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GNDU Question Paper-2021
Ba/Bsc
1
st
Semester (Batch 2024-28) (CBGS)
CHEMISTRY
(Organic Chemistry-I)
Time Allowed: Three Hours Max. Marks:35
Note: Attempt Five questions in all, selecting at least One question from each section. The
Fifth question may be attempted from any section. All questions carry equal marks.
SECTION-A
1. (a) An aqueous solution of tropyllium bromide on treatment with AgNO3 4 gives
precipitate of AgBr. Explain.
(b) Why isopropyl free radical is more stable than n-propyl radical ?
2. (a) Arrange the following free radicals in order of their increasing stability and justify :
CH3, (CH3)3C, (CH3)2CH, C6H5CH2
(b) How do you explain the o, p-directing nature of-CH3 group, though it lacks electron
pair?
SECTION B
3. (a) Using 1-methylcyclohexene as an example, discuss its ozonolysis with suitable
mechanism.
(b) Acetylene forms metal acetylide but dimethylacetylene does not form such
derivatives, why ?
4. (a) Chlorination of n-butane in the presence of light gives a mixture of 28% of 1-
chlorobutane and 72% of 2-chlorobutane while bromination gives 98% of 2-bromobutane
and 2% of 1-bromobutane, explain.
(b) Sketch the mechanism of epoxidation reaction.
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SECTION C
5. (a) Draw energy profile diagram for S
N
1 and S
N
2 reactions.
(b) 1-chlorobut-2-ene reacts with KCN to give a mixture of isomeric products. Give the
structures and suitable mechanisms for the isomeric products.
6. Discuss Baeyer's strain theory. How it is used to explain the reactivity of cyclopropane
and cyclobutane rings ? Also discuss its limitations.
SECTION D
7. Predict the product/products in the following reactions :
(a)
CH3
|
_____|_____
/ \
| |
| |NH2
\ /
\___________/
HNO3 / H2SO4
↓
Product ?
Compound: 2-Methylaniline (o-Toluidine)
(b)
CH3
|
_____|_____
/ \
| |
| |
\ /
\_____ _____/
|
NH2
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Br2 / FeBr3
↓
Product ?
Compound: 3-Methylaniline (m-Toluidine)
(c)
CH3
|
_____|_____
/ \
| |
| |
\ /
\_____ _____/
|
NH2
Br2 / CCl4
↓
Product ?
Compound: 4-Methylaniline (p-Toluidine)
More Accurate Skeletal Structures
(a) o-Toluidine (2-Methylaniline)
CH3
|
/¯¯¯¯¯\
| |
| |NH2
\_____/
(b) m-Toluidine (3-Methylaniline)
CH3
|
/¯¯¯¯¯\
| |
NH2 |
\_____/
(c) p-Toluidine (4-Methylaniline)
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CH3
|
/¯¯¯¯¯\
| |
| |
\_____/
|
NH2
These structures match the compounds shown in the question paper:
(a) o-Toluidine (2-Methylaniline)
(b) m-Toluidine (3-Methylaniline)
(c) p-Toluidine (4-Methylaniline)
8.(a) Why does side-chain halogenation in alkyl benzene take place preferentially at a-
position to the aromatic ring?
(b) Explain what are anti-aromatic compounds. Give two examples.
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GNDU Answer Paper-2021
Bachelor of Computer Application (BCA) (Hons.)
1
st
Semester (Batch 2024-28) (CBGS)
CHEMISTRY
(Organic Chemistry-I)
Time Allowed: Three Hours Max. Marks:35
Note: Attempt Five questions in all, selecting at least One question from each section. The
Fifth question may be attempted from any section. All questions carry equal marks.
SECTION-A
1. (a) An aqueous solution of tropyllium bromide on treatment with AgNO3 4 gives
precipitate of AgBr. Explain.
(b) Why isopropyl free radical is more stable than n-propyl radical ?
Ans: 1. (a) An aqueous solution of tropyllium bromide on treatment with AgNO₃ gives a
precipitate of AgBr. Explain.
Simple Explanation
This question is based on the concepts of ionic compounds, tropyllium ion, and the reaction
of silver nitrate (AgNO₃) with bromide ions (Br⁻).
Step 1: What is Tropyllium Bromide?
Tropyllium bromide is a salt made up of two ions:
Tropyllium ion (C₇H₇⁺) a positively charged ion (cation)
Bromide ion (Br⁻) a negatively charged ion (anion)
It can be written as:
C₇H₇⁺ Br⁻
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Unlike many organic compounds that are covalent, tropyllium bromide behaves like an
ionic salt because the positive charge is present on the tropyllium ion and the negative
charge is on bromide.
Step 2: What happens in water?
When tropyllium bromide is dissolved in water, it breaks into separate ions.
C₇H₇Br → C₇H₇⁺ + Br⁻
Now the bromide ions are free in the solution.
Step 3: What is the role of AgNO₃?
Silver nitrate (AgNO₃) is commonly used to test for halide ions such as chloride, bromide,
and iodide.
In water,
AgNO₃ → Ag⁺ + NO₃⁻
The silver ions (Ag⁺) immediately react with bromide ions (Br⁻).
Ag⁺ + Br⁻ → AgBr↓
The downward arrow (↓) shows that silver bromide (AgBr) is insoluble in water and forms a
pale yellow precipitate.
Diagram
Tropyllium Bromide in Water
Water
│
▼
C₇H₇Br
│
▼
C₇H₇⁺ Br⁻
│
│ + Ag⁺ (from AgNO₃)
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▼
AgBr ↓
(Pale Yellow Precipitate)
Why does this happen?
The important point is that bromide exists as a free ion in solution. Since AgNO₃ reacts only
with free bromide ions, it produces insoluble AgBr.
If bromine had been directly attached to carbon through a covalent bond, AgBr would not
form so easily. Therefore, the formation of AgBr proves that tropyllium bromide is ionic in
nature.
Additional Concept: Why is the Tropyllium Ion Stable?
The tropyllium ion has 7 carbon atoms arranged in a ring.
C
/ \
C C
| |
C C
\ /
C
The positive charge is not fixed on one carbon atom. Instead, it is spread over the entire
ring through resonance.
This spreading of charge makes the ion very stable.
It also contains 6 π-electrons, which satisfies Hückel's rule (4n + 2, where n = 1). Therefore,
the tropyllium ion is aromatic, and aromatic compounds are exceptionally stable.
Conclusion
Tropyllium bromide behaves as an ionic compound. In water, it dissociates into tropyllium
ions and bromide ions. When AgNO₃ is added, the free Br⁻ ions react with Ag⁺ ions to form
an insoluble pale yellow precipitate of AgBr. This reaction confirms the presence of
bromide ions in solution and demonstrates the ionic nature of tropyllium bromide.
1. (b) Why is isopropyl free radical more stable than n-propyl free radical?
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Simple Explanation
This question is based on the concept of free radicals and their stability.
A free radical is an atom or molecule that contains one unpaired electron. Because of this
unpaired electron, free radicals are highly reactive.
Example:
CH₃
The dot () represents the unpaired electron.
Structure of n-Propyl Radical
The n-propyl radical is formed by removing one hydrogen atom from the end carbon of
propane.
CH₃CH₂CH₂
The radical carbon is attached to only one alkyl group.
So, it is called a primary (1°) free radical.
Structure of Isopropyl Radical
The isopropyl radical is formed by removing one hydrogen atom from the middle carbon of
propane.
CH₃
|
CH₃ C
|
H
The radical carbon is attached to two alkyl groups.
Hence, it is called a secondary (2°) free radical.
Diagram Showing the Difference
n-Propyl Radical
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CH₃ CH₂ CH₂
↑
One alkyl group
(Primary Radical)
Isopropyl Radical
CH₃
|
CH₃ C
|
H
↑
Two alkyl groups
(Secondary Radical)
Why are Alkyl Groups Important?
Alkyl groups (like CH₃) donate a small amount of electron density toward the radical carbon
through the +I (electron-releasing inductive) effect.
This reduces the electron deficiency of the radical carbon and increases its stability.
The more alkyl groups attached to the radical carbon, the greater the stabilization.
Hyperconjugation
Another important reason is hyperconjugation.
Hyperconjugation is the overlap of nearby CH σ-bonds with the orbital containing the
unpaired electron.
This spreads the unpaired electron over several atoms instead of keeping it concentrated on
one carbon.
More hyperconjugation means greater stability.
Since the isopropyl radical has more adjacent CH bonds, it experiences more
hyperconjugation than the n-propyl radical.
Stability Order of Free Radicals
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3° Radical
↓
2° Radical
↓
1° Radical
↓
Methyl Radical
or
Tertiary > Secondary > Primary > Methyl
Since isopropyl is a secondary radical and n-propyl is a primary radical, the isopropyl
radical is naturally more stable.
Conclusion
The isopropyl free radical is more stable than the n-propyl free radical because the radical
carbon is attached to two alkyl groups, which stabilize the unpaired electron through the +I
(electron-donating inductive) effect and greater hyperconjugation. In contrast, the n-
propyl radical has only one alkyl group, so it receives less stabilization. Therefore, the
stability order is:
Isopropyl radical > n-Propyl radical.
2. (a) Arrange the following free radicals in order of their increasing stability and justify :
CH3, (CH3)3C, (CH3)2CH, C6H5CH2
(b) How do you explain the o, p-directing nature of-CH3 group, though it lacks electron
pair?
Ans: 2. (a) Arrange the following free radicals in order of their increasing stability and
justify:
CH₃, (CH₃)₃C, (CH₃)₂CH, C₆H₅CH₂
Simple Explanation
Before arranging the radicals, let us first understand what a free radical is.
A free radical is an atom or a group of atoms that contains one unpaired electron ().
Because of this unpaired electron, free radicals are highly reactive and always try to become
stable by pairing their electron.
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The stability of a free radical depends on how well the unpaired electron is shared or spread
out. If the electron is spread over many atoms, the radical becomes more stable.
There are two important concepts involved:
1. Hyperconjugation
Alkyl groups (CH₃) push a small amount of electron density toward the carbon having the
unpaired electron. This sharing of electron density is called hyperconjugation. More alkyl
groups mean greater hyperconjugation and therefore greater stability.
2. Resonance
If the unpaired electron can spread over several atoms through alternating double bonds,
the radical becomes even more stable. This is called resonance.
Structures
Methyl Radical
CH3
Secondary Radical
CH3
|
CH3 C H
Tertiary Radical
CH3
|
CH3 C CH3
Benzyl Radical
C6H5 CH2
In the benzyl radical (C₆H₅CH₂), the unpaired electron is not confined to one carbon. It
spreads over the benzene ring through resonance.
Benzene Ring
_________
/ \
| |
\_________/
|
CH2
Because the electron is delocalized over many atoms, the benzyl radical is more stable than
even the tertiary radical.
Increasing Order of Stability
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CH3 < (CH3)2CH < (CH3)3C < C6H5CH2
Justification
CH₃ (Methyl radical): No alkyl group is attached, so there is almost no
hyperconjugation. It is the least stable.
(CH₃)₂CH (Secondary radical): Two alkyl groups provide hyperconjugation, making it
more stable.
(CH₃)₃C (Tertiary radical): Three alkyl groups provide maximum hyperconjugation
among alkyl radicals, so it is very stable.
C₆H₅CH₂ (Benzyl radical): Stability is highest because the unpaired electron is
delocalized over the benzene ring by resonance.
Final Answer:
CH₃ < (CH₃)₂CH < (CH₃)₃C < C₆H₅CH₂
2. (b) How do you explain the ortho, para-directing nature of the CH₃ group, though it
lacks an electron pair?
Simple Explanation
At first glance, this question seems confusing because the CH₃ (methyl) group has no lone
pair of electrons. Normally, groups having lone pairs donate electrons to the benzene ring
and direct new substituents to the ortho (o) and para (p) positions.
Then why does CH₃ also behave as an ortho-para directing group?
The answer lies in hyperconjugation and the +I (electron-releasing inductive) effect.
What is the +I Effect?
The carbon atoms in the methyl group slightly push electron density toward the benzene
ring through sigma (σ) bonds. This is called the positive inductive (+I) effect.
What is Hyperconjugation?
The CH bonds of the methyl group overlap with the benzene ring and donate a small
amount of electron density. This process is called hyperconjugation.
Because of hyperconjugation, the ortho and para positions become richer in electrons than
the meta position.
Diagram
CH3
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|
2 6
/ \
3 1
| |
4 5
\_________/
2 & 6 = Ortho
4 = Para
3 & 5 = Meta
The methyl group increases electron density mainly at the ortho and para positions.
When an electrophile (E⁺) attacks the ring, these positions are more attractive because they
have higher electron density.
CH3
|
Electron density
↑ ↑
Ortho Para
Therefore, substitution occurs mainly at the ortho and para positions.
Why not Meta?
The meta position does not receive as much extra electron density from hyperconjugation.
Therefore, electrophiles attack the ortho and para positions more easily.
Conclusion
Although the CH₃ group has no lone pair, it donates electrons through the +I effect and
hyperconjugation. These effects increase the electron density at the ortho and para
positions of the benzene ring. As a result, the methyl group acts as an electron-donating,
ortho-para directing group during electrophilic aromatic substitution reactions.
SECTION B
3. (a) Using 1-methylcyclohexene as an example, discuss its ozonolysis with suitable
mechanism.
(b) Acetylene forms metal acetylide but dimethylacetylene does not form such
derivatives, why ?
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Ans: 3. (a) Ozonolysis of 1-Methylcyclohexene (with Mechanism)
Ozonolysis is a very useful reaction in organic chemistry. The word ozonolysis means
breaking a double bond (C=C) with the help of ozone (O₃). You can think of ozone as a
special "chemical scissors" that cuts the double bond into two smaller parts.
What is 1-Methylcyclohexene?
1-Methylcyclohexene is a six-membered ring (cyclohexene) that contains:
One double bond (C=C).
One methyl group (CH₃) attached to carbon number 1.
A simple structure is shown below:
CH3
|
C = C
/ \
| |
| |
\ /
\___/
The double bond is the most reactive part of the molecule.
Step 1: Attack of Ozone (O₃)
When ozone is passed through 1-methylcyclohexene, it attacks the double bond because
the double bond contains electrons that easily react with ozone.
The double bond and ozone first combine to form an unstable compound called the
Molozonide (Primary Ozonide).
1-Methylcyclohexene + O3
↓
Molozonide
(unstable intermediate)
Step 2: Formation of Ozonide
The molozonide is very unstable. It quickly breaks apart and rearranges itself to form a more
stable compound called an ozonide.
Molozonide
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↓ Rearrangement
Ozonide
This ozonide still contains oxygen atoms joining the two carbon atoms.
Step 3: Hydrolysis (Work-up)
The ozonide is then treated with zinc and water (Zn/H₂O) or dimethyl sulfide (DMS).
This step breaks the ozonide completely and produces carbonyl compounds (aldehydes or
ketones).
For 1-methylcyclohexene, the ring opens because the double bond is cut.
The products are:
One ketone group (C=O) on the carbon attached to the methyl group.
One aldehyde group (CHO) on the other carbon.
The product formed is:
CH3CO(CH2)4CHO
which is called 6-oxoheptanal.
Overall Reaction
1-Methylcyclohexene
|
O3
|
Ozonide
|
Zn/H2O
↓
CH3CO(CH2)4CHO
(6-Oxoheptanal)
Why is Ozonolysis Important?
Ozonolysis helps chemists:
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Find the position of double bonds.
Prepare aldehydes and ketones.
Determine the structure of unknown compounds.
Break large molecules into smaller useful compounds.
(b) Why does Acetylene form Metal Acetylides but Dimethylacetylene does not?
This question is based on the acidic nature of alkynes.
What is Acetylene?
Acetylene (Ethyne) is the simplest alkyne.
Formula:
HC≡CH
Notice that both ends have hydrogen atoms.
These hydrogen atoms are called terminal hydrogen atoms.
What is Dimethylacetylene?
Dimethylacetylene is also known as 2-Butyne.
Its structure is:
CH3C≡CCH3
Here, both hydrogen atoms have been replaced by methyl (CH₃) groups.
There is no hydrogen attached directly to the triple-bond carbon.
Why is Acetylene Acidic?
The carbon atoms of the triple bond are sp hybridized.
An sp-hybridized carbon has 50% s-character, which holds electrons closer to the nucleus.
This makes the carbon atom more electronegative and weakens the CH bond.
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As a result, acetylene can lose a hydrogen ion (H⁺).
HC≡CH
↓ loses H⁺
HC≡C⁻
The negatively charged ion formed is called the acetylide ion.
This ion is quite stable.
Formation of Metal Acetylides
When acetylene reacts with sodium metal or ammoniacal silver nitrate, the hydrogen atom
is replaced by a metal.
Example:
HC≡CH + 2Na
↓
NaC≡CNa + H2
or
HC≡CH + AgNO3 + NH4OH
↓
AgC≡CAg
These compounds are called metal acetylides.
Why Doesn't Dimethylacetylene Form Metal Acetylides?
Dimethylacetylene has the structure:
CH3C≡CCH3
There is no terminal hydrogen attached to the triple-bond carbon.
Since there is no acidic hydrogen, it cannot lose H⁺.
Without losing H⁺, an acetylide ion cannot form.
Therefore, metal atoms cannot replace hydrogen.
Hence, dimethylacetylene does not form metal acetylides.
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Easy Comparison
Acetylene
Dimethylacetylene
HC≡CH
CH₃C≡CCH₃
Has terminal hydrogen
No terminal hydrogen
Acidic in nature
Not acidic
Forms acetylide ion
Cannot form acetylide ion
Forms metal acetylides
Does not form metal acetylides
Easy Trick to Remember
Terminal alkyne = Hydrogen at the end = Forms metal acetylide.
Internal alkyne = No end hydrogen = No metal acetylide.
Since acetylene is a terminal alkyne, it forms metal acetylides. Dimethylacetylene (2-
butyne) is an internal alkyne, so it does not form metal acetylides.
Conclusion
Ozonolysis is a reaction in which ozone breaks the carboncarbon double bond of an
alkene into carbonyl compounds (aldehydes and ketones). In 1-methylcyclohexene, ozone
first forms an unstable molozonide, then a stable ozonide, and finally, after treatment with
Zn/H₂O, the ring opens to produce 6-oxoheptanal. In the second part, acetylene forms
metal acetylides because it contains acidic terminal hydrogen atoms attached to sp-
hybridized carbon atoms. In contrast, dimethylacetylene has no terminal hydrogen, so it
cannot lose H⁺ and therefore does not form metal acetylides. These reactions are
important for identifying organic compounds and understanding the chemical behavior of
alkenes and alkynes.
4. (a) Chlorination of n-butane in the presence of light gives a mixture of 28% of 1-
chlorobutane and 72% of 2-chlorobutane while bromination gives 98% of 2-bromobutane
and 2% of 1-bromobutane, explain.
(b) Sketch the mechanism of epoxidation reaction.
Ans: 4. (a) Why does chlorination and bromination of n-butane give different products?
When n-butane (CH₃CH₂CH₂CH₃) reacts with chlorine (Cl₂) or bromine (Br₂) in the
presence of light (hν), a free radical substitution reaction takes place. In this reaction, one
hydrogen atom of butane is replaced by a chlorine or bromine atom.
Structure of n-Butane
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CH3 CH2 CH2 CH3
C1 C2 C3 C4
There are two different types of hydrogen atoms in n-butane:
Primary (1°) hydrogen: Attached to carbon atoms C1 and C4.
Secondary (2°) hydrogen: Attached to carbon atoms C2 and C3.
Primary Carbon Secondary Carbon Secondary Carbon Primary Carbon
CH3 CH2 CH2 CH3
↑ ↑ ↑ ↑
1° H 2° H 2° H 1° H
When a hydrogen is replaced:
From a primary carbon, 1-chlorobutane or 1-bromobutane is formed.
From a secondary carbon, 2-chlorobutane or 2-bromobutane is formed.
Why does chlorination give both products?
During chlorination, chlorine radicals are very reactive. They attack both primary and
secondary hydrogens quite easily.
Although secondary hydrogens form a more stable secondary free radical, chlorine is so
reactive that it is not very selective.
Therefore:
28% 1-chlorobutane
72% 2-chlorobutane
This means chlorine prefers secondary hydrogen, but it also reacts considerably with
primary hydrogen.
Why does bromination give almost only 2-bromobutane?
Bromine radicals are less reactive but much more selective than chlorine radicals.
Before bromine removes a hydrogen atom, it "chooses" the position that gives the most
stable free radical.
A secondary free radical is much more stable than a primary free radical because it is
stabilized by nearby carbon atoms.
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Therefore bromination mainly attacks secondary hydrogen.
Result:
98% 2-bromobutane
2% 1-bromobutane
Easy Comparison
Chlorination
Bromination
Very fast reaction
Slower reaction
Less selective
Highly selective
Produces both products
Produces mainly one product
72% secondary product
98% secondary product
Simple Trick to Remember
Chlorine is in a hurry, so it attacks almost everywhere.
Bromine is careful, so it attacks only the best position.
(b) Mechanism of Epoxidation Reaction
What is Epoxidation?
Epoxidation is the reaction in which an alkene (double bond) reacts with a peroxy acid
(RCO₃H) to form an epoxide.
An epoxide is a three-membered ring containing one oxygen atom.
Example:
CH2 = CH2 + RCO3H
↓
Epoxidation
O
/ \
CH2 CH2
This three-membered ring is called an epoxide or oxirane.
Mechanism of Epoxidation
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The reaction occurs in one single step (concerted mechanism).
Step 1
The oxygen atom from the peroxy acid moves towards the double bond.
At the same time:
The double bond attacks the oxygen.
The OO bond breaks.
A molecule of carboxylic acid is formed.
Everything happens simultaneously.
Simple Diagram
Alkene
CH2 = CH2
+
RCO3H
|
O
/ \
O H
↓
O
/ \
CH2 CH2
+ RCO2H
Reaction Scheme
RCO3H
CH2 = CH2 -----------> Epoxide + RCO2H
Why is this reaction important?
Epoxides are very useful compounds because they can be converted into:
Alcohols
Glycols
Medicines
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Plastics
Industrial chemicals
They are important intermediates in many organic synthesis reactions.
Characteristics of Epoxidation
It occurs with alkenes.
Uses peroxy acid (RCO₃H) such as mCPBA or peracetic acid.
Happens in one step.
No free radical is formed.
Produces a three-membered oxygen ring (epoxide).
Difference Between Chlorination and Bromination
Property
Chlorination
Bromination
Reactivity
Very high
Lower
Selectivity
Low
Very high
Main Product
72% 2-chlorobutane
98% 2-bromobutane
Reason
Attacks many hydrogens
Prefers the most stable radical
Exam Conclusion
The chlorination of n-butane produces 28% 1-chlorobutane and 72% 2-chlorobutane
because chlorine radicals are highly reactive and only moderately selective. In contrast,
bromination produces 98% 2-bromobutane and only 2% 1-bromobutane because bromine
radicals are less reactive but highly selective, favoring the formation of the more stable
secondary free radical. Epoxidation is a one-step reaction in which an alkene reacts with a
peroxy acid to form a three-membered epoxide ring, an important intermediate used in the
preparation of many valuable organic compounds.
SECTION C
5. (a) Draw energy profile diagram for S
N
1 and S
N
2 reactions.
(b) 1-chlorobut-2-ene reacts with KCN to give a mixture of isomeric products. Give the
structures and suitable mechanisms for the isomeric products.
Ans: What is SN1 Reaction?
SN1 stands for Substitution Nucleophilic Unimolecular.
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S = Substitution (one atom is replaced by another)
N = Nucleophilic (the attacking species is a nucleophile, which has extra electrons)
1 = One molecule is involved in the slow step
In the SN1 reaction, the leaving group leaves first, forming a carbocation (positively
charged carbon). After that, the nucleophile attacks the carbocation.
So, the reaction takes place in two steps.
Energy Profile Diagram of SN1
Energy
^
| TS1
| /\
| / \
| / \____
| / \ TS2
| / \ /\
| Reactants _______/ \__/ \______ Products
| Carbocation
+-------------------------------------------------------->
Reaction Progress
Explanation
Reactants are the starting substances.
The first peak (TS1) is the highest, because breaking the CX bond requires
maximum energy.
Between the two peaks, a carbocation intermediate is formed.
The second peak (TS2) represents the attack of the nucleophile.
Since there are two peaks, SN1 is a two-step reaction.
What is SN2 Reaction?
SN2 stands for Substitution Nucleophilic Bimolecular.
2 = Two species are involved in the slow step (substrate + nucleophile)
In SN2, the nucleophile attacks the carbon at the same time the leaving group leaves.
There is no carbocation intermediate.
The reaction happens in only one step.
Energy Profile Diagram of SN2
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Energy
^
| TS
| /\
| / \
| Reactants ________/ \________ Products
|
+------------------------------------------------> Reaction
Progress
Explanation
There is only one peak, which is called the Transition State (TS).
No intermediate is formed.
Bond breaking and bond formation occur simultaneously.
Therefore, SN2 is a single-step reaction.
Difference Between SN1 and SN2 Energy Diagrams
SN1
SN2
Two-step reaction
One-step reaction
Two energy peaks
One energy peak
Carbocation intermediate formed
No intermediate formed
First step is slow
Entire reaction occurs in one step
5. (b) Reaction of 1-Chlorobut-2-ene with KCN
Given Reaction
CH2ClCH=CHCH3 + KCN
|
KCN
Here,
1-Chlorobut-2-ene is an allylic halide because chlorine is attached to a carbon next
to a double bond.
KCN (Potassium Cyanide) provides the CN⁻ (cyanide ion), which acts as a
nucleophile.
Because the compound is an allylic halide, the positive charge formed during substitution
can be shared (delocalized) over two carbon atoms by resonance. Therefore, the
nucleophile can attack at two different positions, giving two isomeric products.
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Step 1: Formation of Allylic Intermediate
When chlorine leaves,
CH2ClCH=CHCH3
|
Cl leaves
CH2⁺CH=CHCH3
This allylic carbocation is resonance stabilized.
Resonance Structures
CH2⁺CH=CHCH3
↔
CH2=CHCH⁺CH3
The positive charge is shared between Carbon-1 and Carbon-3.
Step 2: Attack of CN⁻
Since the positive charge exists at two different carbons, CN⁻ can attack either carbon.
Product 1
Attack at Carbon-1
CNCH2CH=CHCH3
Name: 1-Cyanobut-2-ene
Product 2
Attack at Carbon-3
CH2=CHCH(CN)CH3
Name: 3-Cyanobut-1-ene
Mechanism
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Mechanism for Product 1
CH2ClCH=CHCH3
|
Cl leaves
CH2⁺CH=CHCH3
CN⁻ attacks C-1
↓
CNCH2CH=CHCH3
Mechanism for Product 2
CH2ClCH=CHCH3
↓
Resonance
CH2=CHCH⁺CH3
↓
CN⁻ attacks C-3
↓
CH2=CHCH(CN)CH3
Why Are Two Products Formed?
Normally, a nucleophile attacks only one carbon. However, in an allylic halide, the positive
charge formed after the leaving group departs is stabilized by resonance. This means the
positive charge is spread over two different carbon atoms, making both positions
electrophilic (electron-deficient). As a result, the cyanide ion (CN⁻) can attack either carbon,
producing two different structural isomers.
Key Points to Remember
SN1 reaction occurs in two steps, forms a carbocation, and its energy profile has
two peaks.
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SN2 reaction occurs in one step, has one transition state, and its energy profile has
one peak.
1-Chlorobut-2-ene is an allylic halide, so the intermediate is resonance stabilized.
The CN⁻ ion from KCN can attack at two different resonance positions, producing
two isomeric nitriles:
1. CNCH₂CH=CHCH₃ (1-Cyanobut-2-ene)
2. CH₂=CHCH(CN)CH₃ (3-Cyanobut-1-ene)
6. Discuss Baeyer's strain theory. How it is used to explain the reactivity of cyclopropane
and cyclobutane rings ? Also discuss its limitations.
Ans: Baeyer's Strain Theory Explained in Simple Words
Baeyer's Strain Theory was proposed by the German chemist Adolf von Baeyer in 1885. This
theory explains why some cyclic (ring-shaped) organic compounds are more stable than
others. According to Baeyer, the stability of a ring depends on the strain (stress) present in
it. If the ring is highly strained, it becomes less stable and more reactive.
1. What is Ring Strain?
Imagine you are making a circle using a flexible wire. A large circle is easy to make, but if you
try to bend the wire into a very small triangle or square, you have to force it. The wire
becomes stressed.
The same thing happens with carbon atoms in cyclic compounds.
A carbon atom with sp³ hybridization naturally prefers a bond angle of about 109.5°. This is
called the tetrahedral angle.
When carbon atoms form small rings, they cannot maintain this ideal angle. The bond
angles become smaller, creating angle strain (Baeyer strain).
2. Main Idea of Baeyer's Strain Theory
According to Baeyer:
Carbon rings are assumed to be flat (planar).
The greater the difference between the actual bond angle and 109.5°, the greater
the strain.
More strain means:
o Lower stability
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o Higher reactivity
3. Cyclopropane (Three-membered Ring)
Structure:
C
/ \
C---C
In cyclopropane:
Ring shape = Triangle
Bond angle = 60°
Normal sp³ angle = 109.5°
Difference:
109.5° − 60° = 49.5°
This is a very large difference, so cyclopropane has very high angle strain.
Why is Cyclopropane Highly Reactive?
Because the carbon atoms are forced into an unnatural arrangement.
The CC bonds become weak and bent (often called banana bonds). These weak bonds
break easily during chemical reactions.
Therefore:
Very unstable
Highly reactive
Easily undergoes ring-opening reactions
4. Cyclobutane (Four-membered Ring)
Structure:
C ----- C
| |
C ----- C
In cyclobutane:
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Ring shape = Square
Bond angle = 90°
Difference:
109.5° − 90° = 19.5°
Cyclobutane has less strain than cyclopropane, but still has considerable strain.
Why is Cyclobutane Reactive?
Although its strain is smaller than cyclopropane, the bond angles are still smaller than the
ideal tetrahedral angle.
Hence:
Less stable than larger rings
More reactive than cyclopentane and cyclohexane
Can also undergo ring-opening reactions
5. Comparison of Cyclopropane and Cyclobutane
Compound
Ring Shape
Bond Angle
Angle Strain
Stability
Reactivity
Cyclopropane
Triangle
60°
Very High
Very Low
Very High
Cyclobutane
Square
90°
Moderate
Low
High
6. Limitations of Baeyer's Strain Theory
Although Baeyer's theory was very important, later research showed that it has several
limitations.
(i) Rings are not always planar
Baeyer assumed that all cyclic compounds are flat.
Actually:
Cyclobutane is slightly folded.
Cyclopentane has an envelope shape.
Cyclohexane has a chair conformation.
These shapes reduce strain and increase stability.
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(ii) Only angle strain was considered
Baeyer discussed only bond-angle strain.
He ignored:
Torsional strain (repulsion due to eclipsed bonds)
Steric strain (crowding between atoms)
These also affect stability.
(iii) Could not explain the stability of cyclohexane
According to Baeyer:
Cyclohexane should have strain because a flat hexagon has bond angles of 120°.
But experimentally, cyclohexane is one of the most stable cyclic compounds because it
adopts a chair shape, where bond angles become almost 109.5°.
(iv) Incorrect prediction for large rings
Baeyer predicted that rings larger than six members should be very unstable.
However, many large rings (such as cyclooctane and cyclodecane) exist and are reasonably
stable because they bend and twist to reduce strain.
Conclusion
Baeyer's Strain Theory states that the stability of cyclic compounds depends on the angle
strain produced when bond angles differ from the ideal tetrahedral angle of 109.5°. Small
rings such as cyclopropane (60°) and cyclobutane (90°) experience significant angle strain,
making them less stable and more reactive, especially toward ring-opening reactions.
Although the theory successfully explains the high reactivity of these small rings, it has
important limitations because it assumes that all rings are planar and considers only angle
strain. Modern conformational analysis has shown that many cyclic compounds adopt non-
planar shapes to minimize strain, making them far more stable than Baeyer originally
predicted. Despite its limitations, Baeyer's Strain Theory remains a fundamental concept for
understanding the behavior of cyclic organic compounds.
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SECTION D
7. Predict the product/products in the following reactions :
(a)
CH3
|
_____|_____
/ \
| |
| |NH2
\ /
\___________/
HNO3 / H2SO4
↓
Product ?
Compound: 2-Methylaniline (o-Toluidine)
(b)
CH3
|
_____|_____
/ \
| |
| |
\ /
\_____ _____/
|
NH2
Br2 / FeBr3
↓
Product ?
Compound: 3-Methylaniline (m-Toluidine)
(c)
CH3
|
_____|_____
/ \
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| |
| |
\ /
\_____ _____/
|
NH2
Br2 / CCl4
↓
Product ?
Compound: 4-Methylaniline (p-Toluidine)
More Accurate Skeletal Structures
(a) o-Toluidine (2-Methylaniline)
CH3
|
/¯¯¯¯¯\
| |
| |NH2
\_____/
(b) m-Toluidine (3-Methylaniline)
CH3
|
/¯¯¯¯¯\
| |
NH2 |
\_____/
(c) p-Toluidine (4-Methylaniline)
CH3
|
/¯¯¯¯¯\
| |
| |
\_____/
|
NH2
These structures match the compounds shown in the question paper:
(a) o-Toluidine (2-Methylaniline)
(b) m-Toluidine (3-Methylaniline)
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(c) p-Toluidine (4-Methylaniline)
Ans: 7. Predict the Product(s) in the Following Reactions Simple Explanation
This question tests your understanding of electrophilic aromatic substitution (EAS) and the
directing effects of substituents already present on a benzene ring.
Before solving the reactions, remember two important rules:
1. NH₂ (Amino group) is a very strong activating group. It donates electrons to the
benzene ring, making the ring highly reactive. It directs new groups to the ortho (o)
and para (p) positions.
2. CH₃ (Methyl group) is also an activating group, but it is weaker than NH₂. It also
directs incoming groups to the ortho and para positions.
Since NH₂ is stronger than CH₃, the amino group usually controls where the new atom or
group enters the ring.
Position Names on a Benzene Ring
2 (Ortho)
|
1 -------- 3
(NH2) |
| |
6 -------- 4 (Para)
|
5 (Ortho)
Ortho (o) = Adjacent to NH₂
Meta (m) = One carbon away
Para (p) = Opposite NH₂
(a) o-Toluidine (2-Methylaniline) + HNO₃/H₂SO₄
CH3
|
/¯¯¯¯¯\
| |
| |NH2
\_____/
HNO3 / H2SO4
What happens?
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A mixture of HNO₃ and H₂SO₄ is called the nitrating mixture. It produces the nitronium ion
(NO₂⁺), which replaces one hydrogen atom on the benzene ring.
Both NH₂ and CH₃ direct substitution to ortho and para positions, but NH₂ has a much
stronger effect.
However, one ortho position is already occupied by the methyl group. Therefore, nitration
mainly occurs at the remaining available positions favored by NH₂.
Major Product
The NO₂ group enters mainly at the para position to NH₂ because it experiences less
crowding (steric hindrance).
CH3
|
/¯¯¯¯¯\
| |
NO2 NH2
\_____/
Product: 4-Nitro-2-methylaniline (Major)
Small amounts of other ortho products may also form.
(b) m-Toluidine (3-Methylaniline) + Br₂/FeBr₃
CH3
|
/¯¯¯¯¯\
| |
NH2 |
\_____/
Br2 / FeBr3
What happens?
Here, Br₂ is used with FeBr₃, which generates the electrophile Br⁺.
Normally, NH₂ is so reactive that it does not even need a catalyst, but the question
specifically includes FeBr₃, indicating bromination under catalytic conditions.
Again, NH₂ directs bromine mainly to ortho and para positions.
Major Product
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The bromine atom enters the para position to NH₂, where there is less steric hindrance.
CH3
|
/¯¯¯¯¯\
| |
NH2 Br
\_____/
Product: 4-Bromo-3-methylaniline (Major)
Minor ortho-substituted products may also be formed.
(c) p-Toluidine (4-Methylaniline) + Br₂/CCl₄
CH3
|
/¯¯¯¯¯\
| |
| |
\_____/
|
NH2
What happens?
Here bromination is carried out in CCl₄ (carbon tetrachloride), a non-polar solvent.
The amino group strongly activates the ring. Since the para position is already occupied by
the CH₃ group, bromine enters one of the ortho positions.
Both ortho positions are equivalent because the molecule is symmetrical.
Product
CH3
|
/¯¯¯¯¯\
Br |
| |
\_____/
|
NH2
Product: 2-Bromo-4-methylaniline (same as 6-bromo-4-methylaniline because both
positions are equivalent).
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Key Concepts Asked in the Question
1. Activating Groups
Groups like NH₂ and CH₃ increase the electron density of the benzene ring, making it react
faster.
2. OrthoPara Directing Groups
Both NH₂ and CH₃ direct new substituents to the ortho and para positions.
3. Stronger Directing Effect
The NH₂ group is much stronger than CH₃, so it usually decides where the incoming group
will be attached.
4. Electrophilic Aromatic Substitution (EAS)
In these reactions, an electrophile such as NO₂⁺ or Br⁺ replaces a hydrogen atom on the
benzene ring while the aromatic nature of benzene is preserved.
5. Steric Hindrance
When a position near an existing substituent is crowded, the electrophile prefers a less
crowded position. This is why para substitution is often the major product when available.
Final Answers
Part
Reagent
Major Product
(a)
HNO₃ /
H₂SO₄
4-Nitro-2-methylaniline (major; minor ortho product possible)
(b)
Br₂ / FeBr₃
4-Bromo-3-methylaniline (major; minor ortho product possible)
(c)
Br₂ / CCl₄
2-Bromo-4-methylaniline (≡ 6-bromo-4-methylaniline due to
symmetry)
Exam Tip
Whenever you see NH₂ on a benzene ring, think "strong ortho-para director." If another
group such as CH₃ is also present, first check which positions are ortho and para to NH₂,
then choose the least crowded (usually para) position as the major product. This simple
strategy helps solve most substitution questions involving aniline and toluidine derivatives.
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8.(a) Why does side-chain halogenation in alkyl benzene take place preferentially at a-
position to the aromatic ring?
(b) Explain what are anti-aromatic compounds. Give two examples.
Ans: 8.(a) Why does side-chain halogenation in alkyl benzene take place preferentially at
the α-position to the aromatic ring?
Simple Explanation
To understand this question, first we need to know what alkyl benzene is.
An alkyl benzene is a compound in which a benzene ring is attached to an alkyl group (such
as CH₃, CH₂CH₃, etc.). A common example is toluene (C₆H₅CH₃).
Diagram of Toluene
Benzene Ring
______________
/ \
| |
\______________/
|
CH3
↑
α (Alpha) Carbon
The carbon directly attached to the benzene ring is called the α (alpha) carbon or benzylic
carbon.
What is Side-Chain Halogenation?
Halogenation means replacing a hydrogen atom with a halogen atom like chlorine (Cl) or
bromine (Br).
In side-chain halogenation, the hydrogen atom is removed from the alkyl side chain, not
from the benzene ring.
Example:
C6H5CH3 + Cl2 → C6H5CH2Cl + HCl
Toluene Benzyl chloride
Here, one hydrogen from the CH₃ group is replaced by chlorine.
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Why Does Halogenation Occur at the α-Position?
The reaction mainly occurs at the α-carbon because the intermediate formed during the
reaction is very stable.
When a hydrogen atom is removed from the α-carbon, a benzyl free radical is formed.
C6H5CH3
|
Remove H
↓
C6H5CH2
This radical is called a benzyl radical.
The benzyl radical is highly stable because the unpaired electron is shared (delocalized)
over the benzene ring. This sharing of electrons is known as resonance.
Because the radical is stabilized by resonance, it forms easily, making the reaction occur
mainly at the α-position.
If hydrogen is removed from any other carbon farther away from the benzene ring, the
radical formed is not resonance-stabilized and is much less stable. Therefore, such reactions
are less favorable.
Key Points
The α-carbon is the carbon attached directly to the benzene ring.
Side-chain halogenation replaces hydrogen from the alkyl side chain.
The benzyl radical formed at the α-carbon is stabilized by resonance.
Greater stability means the reaction occurs more easily at this position.
Hence, halogenation takes place preferentially at the α-position.
8.(b) Explain Anti-Aromatic Compounds. Give Two Examples.
Simple Explanation
Many students know about aromatic compounds like benzene, which are very stable. But
there is another class called anti-aromatic compounds, which are exactly the opposite.
An anti-aromatic compound is a cyclic (ring-shaped), flat molecule that has 4n π (pi)
electrons (where n = 0, 1, 2, 3...). These compounds are less stable because their π-
electrons create extra electron repulsion instead of stability.
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Conditions for an Anti-Aromatic Compound
A compound is anti-aromatic only if all four conditions are satisfied:
1. It must be cyclic (ring-shaped).
2. It must be planar (flat).
3. It must have continuous conjugation (alternating single and double bonds).
4. It must contain 4n π electrons.
Examples of 4n π electrons:
4 electrons (n = 1)
8 electrons (n = 2)
12 electrons (n = 3)
If these conditions are fulfilled, the compound becomes anti-aromatic.
Difference Between Aromatic and Anti-Aromatic
Aromatic Compound
Anti-Aromatic Compound
Very stable
Unstable
Has (4n + 2) π electrons
Has 4n π electrons
Lower energy
Higher energy
Example: Benzene
Example: Cyclobutadiene
Example 1: Cyclobutadiene (C₄H₄)
C == C
| |
C == C
Ring-shaped ✔
Planar ✔
Conjugated ✔
Has 4 π electrons ✔
Therefore, it is anti-aromatic.
Example 2: Cyclooctatetraene (Ideal Planar Form)
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Eight-membered ring
with four double bonds
C = C C = C
| |
C = C C = C
If this molecule were perfectly planar, it would contain 8 π electrons, satisfying the 4n rule
(n = 2), so it would be anti-aromatic.
Note: In reality, cyclooctatetraene bends into a tub-shaped structure to avoid anti-
aromaticity, making it non-aromatic. However, its hypothetical planar form is often
discussed as an anti-aromatic example in textbooks.
Easy Way to Remember
Think of aromatic and anti-aromatic compounds like roads:
Aromatic compounds are like a smooth highway, where electrons move freely,
making the molecule very stable.
Anti-aromatic compounds are like a traffic jam, where electrons interfere with each
other, making the molecule unstable.
Conclusion
Side-chain halogenation in alkyl benzene occurs mainly at the α-position because the benzyl
radical formed there is stabilized by resonance with the benzene ring, making the reaction
energetically favorable. Anti-aromatic compounds are cyclic, planar, conjugated molecules
with 4n π electrons. Instead of gaining stability like aromatic compounds, they become
highly unstable due to unfavorable electron arrangement. Cyclobutadiene (C₄H₄) and the
hypothetical planar form of cyclooctatetraene (C₈H₈) are classic examples of anti-aromatic
compounds. Understanding the role of resonance, electron count, and molecular structure
makes these concepts much easier to remember for examinations.
This paper has been carefully prepared for educational purposes. If you notice any mistakes or
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