorine or bromine are added.
• Hydrohalogenation, such as in the addition of HBr, leading to a halogenated alkane. The presence of peroxides can significantly change the outcome of these reactions.
• Hydration, where water is added, leading to alcohol formation. The choice of reagents and conditions, like those involving mercury salts or acid catalysts, determines the specifics of these reactions.

Understanding where and how these reactions occur is crucial in predicting the products for given conditions, especially in more complex alkenes like 3-ethyl-2-pentene.

Markovnikov and anti-Markovnikov Addition

In hydrohalogenation reactions, the concepts of Markovnikov and anti-Markovnikov additions help predict where different atoms will add on to the double bond. Markovnikov's rule states that, in the addition of a protic acid (HX) to an alkene, the hydrogen atom (H) will attach to the less substituted carbon, while the halide (X) will attach to the more substituted carbon.
However, in the presence of peroxides (ROOR), an anti-Markovnikov addition occurs, resulting in the halide attaching to the less substituted carbon. This occurs due to a radical chain mechanism initiated by the peroxide.

• Markovnikov's Rule: Leads to more branched alkyl halides.
• Anti-Markovnikov Addition: Occurs in the presence of peroxides, resulting in products that might appear unexpected depending on one's assumptions about the reagents.

This polarity reversal is valuable in organic synthesis, allowing chemists to selectively target specific positions on an alkene.

Halohydrin Formation

Halohydrin formation is a reaction where a halogen and a hydroxyl group (OH) are added across an alkene's double bond. When an alkene reacts with a halogen in water, the water acts as a nucleophile and attacks one end of the three-member intermediate ring formed by the halogen, opening it up.
This reaction follows Markovnikov's rule, where the hydroxyl group attaches to the less substituted carbon, while the halogen ends up on the more substituted carbon.
The general reaction pathway involves:

• Formation of a halonium ion intermediate, where the halogen forms a temporary three-membered ring with the alkene.
• Attack of the intermediate by water, opening the ring to form a halohydrin.

For example, when Br₂ reacts with water in alkenes like 3-ethyl-2-pentene, bromine attaches to the more substituted carbon, and OH, from water, to the less substituted carbon, creating a versatile intermediate that can further react in organic syntheses.

Oxymercuration-Demercuration

Oxymercuration-demercuration is a two-step process used to add water across an alkene's double bond, with no carbocation rearrangements. This method involves:

• Oxymercuration: In this step, the alkene reacts with mercuric acetate ( Hg(OAc)₂ ) in water, forming a mercurinium ion intermediate. Water acts as a nucleophile attacking the more substituted carbon, aligning with Markovnikov's rule.
• Demercuration: The mercury is removed using sodium borohydride (NaBH₄), reducing the complex to yield an alcohol.

The significant benefit of this reaction is that it achieves the hydration of alkenes without rearrangement, providing high regioselectivity without forming undesired carbocation intermediates. Oxymercuration-demercuration is often preferred when control over the position of the hydroxyl group is crucial, especially in complex alkenes, due to its ability to prevent unwanted shifts or rearrangements by avoiding the formation of unstable intermediates.