Question

When pent-2-ene is treated with HI in presence of a peroxy acids the major product obtained is (a) 2 - Iodo-pentane (b) 3 -Iodo-pentane (c) Both of these (d) None of these

Short answer

(b) 3 -Iodo-pentane

Step-by-step solution

Understanding the Reaction

The problem states that pent-2-ene is treated with HI in the presence of peroxy acids. This means we're dealing with an anti-Markovnikov addition reaction due to the presence of peroxy acids, which causes the iodine atom to add to the less substituted carbon atom of the alkene.

Identifying the Positions in Pent-2-ene

Pent-2-ene has a double bond between the second and third carbon atoms: CH3-CH=CH-CH2-CH3. The position of the double bond is crucial because it determines where the additions will occur.

Applying the Anti-Markovnikov Rule

In an anti-Markovnikov addition, the iodine (I) atom adds to the less substituted carbon. For pent-2-ene, the less substituted carbon of the double bond is the terminal carbon, which is carbon number 3 in this case (CH3-CH=CH-CH2-CH3), since it has more hydrogen atoms.

Determine the Major Products

With iodine adding to the less substituted carbon (the third carbon) and the hydrogen adding to the more substituted carbon, the major product of the reaction is 3-iodo-pentane.

Key concepts

Alkene Reactions

Alkene reactions are a vital part of organic chemistry, given their ability to form various products through different types of addition reactions. An alkene, like pent-2-ene, features a carbon-carbon double bond, which serves as the primary site for these reactions. The double bond allows alkenes to undergo addition reactions, where atoms or groups of atoms are added across the bond.

This adds versatility to alkenes, as they can form a range of different compounds depending on the conditions and reagents used in the reaction. Some common types of alkene reactions include hydrogenation, halogenation, and hydrohalogenation. Each type of reaction has its own set of rules and outcomes.

In the case of pent-2-ene, treating it with hydrogen iodide (HI) and peroxy acids leads to an anti-Markovnikov addition. During this process, the peroxy acids play a crucial role in affecting the direction of the reaction—determining which carbon atom the iodine will attach to, ultimately giving us the major product.

Organic Chemistry Mechanisms

Understanding organic chemistry mechanisms is crucial in predicting the outcomes of reactions like that of pent-2-ene with HI and peroxy acids. An organic chemistry mechanism is a step-by-step explanation illustrating how a reaction proceeds, showcasing each intermediate state and the final product.

An important aspect of mechanisms is the concept of "substitution" levels. In a simple alkene mechanism, we focus on which carbon atoms are more or less substituted, meaning how many other carbon atoms are bonded to it. The degree of substitution affects which atom or group will most likely attach to a specific carbon.

In pent-2-ene's reaction with HI and peroxy acids, we apply the anti-Markovnikov rule. This rule helps predict that the iodine will add to the less substituted carbon of the alkene’s double bond. Here, the less substituted position is carbon 3. Understanding such mechanisms aids in foreseeing reaction products and is essential for mastering organic chemistry.

Halogenation of Alkenes

Halogenation is a key transformation for alkenes, involving the addition of halogens like chlorine or iodine. This reaction is essential for converting simple alkenes into more complex compounds. Specifically, the halogenation we are focusing on is hydrohalogenation, which deals with the addition of a hydrogen halide (such as HI) to an alkene like pent-2-ene.

In the presence of peroxy acids, we observe something unique called the anti-Markovnikov addition. Typically, hydrohalogenation might lead to Markovnikov products, where the halogen attaches to the more substituted carbon; however, peroxy acids induce the opposite effect.

This reaction mechanism guides the halogen, iodine in this case, to attach itself to the less substituted carbon. For pent-2-ene, it means iodine binds to carbon 3, forming 3-iodo-pentane as the major product. Understanding halogenation and its nuances, like Markovnikov and anti-Markovnikov orientations, is vital for successful applications in synthesis and transformation in organic chemistry.