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GNDU Question Paper-2022

Bachelor of Computer Application (BCA) (Hons.)

Semester (Batch 2024-28) (CBGS)

CHEMISTRY

(Organic ChemistryI)

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.

SECTIONA

1. (a) What are carbenes and nitrenes ? Discuss different aspects that affect the stability of

carbanion.

(b) Discuss resonance and hyperconjugation effects with examples.

2. (a) What are van der Waals interactions ? Discuss with examples. How are they

different from H-bonding interactions ?

(b) What is electrophile and nucleophile ? Discuss any chemical reaction where

electrophile and nucleophile are used.

SECTIONB

3. (a) Write plausible product(s) of the following reaction with mechanism :

(i)

CH3CHCH2CH3

N+(CH3)3 Cl−

Ag2O

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───────────────→

H2O, heat

(ii)

H3CC≡CCH3

BH3

────────────→

H2O2, OH−

(iii)

H H

\ /

C==C

/ \

H3C CH3

──────────────→

(b) What is Corey-House reaction ?

4. (a) Out of cis-4-tert-butylcyclohexyl bromide and trans-4-tert-butylcyclohexyl bromide,

which will react faster with NaOH ? Support your answer with explanation and

mechanism.

(b) How will you synthesize trans-3-hexane from 3-hexyne ? Explain the mechanism.

SECTIONC

5. Discuss the mechanism of SN2 reaction with emphasis on energy profile, kinetics and

stereochemistry.

6. (a) What is Baeyers strain theory ? What are its limitations ? Discuss the conformation

of cyclopropane and cyclobutane.

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(b) Discuss nomenclature of alkyl halide with example.

SECTIOND

7. (a) Among OH, NO₂, CN, CH₃, CHO and Cl, which groups are ortho/para directing

and meta directing ? Support your answer with reasoning.

(b) Discuss chemical reaction of alkylbenzene with examples.

8. (a) What is Huckels rule ? Based on this, predict whether cycloheptatriene,

cyclopropenyl cation, cyclopentadienyl anion and cyclooctatetraene are aromatic or not.

(b) Discuss the mechanism of Friedel Craft acylation.

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GNDU Answer Paper-2022

Bachelor of Computer Application (BCA) (Hons.)

Semester (Batch 2024-28) (CBGS)

CHEMISTRY

(Organic ChemistryI)

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.

SECTIONA

1. (a) What are carbenes and nitrenes ? Discuss different aspects that affect the stability of

carbanion.

(b) Discuss resonance and hyperconjugation effects with examples.

Ans: Organic chemistry often looks difficult because it introduces many short-lived particles

and special effects that are invisible to our eyes. But if we understand them by comparing

them with real-life situations, they become much easier to remember. Let us learn each

concept one by one in simple language.

(a) Carbenes

A carbene is a neutral (uncharged) reactive species in which a carbon atom has only six

electrons in its outer shell instead of the normal eight. Since carbon wants to complete its

octet, it becomes highly reactive.

General Formula: R₂C:

Here, the symbol (:) represents two non-bonding electrons (a lone pair).

Example

CH2

This is called Methylene (:CH₂), the simplest carbene.

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Types of Carbenes

1. Singlet Carbene

o Electrons are paired.

o More reactive.

o Usually less stable.

2. Triplet Carbene

o Electrons are unpaired.

o Behaves like a free radical.

o More stable than singlet carbene.

Nitrenes

A nitrene is similar to a carbene, but instead of carbon, it contains a nitrogen atom.

General Formula: RN:

Nitrenes are also highly reactive because nitrogen has only six electrons around it.

Example

R N:

Nitrenes participate in many reactions such as insertion into CH bonds and ring formation.

Stability of Carbanion

A carbanion is a negatively charged carbon atom.

General Formula

R₃C⁻

The carbon atom has one extra pair of electrons, giving it a negative charge.

Example

CH3⁻

Since like charges repel, anything that spreads or reduces this negative charge makes the

carbanion more stable.

Factors Affecting Stability of Carbanion

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1. Electron-Withdrawing Groups (−I Effect)

Groups like NO₂, CN, COOH, Cl pull electrons away from carbon.

This reduces the negative charge on carbon and increases stability.

Example

CH2NO2⁻ > CH3⁻

The nitro group stabilizes the carbanion.

2. Electron-Donating Groups (+I Effect)

Alkyl groups donate electrons.

They increase the electron density on carbon, making the negative charge stronger.

Therefore, alkyl groups decrease stability.

Order

CH3⁻ > CH3CH2⁻ > (CH3)2CH⁻ > (CH3)3C⁻

Least Stable ----------------------> Most Unstable

Thus,

CH3⁻ is the most stable.

3. Resonance Effect

If the negative charge can spread over two or more atoms, stability increases greatly.

Example:

CH2 = CH CH2⁻

↕

CH2⁻ CH = CH2

The negative charge is shared between different atoms, so no single atom carries the full

charge.

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4. Hybridization

Greater s-character means electrons stay closer to the nucleus.

Therefore,

sp > sp² > sp³

Example

HC≡C⁻ > CH2=CH⁻ > CH3⁻

The sp-hybridized carbanion is the most stable.

(b) Resonance Effect

Sometimes a single Lewis structure cannot correctly represent a molecule.

Instead, the electrons move between atoms while the atoms themselves remain fixed.

This phenomenon is called resonance.

The actual molecule is a mixture of all possible resonance structures.

Example: Benzene

⌬

Benzene is often represented by two alternating double-bond structures.

The π-electrons are actually spread equally over all six carbon atoms.

Because of this electron delocalization, benzene is much more stable than expected.

Example: Carbonate Ion

O C O⁻

↕

O⁻ C = O

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The negative charge is shared equally among the oxygen atoms.

Importance of Resonance

Increases stability.

Lowers the energy of molecules.

Equalizes bond lengths.

Spreads the charge over several atoms.

Hyperconjugation Effect

Hyperconjugation is also called "no-bond resonance."

In this effect, electrons from a CH bond of an alkyl group move toward an adjacent double

bond or positively charged carbon.

This electron donation stabilizes the molecule.

Simple Diagram

CH3 CH = CH2

↑

CH electrons help stabilize

the double bond.

Example

(CH3)3C⁺

The tertiary carbocation is more stable because many neighboring CH bonds donate

electron density through hyperconjugation.

More alkyl groups mean more hyperconjugation and greater stability.

Difference Between Resonance and Hyperconjugation

Resonance

Hyperconjugation

Involves movement of π-electrons or lone pairs

Involves movement of σ (CH) electrons

Requires conjugated double bonds or lone pairs

Requires adjacent CH bonds

Gives strong stabilization

Gives moderate stabilization

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Forms resonance structures

Called "no-bond resonance"

Conclusion

Carbenes and nitrenes are highly reactive intermediates because they have incomplete

octets. A carbanion carries a negative charge, and its stability increases when the charge is

reduced or spread out by electron-withdrawing groups, resonance, and higher s-character

(sp hybridization). Resonance stabilizes molecules by delocalizing electrons over multiple

atoms, while hyperconjugation stabilizes molecules through the donation of electrons from

neighboring CH bonds. Understanding these concepts helps explain why some organic

molecules are more stable and react differently than others, making them fundamental

topics in organic chemistry.

2. (a) What are van der Waals interactions ? Discuss with examples. How are they

different from H-bonding interactions ?

(b) What is electrophile and nucleophile ? Discuss any chemical reaction where

electrophile and nucleophile are used.

Ans: Atoms and molecules are always attracting or repelling each other. Some attractions

are very strong, while others are weak. Van der Waals interactions are weak attractive

forces that exist between atoms or molecules, even if they are not chemically bonded

together. These forces help molecules stay close to one another and play an important role

in the properties of solids, liquids, proteins, DNA, and many biological molecules.

Imagine that two balloons are placed close together. Even without tying them, they may

slightly attract each other because of tiny electrical changes. Similarly, the electrons inside

molecules keep moving, creating temporary positive and negative regions. These temporary

charges attract nearby molecules, forming van der Waals interactions.

Examples of van der Waals interactions

Geckos climbing walls: A gecko can walk on walls because millions of tiny hairs on its

feet create van der Waals attractions with the wall surface.

Liquefied gases: Noble gases like helium and neon remain together in liquid form

only because of these weak interactions.

Iodine crystals: Iodine molecules are held together mainly by van der Waals forces.

Simple Diagram

Molecule A Molecule B

(δ+) ---- Weak Attraction ---- (δ−)

van der Waals Interaction

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Difference between van der Waals interactions and Hydrogen Bonding

Van der Waals Interactions

Hydrogen Bonding

Very weak attraction

Much stronger attraction

Present between almost all

molecules

Present only when Hydrogen is bonded with Nitrogen (N),

Oxygen (O), or Fluorine (F)

Caused by temporary

dipoles

Caused by highly polar bonds

Found in gases, liquids, and

solids

Found in water, alcohols, DNA, proteins, etc.

Hydrogen Bond Example

H O ····· H O

Hydrogen Bond

Water molecules stick together because of hydrogen bonding, which is why water has a high

boiling point and surface tension.

Conclusion:

Van der Waals interactions are weak forces that occur between all molecules, whereas

hydrogen bonding is a much stronger and more specific type of intermolecular attraction

found in molecules containing hydrogen attached to nitrogen, oxygen, or fluorine.

2. (b) What is Electrophile and Nucleophile? Discuss any chemical reaction where

electrophile and nucleophile are used.

Chemical reactions happen because some particles want to accept electrons, while others

want to donate electrons. These particles are called electrophiles and nucleophiles.

Electrophile

The word electrophile means "electron-loving." An electrophile is an atom, ion, or molecule

that accepts a pair of electrons because it is electron-deficient. Most electrophiles carry a

positive charge or have a partial positive charge.

Examples:

H⁺ (Hydrogen ion)

NO₂⁺ (Nitronium ion)

BF₃

AlCl₃

Think of an electrophile as a person who is looking for electrons because it has too few.

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Nucleophile

The word nucleophile means "nucleus-loving." A nucleophile is an atom, ion, or molecule

that donates a pair of electrons because it has extra electrons or lone pairs.

Examples:

OH⁻ (Hydroxide ion)

Cl⁻ (Chloride ion)

NH₃ (Ammonia)

H₂O (Water)

A nucleophile can be imagined as a person who has extra electrons and is ready to share

them.

Example Reaction (Nucleophilic Substitution)

One of the simplest reactions involving both electrophiles and nucleophiles is:

CH₃Br + OH⁻ → CH₃OH + Br⁻

Step-by-Step Explanation

1. CH₃Br (Methyl bromide) contains a carbon atom attached to bromine.

2. Bromine attracts electrons more strongly than carbon, making the carbon atom

slightly positive (electrophilic).

3. OH⁻ has an extra pair of electrons, so it acts as a nucleophile.

4. The hydroxide ion attacks the carbon atom and donates its electron pair.

5. Bromine leaves with the bonding electrons, forming Br⁻.

6. The final product formed is CH₃OH (Methanol).

Diagram

OH⁻

↓

CH₃ Br ---------> CH₃ OH + Br⁻

Nucleophile Product

attacks the

Electrophilic Carbon

Another Everyday Example

When hydrochloric acid (HCl) dissolves in water:

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H⁺ + H₂O → H₃O⁺

H⁺ is the electrophile because it accepts electrons.

H₂O is the nucleophile because the oxygen atom donates a lone pair of electrons.

Difference between Electrophile and Nucleophile

Electrophile

Nucleophile

Electron-loving

Electron-donating

Accepts electron pair

Donates electron pair

Electron-deficient

Electron-rich

Usually positive or partially positive

Usually negative or contains lone pairs

Examples: H⁺, NO₂⁺, BF₃

Examples: OH⁻, Cl⁻, NH₃

Conclusion

Electrophiles and nucleophiles are fundamental concepts in organic chemistry. Electrophiles

accept electrons, while nucleophiles donate electrons. Their interaction drives many

important chemical reactions, such as substitution and addition reactions. Understanding

these concepts helps explain how new chemical bonds are formed and why different

substances react with each other.

SECTIONB

3. (a) Write plausible product(s) of the following reaction with mechanism :

(i)

CH3CHCH2CH3

N+(CH3)3 Cl−

Ag2O

───────────────→

H2O, heat

(ii)

H3CC≡CCH3

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BH3

────────────→

H2O2, OH−

(iii)

H H

\ /

C==C

/ \

H3C CH3

──────────────→

(b) What is Corey-House reaction ?

Ans: (i) Hofmann Elimination

Reaction:

CH3

CH3CHCH2CH3

N+(CH3)3 Cl−

Ag2O

H2O, Heat

Step 1: Formation of Quaternary Ammonium Hydroxide

Silver oxide (Ag₂O) reacts with water and replaces Cl⁻ with OH⁻.

RN+(CH3)3 Cl−

Ag2O / H2O

↓

RN+(CH3)3 OH−

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Step 2: Elimination (Heating)

On heating, the OH⁻ removes a β-hydrogen (hydrogen on the carbon next to the carbon

attached to nitrogen). At the same time, the trimethylamine group leaves.

This forms an alkene.

Major Product

According to Hofmann Rule, the less substituted alkene is formed as the major product.

CH2=CHCH2CH3

(1-Butene)

By-products:

Trimethylamine (N(CH₃)₃)

Water

Why does this happen?

The bulky trimethylammonium group prefers removing the hydrogen from the less crowded

carbon. Therefore, 1-butene is formed instead of 2-butene.

(ii) HydroborationOxidation of Alkyne

Reaction

CH3C≡CCH3

BH3

↓

H2O2 / OH−

This reaction occurs in two steps.

Step 1: Hydroboration

BH₃ adds across the triple bond.

CH3C≡CCH3

BH3

An organoborane intermediate is formed.

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Step 2: Oxidation

Hydrogen peroxide in alkaline medium replaces boron with OH.

Initially an enol is formed.

CH3C(OH)=CHCH3

Enols are unstable.

They immediately rearrange (tautomerize) into a ketone.

Final Product

CH3COCH2CH3

(2-Butanone)

Concept

Hydroboration-oxidation converts an alkyne into a carbonyl compound.

For this symmetrical alkyne (2-butyne), the final product is 2-butanone.

(iii) Ozonolysis

Reaction

H H

\ /

C==C

/ \

CH3 CH3

This compound is cis-2-butene.

Step 1

Ozone (O₃) attacks the double bond.

Step 2

The C=C bond breaks completely.

Each carbon of the double bond becomes a carbonyl group.

CH3CH=CHCH3

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↓

CH3CHO + CH3CHO

Products

Two molecules of ethanal (acetaldehyde) are formed.

(b) What is CoreyHouse Reaction?

The CoreyHouse reaction is an important method for preparing higher alkanes by joining

two carbon chains together.

General Reaction

R2CuLi + R'X

↓

RR'

where

R₂CuLi = Lithium dialkyl cuprate (Gilman reagent)

R'X = Alkyl halide

The two alkyl groups combine to form a new carbon-carbon bond.

Example

(CH3)2CuLi + CH3CH2Br

↓

CH3CH2CH3

Product = Propane

Mechanism (Simple)

Step 1

R2CuLi

R'X

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↓

Copper attacks the alkyl halide.

↓

Step 2

R group replaces X.

↓

RR' + RCu + LiX

Important Points to Remember

Hofmann elimination produces the less substituted alkene because the bulky

leaving group favors removal of a β-hydrogen from the less crowded carbon.

Hydroborationoxidation converts an alkyne into a carbonyl compound through an

unstable enol intermediate that changes into a ketone (or aldehyde, depending on

the alkyne).

Ozonolysis breaks a carboncarbon double bond into two smaller carbonyl

compounds, making it useful for identifying the position of double bonds.

CoreyHouse reaction is a powerful carboncarbon bond-forming reaction. It uses a

Gilman reagent (R₂CuLi) and an alkyl halide (R'X) to synthesize larger alkanes.

Exam Tip

To score well, remember these key ideas:

Ag₂O/H₂O + Heat → Hofmann elimination → Less substituted alkene (Hofmann

product).

BH₃ followed by H₂O₂/OH⁻ → Hydroborationoxidation → Ketone or aldehyde.

O₃ → Ozonolysis → Splits the double bond into carbonyl compounds.

CoreyHouse reaction → Joins two carbon chains using a Gilman reagent to form a

new CC bond.

If you remember what each reagent does instead of memorizing products, these reactions

become much easier to solve in examinations.

4. (a) Out of cis-4-tert-butylcyclohexyl bromide and trans-4-tert-butylcyclohexyl bromide,

which will react faster with NaOH ? Support your answer with explanation and

mechanism.

(b) How will you synthesize trans-3-hexane from 3-hexyne ? Explain the mechanism.

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Ans: Simple Explanation

To answer this question, we first need to understand three important concepts:

1. Cyclohexane Chair Form

2. Axial and Equatorial Positions

3. SN2 Reaction Mechanism

Cyclohexane is not a flat ring. It exists in a chair-shaped structure, which is more stable. In

this chair form, every carbon has two possible positions:

Axial (vertical)

Equatorial (sideways)

Axial

C----C

/ \

C C

\ /

C----C

Equatorial

The tert-butyl (-C(CH₃)₃) group is very bulky. Because it is large, it strongly prefers the

equatorial position, where it has more space and experiences less crowding.

Step 1: Identify the Position of Bromine

Since tert-butyl stays in the equatorial position:

In cis-4-tert-butylcyclohexyl bromide, bromine is forced into the axial position.

In trans-4-tert-butylcyclohexyl bromide, bromine remains in the equatorial

position.

cis Isomer

tert-butyl → Equatorial

Br → Axial ✔

trans Isomer

tert-butyl → Equatorial

Br → Equatorial ✘

Step 2: Reaction with NaOH

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NaOH provides the OH⁻ ion, which is a strong nucleophile.

The reaction mainly follows the SN2 mechanism.

In an SN2 reaction:

The nucleophile attacks from the back side of the carbon attached to bromine.

At the same time, bromine leaves.

OH⁻ → C Br

Back-side attack

↓

C OH + Br⁻

For this backside attack to happen easily, the leaving group (Br) should be in the axial

position, where it is more exposed.

Therefore:

cis isomer (Br axial) → Backside attack is easy → Reacts faster

trans isomer (Br equatorial) → Backside attack is difficult → Reacts slower

Mechanism

Cyclohexane + OH⁻

↓ (Back-side attack)

Cyclohexane + Br⁻

Final Answer (a)

cis-4-tert-butylcyclohexyl bromide reacts faster with NaOH because bromine occupies the

axial position, allowing the OH⁻ ion to attack from the back side easily through the SN2

mechanism. In the trans isomer, bromine is equatorial, making the SN2 attack difficult and

therefore the reaction is slower.

4. (b) Synthesis of trans-3-hexene from 3-hexyne

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Understanding the Question

The compound 3-hexyne contains a triple bond (C≡C).

CH₃CH₂C≡CCH₂CH₃

The question asks us to convert this into trans-3-hexene, which contains a double bond.

CH₃CH₂CH=CHCH₂CH₃

Notice that:

Triple bond → Double bond

We must stop at the alkene stage.

The product should specifically be the trans isomer.

Reagent Used

To obtain the trans alkene, we use:

Sodium (Na) or Lithium (Li) in liquid ammonia (NH₃)

This reaction is called the Dissolving Metal Reduction.

Reaction

CH₃CH₂C≡CCH₂CH₃

Na / Liquid NH₃

↓

CH₃CH₂CH=CHCH₂CH₃

(trans)

Why is the Product Trans?

The reaction adds two hydrogen atoms to opposite sides of the triple bond.

This opposite-side addition is called anti addition.

Before

C ≡ C

↓

H added from one side

H added from opposite side

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↓

Trans Alkene

Mechanism (Simplified)

Step 1: Electron Transfer

Sodium donates one electron to the alkyne.

C≡C + e⁻

↓

Radical Anion

Step 2: Protonation

Liquid ammonia provides one hydrogen atom.

Radical Anion

↓

Vinyl Radical

Step 3: Second Electron Transfer

Another sodium atom gives one more electron.

Vinyl Radical

↓

Vinyl Anion

Step 4: Final Protonation

Another hydrogen from ammonia is added.

Vinyl Anion

↓

Trans-3-Hexene

Simple Flow Diagram

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3-Hexyne

CH₃CH₂C≡CCH₂CH₃

│

│ Na / Liquid NH₃

▼

Radical Intermediate

│

▼

Second Electron + H

│

▼

Trans-3-Hexene

CH₃CH₂CH=CHCH₂CH₃

Why Not Use Lindlar's Catalyst?

A common point of confusion is choosing the correct reagent.

Lindlar's catalyst gives the cis alkene because hydrogen atoms are added from the

same side (syn addition).

Na/Liquid NH₃ gives the trans alkene because hydrogen atoms are added from

opposite sides (anti addition).

Final Answer (b)

Trans-3-hexene is synthesized from 3-hexyne by treating it with sodium (or lithium) in

liquid ammonia. The reaction follows a dissolving metal reduction mechanism, where

electrons and protons are added step by step. Since hydrogen atoms are added from

opposite sides (anti addition), the product formed is trans-3-hexene rather than the cis

isomer. This method is highly selective and is the standard laboratory procedure for

preparing trans alkenes from alkynes.

SECTIONC

5. Discuss the mechanism of SN2 reaction with emphasis on energy profile, kinetics and

stereochemistry.

Ans: The SN2 reaction is one of the most important reactions in Organic Chemistry. SN2

stands for Substitution Nucleophilic Bimolecular reaction.

Substitution means one group replaces another.

Nucleophilic means the reaction is carried out by a nucleophile (an electron-rich

species that donates electrons).

Bimolecular means two molecules participate in the rate-determining step, so the

reaction rate depends on both.

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Think of it like this:

Imagine a person (the leaving group) is standing in front of a chair attached to a table (the

carbon atom). Another person (the nucleophile) wants that chair. Instead of waiting, the

new person pushes from the back side while the first person leaves from the front. Both

events happen at the same time. This is exactly how an SN2 reaction occurs.

General Reaction

Br + OH

−

→ CH

OH + Br

−

Here:

CH₃Br = Substrate (alkyl halide)

OH⁻ = Nucleophile

Br⁻ = Leaving group

CH₃OH = Product

Mechanism of SN2 Reaction

The SN2 reaction occurs in only one step.

Step 1: Backside Attack

The nucleophile attacks the carbon atom from the opposite side of the leaving group.

At the same time:

New bond begins to form.

Old bond begins to break.

There is no intermediate in this reaction.

Diagram

Before Reaction

CH3 --------> attacked by OH⁻ from backside

OH⁻

↑

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Transition State

/ \

OH·····

(Both bonds are partially formed)

After Reaction

CH3

+ Br⁻

The carbon is briefly connected to both the nucleophile and leaving group in the transition

state.

Energy Profile of SN2 Reaction

Since the reaction occurs in one single step, there is only one energy barrier and one

transition state.

Energy

| Transition State

| /\

| / \

|___________________/ \____________

Reactants Products

----------------------> Reaction Progress

Explanation

Reactants start at a certain energy.

Energy increases until the transition state is reached.

This point has the highest energy.

After crossing this point, energy decreases and products are formed.

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Important Points

Only one transition state.

No intermediate.

Reaction is generally fast for methyl and primary alkyl halides.

Kinetics of SN2 Reaction

Kinetics means the study of reaction speed.

The rate law for SN2 reaction is:

Rate = k [Substrate][Nucleophile]

This means the reaction depends on both:

1. Concentration of substrate

2. Concentration of nucleophile

If either concentration doubles, the reaction rate increases accordingly.

This is why SN2 is called a bimolecular reaction.

Factors Affecting the Rate

Strong nucleophile → Faster reaction

Better leaving group → Faster reaction

Less crowded carbon atom → Faster reaction

Order of reactivity:

Methyl > Primary > Secondary >> Tertiary

Tertiary alkyl halides react very slowly because bulky groups block the backside attack.

Stereochemistry of SN2 Reaction

One of the most important characteristics of SN2 is its stereochemistry.

Since the nucleophile attacks from the back side, the arrangement of atoms around the

carbon atom becomes completely inverted.

This is called the Walden Inversion or Inversion of Configuration.

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Diagram

Before Reaction

/ | \

A B D

Backside Attack

↑

Nu⁻

After Reaction

/ | \

D B A

(Configuration Inverted)

It is similar to turning an umbrella inside outthe overall shape remains, but its orientation

flips.

Characteristics of SN2 Reaction

Takes place in one step.

No intermediate is formed.

One transition state is present.

Shows backside attack.

Produces Walden inversion.

Rate depends on both substrate and nucleophile.

Favoured by methyl and primary alkyl halides.

Favoured by strong nucleophiles and good leaving groups.

Conclusion

The SN2 reaction is a single-step nucleophilic substitution reaction in which a nucleophile

attacks the carbon atom from the back side while the leaving group departs simultaneously.

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Because both events occur together, no intermediate is formed and only one transition

state appears in the energy profile. The reaction follows the rate law Rate =

k[Substrate][Nucleophile], showing that its speed depends on the concentrations of both

reactants. A unique feature of SN2 is Walden inversion, where the spatial arrangement

around the carbon atom is completely reversed after the reaction. These characteristics

make SN2 reactions easy to identify and highly important in understanding organic reaction

mechanisms.

6. (a) What is Baeyers strain theory ? What are its limitations ? Discuss the conformation

of cyclopropane and cyclobutane.

(b) Discuss nomenclature of alkyl halide with example.

Ans: Organic chemistry contains many compounds that have carbon atoms joined together

to form rings. These are called cyclic compounds. The stability of these rings depends on the

angles between the carbon atoms. To explain why some rings are more stable than others,

Adolf von Baeyer proposed the Baeyer's Strain Theory in 1885.

Baeyer's Strain Theory

According to Baeyer's Strain Theory, carbon atoms prefer a bond angle of 109.5°, which is

the normal tetrahedral angle of an sp³ hybridized carbon atom.

If the bond angle in a ring is different from 109.5°, the molecule experiences angle strain.

The greater the difference, the greater the strain, and therefore the less stable the ring

becomes.

For example:

Cyclopropane has bond angles of 60°, which is much smaller than 109.5°.

Cyclobutane has bond angles close to 90°.

Cyclopentane has bond angles around 108°, very close to the ideal angle, making it

more stable.

Diagram: Ring Angles

Cyclopropane Cyclobutane Cyclopentane

C C------C C

/ \ | | / \

C---C C------C C C

Angle = 60° Angle ≈ 90° Angle ≈108°

The closer the bond angle is to 109.5°, the lower the strain and the greater the stability.

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Limitations of Baeyer's Strain Theory

Although Baeyer's theory was very important, later research showed that it has several

limitations.

1. It assumed all ring molecules are flat (planar).

o In reality, most cyclic compounds bend or fold in space to reduce strain.

2. It considered only angle strain.

o It ignored other types of strain such as torsional strain (repulsion caused by

eclipsing bonds) and steric strain (crowding of atoms).

3. It failed to explain the stability of larger rings.

o Rings such as cyclohexane are actually very stable because they adopt non-

planar shapes.

4. It could not explain different conformations like the chair and boat forms of

cyclohexane.

Because of these limitations, modern conformational analysis provides a much better

explanation of ring stability.

Conformation of Cyclopropane

Cyclopropane consists of three carbon atoms forming a triangular ring.

Structure

/ \

C---C

Bond angle = 60°

Ideal angle = 109.5°

Since 60° is much smaller than 109.5°, cyclopropane has very high angle strain.

In addition, all the CH bonds overlap with each other (called eclipsed bonds), creating

torsional strain.

Because of these two types of strain:

Cyclopropane is highly unstable.

It is more reactive than larger cycloalkanes.

The ring remains almost planar because it is too small to bend significantly.

Conformation of Cyclobutane

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Cyclobutane contains four carbon atoms.

Planar Structure

C------C

| |

C------C

If cyclobutane remained flat:

Bond angles would be about 90°.

It would suffer from angle strain and torsional strain because the hydrogen atoms

would eclipse each other.

To reduce this strain, cyclobutane does not stay flat.

Folded (Puckered) Conformation

/ \

C C

\ /

In this folded shape:

Some eclipsing interactions are reduced.

Torsional strain decreases.

The molecule becomes more stable than the completely flat form.

However, cyclobutane still has considerable angle strain, so it is less stable than

cyclopentane and cyclohexane.

(b) Nomenclature of Alkyl Halides with Example

What are Alkyl Halides?

Alkyl halides (also called haloalkanes) are organic compounds in which one or more

hydrogen atoms of an alkane are replaced by halogen atoms such as:

Fluorine (F)

Chlorine (Cl)

Bromine (Br)

Iodine (I)

General Formula:

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R X

R = Alkyl group

X = F, Cl, Br or I

Rules for Nomenclature (IUPAC)

Rule 1: Choose the longest carbon chain

This chain becomes the parent alkane.

Rule 2: Number the carbon chain

Number from the end nearest to the halogen atom.

Rule 3: Name the halogen as a prefix

Fluoro (F)

Chloro (Cl)

Bromo (Br)

Iodo (I)

Rule 4: Write the complete name

Position + Halogen Prefix + Parent Alkane

Examples

Example 1

CH3CH2Cl

Two carbon atoms → Ethane

Chlorine on carbon 1

Name: Chloroethane

Example 2

CH3CH(Br)CH3

Three carbon atoms → Propane

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Bromine on carbon 2

Name: 2-Bromopropane

Example 3

CH3CH2CH2I

Three carbon atoms → Propane

Iodine on carbon 1

Name: 1-Iodopropane

Summary

Baeyer's Strain Theory explains that cyclic compounds become unstable when their bond

angles differ significantly from the ideal tetrahedral angle of 109.5°. Cyclopropane has

severe angle and torsional strain because of its 60° bond angles, making it highly reactive.

Cyclobutane reduces some of its strain by adopting a folded (puckered) conformation

instead of remaining flat. Although Baeyer's theory laid the foundation for understanding

ring stability, it has important limitations because it assumes all rings are planar and ignores

other types of strain. Alkyl halides are compounds in which a halogen replaces a hydrogen

atom in an alkane, and they are named by selecting the longest carbon chain, numbering it

correctly, and using prefixes such as chloro, bromo, fluoro, or iodo before the parent alkane

name. This systematic naming method makes it easy to identify the structure of the

compound.

SECTIOND

7. (a) Among OH, NO₂, CN, CH₃, CHO and Cl, which groups are ortho/para directing

and meta directing ? Support your answer with reasoning.

(b) Discuss chemical reaction of alkylbenzene with examples.

Ans: Benzene is a ring made of six carbon atoms with alternating double bonds. When one

atom or group (called a substituent) is already attached to the benzene ring, it influences

where the next incoming group will attach during an electrophilic substitution reaction.

The positions on the benzene ring are:

2 (Ortho)

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1 3 (Meta)

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6 4 (Para)

\ /

If the first substituent is at position 1:

Ortho (o-) = Positions 2 and 6

Meta (m-) = Positions 3 and 5

Para (p-) = Position 4

Ortho/Para Directing Groups

These groups increase the electron density of the benzene ring by donating electrons.

Because of this, the ortho and para positions become more attractive for the incoming

electrophile.

The ortho/para directing groups are:

OH (Hydroxyl)

CH₃ (Methyl)

Cl (Chlorine)

Although Cl withdraws electrons by induction, it has lone pairs that can donate electrons

by resonance. Therefore, it still directs new groups to the ortho and para positions, though

it slows the reaction.

Meta Directing Groups

These groups withdraw electrons from the benzene ring, making the ortho and para

positions less stable during the reaction. Therefore, the incoming group prefers the meta

position.

The meta directing groups are:

NO₂ (Nitro)

CN (Cyano)

CHO (Aldehyde)

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Electron Effect

Directing Position

Examples

Electron donating

Ortho & Para

OH, CH₃, Cl

Electron withdrawing