Problem 10 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

An ether (A), \(\mathrm{C}_{5} \mathrm{H}_{12} \mathrm{O}\), when heated with excess of hot concentrated HI produced two alkyl halides which when treated with NaOH yielded compounds (B) and (C). Oxidation of (B) and (C) gave a propanone and an ethanoic acid respectively. The IUPAC name of the ether (A) is (a) 2 -ethoxypropane (b) ethoxypropane (c) methoxybutane (d) 2 -methoxybutane

Step-by-Step Solution

Answer

The correct IUPAC name of the ether (A) is 2-ethoxypropane.

Step 1: Analyze the Decomposition Reaction

When the ether \(A\) reacts with excess HI, it cleaves to form two alkyl halides. Since \(A\) is \(\mathrm{C}_{5} \mathrm{H}_{12} \mathrm{O}\), this molecule can be split into different combinations of alkyl groups that will form alkyl iodides. Possible decomposition pathways give: \(\text{alkyl halides} = \text{ethyl iodide and propyl iodide}\) or \(\text{methyl iodide and butyl iodide}\).

Step 2: Focus on the Resulting Alkyl Halides

The problem states that one alkyl iodide turns into propanone when oxidized, and the other turns into ethanoic acid. This means the propyl group corresponds to the ketone in the oxidation product (propanone), so the ether cannot have a methyl group together with a butyl group since neither would yield propanone upon oxidation. Hence, the ether likely cleaves into ethyl iodide and propyl iodide.

Step 3: Analyzing the Products of Hydrolysis with NaOH

Upon treatment with NaOH, the \(\text{ethyl iodide}\) gives ethanol (\(\text{B}\)), and \(\text{propyl iodide}\) gives propan-1-ol (\(\text{C}\)). When these alcohols are oxidized, ethanol forms ethanoic acid and propan-1-ol forms propanone, matching the problem statement.

Step 4: Conclude the Structure and IUPAC Name

Given that the alkyl iodides are ethyl and propyl, and the oxidation products match the given conditions, the ether must be a combination of these two. This indicates the ether is 2-ethoxypropane, where the ether is linked between an ethyl group and a propyl group's second carbon.

Key Concepts

Oxidation Reactions

In organic chemistry, oxidation reactions are essential for transforming certain functional groups into others. Here, we will explore the oxidation reactions relevant to the problem involving the ether and its cleavage products. When we talk about oxidation in organic chemistry, it usually involves increasing the number of bonds to oxygen or decreasing the number of hydrogen atoms.
For example, alcohols can be oxidized to ketones or carboxylic acids depending on their structure:

• Primary alcohols, like ethanol, are oxidized to form aldehydes and further to carboxylic acids, such as ethanoic acid.
• Secondary alcohols, such as propan-1-ol, are oxidized to form ketones, like propanone.

In the given problem, ethyl iodide is converted into ethanol, which upon oxidation yields ethanoic acid. Similarly, propyl iodide gives propan-1-ol, which oxidizes to form propanone. This chemistry plays a crucial role in determining the types of alkyl halides that result from the ether's cleavage, contributing to understanding the ether's structural possibilities.

Alkyl Halides

Alkyl halides are organic compounds that have one or more halogens (like iodine) bonded to an alkyl group. In the context of the given problem involving ethers, the breakdown using HI forms these alkyl halides. This happens because HI is a strong hydrohalic acid capable of cleaving ethers into corresponding alkyl halides and water.
Understanding alkyl halide formation helps in analyzing the ether cleavage:

• With molecular formula \(C_{5}H_{12}O\), the ether splits with HI to possibly form ethyl iodide and propyl iodide.
• These halides are critical because upon treatment with NaOH, they revert to corresponding alcohols which are then oxidized to the specified products.

Alkyl halides in this problem enable the transformation of an ether into usable products, facilitating the determination of the ether's structure and name.

IUPAC Nomenclature

IUPAC nomenclature provides a systematic way to name organic compounds based on their structure. For ethers, the name usually includes the names of the alkyl groups attached to the oxygen atom.
Let's dive into naming ethers using IUPAC rules. An ether consists of two alkyl or aryl groups connected by an oxygen atom. According to IUPAC:

• The smaller alkyl group becomes the alkoxy substituent.
• The larger alkyl group is considered the parent chain.

In our problem, after analyzing the ether and its cleavage products, we concluded it must be 2-ethoxypropane. Here's why:

• "Ethoxy" indicates an ethyl group linked to oxygen, suggesting it acts as a substituent.
• "Propane" is the main chain, indicating a three-carbon backbone derived from the propyl group.

Therefore, the ether is named 2-ethoxypropane because the ethyl group links to the second carbon of the propane chain. IUPAC nomenclature thus helps clearly discern the compound's structure, aiding in the correct identification and understanding of its chemistry.