Problem 10 - A Complete Resource Book in Chemistry for JEE Main

Question

Isopropyl bromide on Wurtz reaction gives (a) hexane (b) propane (c) 2,3 -dimethyl butane (d) neo-hexane

Step-by-Step Solution

Answer

The product is 2,3-dimethylbutane (c).

Step 1: Understanding Wurtz Reaction

The Wurtz reaction involves the coupling of two alkyl halide molecules in the presence of sodium metal and an ether solvent. The reaction leads to the formation of a new carbon-carbon bond, resulting in an alkane that is double the carbon chain length of the original alkane halide.

Step 2: Analyzing Isopropyl Bromide

Isopropyl bromide, also known as 2-bromopropane, has the structure C3H7Br. This compound consists of a three-carbon chain with a bromine atom attached to the second carbon.

Step 3: Applying the Wurtz Reaction to Isopropyl Bromide

When two molecules of isopropyl bromide undergo the Wurtz reaction, the bromine atoms are removed, and the remaining alkyl groups are joined together by sodium to form a single molecule.

Step 4: Predicting the Product

Two isopropyl groups (C3H7-) will combine together to form 2,3-dimethylbutane (C6H14) as the new carbon-carbon bonds form between the isopropyl groups.

Key Concepts

Isopropyl bromide

Isopropyl bromide, more formally known as 2-bromopropane, is an organic compound with the molecular formula C\(_3\)H\(_7\)Br. This molecule belongs to the family of alkyl halides, where an alkyl group is attached to a halogen atom—in this case, bromine. The structure of isopropyl bromide features a chain of three carbon atoms. The bromine atom is bonded to the middle carbon, giving it a secondary alkyl halide characteristic. This position makes isopropyl bromide particularly interesting for reactions that involve removing the bromine atom, as it provides unique opportunities for forming new bonds in organic synthesis.

Alkyl halide coupling

Alkyl halide coupling is a fundamental reaction in organic chemistry, particularly for creating larger and more complex organic molecules. The Wurtz reaction is a primary example where alkyl halide coupling comes into play. This reaction involves two molecules of an alkyl halide reacting together, facilitated by sodium metal.

• The halide atoms are replaced during the reaction.
• Carbon-carbon bonds form between the remaining alkyl groups.
• Sodium acts as a reducing agent, allowing the bond formation.

Through coupling reactions, chemists can build larger carbon-based structures from simpler molecules, significantly expanding the methodology for constructing varied hydrocarbons.

Sodium in ether solvent

In the Wurtz reaction, sodium metal and an ether solvent play critical roles. Sodium, a highly reactive metal, serves as a reducing agent. It aids in breaking the carbon-halogen bonds in the alkyl halide molecules.

• Ether functions as the solvent providing a medium for the reaction.
• Its non-polar nature prevents it from reacting actively with sodium.
• It also stabilizes the reaction intermediates, supporting efficient coupling.

The use of ether as a solvent is crucial because it provides an inert environment. This helps avert side reactions and focuses the reaction towards the desired product, improving the yield of the carbon-carbon coupled product.

Carbon-carbon bond formation

Carbon-carbon bond formation is a cornerstone of building organic molecules. In the context of the Wurtz reaction, the goal is to effectively join two alkyl chains through a new carbon-carbon bond forming process.

• Two alkyl bromide molecules first interact with sodium.
• The reactive bromide groups are removed.
• Remaining carbon atoms directly bond, doubling the alkane length.

For isopropyl bromide, this reaction specifically results in forming 2,3-dimethylbutane. As both alkyl chains from the isopropyl groups unite, the product showcases the power of utilizing the Wurtz reaction for synthesizing higher alkanes from simpler precursors. This ability to form stable, extended carbon chains is fundamental in organic compound synthesis.