Unit I Short-Answer Practice: Benzene and Its Derivatives

Practice important 2-mark, 3-mark, and 5-mark university questions with concise model answers, examiner expectations, essential keywords, and common mistakes.

How to Practice Theory Questions

Read the question and first prepare your own answer. Then open the model answer, compare the structure and keywords, and rewrite the answer within the suggested word limit.

Part A: Important 2-Mark Questions and Model Answers

Write each answer in approximately 40–60 words. Include the essential definition, scientific reason, equation, or example required by the question.

1. Calculate the degree of unsaturation of benzene.

Degree of unsaturation = (2C + 2 − H) ÷ 2

[ means 2× the number of carbon atoms, The additional +2 completes the maximum hydrogen count for a saturated open-chain hydrocarbon. The entire expression is divided by 2 because the loss of every two hydrogen atoms produces one degree of unsaturation.]

For benzene, C₆H₆:
= (2 × 6 + 2 − 6) ÷ 2
= 8 ÷ 2 = 4

Therefore, benzene has four degrees of unsaturation.

One degree of unsaturation represents either: one ring, or one double bond, or one equivalent unit of unsaturation.

Benzene contains:

 

  • one six-membered ring = 1 degree of unsaturation
  • three π bonds associated with its three carbon–carbon double bonds = 3 degrees of unsaturation

Therefore: 1 ring+3 π bonds=4 degrees of unsaturation

According to Hückel’s rule, a cyclic, planar, fully conjugated molecule is aromatic when it contains (4n + 2) π electrons, where n is 0, 1, 2, 3, and so on. Benzene contains six π electrons; therefore, n = 1 and it is aromatic.

Keywords: cyclic, planar, conjugated, 4n + 2 π electrons.

All six carbon–carbon bonds in benzene are equal because the six π electrons remain delocalised over the entire ring. Benzene represents a resonance hybrid rather than separate alternating single and double bonds. Consequently, every C–C bond has an intermediate bond order and a length of approximately 1.39 Å.

Keywords: delocalisation, resonance hybrid, equal bond length, 1.39 Å.

The Kekulé structure fails to explain why all six carbon–carbon bonds in benzene have equal length. It also predicts two different ortho-disubstituted derivatives because the fixed structure contains alternating single and double bonds, whereas only one ortho derivative is observed experimentally.

Keywords: equal bond length, ortho derivatives, fixed alternating bonds.

Benzene generally undergoes substitution because substitution temporarily disrupts aromaticity but restores it in the final product. In contrast, an addition reaction permanently destroys the delocalised π-electron system and removes aromatic stabilisation. Therefore, substitution is energetically more favourable.

Keywords: aromaticity, delocalised π system, aromatic stabilisation.

An electrophile is an electron-deficient species that accepts an electron pair from an electron-rich molecule. In electrophilic aromatic substitution, the benzene ring donates π electrons to the electrophile. Examples include the nitronium ion, NO₂⁺, and the acylium ion, RCO⁺.

Keywords: electron-deficient, electron-pair acceptor, NO₂⁺, RCO⁺.

A sigma complex is a resonance-stabilised carbocation intermediate formed when an electrophile creates a sigma bond with one carbon atom of benzene. During its formation, aromaticity is temporarily lost. Deprotonation then restores the aromatic ring and produces the substituted benzene.

Keywords: carbocation intermediate, sigma bond, aromaticity lost, deprotonation.

Halogens deactivate the benzene ring through their strong negative inductive effect, which withdraws electron density. However, their lone pairs donate electron density through resonance and stabilise the ortho and para sigma complexes. Therefore, halogens are deactivating but ortho–para directing.

Keywords: −I effect, resonance donation, lone pair, ortho–para direction.

Friedel–Crafts acylation forms a resonance-stabilised acylium ion that does not undergo rearrangement. Moreover, the acyl group deactivates the aromatic ring and prevents repeated substitution. Therefore, acylation gives a more controlled product than Friedel–Crafts alkylation.

Keywords: acylium ion, no rearrangement, ring deactivation, controlled substitution.

Activating substituents increase the rate of electrophilic aromatic substitution by donating electron density to the benzene ring. Deactivating substituents decrease the reaction rate by withdrawing electron density. For example, –OH activates the ring, whereas –NO₂ strongly deactivates it.

Keywords: electron donation, electron withdrawal, reaction rate, –OH, –NO₂.

Part B: Three-Mark Questions and Model Answers

Practice these important three-mark questions using brief explanations, essential scientific terms, and suitable examples. Each model answer presents the points generally expected in a university examination.

1. Calculate the degree of unsaturation of benzene and explain what the calculated value represents.

Benzene has the molecular formula C₆H₆.

A saturated open-chain hydrocarbon containing six carbon atoms would have the molecular formula C₆H₁₄. Therefore, benzene contains eight hydrogen atoms fewer than the corresponding saturated hydrocarbon.

Every degree of unsaturation represents a deficiency of two hydrogen atoms. Therefore:

Degree of unsaturation = Hydrogen deficiency divided by 2

Degree of unsaturation = 8 divided by 2 = 4

Thus, benzene has four degrees of unsaturation. These four degrees correspond to one ring and three double bonds.

Keywords: C₆H₆, hydrogen deficiency, four degrees of unsaturation, one ring and three double bonds.

Hückel’s rule states that a cyclic, planar and completely conjugated compound is aromatic when it contains 4n + 2 pi electrons. In this expression, n represents a whole number beginning with zero.

Benzene is cyclic and planar. Moreover, each carbon atom contains an unhybridized p orbital, producing continuous conjugation throughout the ring.

Benzene contains six pi electrons. When the value of n is 1:

4 multiplied by 1 plus 2 equals 6.

Therefore, benzene satisfies Hückel’s rule and shows aromatic character.

Keywords: cyclic, planar, continuous conjugation, six pi electrons, 4n plus 2 rule and aromaticity.

Each carbon atom in benzene undergoes sp2 hybridization and possesses one unhybridized p orbital. The six parallel p orbitals overlap continuously around the ring.

Consequently, the six pi electrons become delocalized over all six carbon atoms. They are not confined between particular pairs of carbon atoms.

Therefore, every carbon–carbon bond has the same bond character and the same length of approximately 1.39 angstrom. This value lies between the length of a normal carbon–carbon single bond and a carbon–carbon double bond.

The equality of the bonds supports the resonance-hybrid structure of benzene rather than a fixed Kekulé structure.

Keywords: sp2 hybridization, p-orbital overlap, electron delocalization, resonance hybrid, equal bond length and 1.39 angstrom.

The Kekulé structure represents benzene as a six-membered ring containing alternating single and double bonds. However, it has several limitations.

First, it predicts two different types of carbon–carbon bonds. Experimental evidence shows that all six carbon–carbon bonds have an identical length of approximately 1.39 angstrom.

Second, a fixed Kekulé structure predicts more than one form for some disubstituted derivatives. However, only one ortho, one meta and one para disubstituted derivative are obtained for identical substituents.

Third, a structure containing three ordinary double bonds should readily undergo addition reactions. Benzene mainly undergoes substitution reactions and resists addition under ordinary conditions.

Thus, the fixed Kekulé structure cannot explain bond equality, the number of derivatives, or the unusual stability of benzene.

Keywords: equal bond lengths, disubstituted derivatives, substitution reaction, resonance and aromatic stability.

Benzene contains six delocalized pi electrons distributed over the entire cyclic structure. This delocalization gives benzene considerable aromatic stability.

During an electrophilic substitution reaction, aromaticity is temporarily lost when the electrophile forms a bond with the ring. However, removal of a proton restores the delocalized electron system and aromaticity.

An addition reaction would permanently break the continuous cyclic conjugation and destroy aromatic stabilization. The resulting addition product would be less stable than benzene.

Therefore, benzene generally undergoes substitution reactions because substitution allows it to retain its aromatic character.

Keywords: delocalized pi electrons, aromatic stability, temporary loss of aromaticity, deprotonation and restoration of aromaticity.

A sigma complex is the positively charged intermediate formed when the pi electrons of benzene attack an electrophile during electrophilic aromatic substitution.

In this intermediate, the electrophile and a hydrogen atom remain attached to the same carbon. This carbon temporarily changes from sp2 to sp3 hybridization, and the ring loses its aromaticity.

The positive charge becomes delocalized over three carbon atoms through resonance. Therefore, the sigma complex is also called an arenium ion or Wheland intermediate.

Its formation requires considerable energy because aromaticity is temporarily lost. Consequently, formation of the sigma complex is generally the slow and rate-determining step of electrophilic aromatic substitution.

Keywords: arenium ion, Wheland intermediate, positively charged intermediate, resonance stabilization, temporary loss of aromaticity and rate-determining step.

During nitration, concentrated nitric acid reacts with concentrated sulfuric acid. Sulfuric acid acts as the stronger acid and protonates nitric acid.

The protonated nitric acid then loses a molecule of water and produces the nitronium ion.

The nitronium ion is a positively charged, electron-deficient species. Therefore, it acts as the active electrophile during nitration.

The pi electrons of benzene attack the nitrogen atom of the nitronium ion and form a sigma complex. Subsequent removal of a proton restores aromaticity and produces nitrobenzene.

Keywords: concentrated nitric acid, concentrated sulfuric acid, protonation, loss of water, nitronium ion, electrophile and nitrobenzene.

Friedel–Crafts alkylation introduces an alkyl group into the benzene ring. It commonly uses an alkyl halide and anhydrous aluminium chloride. The active electrophile behaves like a carbocation.

Friedel–Crafts acylation introduces an acyl group into the benzene ring. It uses an acyl chloride or acid anhydride with anhydrous aluminium chloride. The active electrophile is the resonance-stabilized acylium ion.

Alkylation may produce carbocation rearrangement and polyalkylation. In contrast, the acylium ion generally does not rearrange, and the acyl group deactivates the ring. Therefore, polyacylation usually does not occur.

Consequently, Friedel–Crafts acylation often provides a more controlled method of introducing a carbon chain into benzene.

Keywords: alkyl group, acyl group, aluminium chloride, carbocation, acylium ion, rearrangement and polyalkylation.

Halogens withdraw electron density from the benzene ring through their negative inductive effect. This withdrawal reduces the overall reactivity of the ring towards electrophiles. Therefore, halogens deactivate the benzene ring.

However, each halogen contains lone-pair electrons. These lone pairs interact with the pi-electron system of the ring through resonance.

During ortho and para substitution, the halogen donates electron density by resonance and stabilizes the corresponding sigma complexes. The meta sigma complex does not receive the same resonance stabilization.

Therefore, halogens are deactivating because of their inductive effect but direct incoming electrophiles mainly towards the ortho and para positions because of resonance donation.

Keywords: negative inductive effect, lone-pair electrons, resonance donation, deactivation, ortho–para direction and sigma-complex stabilization.

Activating groups increase the electron density of the benzene ring and make it more reactive towards electrophilic substitution. Most activating groups direct an incoming electrophile towards the ortho and para positions.

Examples include the hydroxyl group, amino group, alkoxy group and alkyl group. These groups release electron density through resonance or an electron-donating inductive effect.

Deactivating groups withdraw electron density from the benzene ring and reduce its reactivity towards electrophiles. Most deactivating groups direct further substitution towards the meta position.

Examples include the nitro group, cyano group, carboxyl group, aldehyde group and sulfonic acid group.

Halogens form an important exception. They deactivate the ring through their negative inductive effect but direct substitution towards the ortho and para positions through resonance donation.

Keywords: activating group, deactivating group, electron donation, electron withdrawal, ortho–para director, meta director and halogen exception.

Part C: Long-Answer Questions and Model Answers

Practice these important long-answer questions using brief explanations, essential scientific terms, and suitable examples. Each model answer presents the points generally expected in a university examination.

1. Discuss the structure of benzene and explain the experimental evidence supporting its cyclic and aromatic structure.

Benzene is an aromatic hydrocarbon with the molecular formula C₆H₆. Its unusual stability and chemical behaviour cannot be explained by considering it an ordinary cyclic polyunsaturated compound.

1. Molecular formula and hydrogen deficiency

The molecular formula of benzene is C₆H₆. The corresponding saturated open-chain hydrocarbon containing six carbon atoms is hexane, C₆H₁₄.

Therefore:

Hydrogen deficiency = 14 − 6 = 8 hydrogen atoms

Since one degree of unsaturation represents a deficiency of two hydrogen atoms:

Degree of unsaturation = 8 ÷ 2 = 4

Thus, benzene has four degrees of unsaturation. These correspond to one ring and three double bonds.

2. Kekulé structure of benzene

Kekulé proposed that benzene consists of:

• A six-membered carbon ring
• One hydrogen atom attached to each carbon atom
• Three alternating carbon–carbon single and double bonds

This structure explains the molecular formula C₆H₆ and the tetravalency of carbon.

3. Formation of only one monosubstituted derivative

Benzene produces only one monosubstituted derivative when any one hydrogen atom is replaced by another atom or group.

Examples:

C₆H₆ + Cl₂ → C₆H₅Cl + HCl
Condition: Anhydrous ferric chloride, FeCl₃

Benzene (C₆H₆) forms chlorobenzene (C₆H₅Cl).

C₆H₆ + HNO₃ → C₆H₅NO₂ + H₂O
Condition: Concentrated H₂SO₄, 50–60°C

Benzene (C₆H₆) forms nitrobenzene (C₆H₅NO₂).

The formation of only one monosubstituted product indicates that all six carbon atoms and all six hydrogen atoms in benzene are equivalent.

4. Formation of three disubstituted derivatives

When two hydrogen atoms of benzene are replaced, three positional isomers may form:

• Ortho derivative: substituents occupy positions 1 and 2
• Meta derivative: substituents occupy positions 1 and 3
• Para derivative: substituents occupy positions 1 and 4

For example, dimethylbenzene has the molecular formula C₈H₁₀ and exists as:

• Ortho-xylene or 1,2-dimethylbenzene, C₆H₄(CH₃)₂
• Meta-xylene or 1,3-dimethylbenzene, C₆H₄(CH₃)₂
• Para-xylene or 1,4-dimethylbenzene, C₆H₄(CH₃)₂

The formation of three disubstituted derivatives supports a symmetrical six-membered cyclic structure.

5. Equal carbon–carbon bond lengths

A fixed Kekulé structure should contain:

• Three carbon–carbon single bonds, approximately 1.54 angstrom
• Three carbon–carbon double bonds, approximately 1.34 angstrom

However, X-ray diffraction shows that all six carbon–carbon bonds in benzene have the same length of approximately 1.39 angstrom.

Therefore, benzene does not contain three permanently localized single bonds and three permanently localized double bonds. Instead, its six pi electrons remain delocalized over the entire ring.

6. Heat of hydrogenation and resonance energy

Cyclohexene (C₆H₁₀), which contains one isolated carbon–carbon double bond, releases approximately 120 kilojoules per mole during hydrogenation.

If benzene contained three ordinary double bonds, its expected heat of hydrogenation would be approximately:

3 × 120 = 360 kilojoules per mole

However, the experimentally observed heat of hydrogenation of benzene is approximately 208 kilojoules per mole.

Hydrogenation reaction:

C₆H₆ + 3H₂ → C₆H₁₂
Condition: Nickel, platinum or palladium catalyst; elevated temperature and pressure

Benzene (C₆H₆) forms cyclohexane (C₆H₁₂).

The difference between the expected and observed values is approximately:

360 − 208 = 152 kilojoules per mole

This difference represents the resonance energy or aromatic stabilization energy of benzene.

7. Substitution rather than addition

Ordinary alkenes readily undergo addition reactions. However, benzene generally undergoes substitution reactions because substitution allows the aromatic system to be restored.

Example of substitution: C₆H₆ + Br₂ → C₆H₅Br + HBr

Condition: Anhydrous FeBr₃

Benzene (C₆H₆) forms bromobenzene (C₆H₅Br).

Addition occurs only under vigorous conditions.

Example: C₆H₆ + 3Cl₂ → C₆H₆Cl₆

Condition: Ultraviolet light or sunlight

Benzene (C₆H₆) forms benzene hexachloride, C₆H₆Cl₆.

Substitution under ordinary conditions and addition under vigorous conditions demonstrate the aromatic stability of benzene.

8. Resistance to oxidation

Benzene does not readily decolourise bromine water or dilute alkaline potassium permanganate, KMnO₄, under ordinary conditions. This behaviour distinguishes benzene from ordinary alkenes.

Conclusion

Benzene (C₆H₆) is a planar, cyclic and conjugated molecule. It contains six equivalent carbon atoms and six delocalized pi electrons. Equal carbon–carbon bond lengths, resonance energy, substitution reactions, and the formation of one monosubstituted and three disubstituted derivatives support its aromatic structure.

Examination keywords: C₆H₆, four degrees of unsaturation, six-membered ring, equivalent carbon atoms, 1.39 angstrom bond length, delocalized pi electrons, resonance energy, substitution reaction and aromatic stability.

Benzene (C₆H₆) is a planar, cyclic, conjugated and aromatic hydrocarbon. Its exceptional stability can be explained through sp² hybridization, continuous p-orbital overlap, pi-electron delocalization, resonance and Hückel’s rule.

1. Hybridization of carbon atoms

Each of the six carbon atoms in benzene undergoes sp² hybridization.

Every carbon atom contains:

• Three sp²-hybridized orbitals
• One unhybridized p orbital
• One electron in the unhybridized p orbital

Two sp² orbitals form carbon–carbon sigma bonds with neighbouring carbon atoms. The third sp² orbital forms a carbon–hydrogen sigma bond.

Therefore, benzene contains:

• Six carbon–carbon sigma bonds
• Six carbon–hydrogen sigma bonds
• A total of twelve sigma bonds

2. Planar hexagonal structure

The three sp² orbitals of each carbon atom lie in the same plane and maintain bond angles of approximately 120 degrees.

Therefore, all six carbon atoms and all six hydrogen atoms lie in the same plane. Benzene consequently possesses a planar hexagonal structure.

3. Formation of the delocalized pi-electron system

Each carbon atom contains one unhybridized p orbital perpendicular to the plane of the ring.

The six parallel p orbitals overlap sideways with one another around the complete ring. This continuous overlap produces two delocalized pi-electron clouds:

• One pi-electron cloud above the ring
• One pi-electron cloud below the ring

The six pi electrons are not confined between particular pairs of carbon atoms. Instead, they remain delocalized over all six carbon atoms.

4. Resonance in benzene

Benzene can be represented by two equivalent Kekulé contributing structures. These structures differ only in the positions assigned to the three double bonds.

The actual benzene molecule does not rapidly alternate between these two structures. Instead, it exists as a single resonance hybrid.

The resonance hybrid is commonly represented as a regular hexagon containing a circle. The circle represents the delocalization of six pi electrons over the entire ring.

5. Consequences of resonance

Pi-electron delocalization produces the following characteristics:

• All six carbon–carbon bonds become equivalent.
• Every carbon–carbon bond has a length of approximately 1.39 angstrom.
• Benzene gains approximately 150–152 kilojoules per mole of additional stability.
• Benzene generally undergoes substitution rather than addition.
• Benzene resists oxidation under ordinary conditions.

6. Conditions required for aromaticity

A compound must satisfy four conditions to show aromatic character:

• It must be cyclic.
• It must be planar.
• It must have continuous conjugation throughout the ring.
• It must contain the required number of pi electrons according to Hückel’s rule.

7. Hückel’s rule

Hückel’s rule states that a cyclic, planar and completely conjugated molecule is aromatic when it contains 4n + 2 pi electrons.

Here, n represents a whole number such as 0, 1, 2 or 3.

Benzene contains six pi electrons.

For n = 1:

4n + 2 = (4 × 1) + 2 = 6 pi electrons

Therefore, benzene satisfies Hückel’s rule and is aromatic.

8. Aromatic stability demonstrated by hydrogenation

Benzene undergoes complete catalytic hydrogenation according to the following reaction:

C₆H₆ + 3H₂ → C₆H₁₂
Condition: Nickel, platinum or palladium catalyst; elevated temperature and pressure

Benzene (C₆H₆) forms cyclohexane (C₆H₁₂).

The observed heat released during this reaction is substantially lower than expected for a hypothetical compound containing three isolated double bonds. This difference confirms the additional aromatic stabilization of benzene.

Conclusion

Benzene (C₆H₆) contains six sp²-hybridized carbon atoms and six continuously overlapping p orbitals. Its six delocalized pi electrons satisfy the 4n + 2 rule when n equals one. Therefore, benzene is planar, cyclic, completely conjugated and aromatic.

Examination keywords: C₆H₆, sp² hybridization, 120-degree bond angle, p-orbital overlap, six pi electrons, resonance hybrid, 1.39 angstrom, 4n + 2 rule and aromatic stability.

Electrophilic aromatic substitution is the characteristic reaction of benzene (C₆H₆). In this reaction, an electrophile replaces a hydrogen atom of the aromatic ring without causing permanent loss of aromaticity.

General reaction: C₆H₆ + E⁺ → C₆H₅E + H⁺

Here, E⁺ represents an electrophile and C₆H₅E represents the substituted benzene product.

The mechanism occurs in three principal stages.

1. Generation of the electrophile

A reagent interacts with a strong acid or Lewis acid catalyst to produce an electron-deficient electrophile.

Important examples include:

Nitration:

HNO₃ + H₂SO₄ → NO₂⁺ + HSO₄⁻ + H₂O

Nitric acid (HNO₃) and sulfuric acid (H₂SO₄) generate the nitronium ion (NO₂⁺).

Chlorination:

Cl₂ + FeCl₃ → Cl⁺ + FeCl₄⁻

Chlorine (Cl₂) and ferric chloride (FeCl₃) generate an electrophilic chlorine species.

Bromination:

Br₂ + FeBr₃ → Br⁺ + FeBr₄⁻

Bromine (Br₂) and ferric bromide (FeBr₃) generate an electrophilic bromine species.

Friedel–Crafts alkylation:

CH₃Cl + AlCl₃ → CH₃⁺ + AlCl₄⁻

Methyl chloride (CH₃Cl) and aluminium chloride (AlCl₃) generate a methyl electrophile.

Friedel–Crafts acylation:

CH₃COCl + AlCl₃ → CH₃CO⁺ + AlCl₄⁻

Acetyl chloride (CH₃COCl) and aluminium chloride (AlCl₃) generate the acylium ion (CH₃CO⁺).

2. Formation of the sigma complex

The electron-rich pi system of benzene attacks the electrophile. A new carbon–electrophile sigma bond forms.

During this step:

• One ring carbon temporarily changes from sp² to sp³ hybridization.
• The attacked carbon carries both hydrogen and the electrophile.
• Aromaticity is temporarily lost.
• A positively charged intermediate forms.

This positively charged intermediate is called:

• Sigma complex
• Arenium ion
• Wheland intermediate

The positive charge becomes delocalized over three carbon atoms through resonance. However, the intermediate remains less stable than aromatic benzene.

Because aromaticity is lost during sigma-complex formation, this step requires the highest activation energy. Therefore, it generally represents the slow, rate-determining step.

3. Deprotonation and restoration of aromaticity

A base removes the proton from the carbon atom bonded to the electrophile. The electrons from the carbon–hydrogen bond return to the ring and restore the delocalized pi-electron system.

As a result:

• Aromaticity is restored.
• Substituted benzene forms.
• The catalyst is regenerated.

Example: chlorination of benzene

Overall reaction:

C₆H₆ + Cl₂ → C₆H₅Cl + HCl
Condition: Anhydrous FeCl₃

Benzene (C₆H₆) forms chlorobenzene (C₆H₅Cl).

Electrophile generation:

Cl₂ + FeCl₃ → Cl⁺ + FeCl₄⁻

Final deprotonation:

Sigma complex + FeCl₄⁻ → C₆H₅Cl + HCl + FeCl₃

Ferric chloride (FeCl₃) is regenerated.

Why substitution is preferred over addition?

Substitution temporarily disturbs aromaticity, but the final deprotonation step restores the aromatic pi-electron system.

In contrast, addition would permanently destroy cyclic conjugation and aromatic stabilization. Therefore, benzene generally undergoes electrophilic substitution instead of addition.

Common electrophilic aromatic substitution reactions

• Nitration produces nitrobenzene (C₆H₅NO₂).
• Sulphonation produces benzenesulphonic acid (C₆H₅SO₃H).
• Chlorination produces chlorobenzene (C₆H₅Cl).
• Bromination produces bromobenzene (C₆H₅Br).
• Friedel–Crafts alkylation produces alkylbenzene.
• Friedel–Crafts acylation produces aromatic ketone.

Conclusion

Electrophilic aromatic substitution involves electrophile generation, attack of benzene on the electrophile, formation of a resonance-stabilized sigma complex, removal of a proton and restoration of aromaticity.

Examination keywords: electrophile, Lewis acid, pi-electron attack, sigma complex, arenium ion, resonance stabilization, rate-determining step, deprotonation, catalyst regeneration and restoration of aromaticity.

Nitration of benzene is an electrophilic aromatic substitution reaction in which one hydrogen atom of benzene (C₆H₆) is replaced by a nitro group, NO₂.

Overall reaction: C₆H₆ + HNO₃ → C₆H₅NO₂ + H₂O

Condition: Concentrated H₂SO₄, 50–60°C

Benzene (C₆H₆) reacts with concentrated nitric acid (HNO₃) in the presence of concentrated sulfuric acid (H₂SO₄) to produce nitrobenzene (C₆H₅NO₂) and water (H₂O).

The mixture of concentrated nitric acid and concentrated sulfuric acid is called the nitrating mixture.

Step 1: Formation of the nitronium ion

Concentrated sulfuric acid acts as the stronger acid and protonates nitric acid.

HNO₃ + H₂SO₄ → H₂NO₃⁺ + HSO₄⁻

Nitric acid (HNO₃) accepts a proton and forms protonated nitric acid (H₂NO₃⁺).

The protonated nitric acid then loses water:

H₂NO₃⁺ → NO₂⁺ + H₂O

Therefore, the overall electrophile-generation reaction is:

HNO₃ + H₂SO₄ → NO₂⁺ + HSO₄⁻ + H₂O

The nitronium ion (NO₂⁺) is the active electrophile responsible for nitration.

Step 2: Attack of benzene on the electrophile

The electron-rich pi system of benzene attacks the positively charged nitrogen atom of the nitronium ion.

C₆H₆ + NO₂⁺ → [C₆H₆NO₂]⁺

The product [C₆H₆NO₂]⁺ represents the positively charged sigma complex.

During this step:

• A carbon–nitrogen sigma bond forms.
• One ring carbon temporarily changes from sp² to sp³ hybridization.
• The attacked carbon contains both hydrogen and the nitro group.
• Aromaticity is temporarily lost.
• A positively charged sigma complex forms.

The positive charge becomes delocalized over three carbon atoms through resonance.

Because the ring temporarily loses aromaticity, formation of the sigma complex is the slow and rate-determining step.

Step 3: Deprotonation and restoration of aromaticity

The hydrogen sulfate ion (HSO₄⁻) removes the proton from the carbon atom bearing the nitro group.

[C₆H₆NO₂]⁺ + HSO₄⁻ → C₆H₅NO₂ + H₂SO₄

The electrons of the broken carbon–hydrogen bond return to the ring. Consequently:

• The delocalized pi-electron system is restored.
• Aromaticity returns.
• Nitrobenzene (C₆H₅NO₂) forms.
• Sulfuric acid (H₂SO₄) is regenerated.

Role of sulfuric acid

Concentrated sulfuric acid performs the following functions:

• It protonates nitric acid.
• It helps generate the nitronium ion (NO₂⁺).
• It provides a strongly acidic reaction medium.
• It is regenerated during the final step.

Reaction conditions

Reagent: Concentrated nitric acid, HNO₃

Catalyst and acidic medium: Concentrated sulfuric acid, H₂SO₄

Temperature: Approximately 50–60°C

Electrophile: Nitronium ion, NO₂⁺

Main product: Nitrobenzene, C₆H₅NO₂

By-product: Water, H₂O

Importance of temperature control

The temperature should remain near 50–60°C. Excessive heating may encourage further nitration and produce dinitrobenzene derivatives.

Conclusion

Nitration of benzene involves generation of the nitronium ion (NO₂⁺), electrophilic attack on benzene (C₆H₆), formation of a resonance-stabilized sigma complex, removal of a proton, restoration of aromaticity and formation of nitrobenzene (C₆H₅NO₂).

Examination keywords: benzene, C₆H₆, nitric acid, HNO₃, sulfuric acid, H₂SO₄, nitrating mixture, nitronium ion, NO₂⁺, sigma complex, rate-determining step, deprotonation, nitrobenzene, C₆H₅NO₂ and 50–60°C.

Sulphonation of benzene is an electrophilic aromatic substitution reaction in which one hydrogen atom of benzene (C₆H₆) is replaced by a sulphonic acid group, SO₃H.

Benzene reacts with fuming sulphuric acid or sulphur trioxide dissolved in concentrated sulphuric acid to produce benzenesulphonic acid (C₆H₅SO₃H).

Overall reaction:

C₆H₆ + H₂SO₄ ⇌ C₆H₅SO₃H + H₂O

Condition: Fuming sulphuric acid containing SO₃; heat

Benzene (C₆H₆) reacts with sulphuric acid (H₂SO₄) to produce benzenesulphonic acid (C₆H₅SO₃H) and water (H₂O).

1. Generation of the electrophile

Fuming sulphuric acid contains sulphur trioxide (SO₃), which acts as an electrophile.

In a strongly acidic medium, sulphur trioxide may become protonated:

SO₃ + H₂SO₄ ⇌ SO₃H⁺ + HSO₄⁻

The protonated sulphur trioxide ion (SO₃H⁺) is a stronger electrophile than neutral sulphur trioxide.

Depending on the reaction medium, SO₃ or SO₃H⁺ may act as the active electrophile.

2. Attack of benzene on the electrophile

The pi electrons of benzene attack the sulphur atom of the electrophile. A carbon–sulphur sigma bond forms, producing a positively charged sigma complex.

C₆H₆ + SO₃H⁺ → [C₆H₆SO₃H]⁺

During this step:

• One carbon atom temporarily changes from sp² to sp³ hybridization.
• Aromaticity is temporarily lost.
• The positive charge becomes delocalized over three carbon atoms.
• A resonance-stabilized sigma complex forms.

Formation of the sigma complex is generally the slow and rate-determining step.

3. Deprotonation and restoration of aromaticity

The hydrogen sulphate ion (HSO₄⁻) removes the proton from the carbon atom bearing the sulphonic acid group.

[C₆H₆SO₃H]⁺ + HSO₄⁻ → C₆H₅SO₃H + H₂SO₄

The electrons from the carbon–hydrogen bond return to the ring, aromaticity is restored, and benzenesulphonic acid (C₆H₅SO₃H) forms.

4. Reversible nature of sulphonation

Sulphonation is a reversible reaction.

Sulphonation:

C₆H₆ + H₂SO₄ ⇌ C₆H₅SO₃H + H₂O

Desulphonation:

C₆H₅SO₃H + H₂O → C₆H₆ + H₂SO₄

Condition for desulphonation: Dilute aqueous acid or steam; heat

Heating benzenesulphonic acid with dilute acid or steam removes the SO₃H group and regenerates benzene.

5. Importance of sulphonation

The sulphonic acid group can serve as:

• A water-solubilizing group
• A temporary blocking group in aromatic synthesis
• An intermediate in the preparation of phenols, dyes, detergents and pharmaceutical compounds

Conclusion

Sulphonation of benzene involves electrophile generation, attack of benzene on SO₃ or SO₃H⁺, formation of a sigma complex, deprotonation and restoration of aromaticity. The reaction is reversible, and heating benzenesulphonic acid with steam regenerates benzene.

Examination keywords: benzene, C₆H₆, fuming H₂SO₄, SO₃, SO₃H⁺, sigma complex, benzenesulphonic acid, C₆H₅SO₃H, reversible reaction and desulphonation.

Halogenation of benzene is an electrophilic aromatic substitution reaction in which one hydrogen atom of benzene (C₆H₆) is replaced by chlorine or bromine.

Because chlorine and bromine are not sufficiently electrophilic by themselves, a Lewis acid catalyst such as anhydrous ferric chloride (FeCl₃), ferric bromide (FeBr₃) or aluminium chloride (AlCl₃) is required.

A. Chlorination of benzene

Overall reaction:

C₆H₆ + Cl₂ → C₆H₅Cl + HCl

Condition: Anhydrous FeCl₃ or AlCl₃

Benzene (C₆H₆) reacts with chlorine (Cl₂) to form chlorobenzene (C₆H₅Cl) and hydrogen chloride (HCl).

Step 1: Generation of the electrophile

Chlorine interacts with anhydrous ferric chloride:

Cl₂ + FeCl₃ ⇌ Cl⁺ + FeCl₄⁻

Ferric chloride polarizes the chlorine molecule and produces a strongly electrophilic chlorine species.

Step 2: Formation of the sigma complex

The pi electrons of benzene attack the electrophilic chlorine species:

C₆H₆ + Cl⁺ → [C₆H₆Cl]⁺

The resulting sigma complex temporarily loses aromaticity. Its positive charge becomes delocalized over three carbon atoms through resonance.

Step 3: Deprotonation

The tetrachloroferrate ion (FeCl₄⁻) removes the proton:

[C₆H₆Cl]⁺ + FeCl₄⁻ → C₆H₅Cl + HCl + FeCl₃

Chlorobenzene (C₆H₅Cl) forms, aromaticity returns, and FeCl₃ is regenerated.

B. Bromination of benzene

Overall reaction:

C₆H₆ + Br₂ → C₆H₅Br + HBr

Condition: Anhydrous FeBr₃

Benzene (C₆H₆) reacts with bromine (Br₂) to form bromobenzene (C₆H₅Br) and hydrogen bromide (HBr).

Electrophile generation:

Br₂ + FeBr₃ ⇌ Br⁺ + FeBr₄⁻

Sigma-complex formation:

C₆H₆ + Br⁺ → [C₆H₆Br]⁺

Deprotonation:

[C₆H₆Br]⁺ + FeBr₄⁻ → C₆H₅Br + HBr + FeBr₃

Ferric bromide (FeBr₃) is regenerated.

Role of the Lewis acid catalyst

The Lewis acid:

• Accepts electron density from the halogen molecule.
• Polarizes the halogen–halogen bond.
• Generates a stronger electrophilic halogen species.
• Is regenerated after deprotonation.

Difference from addition reaction

Under normal catalytic conditions, benzene undergoes substitution and retains aromaticity.

However, benzene undergoes photochemical addition with chlorine under ultraviolet light:

C₆H₆ + 3Cl₂ → C₆H₆Cl₆

Condition: Ultraviolet light or sunlight

The product is benzene hexachloride (C₆H₆Cl₆), also called hexachlorocyclohexane.

Conclusion

Chlorination and bromination of benzene involve Lewis acid-assisted electrophile generation, sigma-complex formation, deprotonation and regeneration of aromaticity.

Examination keywords: C₆H₆, Cl₂, Br₂, FeCl₃, FeBr₃, Lewis acid, sigma complex, chlorobenzene, C₆H₅Cl, bromobenzene, C₆H₅Br and catalyst regeneration.

Friedel–Crafts alkylation is an electrophilic aromatic substitution reaction in which an alkyl group replaces a hydrogen atom of benzene (C₆H₆).

An alkyl halide reacts with benzene in the presence of an anhydrous Lewis acid catalyst, usually aluminium chloride (AlCl₃).

Example: Methylation of benzene

Overall reaction:

C₆H₆ + CH₃Cl → C₆H₅CH₃ + HCl

Condition: Anhydrous AlCl₃

Benzene (C₆H₆) reacts with chloromethane or methyl chloride (CH₃Cl) to produce methylbenzene or toluene (C₇H₈) and hydrogen chloride (HCl).

Step 1: Generation of the electrophile

CH₃Cl + AlCl₃ ⇌ CH₃⁺ + AlCl₄⁻

Aluminium chloride accepts electron density from chlorine and strongly polarizes the carbon–chlorine bond. The resulting methyl electrophile behaves as a carbocation-like species.

Step 2: Formation of the sigma complex

The pi electrons of benzene attack the methyl electrophile:

C₆H₆ + CH₃⁺ → [C₆H₆CH₃]⁺

A carbon–carbon sigma bond forms. Aromaticity is temporarily lost, and the positive charge becomes delocalized through resonance.

Step 3: Deprotonation

The tetrachloroaluminate ion removes a proton:

[C₆H₆CH₃]⁺ + AlCl₄⁻ → C₆H₅CH₃ + HCl + AlCl₃

Toluene (C₇H₈) forms, aromaticity returns, and aluminium chloride is regenerated.

Example: Ethylation of benzene

C₆H₆ + C₂H₅Cl → C₆H₅C₂H₅ + HCl

Condition: Anhydrous AlCl₃

Benzene forms ethylbenzene (C₈H₁₀).

Important limitations

1. Polyalkylation

An alkyl group releases electron density and activates the aromatic ring. Therefore, the first alkylated product may react more readily than benzene and undergo further alkylation.

For example:

C₆H₅CH₃ + CH₃Cl → C₆H₄(CH₃)₂ + HCl

Condition: Anhydrous AlCl₃

Toluene (C₇H₈) may form dimethylbenzene or xylene (C₈H₁₀).

Using excess benzene may reduce polyalkylation.

2. Carbocation rearrangement

Primary alkyl electrophiles may rearrange into more stable secondary or tertiary carbocations. Therefore, the expected straight-chain alkylbenzene may not form as the principal product.

3. Failure with strongly deactivated rings

Aromatic compounds containing strongly electron-withdrawing groups generally do not undergo Friedel–Crafts alkylation.

Examples of strongly deactivating groups include:

• Nitro group, NO₂
• Sulphonic acid group, SO₃H
• Cyano group, CN
• Carboxyl group, COOH

4. Failure with amino groups

Aniline (C₆H₅NH₂) generally does not undergo normal Friedel–Crafts alkylation because its amino group forms a Lewis acid–base complex with AlCl₃.

5. Vinyl and aryl halides are unsuitable

Vinyl halides and aryl halides do not readily generate the required carbocations. Therefore, they generally do not act as effective alkylating agents in this reaction.

Conclusion

Friedel–Crafts alkylation introduces an alkyl group into benzene through electrophile generation, sigma-complex formation and deprotonation. However, polyalkylation, carbocation rearrangement and failure with strongly deactivated aromatic compounds limit its synthetic usefulness.

Examination keywords: benzene, C₆H₆, alkyl halide, anhydrous AlCl₃, carbocation, sigma complex, toluene, C₇H₈, polyalkylation, rearrangement and deactivated aromatic ring.

Friedel–Crafts acylation is an electrophilic aromatic substitution reaction in which an acyl group, RCO, replaces a hydrogen atom of benzene (C₆H₆).

An acyl chloride or acid anhydride reacts with benzene in the presence of anhydrous aluminium chloride (AlCl₃).

Example: Acetylation of benzene

Overall reaction:

C₆H₆ + CH₃COCl → C₆H₅COCH₃ + HCl

Condition: Anhydrous AlCl₃, followed by hydrolysis

Benzene (C₆H₆) reacts with acetyl chloride or ethanoyl chloride (CH₃COCl) to produce acetophenone (C₆H₅COCH₃), whose molecular formula is C₈H₈O.

Step 1: Formation of the acylium ion

CH₃COCl + AlCl₃ → CH₃CO⁺ + AlCl₄⁻

The electrophile is the acylium ion (CH₃CO⁺).

The acylium ion is stabilized by resonance:

CH₃–C≡O⁺ ↔ CH₃–C⁺=O

Because of resonance stabilization, the acylium ion generally does not undergo rearrangement.

Step 2: Formation of the sigma complex

The pi electrons of benzene attack the positively charged carbon atom of the acylium ion:

C₆H₆ + CH₃CO⁺ → [C₆H₆COCH₃]⁺

A carbon–carbon bond forms, aromaticity is temporarily lost, and a resonance-stabilized sigma complex develops.

Step 3: Deprotonation

The tetrachloroaluminate ion removes a proton:

[C₆H₆COCH₃]⁺ + AlCl₄⁻ → C₆H₅COCH₃·AlCl₃ + HCl

The aromatic system is restored. The ketone product initially forms a complex with aluminium chloride.

Step 4: Hydrolysis of the complex

C₆H₅COCH₃·AlCl₃ + H₂O → C₆H₅COCH₃ + aluminium-containing hydrolysis products

Acidic hydrolysis releases acetophenone (C₈H₈O).

Comparison with Friedel–Crafts alkylation

1. Group introduced

Alkylation introduces an alkyl group, while acylation introduces an acyl group.

2. Electrophile

Alkylation involves a carbocation or carbocation-like electrophile. Acylation involves a resonance-stabilized acylium ion.

3. Rearrangement

Carbocations may rearrange during alkylation. Acylium ions generally do not rearrange during acylation.

4. Multiple substitution

An alkyl group activates the benzene ring. Therefore, polyalkylation may occur.

An acyl group withdraws electron density and deactivates the aromatic ring. Therefore, polyacylation generally does not occur.

5. Synthetic control

Friedel–Crafts acylation usually provides better control and produces a single principal product.

Reduction of the acyl product

The carbonyl group of an aromatic ketone may subsequently be reduced to obtain an alkylbenzene without carbocation rearrangement.

Acetophenone (C₈H₈O) may be reduced to ethylbenzene (C₈H₁₀).

C₆H₅COCH₃ → C₆H₅CH₂CH₃

Suitable reduction methods include Clemmensen reduction or Wolff–Kishner reduction.

Conclusion

Friedel–Crafts acylation introduces an acyl group through formation of an acylium ion, sigma-complex formation, deprotonation and hydrolysis. It avoids rearrangement and polyacylation, making it more controlled than Friedel–Crafts alkylation.

Examination keywords: benzene, C₆H₆, acetyl chloride, CH₃COCl, anhydrous AlCl₃, acylium ion, sigma complex, acetophenone, C₈H₈O, hydrolysis, no rearrangement and no polyacylation.

A substituent already attached to the benzene ring affects both the rate and position of further electrophilic aromatic substitution.

The substituent may:

• Activate or deactivate the benzene ring.
• Direct the incoming electrophile towards the ortho, meta or para position.

1. Activating groups

Activating groups increase electron density in the aromatic ring. Therefore, the substituted benzene reacts faster than benzene.

Most activating groups direct the incoming electrophile towards the ortho and para positions.

Important activating groups include:

• Amino group, NH₂
• Hydroxyl group, OH
• Alkoxy group, OR
• Alkyl group, R

Examples:

Aniline (C₆H₅NH₂) contains the amino group.

Phenol (C₆H₅OH) contains the hydroxyl group.

Toluene (C₆H₅CH₃) contains the methyl group.

The amino and hydroxyl groups donate electron density by resonance. Alkyl groups release electron density mainly through the positive inductive effect and hyperconjugation.

2. Ortho–para direction by an activating group

Consider nitration of toluene (C₇H₈):

C₆H₅CH₃ + HNO₃ → CH₃C₆H₄NO₂ + H₂O

Condition: Concentrated H₂SO₄; controlled temperature

Toluene mainly produces:

• Ortho-nitrotoluene, C₇H₇NO₂
• Para-nitrotoluene, C₇H₇NO₂

The methyl group increases electron density and stabilizes the ortho and para sigma complexes.

3. Deactivating groups

Deactivating groups withdraw electron density from the aromatic ring. Therefore, the substituted benzene reacts more slowly than benzene.

Most deactivating groups direct the incoming electrophile towards the meta position.

Important deactivating groups include:

• Nitro group, NO₂
• Sulphonic acid group, SO₃H
• Cyano group, CN
• Carboxyl group, COOH
• Aldehyde group, CHO
• Acyl group, COR
• Ester group, COOR

These groups withdraw electron density through negative inductive or negative resonance effects.

4. Meta direction by a deactivating group

Consider nitration of nitrobenzene (C₆H₅NO₂):

C₆H₅NO₂ + HNO₃ → C₆H₄(NO₂)₂ + H₂O

Condition: Concentrated H₂SO₄; heat

The principal product is meta-dinitrobenzene (C₆H₄N₂O₄).

The nitro group withdraws electron density strongly. The sigma complexes produced by ortho and para attack contain particularly unstable resonance contributors in which positive charge develops near the electron-withdrawing group.

The meta sigma complex avoids these highly destabilized contributors. Therefore, substitution occurs mainly at the meta position.

5. Halogen exception

Halogens such as fluorine, chlorine, bromine and iodine are deactivating but ortho–para directing.

For example, chlorobenzene (C₆H₅Cl) is less reactive than benzene because chlorine withdraws electron density through its negative inductive effect.

However, chlorine contains lone-pair electrons and donates electron density through resonance. This donation stabilizes the ortho and para sigma complexes.

Therefore, chlorine deactivates the ring but directs further substitution towards the ortho and para positions.

Nitration of chlorobenzene:

C₆H₅Cl + HNO₃ → ClC₆H₄NO₂ + H₂O

Condition: Concentrated H₂SO₄; heat

The main products are:

• Ortho-nitrochlorobenzene, C₆H₄ClNO₂
• Para-nitrochlorobenzene, C₆H₄ClNO₂

6. Steric effect on orientation

Although a substituent may direct electrophiles towards both ortho and para positions, a bulky incoming electrophile or bulky substituent may reduce ortho substitution.

The para product may predominate because it experiences less steric crowding.

Summary of directing effects

Strongly activating, ortho–para directing:

NH₂, NHR, NR₂, OH and OR

Moderately activating, ortho–para directing:

NHCOR and OCOR

Weakly activating, ortho–para directing:

R groups such as CH₃

Deactivating but ortho–para directing:

F, Cl, Br and I

Deactivating and meta directing:

NO₂, SO₃H, CN, CHO, COR, COOH and COOR

Conclusion

Electron-donating groups usually activate the benzene ring and direct substitution towards the ortho and para positions. Electron-withdrawing groups usually deactivate the ring and direct substitution towards the meta position. Halogens form an important exception because they deactivate the ring but direct substitution towards the ortho and para positions.

Examination keywords: activating group, deactivating group, electron donation, electron withdrawal, inductive effect, resonance effect, sigma-complex stability, ortho–para director, meta director, steric effect and halogen exception.

DDT, BHC, saccharin and chloramine-T are important aromatic or cyclic organic compounds commonly discussed in pharmaceutical organic chemistry. Their structures, molecular formulas and uses are given below.

1. DDT (Full name: Dichlorodiphenyltrichloroethane)

Molecular formula: C₁₄H₉Cl₅

IUPAC name: 1,1,1-Trichloro-2,2-bis(4-chlorophenyl)ethane

Condensed structural representation:

(p-ClC₆H₄)₂CHCCl₃

Structural features

DDT contains two para-chlorophenyl rings attached to a carbon atom. This carbon is also connected to a trichloromethyl group, CCl₃.

Preparation

DDT is prepared by the condensation of chlorobenzene (C₆H₅Cl) with chloral or trichloroacetaldehyde (CCl₃CHO) in the presence of concentrated sulphuric acid (H₂SO₄).

2C₆H₅Cl + CCl₃CHO → C₁₄H₉Cl₅ + H₂O

Condition: Concentrated H₂SO₄

Two molecules of chlorobenzene react with one molecule of chloral to produce DDT and water.

Uses

• DDT was widely used as a contact insecticide against mosquitoes, flies, lice and agricultural pests.
• It played an important role in controlling malaria-transmitting mosquitoes.
• It was used to control insects responsible for typhus and other vector-borne diseases.
• Its environmental use is now banned or strictly restricted in many countries because it persists in the environment, accumulates in living organisms and undergoes biomagnification.
• Any current public-health use remains subject to strict regulatory control.

Important caution: DDT should be described as a historically important but environmentally persistent insecticide, not as a generally recommended household pesticide.

2. BHC (Full name: Benzene hexachloride)

Preferred chemical name: 1,2,3,4,5,6-Hexachlorocyclohexane

Molecular formula: C₆H₆Cl₆

BHC is also called hexachlorocyclohexane or HCH.

Structural features

BHC contains a six-membered cyclohexane ring. Each carbon atom of the ring carries one hydrogen atom and one chlorine atom.

Therefore, each of the six carbon atoms is attached to:

• Two neighbouring carbon atoms
• One hydrogen atom
• One chlorine atom

Preparation

BHC forms by the photochemical addition of chlorine to benzene.

C₆H₆ + 3Cl₂ → C₆H₆Cl₆

Condition: Ultraviolet light or sunlight

Benzene (C₆H₆) reacts with three molecules of chlorine (Cl₂) to produce benzene hexachloride (C₆H₆Cl₆).

This is an addition reaction. Therefore, the aromatic character of benzene is lost.

Isomers

BHC exists as several stereoisomers. Its gamma isomer is called gamma-hexachlorocyclohexane or lindane.

Uses

• BHC was historically used as an agricultural insecticide against soil and crop pests.
• Lindane was used for controlling lice and scabies under medical supervision.
• It was used for controlling mosquitoes, flies, beetles and other insects.
• Its use is now prohibited or restricted in many regions because of toxicity, environmental persistence and possible neurological effects.

Important caution: Lindane should not be presented as a routine first-choice treatment because safer alternatives may be available.

3. Saccharin (Chemical name: Benzoic sulfimide)

Molecular formula: C₇H₅NO₃S

Structural description

Saccharin contains:

• A benzene ring
• A five-membered heterocyclic ring fused with the benzene ring
• One carbonyl group, C=O
• One sulphonyl group, SO₂
• One imide nitrogen bearing hydrogen

The carbonyl carbon and sulphonyl sulphur form part of the five-membered ring. The nitrogen atom lies between them within the heterocyclic ring.

Uses

• Saccharin is a non-nutritive artificial sweetening agent.
• It is approximately 300–500 times sweeter than sucrose.
• It is used in sugar-free beverages, tabletop sweeteners, confectionery and processed foods.
• It is used in pharmaceutical syrups, chewable tablets, lozenges and oral-care products to improve taste.
• It provides sweetness with negligible caloric contribution.
• It may be used in products intended for people who need to control sugar intake.
• Sodium saccharin (C₇H₄NNaO₃S) is more water-soluble and is commonly used in commercial products.

Limitation

Saccharin may produce a bitter or metallic aftertaste, particularly at higher concentrations. Therefore, manufacturers often combine it with other sweetening agents.

4. Chloramine-T (Chemical name: Sodium N-chloro-p-toluenesulphonamide)

Molecular formula of anhydrous chloramine-T: C₇H₇ClNNaO₂S

It is commonly available as chloramine-T trihydrate:

C₇H₇ClNNaO₂S·3H₂O

Condensed structural representation:

p-CH₃C₆H₄SO₂NClNa

Structural features

Chloramine-T contains:

• A benzene ring
• A methyl group, CH₃, at the para position
• A sulphonamide-derived group, SO₂NClNa
• Active chlorine attached to nitrogen

In water, chloramine-T releases active chlorine-containing species. Therefore, it acts as a disinfecting and oxidizing agent.

Uses

• Chloramine-T is used as a disinfectant for surfaces, utensils and equipment.
• It has been used for disinfecting water under properly controlled conditions.
• It acts as an antiseptic and deodorizing agent in suitable diluted preparations.
• It is used as an oxidizing and chlorinating reagent in chemical laboratories.
• It is employed in analytical chemistry, including certain oxidation and iodometric procedures.
• It may be used in pharmaceutical and healthcare environments according to approved concentrations and safety procedures.

Safety

Chloramine-T solutions must be prepared at the recommended concentration. Concentrated material may irritate the skin, eyes and respiratory system. It should not be mixed indiscriminately with acids or incompatible cleaning chemicals.

Conclusion

DDT (C₁₄H₉Cl₅) and BHC (C₆H₆Cl₆) are historically important chlorinated insecticides whose present use is restricted because of environmental and toxicological concerns. Saccharin (C₇H₅NO₃S) is an artificial sweetening agent, while chloramine-T (C₇H₇ClNNaO₂S·3H₂O) acts as a disinfectant, oxidizing agent and chlorinating reagent.

Examination keywords: DDT, C₁₄H₉Cl₅, chloral, BHC, C₆H₆Cl₆, ultraviolet light, lindane, saccharin, C₇H₅NO₃S, artificial sweetener, chloramine-T, active chlorine, disinfectant and environmental persistence.