Problem 128 - A Complete Resource Book in Chemistry for JEE Main

Question

Identify the set of reagents/reaction conditions 'X' and ' \(\mathrm{Y}\) ' in the following set of transformations: CCCBr (a) \(\mathrm{X}=\) dilute aqueous \(\mathrm{NaOH}, 20^{\circ} \mathrm{C}\) \(\mathrm{Y}=\) HBr/acetic acid, \(20^{\circ} \mathrm{C}\) (b) \(\mathrm{X}=\) concentrated alcoholic \(\mathrm{NaOH}, 80^{\circ} \mathrm{C}\) \(\mathrm{Y}=\mathrm{HBr} /\) acetic acid, \(20^{\circ} \mathrm{C}\) (c) \(\mathrm{X}=\) dilute aqueous \(\mathrm{NaOH}, 20^{\circ} \mathrm{C}\) \(\mathrm{Y}=\mathrm{Br}_{2} / \mathrm{CHCl}_{3}, 0^{\circ} \mathrm{C}\) (d) \(\mathrm{X}=\) concentrated alcoholic \(\mathrm{NaOH}, 80^{\circ} \mathrm{C}\) \(\mathrm{Y}=\mathrm{Br}_{2} / \mathrm{CHCl}_{3}, 0^{\circ} \mathrm{C}\)

Step-by-Step Solution

Answer

Option (b) is likely correct with these conditions for transformations.

Step 1: Identify Reagent X in Transformation

We start by determining the role of the first reagent, \(\mathrm{X}\), in the transformation of \(\mathrm{CCCBr}\). The options given are:(a) Dilute aqueous \(\mathrm{NaOH}\), \(20^{\circ} \mathrm{C}\): Provides a nucleophilic environment for substitution reactions.(b) Concentrated alcoholic \(\mathrm{NaOH}\), \(80^{\circ} \mathrm{C}\): Typically used for elimination reactions to form alkenes.(c) Dilute aqueous \(\mathrm{NaOH}\), \(20^{\circ} \mathrm{C}\): Again, primarily for nucleophilic substitution but under mild conditions.(d) Concentrated alcoholic \(\mathrm{NaOH}\), \(80^{\circ} \mathrm{C}\): Promotes elimination reactions to produce alkenes.

Step 2: Determine the Resulting Product after Reagent X

For cases (a) and (c) with dilute aqueous \(\mathrm{NaOH}\), \(\mathrm{CCCBr}\) undergoes nucleophilic substitution, potentially forming \(\mathrm{CCC(OH)}\), an alcohol.For cases (b) and (d), in concentrated alcoholic \(\mathrm{NaOH}\) at high temperature, elimination leads to the formation of an alkene, \(\mathrm{C=C}\) from \(\mathrm{CCCBr}\).

Step 3: Identify Reagent Y in the Transformation of Resulting Product

Next, we identify the role of the second reagent, \(\mathrm{Y}\), given different conditions.(a) and (b) use \(\mathrm{HBr}\) in acetic acid. After forming an alcohol from (a), \(\mathrm{HBr}\) would convert it back into a bromoalkane via substitution. In (b), for a resulting alkene, \(\mathrm{HBr}\) adds across the double bond to form a bromoalkane.(c) and (d) with \(\mathrm{Br}_{2} / \mathrm{CHCl}_{3}\), involve halogenation. (c) forms dibromo compounds from alcohols, while (d) brominates alkenes to form dibromoalkanes.

Step 4: Select the Correct Set of Reagents X and Y

Analyzing the transformations:- **(a)** As \(\mathrm{X}\) leads to an alcohol and \(\mathrm{Y}\) turns it back to an alkyl bromide; it describes a substitution to alkane and then back to bromide.- **(b)** \(\mathrm{X}\) results in an alkene, and \(\mathrm{HBr}\) then leads to the addition across the double bond, which describes forming an alkyl bromide from an alkene effectively.- **(c)** Likely incorrect as bromination of alcohol is not direct.- **(d)** Alkene from \(\mathrm{X}\) brominated in \(\mathrm{Y}\), making \(\mathrm{(d)}\) suitable if adding bromine across the double bond of alkene.

Key Concepts

Nucleophilic Substitution

Nucleophilic substitution is a fundamental reaction in organic chemistry where a nucleophile, a chemical species with a free pair of electrons or a pi bond, replaces a leaving group in a molecule. In our given example, this transformation is observed when dilute aqueous \(\mathrm{NaOH}\) is used under \(20^\circ \mathrm{C}\). This environment is mild and suitable for the hydroxide ions from \(\mathrm{NaOH}\) to replace the bromine atom attached to the carbon on the alkyl group, transforming the bromoalkane to an alcohol (like \(\mathrm{CCC(OH)}\)).

The key here is the strong nucleophilic nature of the \(\mathrm{OH}^-\) ion. Its high electron density allows it to attack the partially positive carbon atom bonded to the bromine, which is a good leaving group due to its size and ability to stabilize the negative charge after departure. This kind of reaction is generally classified as \(\mathrm{S_N2}\) or \(\mathrm{S_N1}\) depending on the kinetics and conditions, with \(\mathrm{S_N2}\) being more likely in primary bromoalkanes and \(\mathrm{S_N1}\) in tertiary.

Elimination Reactions

Elimination reactions involve the removal of elements from a molecule, typically leading to the formation of a double bond or an alkene. In this case, concentrated alcoholic \(\mathrm{NaOH}\) at a high temperature of \(80^\circ \mathrm{C}\) is used. This condition favors elimination (\(\mathrm{E2}\) mechanism) over substitution.

During elimination, the hydroxide ions act as a base rather than a nucleophile. They abstract a hydrogen atom adjacent to the carbon attached to the bromine, leading to the formation of a double bond as the bromide ion leaves. This results in an alkene through a concerted mechanism, where the formation of \(\mathrm{C=C}\) double bond happens simultaneously with the loss of the bromide ion.

This pathway is often chosen when the goal is to synthesize olefins from alkyl halides, as it efficiently removes a \(\mathrm{HX}\) (in this case, \(\mathrm{HBr}\)) and forms a more stable carbon-carbon double bond.

Halogenation

Halogenation is a crucial reaction where a molecule becomes saturated with halogen atoms. In the context of our bromoalkane transformations, \(\mathrm{Br}_2\) in \(\mathrm{CHCl}_3\) under \(0^\circ \mathrm{C}\) is used. When \(\mathrm{Br}_2\) reacts with an alkene, it leads to the addition of bromine atoms across the double bond, forming dibromo alkanes.

This reaction is generally straightforward and involves the formation of a cyclic bromonium ion intermediate, which then leads to the anti-addition of bromine across the double bond. This stereospecific reaction ensures that bromine atoms add to opposite sides, which can affect the molecule's overall spatial configuration. With an alkene as the starting material, this halogenation process is highly regioselective and results in vicinal dibromides. This is a typical reaction for activating alkenes and adding functional diversity to them.

Reagent Identification

Identifying appropriate reagents is key to executing desired organic transformations. Understanding the role of each reagent and their specific reaction conditions allows chemists to predict the outcome of complex synthetic paths.

- **Aqueous \(\mathrm{NaOH}\):** Used for performing nucleophilic substitution to swap an \(\mathrm{X}\) group such as bromine with hydroxide.
- **Alcoholic \(\mathrm{NaOH}\):** When concentrated and heated, it shifts reaction favorability towards elimination, producing alkenes.
- **\(\mathrm{HBr}\) / Acetic Acid:** Suitable for adding a bromine atom across double or converting alcohols back to alkyl bromides through substitution.
- **\(\mathrm{Br}_2\) / \(\mathrm{CHCl}_3\):** Employed for adding bromine across carbon-carbon double bonds in alkenes, leading to dibromo compounds.

Through careful reagent selection and understanding conditions, one can induce the precise chemical changes desired in organic synthesis.