Problem 72 - A Complete Resource Book in Chemistry for JEE Main

Question

\(\mathrm{CH}_{3}-\mathrm{CH}_{2}-\mathrm{CH}_{2}-\mathrm{Cl} \stackrel{\text { alc. } \mathrm{KOH}}{\longrightarrow}(\mathrm{X}) \stackrel{\mathrm{HBr}}{\longrightarrow}(\mathrm{Y}) \stackrel{\mathrm{Na} \text { /ether }}{\longrightarrow}(\mathrm{Z})\) In the above sequence of reaction, the product (Z) is (a) propane (b) hexane (c) 2,3 -dimethylbutane (d) allyl bromide

Step-by-Step Solution

Answer

The product (Z) is hexane.

Step 1: Identify the First Reaction

The first reaction involves the elimination of HCl from the given compound using alcoholic KOH, which is a common reagent for dehydrohalogenation. Here, the chlorine atom and a hydrogen atom from the adjacent carbon are removed to form an alkene. Therefore, the product (X) is propene, \( ext{CH}_3- ext{CH}= ext{CH}_2\).

Step 2: Add HBr to Propene

The second reaction involves adding HBr to the alkene (X). In propene, the addition follows Markovnikov's rule, meaning the hydrogen from HBr will attach to the carbon with more hydrogen atoms, resulting in 2-bromopropane, \( ext{CH}_3- ext{CHBr}- ext{CH}_3\), as the product (Y).

Step 3: Perform a Wurtz Reaction

In the third step, compound (Y) undergoes a Wurtz reaction with sodium in ether. This involves coupling of two alkyl halide molecules. When two molecules of 2-bromopropane undergo this reaction, the product formed is hexane, \( ext{CH}_3- ext{(CH}_2)_4- ext{CH}_3\). Thus, the final product (Z) is hexane.

Key Concepts

Dehydrohalogenation

Dehydrohalogenation is a crucial reaction in organic chemistry where an alkyl halide loses its halogen atom along with a hydrogen atom from the adjacent carbon atom. This process results in the formation of an alkene. The term "dehydrohalogenation" itself refers to the removal ("de-") of hydrogen ("hydro-") and a halogen ("-halogenation"). In the given exercise, the starting compound, propyl chloride (\(\text{CH}_3-\text{CH}_2-\text{CH}_2-\text{Cl}\)), undergoes dehydrohalogenation using alcoholic potassium hydroxide (KOH). This preferred reagent promotes elimination over substitution, ensuring the halide leaves and forms a double bond in place, thus forming propene (\(\text{CH}_3-\text{CH}=\text{CH}_2\)). This transformation highlights the importance of choosing the right conditions, as alcoholic KOH is known for favoring the formation of alkenes over alcohols. To recap:

• Dehydrohalogenation removes a hydrogen and a halogen from the molecule.
• This process results in the formation of a new double bond, yielding an alkene.
• Alcoholic KOH is a common dehydrohalogenation agent that encourages elimination reactions.

Understanding dehydrohalogenation is a stepping stone to mastering organic synthesis and modification of aliphatic hydrocarbons.

Markovnikov's Rule

Markovnikov's Rule comes into play when adding acids, like HBr, across the double bond of an alkene. This principle helps predict the major product by stating that the hydrogen atom from the acid will attach to the doubly bonded carbon with more hydrogen atoms already attached to it. In simpler terms, "who ever has, gets more." In the exercise, propene (\(\text{CH}_3-\text{CH}=\text{CH}_2\)) reacts with HBr. Here, the hydrogen from HBr attaches to the terminal carbon which already has two hydrogen atoms. The bromine atom bonds with the other carbon, resulting in 2-bromopropane (\(\text{CH}_3-\text{CHBr}-\text{CH}_3\)).

• Markovnikov's Rule determines the regioselectivity of hydrogen additions.
• Alkenes follow this rule to form more stable, substituted carbon centers in the product.
• This is key in predicting the outcome of electrophilic addition reactions.

By understanding Markovnikov's Rule, students can better anticipate the products of alkene reactions under certain conditions, ensuring they grasp this integral concept in organic chemistry.

Wurtz Reaction

The Wurtz Reaction is a classic method to synthesize higher alkanes from smaller alkyl halides. By using sodium metal in a dry ether solution, two alkyl halides undergo a coupling reaction, forming a new carbon-carbon bond. In the exercise, 2-bromopropane (\(\text{CH}_3-\text{CHBr}-\text{CH}_3\)) is subjected to a Wurtz Reaction with sodium. Two molecules of this compound couple together, resulting in hexane (\(\text{CH}_3-\text{(CH}_2)_4-\text{CH}_3\)) as the final product.

• The Wurtz Reaction couples two identical or different alkyl halides.
• It is useful for lengthening carbon chains and producing even-numbered alkanes.
• This reaction highlights the role of sodium in facilitating bond formation between halides.

Though highly effective, the reaction is limited to symmetric alkanes and can face challenges like side reactions or formation of other products. Nevertheless, mastering the Wurtz Reaction provides a valuable tool for constructing larger alkane structures in synthetic organic chemistry.