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Which separates by sulphonation? (1) 0-xylene (2) m-xylene (3) p-xylene (4) All

Short Answer

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m-xylene and p-xylene separate by sulphonation.

Step by step solution

01

Understanding Sulphonation

Sulphonation is a chemical process where a sulfonic acid group \(\text{-SO}_3\text{H}\) is introduced into an organic compound, usually an aromatic hydrocarbon. The position where the sulfonic acid group attaches can vary depending on the structure of the aromatic compound.
02

Identify the Structures

Examine the structures of the given xylenes. 0-xylene has the substituents in the ortho position (adjacent to each other), m-xylene has them in the meta position (separated by one carbon), and p-xylene has them in the para position (opposite each other).
03

Stability of Products through Sulphonation

Consider the stability of the intermediates formed during sulphonation. Each type of xylene will form different sulfonated products. 0-xylene tends to form multiple products because the sulfonation can occur at different positions due to the adjacent methyl groups.
04

Separation by Sulphonation

The compounds that have distinct positions for substitution will separate more easily. In the case of xylenes, m-xylene (3) and p-xylene (4) form distinct monosulfonated products, making them easier to separate from other isomers and multiple-substituted products.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Chemical Process
Sulphonation is a crucial chemical process in organic chemistry. It involves introducing a sulfonic acid group \(-SO_3H\) into an organic compound. This process typically targets aromatic hydrocarbons, making them more soluble in water and reactive.
The sulfonation of xylenes, which are aromatic hydrocarbons with two methyl groups attached to a benzene ring, specifically aids in generating useful intermediates for various chemical applications.
The position where the sulphonic group attaches determines the product's properties and potential applications.
Aromatic Hydrocarbons
Aromatic hydrocarbons, such as xylenes, have a significant role in the chemical industry. These compounds feature a benzene ring, which provides unique stability and reactivity due to its resonance structure.
Xylenes have three isomers: ortho-xylene, meta-xylene, and para-xylene. The difference between these isomers lies in the position of the two methyl groups attached to the benzene ring:
  • Ortho-xylene (0-xylene) has adjacent methyl groups.
  • Meta-xylene (m-xylene) has methyl groups separated by one carbon.
  • Para-xylene (p-xylene) has methyl groups opposite each other.
Isomer Separation
Separating isomers like xylenes can be quite challenging. However, sulphonation offers a practical solution.
During sulphonation, each xylene isomer reacts to form a monosulfonated product at different positions, creating distinct products.:
  • 0-xylene tends to form multiple products.
  • m-xylene forms one distinct monosulfonated product.
  • p-xylene also forms a distinct monosulfonated product.
Therefore, the process can effectively separate m-xylene and p-xylene due to their unique product formation.
Organic Chemistry
Understanding sulphonation and isomer separation is foundational in organic chemistry. Students and professionals must grasp how the structure of aromatic compounds influences their reactivity and the resultant products.
Steps to study include:
  • Learning the basic properties and structures of aromatic hydrocarbons.
  • Understanding electrophilic aromatic substitution mechanisms, such as sulphonation.
  • Predicting the positions of substitution on different isomers.
  • Applying this knowledge in practical scenarios, such as in industrial processes or laboratory synthesis.
Engineering and Medical Entrance Exams
Sulphonation of xylenes is a common topic in chemistry sections of engineering and medical entrance exams. Here are some tips for tackling related questions:
  • Understand the basic principles of sulphonation and aromatic chemistry.
  • Be able to identify different isomers of aromatic compounds based on substituent positions.
  • Practice predicting the outcomes of chemical reactions, like sulphonation, on various compounds.
  • Review example questions and solutions to build familiarity and confidence with the material.
By mastering these concepts, students can perform better in entrance exams and build a strong foundation for future studies in chemistry.

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Most popular questions from this chapter

When nitrobenzene is treated with \(\mathrm{Br}_{2}\) in the presence of \(\mathrm{FeBr}_{3}\), the major product formed is \(\mathrm{m}\) -bromonitrobenzene. Statement which is related to obtain the m-isomer is (1) the clectron density on meta-carbon is more than that on ortho- and para- positions (2) loss of aromaticity when \(\mathrm{Br}^{+}\) attacks at the orthoand para- positions and not at meta-position (3) casier loss of II" to regain aromaticity from the meta-position than from ortho- and parapositions (4) None of the above

The wrong statement in the following is (1) Sulphonation of benzenc takes place only with hot concentrated sulphuric acid. (2) In the nitration mixture concentrated sulphuric acid is uscd for the formation of nitronium ion. (3) Bccause of unsaturation benzene casily undergoes addition rcactions. (4) Benzene burns with a sooty flame.

Which of the following represents the correct decreasing relative reactivity towards an electrophile, \(E^{19}\) (1) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{3}, \mathrm{p}-\mathrm{CH}_{3} \mathrm{C}_{6} \mathrm{H}_{4} \mathrm{CH}_{3}\), \(\mathrm{p}-\mathrm{CH}_{3} \mathrm{C}_{6} \mathrm{H}_{4} \mathrm{NO}_{2}, \mathrm{p}-\mathrm{NO}_{2} \mathrm{C}_{6} \mathrm{H}_{4} \mathrm{NO}_{2}\) (2) \(\mathrm{p}-\mathrm{CII}_{3} \mathrm{C}_{6} \mathrm{I}_{4} \mathrm{CH}_{3}, \mathrm{p}-\mathrm{CII}_{3} \mathrm{C}_{6} \mathrm{II}_{4} \mathrm{NO}_{2}\) \(\mathrm{p}-\mathrm{NO}_{2} \mathrm{C}_{6} \mathrm{II}_{4} \mathrm{NO}_{2}, \mathrm{C}_{6} \mathrm{I}_{5} \mathrm{CII}_{3}\) (3) \(\mathrm{p}-\mathrm{CII}_{3} \mathrm{C}_{6} \mathrm{H}_{4} \mathrm{NO}_{2}, \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CII}_{3}\), \(\mathrm{p}-\mathrm{CII}_{3} \mathrm{C}_{6} \mathrm{I} \mathrm{I}_{4} \mathrm{CH}_{3}, \mathrm{p}-\mathrm{NO}_{2} \mathrm{C}_{6} \mathrm{I}_{4} \mathrm{NO}_{2}\) (4) \(\mathrm{p}-\mathrm{CII}_{3} \mathrm{C}_{6} \mathrm{II}_{4} \mathrm{CH}_{3}, \mathrm{C}_{6} \mathrm{II}_{5} \mathrm{C} \mathrm{II}_{3}\), \(\mathrm{p}-\mathrm{CII}_{3} \mathrm{C}_{6} \mathrm{H}_{4} \mathrm{NO}_{2}, \mathrm{p}-\mathrm{NO}_{2} \mathrm{C}_{6} \mathrm{I}_{4} \mathrm{NO}_{2}\)

In which of the following new carbon-carbon bond is not formed? (1) Wurtz rcaction (2) Fricdel Crafts reaction (3) Passing benzene through hot iron tubes (4) Sabatier Scnder's reaction

The reaction least likely to oecur is (1) \(\mathrm{C}_{6} \mathrm{H}_{6}+\mathrm{HNO}_{3} \stackrel{\mathrm{II}, \mathrm{so}}{\longrightarrow} \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NO}_{2}\) (2) \(\mathrm{C}_{6} \mathrm{H}_{6}+\mathrm{H}_{2} \mathrm{SO}_{4} \stackrel{\text { Heat }}{\longrightarrow} \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{SO}_{3} \mathrm{H}\) (3) \(\mathrm{C}_{6} \mathrm{H}_{6}+\mathrm{Cl}_{2} \stackrel{\text { UY }}{\longrightarrow} \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{Cl}\) (4) \(\mathrm{C}_{6} \mathrm{H}_{6}+\mathrm{Br}_{2} \longrightarrow \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{Br}\)

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