Test: Aromatic Electrophilic Substitution - JEE MCQ

The correct answer is Option B.
Fuming sulphuric acid, H₂S₂O₇, can be thought of as a solution of SO₃ in sulphuric acid - and so is a much richer source of the SO₃. Sulphur trioxide is an electrophile because it is a highly polar molecule with a fair amount of positive charge on the sulphur.

The correct answer is Option B.
Nitration is electrophilic substitution reaction; in its first step HNO3 takes protons from sulphuric acid and then forms −NO2+ so here HNO3 acts as base.
 
Mechanism of reaction as shown

The reactivity order for electrophilic aromatic substitution is based on the substituents' effects:

• Most reactive: Compound III (toluene, −CH3-​ activates the ring)
• Next: Compound I (benzene, neutral)
• Next: Compound IV (chlorobenzene, mildly deactivating)
• Least reactive: Compound II (nitrobenzene, strongly deactivating)

Correct answer: C: II < IV < I < III

The correct answer is option B

The correct answer is option D

This reaction is a bromination in the presence of Br2​ and FeBr3 a typical electrophilic aromatic substitution.

Structure Analysis:

• The compound has an amide group (−CONH−) attached to a benzene ring. This group is an electron-donating group due to resonance, activating the ring, particularly at the ortho and para positions relative to the −CONH− group.

Expected Product:

• Bromination will most likely occur at the para position (due to steric hindrance at the ortho position).
• Thus, the correct major product should have a bromine atom at the para position to the −CONH− group.

Halonium ion act as electrophile in halogenation. Nitronium ion is used in nitration. Sulphonium ion is used in sulphonation. Acylium ion is used in acylation.

The correct answer is Option A
Toluene is having one methyl group which is an electron-donating group causing a negative charge on the carbon atom of the ring so it is highly reactive towards electrophile which is already electron deficient.
Benzene has a delocalised set of electron clouds which attracts electrophile while chloro and nitro group are electronegative causing positive charge on carbon atoms so are not reactive towards electrophilic substitution.
Cl shows -I effect and −NO₂ shows a strong electron-withdrawing effect. So least reactive
a)IV < III < II < I
 

This reaction is also a bromination with Br2\text{Br}_2Br2​ and FeBr3\text{FeBr}_3FeBr3​ on a heterocyclic aromatic compound with both a methyl and a carbonyl group.

Structure Analysis:

• The methyl group is an electron-donating group and activates the ortho and para positions on the ring.
• The carbonyl group (−CO−)(-CO-)(−CO−) is electron-withdrawing, so it deactivates the ring, especially at the meta positions relative to itself.

Expected Product:

• Bromination will most likely occur at the ortho position relative to the methyl group, which is electron-donating and thus activates the position.

Answer: B

1. Option A: Sulphonation of benzene (C6H6) requires concentrated H2SO4​ at high (boiling) temperatures to produce the sulphonated product (C6H5SO3H).

2. Option B: In the sulphonation reaction, SO3​ acts as the electrophile, attacking the benzene ring to form the sulphonated product.

3. Option C: Sulphonation is a reversible reaction. Adding water can reverse the reaction, converting C6H5SO3H back to benzene.

Since all statements are correct, D: All of the above is the right choice.

The first reaction oxidizes an achiral aromatic compound (A, C₁₀H₁₀) using cold, dilute KMnO₄ to form a racemic mixture (B, C₁₀H₁₂O₂), indicating an alkene in A. Further oxidation cleaves the structure (C₈H₆O₄), and the final steps reduce and heat the compound, ending with C₈H₄O₃.

Based on this, A is most likely naphthalene (Option C).

Reaction Analysis:

1. A to B:

   • A (C₁₀H₁₀) is oxidized by cold, dilute KMnO₄ to form B (C₁₀H₁₂O₂), a racemic mixture. This indicates that A likely contains a double bond, and B has two hydroxyl groups added.

2. B to C:

   • B undergoes further oxidation with hot, aqueous KMnO₄/KOH to form C (C₈H₆O₄), which indicates cleavage of bonds, likely producing a dicarboxylic acid.

3. C to D:

   • C is reduced with H₂/Pd-C, which typically leads to the formation of D (C₁₀H₁₂), suggesting that D is a fully saturated hydrocarbon.

Question: What is correct about D?

• D is likely an achiral compound, as reduction would lead to a simple saturated hydrocarbon without stereocenters.

Correct Answer: Option C: An achiral compound.

Key Points:

• A is likely an aromatic compound, and nitration introduces a nitro group (-NO₂) to the ring.
• Given the likely structure of A (naphthalene or a similar aromatic system), the nitro group will preferentially attach at a specific position based on the substitution pattern of A.

Conclusion:

The major product of nitration will have the nitro group at a position that is favorable for electrophilic substitution.

The correct structure is Option A, where the nitro group is attached at the 1-position of the aromatic ring.

Reaction Sequence Breakdown:

1. A to B:

   • A (C₁₀H₁₄) undergoes nitration with conc. HNO₃ and H₂SO₄ to form a nitro compound (B, C₁₀H₁₃NO₂). This suggests that A is likely an alkylbenzene, where nitration introduces a nitro group onto the aromatic ring.

2. B to C:

   • B is brominated using Br₂ under UV light, resulting in C (C₁₀H₁₂BrNO₂). The presence of UV light indicates that bromination occurs at the benzylic position (on the side chain) rather than the aromatic ring.

3. C to D and E:

   • C undergoes further reactions that involve elimination (using NaNH₂) and oxidation (H₂SO₄) to form a carbonyl compound (E).

Conclusion:

• A (C₁₀H₁₄) is likely ethylbenzene or a similar alkyl-substituted benzene derivative, as the nitration and bromination reactions fit well with such structures.

The correct structure of A is Option B: Ethylbenzene.

Key Information:

• A to B: A (C₁₀H₁₄) undergoes nitration to form B (C₁₀H₁₃NO₂).
• B to C: B is brominated using Br₂ under UV light, suggesting that bromine is added at the benzylic position (not on the aromatic ring), forming C (C₁₀H₁₂BrNO₂).
• C is likely achiral, as no new chiral centers are introduced.
• It is not enantiomeric, as the structure does not suggest chirality.
• Bromination here is not an electrophilic aromatic substitution but rather a free radical substitution at the benzylic position.

Conclusion:

The correct answer is Option C: It is achiral.

Reaction Breakdown:

1. A to B:

   • A (C₁₀H₁₄) undergoes nitration with conc. HNO₃ and H₂SO₄, forming B (C₁₀H₁₃NO₂), a nitro-substituted aromatic compound.

2. B to C:

   • B is brominated under UV light to form C (C₁₀H₁₂BrNO₂), indicating benzylic bromination.

3. C to D:

   • C is treated with KOH and Br₂, leading to the formation of D by elimination reactions.

4. D to E:

   • D is subjected to NaNH₂, followed by oxidation using H₂SO₄, to yield E, a carbonyl compound.

Conclusion:

• The final step indicates the formation of a carbonyl group, which suggests that E is likely a ketone or aldehyde.
• Given the starting material and the steps involved, E is most likely a benzaldehyde derivative (due to the oxidation of the benzylic position).

The structure of E is benzaldehyde or a related structure, depending on the exact substituents in the molecule.

Understanding the Compound

- The compound C6H4Br2(P) is a dibromo benzene. This means it has two bromine atoms attached to the benzene ring.

Understanding Isomerism

- The less polar isomer refers to the geometric arrangement of the bromine atoms on the benzene ring that results in less polarity.
- In the case of dibromo benzene, there are three possible geometric arrangements (ortho, meta, and para) but only ortho-dibromo benzene and para-dibromo benzene are polar. The less polar of the two is the para isomer.

Nitration Reaction

- Nitration is a general class of a chemical process for the introduction of a nitro group into an organic chemical compound.
- The nitration of benzene involves substitution of a hydrogen atom by a nitro group (-NO2) and this is facilitated by a mixture of concentrated nitric acid (HNO3) and sulfuric acid (H2SO4).

Mono Nitration Products

- Considering that the compound is para-dibromo benzene, the nitration can occur at any of the 4 remaining carbon atoms in the benzene ring.
- However, since the question asks for mono nitration products (meaning only one nitro group is added), the number of unique products will depend on the symmetry of the molecule.
- For para-dibromo benzene, the molecule is symmetric. Nitration at any of the 4 remaining positions will result in only 2 unique mono-nitration products due to this symmetry.

So, the answer is 2.

C7H8 is toluene: that’s the only benzenoid isomer possible. To get C7H8O, you must mentally “insert an oxygen atom” somewhere.
There are 5 ways you can mentally insert an oxygen atom into a C-C or C-H bond of toluene.
So the isomers of C7H8O are:
2-methylphenol = o-cresol
3-methylphenol = m-cresol
4-methylphenol = p-cresol
benzyl alcohol = phenylmethanol = (hydroxymethyl)benzene
methoxybenzene = anisole = methyl phenyl ether

The correct answer is:

Analysis of Each Structure:

1. I (Pyrrole): Activated due to the lone pair of electrons on nitrogen, which can donate to the ring.
2. II (Pyridine): Deactivated because nitrogen withdraws electron density from the ring.
3. III (Benzene with two methyl groups): Activated by the electron-donating methyl groups (-CH₃).
4. IV (Benzene with -NHCOCH₃ group): Activated by the -NH group (amide), which donates electrons.
5. V (Benzene with -COCH₃ group): Deactivated by the electron-withdrawing carbonyl group (-C=O).
6. VI (Benzene with -NHCOCH₃ group): Activated by the -NH group (amide).
7. VII (Naphthalene): Activated, similar to a typical aromatic system.
8. VIII (Diphenyl): Activated, typical aromatic system.
9. IX (Cyclohexane ring with alkyl): Not aromatic, so not relevant for activation.
10. X (Benzaldehyde): Deactivated by the electron-withdrawing aldehyde group (-CHO).

Conclusion:

The compounds with activated aromatic rings are I, III, IV, VI, VII, and VIII.

Thus, there are 6 activated aromatic rings for electrophilic substitution reactions.