Question

Find the number of reagents which can oxidise \(\mathrm{CH}_{\mathrm{s}}-\mathrm{C}_{\mathrm{CH}_{3}}^{\frac{1}{\mathrm{O}}}\) (i) \(\mathrm{I}_{2} / \mathrm{NaOH}\) (ii) \(\mathrm{H}_{2} \mathrm{O}_{2}\) (iii) O=C(O)c1cccc(Cl)c1 (iv) \(\mathrm{H}_{2} / \mathrm{Pd}\) (v) \(\mathrm{NOCl}\) (vi) \(\mathrm{NH}_{2}-\mathrm{NH}_{2} / \mathrm{pH}(4.5-6.0)\) (vii) \(\mathrm{LiAlH}_{4}\) (viii) \(\mathrm{NaBH}_{4}\)

Short answer

The number of reagents that can oxidize \(\mathrm{CH}_{\mathrm{s}}-\mathrm{C}_{\mathrm{CH}_{3}}^{\frac{1}{\mathrm{O}}}\) is 2, which are \(\mathrm{H}_{2} \mathrm{O}_{2}\) (Hydrogen peroxide) and \(\mathrm{NOCl}\) (Nitrosyl chloride).

Step-by-step solution

Understand the Compound Structure

First, let's decipher the given compound \(\mathrm{CH}_{\mathrm{s}}-\mathrm{C}_{\mathrm{CH}_{3}}^{\frac{1}{\mathrm{O}}}\). It consists of two carbon atoms, with one of them connected to a sulfanyl group (SH) and the other connected to a methyl group (CH3). The superscript \(\frac{1}{\mathrm{O}}\)" signifies that the carbon atom connected to the sulfanyl group has a positive charge (partial oxidation).

Analyze Each Reagent

Now we'll analyze each reagent to see if it can oxidize the given compound. (i) \(\mathrm{I}_{2} / \mathrm{NaOH}\): Iodine in basic conditions favors a substitution reaction rather than oxidation. It will not oxidize the compound. (ii) \(\mathrm{H}_{2} \mathrm{O}_{2}\): Hydrogen peroxide (H₂O₂) is a strong oxidizing agent. It can oxidize the given compound. (iii) O=C(O)c1cccc(Cl)c1: This is a carboxylic acid, with a benzene ring and a chlorine atom. It is not capable of oxidizing the given compound. (iv) \(\mathrm{H}_{2} / \mathrm{Pd}\): Hydrogen gas with a palladium catalyst typically works as a reducing agent, which does the opposite of oxidizing. Therefore, it will not oxidize the compound. (v) \(\mathrm{NOCl}\): Nitrosyl chloride (NOCl) can act as an oxidizing agent. It can oxidize the given compound. (vi) \(\mathrm{NH}_{2}-\mathrm{NH}_{2} / \mathrm{pH}(4.5-6.0)\): Hydrazine (NH₂NH₂) in mildly acidic conditions acts as a reducing agent, so it will not oxidize the compound. (vii) \(\mathrm{LiAlH}_{4}\): Lithium aluminum hydride (LiAlH₄) is a strong reducing agent, so it will not oxidize the compound. (viii) \(\mathrm{NaBH}_{4}\): Sodium borohydride (NaBH₄) is a mild reducing agent, so it will not oxidize the compound.

Count the Oxidizing Agents

After going through the list of reagents, we have found that there are 2 reagents that can oxidize the given compound: 1. Hydrogen peroxide (H₂O₂) 2. Nitrosyl chloride (NOCl) So the number of reagents which can oxidize \(\mathrm{CH}_{\mathrm{s}}-\mathrm{C}_{\mathrm{CH}_{3}}^{\frac{1}{\mathrm{O}}}\) is 2.

Key concepts

Reagent Analysis

In the context of organic chemistry, reagent analysis is a critical skill that involves understanding and predicting the reactivity of various substances with organic compounds.

For instance, the analysis of reagents concerning their ability to oxidize an organic compound involves examining their chemical nature and the possible reactions they can undergo. Oxidizing agents are substances that can accept electrons during a chemical reaction, leading to the oxidation of the other substance.

When analyzing reagents for oxidation, the two key characteristics to consider are the presence of oxygen or a high oxidation state element and the conditions under which the reagent is used (such as pH, temperature, and presence of metal catalysts). Differentiating between oxidizing and reducing agents is vital, as they have the opposite effect on the organic substrate. Oxidizing agents add oxygen to or remove hydrogen from substances, while reducing agents do the reverse.

Organic Compound Oxidation

The process of organic compound oxidation involves the loss of electrons, and commonly hydrogen atoms, with a simultaneous gain of oxygen atoms or increase in the oxidation state of one or more atoms within the molecule. This type of reaction is fundamental in synthetic organic chemistry because it alters the functionality and reactivity of organic molecules, making them suitable precursors for further chemical transformations.

Oxidative processes can be used to transform, for example, alcohols into ketones or aldehydes, and aldehydes into carboxylic acids. The reagents used in these reactions must be carefully chosen based on the specific type of oxidation desired and the particular functional groups present in the molecule that needs to be oxidized.

Hence, students should practice identifying potential oxidizing agents and predicting the outcomes of their reactions with different organic substrates. As illustrated in the provided exercise, hydrogen peroxide and nitrosyl chloride are two such oxidizing agents capable of oxidizing the given organic compound due to their electron-accepting characteristics.

JEE Main and Advanced Organic Chemistry

The Joint Entrance Examination (JEE) Main and Advanced are competitive exams in India that assess students' knowledge and understanding of various core subjects, including organic chemistry. The questions related to organic chemistry typically require the application of concepts like reagent analysis and organic compound oxidation.

Preparing for JEE organic chemistry involves not just memorizing reactions but also developing a deep comprehension of reaction mechanisms and the ability to predict reaction products. Knowing how and why certain reagents behave as oxidizing agents can have a significant impact on a student's performance in not only academics but also these competitive exams.

For JEE aspirants, a clear grasp of organic oxidation processes is essential. Questions might involve identifying suitable oxidizing agents for a given organic transformation or explaining the mechanism by which an organic compound is oxidized. Therefore, it is critical for JEE aspirants to understand these concepts deeply, recognize patterns in organic reaction behavior, and apply this knowledge to solve complex problems under exam conditions.