Question

Choose the correct option(s): $$ \mathrm{NBS} \stackrel{\mathrm{NaOBr}}{\longrightarrow} \mathrm{P} \stackrel{\mathrm{NaNO}_{2} / \mathrm{HCl}}{\longrightarrow} \mathrm{Q} \stackrel{\Delta}{\longrightarrow} \mathrm{R} $$ (A) (P) on strong heating will produce (R) (B) Formation of (Q) involves \(\mathrm{N}_{2}\) as leaving group (C) Formation of (P) is Hofmann bromamide degradation (D) Rearrangment step is rate determining step to form (P)

Short answer

The correct option is (B): Formation of (Q) involves N₂ as the leaving group.

Step-by-step solution

Formation of compound P

Through the Hofmann bromamide degradation reaction, a primary amine is converted into an amine with one carbon atom less than the starting material. NBS, in the presence of a strong base (NaOBr), reacts to form an intermediate product (compound P). To identify compound P, we must know that NBS is an electrophilic brominating agent, primarily used to brominate allylic and benzylic Carbon atoms.

Formation of compound Q

Compound P undergoes a reaction with the mixture of NaNO₂ and HCl, which transforms the amine group into a diazonium group. This reaction involves the release of N₂ gas (due to N₂ being a good leaving group), forming compound Q.

Formation of compound R

With strong heating, the diazonium group in compound Q may rearrange, causing a loss of N₂ gas, generating an intermediate carbocation, which can then lead to further reaction pathways or rearrangement to form compound R. Now let's analyze the options: (A) (P) on strong heating will produce (R): This statement is incorrect because it's not compound P that produces R upon strong heating; it's compound Q that acts as the substrate for the formation of compound R upon strong heating. (B) Formation of (Q) involves N₂ as leaving group: This statement is correct because during the conversion of compound P to compound Q, the diazonium group leaves in the form of N₂ gas. (C) Formation of (P) is Hofmann bromamide degradation: This statement is incorrect because Hofmann bromamide degradation involves the conversion of an amide to an amine with one carbon atom less, which is not happening here. Instead, NBS brominates compound P by electrophilic addition. (D) Rearrangment step is rate determining step to form (P): This statement is incorrect because compound P is formed through electrophilic addition rather than by rearrangement. In conclusion, the only correct option is (B).

Key concepts

Hofmann Bromamide Degradation

Hofmann bromamide degradation is a classic organic reaction that converts an amide into an amine, which has one less carbon atom than the original amide. This conversion is carried out by treating the amide with bromine in an alkaline solution, typically using sodium hydroxide (NaOH). The process involves several steps, beginning with the formation of an N-bromoamide, followed by its deprotonation, and subsequent rearrangement known as the Hofmann rearrangement. This leads to the formation of an isocyanate intermediate which, upon hydrolysis, provides the final primary amine product.

For students focusing on JEE organic chemistry, understanding the mechanism of this reaction is essential as it not only explains how to remove a carbon atom from an amide but also illustrates the intricacy of nitrogen-containing compound transformations. When tackling problems that involve identifying reaction products, always check if the starting material is an amide and if the reagents align with Hofmann degradation conditions to correctly predict the amine product.

Diazonium Group Reactions

Reactions involving diazonium groups are crucial in the synthesis of various aromatic compounds within organic chemistry. The formation of a diazonium salt typically starts from a primary aromatic amine reacting with nitrous acid, which is often generated in situ from sodium nitrite (NaNO₂) and hydrochloric acid (HCl). The critical feature of diazonium salts is their stability under cold conditions and their propensity to form nitrogen gas (N₂) as a very good leaving group during reactions.

The most common reaction seen in JEE organic chemistry involving diazonium salts is their ability to undergo substitution reactions. Diazonium compounds can participate in Sandmeyer reactions, where they are replaced by a halide or a cyanide ion, or in azo coupling reactions where they form colorful azo compounds. These reactions can be used to deduce the formation of different products based on the choice of the reaction partner. The understanding of diazonium group reactions allows students to navigate the synthesis of complex aromatic molecules systematically.

Electrophilic Addition

Electrophilic addition is a fundamental type of reaction in organic chemistry where an electrophile and a nucleophile come together to form more complex molecules. It is especially relevant in the chemistry of alkenes and alkynes, which contain double or triple bonds, respectively. These unsaturated compounds are characterized by electron-rich regions that are attractive to electrophiles.

During an electrophilic addition reaction to an alkene, the double bond opens up to allow the attachment of two groups. The reaction typically proceeds via the formation of a carbocation intermediate, which is then attacked by a nucleophile. In the context of JEE organic chemistry, students must recognize the role of reagents such as bromine (Br₂) or hydrohalic acids (HX) as electrophiles that add across the double bond of alkenes. It's crucial to understand the Markovnikov's rule, which helps predict the regiochemistry of the electrophilic addition reaction. This rule states that the electrophile will attach to the less substituted carbon of the double bond, leading to the more stable carbocation intermediate. Through mastery of these concepts, students can analyze the outcomes of electrophilic addition reactions and predict the major products with confidence.