Problem 72 - A Complete Resource Book in Chemistry for JEE Main

Question

The final product \(\mathrm{B}\) in the following reaction is \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}=\mathrm{CH}_{2} \underset{\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2}\right)_{2}}{\stackrel{\mathrm{NBS}, \mathrm{hv}}{\mathrm{CH}_{2} \mathrm{OH}}} \mathrm{A} \stackrel{\mathrm{CH}_{3} \mathrm{SNa}}{\longrightarrow} \mathrm{B}\) (a) methyl 1-methylallylsulphide (b) allyl methyl sulphide (c) \(\mathrm{n}\)-butyl methyl sulphide (d) diallyl sulphide

Step-by-Step Solution

Answer

The final product \(\mathrm{B}\) is \(n\)-butyl methyl sulphide, option (c).

Step 1: Understand the Starting Material

The starting material in the reaction is 1-butene, which is represented as \( \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}=\mathrm{CH}_{2} \). This structure is a simple alkene with a terminal double bond.

Step 2: Examine Reaction of 1-Butene with NBS and hv

N-Bromosuccinimide (NBS) in the presence of light (hv) induces allylic bromination. This reaction introduces a bromine atom at the allylic position of 1-butene, resulting in the formation of 3-bromo-1-butene, represented as \( \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CHBrCH}_{2} \).

Step 3: Identify the Use of CH3SNa

In the second step, \( \mathrm{A} \) is reacted with methylsodium thioether (\( \mathrm{CH}_{3} \mathrm{SNa} \)). This leads to a nucleophilic substitution, where the bromine atom in 3-bromo-1-butene is replaced with the methylthio group \( \mathrm{SCH}_{3} \). This substitution results in the formation of \( \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}(\mathrm{SCH}_{3})\mathrm{CH}_{2} \).

Step 4: Determine the Final Product B

The final product \( \mathrm{B} \) is thus \( \mathrm{n} \)-butyl methyl sulfide, noted as \( \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}(\mathrm{SCH}_{3})\mathrm{CH}_{2} \). This matches option \(\mathrm{(c)}\) \(n\)-butyl methyl sulphide.

Key Concepts

Alkene

Alkenes are hydrocarbons that contain at least one carbon-carbon double bond. In the context of this exercise, we are looking at 1-butene, a simple alkene, where the structure is represented as \( \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}=\mathrm{CH}_{2} \). The key features of alkenes include:

• Double bonds: This feature makes alkenes more reactive than alkanes because the electrons in the double bond can readily participate in chemical reactions.
• Unsaturated nature: Unlike alkanes, alkenes can add atoms or groups of atoms due the presence of the double bond, which can open and form new bonds.
• Isomerism: Alkenes can exhibit geometric (cis/trans or E/Z) isomerism based on the relative positions of substituents around the double bond.

In synthesis reactions, alkenes serve as versatile building blocks. Their double bond can undergo a variety of reactions such as polymerization, hydrohalogenation, and halogenation, which are key in the synthesis of complex organic molecules.

Allylic Bromination

Allylic bromination is a reaction where a bromine atom is introduced to an alkene at an allylic position - the carbon atom adjacent to the double bond. This reaction is typically carried out using N-Bromosuccinimide (NBS) and light (hv) to initiate free-radical halogenation. In this exercise:

• 1-butene is transformed into 3-bromo-1-butene through the allylic bromination process.
• The presence of light helps in homolytically breaking the NBS to initiate the radical mechanism, essential for creating allylic bromides.
• This step is crucial because it activates the substrate by making it more reactive for subsequent transformations, like nucleophilic substitutions.

The process involves radical mechanisms where the intermediate - a free radical - is stabilized by delocalization over the allylic position. Allylic bromination is preferred when the goal is to advance in synthetic steps that require creating reactive intermediates like allyl bromides.

Organic Synthesis

Organic synthesis involves constructing complex organic compounds from simpler pre-existing molecules. The steps might involve multiple reactions and transformations that achieve the required functional groups and molecular configurations.

• The transformation discussed in this exercise displays the usefulness of a stepwise approach in organic synthesis. Starting from 1-butene, NBS induced allylic bromination gives an allyl bromide.
• The introduction of a nucleophile like \(\mathrm{CH}_3\mathrm{SNa}\) represents a classic example of nucleophilic substitution, where a nucleophile replaces a leaving group (bromide in this case) resulting in a sulfide product.
• This step-by-step synthesis provides insights into designing pathways for more complex molecule production by strategically modifying the molecular structure.

In organic synthesis, reactions are carefully chosen based on their mechanisms and the kind of intermediates or products desired. One must always consider factors like selectivity and reaction conditions to ensure successful transformation. Through understanding these principles, complex molecules can be synthesized for a wide array of applications in industries such as pharmaceuticals and materials science.