Problem 104 - A Complete Resource Book in Chemistry for JEE Main

Question

Which of the following sequence of reagents can be used for the conversion of \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{2} \mathrm{CH}_{3}\) into \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}=\mathrm{CH}_{2} ?\) (a) \(\mathrm{SOCl}_{2} ; \mathrm{H}_{2} \mathrm{O}\) (b) \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\); alc. \(\mathrm{KOH}\) (c) \(\mathrm{Cl}_{2} / \mathrm{hv} ; \mathrm{H}_{2} \mathrm{O}\) (d) \(\mathrm{SOCl}_{2}\); alc. \(\mathrm{KOH}\)

Step-by-Step Solution

Answer

Options (b) and (c) are suitable for the conversion. Option (c) is more common in practice.

Step 1: Recognize the Starting Material and Target Product

The starting material is ethylbenzene, \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{2} \mathrm{CH}_{3}\), and the target product is styrene, \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}=\mathrm{CH}_{2}\). This transformation involves converting the alkyl side chain's two-carbon moiety into a double-bonded vinyl group.

Step 2: Determine the Required Reaction Type

The conversion of ethylbenzene to styrene involves the creation of a double bond by dehydrohalogenation. This process typically requires halogenation first, followed by removal of the halogen and a hydrogen to form the double bond.

Step 3: Examine Each Reagent Option for Halogenation

For the sequence to work, the initial step must introduce a halogen onto the alkyl chain.- (a) \(\mathrm{SOCl}_{2} ; \mathrm{H}_{2} \mathrm{O}\) is used to convert alcohols to chlorides but does not apply since our starting material is not an alcohol.- (b) \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) would introduce a chlorine via an allylic chlorination with UV light, suitable in some conditions.- (c) \(\mathrm{Cl}_{2} / \mathrm{hv}\) uses light to induce radical chlorination, forming a halogenated intermediate on a saturated carbon.- (d) \(\mathrm{SOCl}_{2}\) is again used for the conversion of \(\mathrm{OH}\) groups to \(\mathrm{Cl}\), which is inappropriate here.

Step 4: Evaluate the Reagent Sequences for Elimination

After halogenation, elimination with alcoholic \(\mathrm{KOH}\) is needed to form the double bond, removing both halogen and a hydrogen atom by forming the alkene.- (b) Halogenation with \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) and subsequent treating with alcoholic \(\mathrm{KOH}\) introduces and then removes chlorine to form the desired alkene, styrene.- (c) Similarly, \(\mathrm{Cl}_{2} / \mathrm{hv}\) can properly chlorinate, followed by dehydrohalogenation with alcoholic \(\mathrm{KOH}\), achieving the desired transformation.

Step 5: Confirm the Correct Reagent Sequence

Options (b) and (c) both work theoretically, but practical conditions may vary; however, standard radical chlorination used in option (c) is more common and reliable for this context. Meanwhile, sequence (d) only applies to alcohols, nor aligns with the process of halogenation followed by dehydrohalogenation for producing an alkene.

Key Concepts

Dehydrohalogenation

Dehydrohalogenation is a critical reaction in organic synthesis used for alkene formation. This process involves the elimination of a hydrogen halide (HX) from an alkyl halide, resulting in the formation of a double bond.

When discussing the transformation of ethylbenzene to styrene, dehydrohalogenation plays a pivotal role. It involves both the removal of a halogen, such as chlorine, and a hydrogen atom from adjacent carbon atoms in the molecule. As a result, a double bond is formed, converting the starting material into the desired alkene.

The reagent typically used for this reaction is alcoholic potassium hydroxide (alc. KOH). This strong base not only aids in removing the halogen as a halide ion but also facilitates the abstraction of a hydrogen atom, leading to the creation of a \\[\text{C=C}\] bond.

• Start with halogenation to introduce a halogen to the alkyl chain.
• Proceed with dehydrohalogenation to form the double bond.

The success of dehydrohalogenation depends on the correct placement of the halogen in the molecule, as well as the choice of base for elimination.

Radical Chlorination

Radical chlorination is essential when introducing a chlorine atom to an alkane, which is necessary for preparing substrates for dehydrohalogenation. In this context, radical chlorination involves using molecular chlorine (\(\text{Cl}_2\)) under UV light conditions, resulting in homolytic cleavage of the \\[\text{Cl-Cl}\] bond to generate chlorine radicals.

The chlorine radicals are highly reactive and capable of abstracting a hydrogen atom from the alkyl chain, particularly in the context of ethylbenzene. This step converts a saturated carbon, like those in the side chain of ethylbenzene, into a chlorinated intermediate by replacing a hydrogen atom with a chlorine atom.

• Chlorination steps require light (\(\text{hv}\)) as an energy source for radical generation.
• This introduces a suitable site for further reaction, such as dehydrohalogenation.

This method is favored for its efficiency and ability to selectively chlorinate the desired position, preparing the molecule for further conversion into alkenes through elimination reactions.

Alkene Formation

Alkene formation is achieved by introducing a double bond into an alkane or alkyl-halide, transforming the molecule into a more reactive, unsaturated compound. The classical method involves a two-step process where halogenation is followed by dehydrohalogenation.

In our example, starting with ethylbenzene, formation of the corresponding alkene, styrene, involves \(\text{C=C}\) bond creation. This is done by first employing radical chlorination to introduce a chlorine atom to the molecule, creating the necessary halogenated intermediate.

• The subsequent dehydrohalogenation step removes the halogen and a hydrogen from adjacent carbons.
• This process effectively forms the new double bond, converting the compound into an alkene.

The strategic use of specific reagents and conditions allows for selective and controlled alkene formation, making it a cornerstone method in synthetic organic chemistry.