Author: Sarvan Kumar

Sarvan Kumar http://sarvankumar.wordpress.com My name is Sarvan Kumar, I am post graduate in chemistry (MSc). besides teaching chemistry at various schools and coaching centres I have been giving home tuitions to students for 10 years. I have helped students score good marks in chemistry not only in board examination but also JEE, NEET, SAT, IGCSE and IB examinations. Nearly 90% of my students have scored more than 95% in their CBSE board examination. Moreover they have also secured a seat in prestigious engineering and medical college. Since every student is different hence my unique style of teach

Nucleophilicity means the ability of a species to donate an electron pair to an electrophile and form a bond. In exams, students must quickly compare nucleophiles using a few core rules.

When we talk about atoms or molecules that have lone-pair electrons, sometimes we call them
bases and sometimes we call them nucleophiles (Table 9.1). What is the difference between a base
and a nucleophile?

Basicity tells us how easily a compound (base) can donate its lone pair of electrons to a proton (H⁺).
A strong base donates its electron pair more easily than a weak base.

Basicity is related to the acid dissociation constant (Ka) of its conjugate acid, which shows how easily that conjugate acid releases a proton. If the conjugate acid releases H⁺ easily, the base is weaker; if it does not release H⁺ easily, the base is stronger.

Nucleophilicity describes how easily a compound (nucleophile) can attack an electron-deficient atom and form a new bond.
It is measured by the rate constant (k), which indicates how fast the nucleophile reacts in a chemical reaction.

Note: Species with a negative charge is a stronger base and a better nucleophile than a
species that has the same attacking atom but is neutral

If the attacking atoms are the same size, stronger bases are better nucleophiles.

⁻NH₂ > HO⁻ > F⁻

If, however, the attacking atoms of the nucleophiles are very different in size, another factor
comes into play: the polarizability

Effect of Solvent on Nucleophilicity (Simple Point-wise) 🧪

Aprotic Polar Solvent
(Example: DMSO, DMF, Acetone)

Solvent molecules do not contain O–H or N–H bonds.
There is no strong hydrogen bonding with nucleophiles.
Therefore nucleophiles remain free and reactive.
Basicity and nucleophilicity follow the same order.
Stronger base → stronger nucleophile.

Example (Halide ions) : $F^- > Cl^- > Br^- > I^-$

Here I⁻ is the weakest base, so it is the poorest nucleophile in aprotic solvents.

2. Protic Polar Solvent
(Example: Water, Alcohol)

Solvent molecules contain O–H or N–H bonds.
They form hydrogen bonding with nucleophiles.
Small ions get strongly solvated (surrounded by solvent molecules).
This reduces their nucleophilicity.

Result:
Basicity and nucleophilicity order becomes opposite.

Example (Halide ions): $I^- > Br^- > Cl^- > F^-$

I⁻ is the best nucleophile in protic solvents because it is large and less solvated.

✅ Quick NEET/JEE Rule

Aprotic solvent: Basicity order = Nucleophilicity order
Protic solvent: Large atom = Better nucleophile

PROBLEM : Indicate whether each of the following solvents is protic or aprotic: a. chloroform (CHCl₃) b. diethyl ether c. acetic acid d. hexane

To classify protic vs aprotic solvents, remember the rule:

Protic solvent: contains H bonded to O, N, or F → can form hydrogen bonding and donate H⁺.
Aprotic solvent: no H attached to O, N, or F.

(a) Chloroform

Structure: CHCl₃

Hydrogen is attached to carbon, not O/N/F.
Cannot donate hydrogen for hydrogen bonding.

Result: Aprotic solvent

(b) Diethyl ether

Structure: C₂H₅–O–C₂H₅

Oxygen is present, but no O–H bond.
Hydrogens are attached only to carbon.

Result: Aprotic solvent

Structure: CH₃–COOH

Contains O–H bond in the –COOH group.
Can donate proton and form hydrogen bonding.

Result: Protic solvent

(d) Hexane

Structure: alkane

Only C–H and C–C bonds.
No O–H, N–H, or F–H bonds.

Result: Aprotic solvent

Relative Nucleophilicity toward CH₃l in Methanol

PROBLEM: Which is a stronger base: RO^–or RS^–?

b. Which is a better nucleophile in an aqueous solution?

c. Which is a better nucleophile in DMSO?

Let RO⁻ (alkoxide) and RS⁻ (thiolate) be compared.

Key idea: Basicity and nucleophilicity are not always the same, especially depending on the solvent.

(a) Which is a stronger base?

Compare conjugate acids:

RO⁻ → conjugate acid Alcohol (pKa ≈ 16)
RS⁻ → conjugate acid Thiol (pKa ≈ 10)

Lower pKa → stronger acid → weaker conjugate base.

Since thiols are more acidic than alcohols, their conjugate base RS⁻ is weaker.

Stronger base: RO⁻

(b) Better nucleophile in aqueous solution (polar protic solvent)

Water is a protic solvent, so small anions are strongly solvated by hydrogen bonding.

RO⁻ (O is smaller) → strongly solvated → nucleophilicity decreases
RS⁻ (S is larger, more polarizable) → less solvated → more reactive

Better nucleophile in water: RS⁻

(c) Better nucleophile in Dimethyl sulfoxide (polar aprotic solvent)

In polar aprotic solvents, anions are not strongly solvated, so nucleophilicity follows basicity.

Since:

RO⁻ is a stronger base than RS⁻

Better nucleophile in DMSO: RO⁻

Which is stronger nucleophile?

(a) Br⁻ or Cl⁻ in H₂O

Water strongly solvates smaller ions.

Br⁻ is larger → less solvated → more reactive.

Better nucleophile: Br⁻

b. CH₃O⁻ or CH₃OH in H₂O

Charged species are stronger nucleophiles than neutral molecules.

Methoxide ion vs Methanol

Better nucleophile: CH₃O⁻

c. CH₃O⁻ or CH₃OH in DMSO

Same logic: anion >> neutral molecule.

Better nucleophile: CH₃O⁻

(d) Br⁻ or Cl⁻ in DMSO

In aprotic solvent → nucleophilicity follows basicity.

Basicity order:
Cl⁻ > Br⁻

Better nucleophile: Cl⁻

(e) HO⁻ or NH₂⁻ in H₂O

Hydroxide ion vs Amide ion

NH₂⁻ is a much stronger base, but in water it reacts with the solvent:

NH₂⁻ + H₂O → NH₃ + HO⁻

So NH₂⁻ cannot exist effectively in water.

Better nucleophile: HO⁻

(f) I⁻ or Br⁻ in H₂O

In protic solvent → larger ion is better nucleophile.

I⁻ > Br⁻

Better nucleophile: I⁻

(g) I⁻ or Br⁻ in DMSO

Aprotic solvent → nucleophilicity follows basicity.

Basicity:
Br⁻ > I⁻

Better nucleophile: Br⁻

(h) HO⁻ or NH₂⁻ in DMSO

DMSO is aprotic → nucleophilicity follows basicity.

NH₂⁻ is stronger base.

Better nucleophile: NH₂⁻

Nucleophilicity Is Affected by Steric Effects

Basicity Comparison

tert-Butoxide (t-BuO⁻) has three methyl groups pushing electron density toward oxygen.

This increases electron density, making it a stronger base.

Stronger base: $(CH_3)_3CO^- \; > \; CH_3CH_2O^-$

Nucleophilicity Comparison

Nucleophilicity depends strongly on steric hindrance.

tert-Butoxide ion is bulky because of three CH₃ groups.
This blocks approach to the electrophilic carbon in SN2 reactions.

Ethoxide is smaller, so it can attack carbon more easily.

PROBLEM :
Rank the following species from best nucleophile to poorest nucleophile in an aqueous solution:

Key Concepts to Arrange Basic Strength (NEET/JEE)

1.Charge on Species (Most Important First)

Species with negative charge are stronger bases than neutral ones.

Example $OH^- > H_2O$ $NH_2^- > NH_3$

Basicity is a measure of how well a compound (a base) shares its lone pair with a proton. The
stronger the base, the better it shares its electrons. Basicity is measured by an equilibrium constant
(the acid dissociation constant, K_a) that indicates the tendency of the conjugate acid of the base to
lose a proton

2. Electronegativity (Across a Period)

If atoms are in the same period, basicity decreases with electronegativity.

Less electronegative atom holds negative charge poorly → stronger base.

3.Hybridization Effect

More s-character stabilizes negative charge, making the base weaker.

4.Resonance

If negative charge is delocalized, the base becomes less basic.

Example:

Acetate ion vs Ethoxide

5.Inductive Effect

Electron-withdrawing groups (–I effect) decrease basicity.

Example

6. Aromaticity

If negative charge contributes to aromatic stabilization, the base becomes weaker.

Example
Cyclopentadienyl anion is stable → weak base.

Quick NEET/JEE Order (Decision Flow)

When comparing bases, check in this order:

1️⃣ Charge
2️⃣ Resonance
3️⃣ Electronegativity
4️⃣ Hybridization
5️⃣ Inductive effect
6️⃣ Aromaticity
7️⃣ Atom size

20 NEET / JEE MCQs – Basic Strength

1

Order of basicity:

CH₃⁻ , NH₂⁻ , OH⁻ , F⁻

A. CH₃⁻ > NH₂⁻ > OH⁻ > F⁻
B. NH₂⁻ > CH₃⁻ > OH⁻ > F⁻
C. OH⁻ > NH₂⁻ > CH₃⁻ > F⁻
D. F⁻ > OH⁻ > NH₂⁻ > CH₃⁻

2

Which is the strongest base?

A. NH₃
B. CH₃NH₂
C. (CH₃)₂NH
D. C₆H₅NH₂

3

Order of basicity:

NH₃ , CH₃NH₂ , (CH₃)₂NH , (CH₃)₃N (aqueous)

A. 2° > 1° > 3° > NH₃
B. 3° > 2° > 1° > NH₃
C. NH₃ > 1° > 2° > 3°
D. 1° > 2° > 3° > NH₃

4

Which is the weakest base?

A. Pyridine
B. Aniline
C. NH₃
D. CH₃NH₂

5

Correct order of basicity:

NH₂⁻ , OH⁻ , F⁻

A. NH₂⁻ > OH⁻ > F⁻
B. F⁻ > OH⁻ > NH₂⁻
C. OH⁻ > NH₂⁻ > F⁻
D. NH₂⁻ > F⁻ > OH⁻

6

Which is the strongest base?

A. HC≡C⁻
B. CH₂=CH⁻
C. CH₃⁻
D. C₆H₅⁻

7

Correct order of basicity:

CH₃NH₂ , C₆H₅NH₂ , NH₃

A. CH₃NH₂ > NH₃ > C₆H₅NH₂
B. NH₃ > CH₃NH₂ > C₆H₅NH₂
C. C₆H₅NH₂ > CH₃NH₂ > NH₃
D. NH₃ > C₆H₅NH₂ > CH₃NH₂

8

Which compound is most basic?

A. Pyrrole
B. Pyridine
C. Aniline
D. Ammonia

9

Which is strongest base?

A. CH₃O⁻
B. C₂H₅O⁻
C. OH⁻
D. F⁻

10

Order of basicity:

sp³ N , sp² N , sp N

A. sp³ > sp² > sp
B. sp > sp² > sp³
C. sp² > sp³ > sp
D. sp³ > sp > sp²

11

Which is least basic?

A. CH₃NH₂
B. NH₃
C. Aniline
D. (CH₃)₂NH

12

Which has maximum basicity?

A. NH₂⁻
B. OH⁻
C. H₂O
D. NH₃

13

Order of basicity:

Pyrrole , Pyridine , Piperidine

A. Piperidine > Pyridine > Pyrrole
B. Pyridine > Piperidine > Pyrrole
C. Pyrrole > Pyridine > Piperidine
D. Pyridine > Pyrrole > Piperidine

14

Which is stronger base?

A. CH₃NH₂
B. C₆H₅NH₂

15

Most basic compound:

A. NH₃
B. CH₃NH₂
C. C₂H₅NH₂
D. (CH₃)₂NH

16

Order of basicity:

NH₃ , PH₃ , AsH₃

A. NH₃ > PH₃ > AsH₃
B. AsH₃ > PH₃ > NH₃
C. PH₃ > NH₃ > AsH₃
D. NH₃ > AsH₃ > PH₃

17

Which is least basic?

A. Pyrrole
B. Pyridine
C. Piperidine
D. NH₃

18

Which is strongest base?

A. NH₂⁻
B. CH₃O⁻
C. OH⁻
D. NH₃

19

Order of basicity:

Aniline , p-methoxyaniline , p-nitroaniline

A. p-OCH₃ > aniline > p-NO₂
B. aniline > p-OCH₃ > p-NO₂
C. p-NO₂ > aniline > p-OCH₃
D. p-OCH₃ > p-NO₂ > aniline

20

Which is strongest base?

A. NH₃
B. CH₃NH₂
C. (CH₃)₂NH
D. (CH₃)₃N

Answer Key

Cracking of Alkanes — JEE Advanced Points

Cracking = breaking of C–C bonds of higher alkanes to give smaller alkanes + alkenes.

Types of Cracking

(A) Thermal Cracking

High temperature (700–900°C)
Free radical mechanism
Produces more alkenes
Less selective

(B) Catalytic Cracking

450–550°C
Zeolite (acidic catalyst)
Carbocation mechanism
More branched alkanes
Higher octane number

Thermal Cracking Example (Free Radical)

Example 1:

$\text{n-Butane} \xrightarrow{800^\circ C} \text{Ethene} + \text{Ethane}$

Example 2:

$\text{n-Pentane} \rightarrow \text{Propene} + \text{Ethane}$

Major product contains more substituted alkene due to radical stability

Example 3:

$\text{n-Decane} \rightarrow \text{Pentane} + \text{Pentene}$

Catalytic Cracking Example (Carbocation Mechanism)

Occurs in presence of zeolite catalyst (acidic sites).

$\text{n-Hexane} \xrightarrow{\text{Zeolite}} \text{2-Methylpentane} + \text{Propene}$

✔ Rearrangement possible
✔ Branched products favored

Example 5:

$\text{n-Octane} \rightarrow \text{Isobutane} + \text{Butene}$

More stable tertiary carbocation pathway preferred.

Aromatisation of Alkanes

Aromatisation is the conversion of higher alkanes (C₆ or more) into aromatic hydrocarbons (benzene and its derivatives) by dehydrogenation and cyclisation at high temperature in presence of catalysts.

n-Octane (C₈) → Xylene (or ethylbenzene)

Also possible under catalytic reforming conditions.

Theoretical Point of View

There is no fixed maximum carbon limit.
Any C₆ or higher straight-chain alkane (C₆⁺) can undergo aromatisation because at least six carbons are required to form a benzene ring.

🔹 Practical / Industrial Point of View

In petroleum refining (catalytic reforming):

✔ Most effective for C₆ to C₁₀ alkanes
✔ Above C₁₀–C₁₂, cracking and side reactions increase
✔ Very long chains prefer cracking rather than clean aromatisation

So practically: $\boxed{\textbf{Industrially important range: C₆–C₁₀ (approx.)}}$

Cyclohexane undergoes multiple dehydrogenations:

$\text{Cyclohexane} \rightarrow \text{Cyclohexene} \rightarrow \text{Cyclohexadiene} \rightarrow \text{Benzene}$

👉 Total 4 molecules of H₂ released.

🔹 Key Concept

✔ First dehydrogenation, then cyclisation, then aromatic stabilization
✔ Metal catalyst helps in removal of hydrogen
✔ Final product gains extra stability due to aromatic resonance

Mechanism of isomerisation of Alkane

Isomerisation of alkane is the process in which a straight-chain alkane (n-alkane) is converted into its branched-chain isomer without changing the molecular formula.

Conditions Required

Catalyst:
- Anhydrous AlCl₃
- HF
- Pt/Al₂O₃
Temperature: 250–400°C (industrial process)
Occurs via carbocation mechanism

Mechanism (Simplified for JEE/NEET)

Formation of carbocation
Hydride shift or methyl shift
Formation of more stable branched carbocation
Deprotonation → Branched alkane

Industrial Importance

Used in petroleum refining
Improves octane number of petrol
Branched alkanes burn more smoothly (less knocking)

Example:
n-Pentane → Isopentane
n-Hexane → Isohexane

The mechanism for butane:

it probably begins by the protonation of butene impurities.

Step 1: Initiation

Propagation:

Termination:

Isomerisation of Pentane (C₅H₁₂)

During catalytic isomerisation of n-pentane (using AlCl₃/HCl or Pt/Al₂O₃), rearrangement proceeds via a carbocation intermediate.

🔹 Possible products:

2,2-Dimethylpropane (Neopentane)

2-Methylbutane (Isopentane)

Major Product → 2-Methylbutane (Isopentane)

✔ Why is it major?

Formation occurs via secondary carbocation, which is reasonably stable.
It requires only one methyl shift from n-pentane.
Formation of neopentane requires further rearrangement and is less favored.
Under equilibrium, moderately branched alkane forms in higher amount.

Isomerisation of n-Hexane — Major Product (Isohexane)

When n-hexane (C₆H₁₄) undergoes catalytic isomerisation (AlCl₃/HCl, Pt/Al₂O₃, zeolites), branched isomers are formed via carbocation rearrangement.

🔹 Possible Isohexanes (Structural Isomers)

1) 2-Methylpentane ✅ Major Product

2) 3-Methylpentane

3) 2,2-Dimethylbutane

4) 2,3-Dimethylbutane

✅ Major Isohexane → 2-Methylpentane

✔ Reason (JEE/NEET Concept):

Forms via secondary carbocation with minimal rearrangement.
Requires only one hydride/methyl shift.
Moderately branched alkane is favored kinetically.
Highly branched (like 2,2-dimethylbutane) needs further rearrangement → less formed.

Practically (Industrial Catalytic Isomerisation)

Commercially important for C₄ to C₇/C₈ alkanes
Used in petroleum refining to increase octane number.
Higher alkanes (> C₈) tend to undergo cracking along with isomerisation.

Free Radical Halogenation (Alkanes) – JEE / NEET Level

Free radical halogenation is a substitution reaction of alkanes in which a hydrogen atom is replaced by a halogen (Cl or Br) in the presence of UV light (hv) or heat.

(X = Cl or Br)

Mechanism:

Step 1

Step 2: Propagation (Chain continues)

Step 3:Termination (Radicals combine)

Order of Reactivity of Halogens

$F_2 > Cl_2 > Br_2 > I_2$

But in practice:

Fluorination → Explosive
Chlorination → Fast, less selective
Bromination → Slow, more selective
Iodination → Not feasible (endothermic)

Selectivity Rule (Very Important for JEE)

Relative Reactivity of Hydrogens

$3^\circ > 2^\circ > 1^\circ$

Because:
✔ Stability of radical
✔ Hyperconjugation
✔ Inductive effect

Energy Profile Concept

✔ Chlorination → Low activation difference
✔ Bromination → Large activation difference
👉 Bromination gives major substituted product (Hammond postulate)

JEE / NEET Special Points

✔ Reaction requires UV light
✔ Radical inhibitor (like O₂) slows reaction
✔ Tertiary hydrogen is most reactive
✔ Allylic and benzylic positions are highly reactive
✔ Reaction gives mixture of products

Allylic Halogenation (Very Important Exception)

When alkene is treated with halogen under radical conditions → substitution occurs at allylic position, not addition.

Example:
Propene + Br₂ (hv) → Allylic bromide

Best reagent: NBS (N-Bromosuccinimide)
Reason:
✔ Allylic radical is resonance stabilized
✔ More stable than normal 3° radical sometimes

Benzylic Halogenation

Example:
Toluene + Br₂ (hv) → Benzyl bromide

Reason:
✔ Benzyl radical is highly resonance stabilized
✔ Very reactive position

Reactivity order (special case): $Benzylic > Allylic > 3^\circ > 2^\circ > 1^\circ$

Energy Profile Concept (JEE Advanced Angle)

• Chlorination → Exothermic
• Bromination → Slightly exothermic
• Iodination → Endothermic

Iodination in Presence of HIO₃ (Why It Works?)

Iodination is carried out in presence of HIO₃ because it oxidizes HI to I₂, removing HI and shifting equilibrium forward, making the reaction feasible.

The Problem (Normal Iodination

$RH + I_2 \xrightarrow{hv} RI + HI$

✔ Propagation step is endothermic
✔ Reaction is reversible
✔ HI formed shifts equilibrium backward
👉 So iodination is normally very slow / does not proceed.

⚡ Role of HIO₃ (Iodic Acid)

When HIO₃ is added: $5HI + HIO_3 \rightarrow 3I_2 + 3H_2O$

What happens?

✔ HI (product) is removed
✔ Equilibrium shifts forward (Le Chatelier principle)
✔ Reaction becomes feasible

Stability of Free Radicals (Including Vinylic Radical) – JEE / NEET

Chlorination – Relative Reactivity of Different Hydrogens (JEE / NEET)

Meaning of the Ratio

3° hydrogen reacts about 5 times faster than 1° hydrogen
2° hydrogen reacts about 3.8 times faster than 1° hydrogen
1° hydrogen ≈ methyl hydrogen (taken as 1)

Why This Order?

Hydrogen abstraction forms a radical: $R-H + Cl^\bullet \rightarrow R^\bullet + HCl$

Since radical stability order is: $3^\circ > 2^\circ > 1^\circ > CH_3^\bullet$

More stable radical → lower activation energy → faster reaction.

But remember
Chlorination is less selective, so differences are moderate.

Comparison with Bromination (Very Important for JEE)

For bromination: $3^\circ : 2^\circ : 1^\circ = 1600 : 80 : 1$

👉 Bromination is highly selective.
👉 Chlorination is fast but less selective.

Numerical 1

Calculate percentage of products formed when propane undergoes chlorination.

Step 1: Identify hydrogens

Propane (CH₃–CH₂–CH₃)

6 primary H
2 secondary H

Step 2: Multiply by reactivity ratio

Primary: $6 × 1 = 6$

Secondary: $2 × 3.8 = 7.6$

Step 3: Total

$6 + 7.6 = 13.6$

Final Answer

✔ 1-chloropropane ≈ 44%
✔ 2-chloropropane ≈ 56%

Order of C–H Bond Dissociation Energy (BDE)

Bond strength decreases as radical stability increases

Question:

Ans: IV > III> 1 > II >V

Corey–House Synthesis (Gilman Coupling)

Corey–House synthesis is a method used to prepare higher alkanes by coupling an alkyl halide with a Gilman reagent (lithium dialkylcuprate, R₂CuLi).

Preparation of Gilman Reagent

Important JEE/NEET Points

✔ Best with primary alkyl halides
✔ Does not work well with tertiary halides (elimination dominates)
✔ Useful for forming unsymmetrical alkanes
✔ Milder and more selective than Wurtz reaction
✔ Works via substitution mechanism

Mechanistic Reason

Proceeds via substitution (SN2-like)
No free radical chain process
Controlled coupling

Rate of Hydrogenation of Alkene

Hydrogenation of alkene is the addition of H₂ across a C=C double bond in presence of a metal catalyst like Ni, Pd, or Pt, forming an alkane.

1. Based on Substitution of Double Bond

2.Cis vs Trans Alkenes

Reason:

Cis isomer is less stable (higher energy)
Therefore hydrogenates faster

3.Cyclic vs Acyclic

cyclopropane reacts faster due to ring strain.

4. Conjugated vs Isolated

Isolated double bond hydrogenates normally.

In conjugated systems, 1,4-addition may occur depending on conditions.

More stable alkene → Lower rate of hydrogenation why?

Hydrogenation is an exothermic reaction.

More stable alkene = lower potential energy
Therefore it releases less heat on hydrogenation
Lower energy difference → smaller driving force
Hence reacts more slowly

1. Which alkene hydrogenates fastest?

A) CH₃–CH=CH₂
B) CH₂=CH₂
C) (CH₃)₂C=CH₂
D) (CH₃)₂C=C(CH₃)₂

Answer: B

2. The rate of hydrogenation is inversely proportional to:

A) Heat of hydrogenation
B) Stability of alkene
C) Surface area of catalyst
D) Pressure of H₂

Answer: B

3. Which has lowest heat of hydrogenation?

A) 1-Butene
B) cis-2-Butene
C) trans-2-Butene
D) 2-Methyl-2-butene

Answer: D

4. Correct order of rate of hydrogenation:

A) Tetra > Tri > Di > Mono
B) Mono > Di > Tri > Tetra
C) Tri > Di > Mono > Tetra
D) Di > Mono > Tri > Tetra

Answer: B

5. Which reacts faster in hydrogenation?

A) trans-2-Butene
B) cis-2-Butene
C) Both equal
D) 2-Methyl-2-butene

Answer: B

6. Hydrogenation of alkene is:

A) Anti addition
B) Syn addition
C) Free radical reaction
D) Electrophilic addition

Answer: B

7. Which alkene is most stable?

A) CH₂=CH₂
B) CH₃–CH=CH₂
C) (CH₃)₂C=CH₂
D) (CH₃)₂C=C(CH₃)₂

Answer: D

8. Which will hydrogenate slowest?

A) 1-Butene
B) 2-Butene
C) 2-Methylpropene
D) 2,3-Dimethyl-2-butene

Answer: D

9. More stable alkene has:

A) Higher heat of hydrogenation
B) Lower heat of hydrogenation
C) Same heat of hydrogenation
D) Zero heat of hydrogenation

Answer: B

10. Strained cyclic alkene hydrogenates faster because:

A) It is more stable
B) It has lower energy
C) Relief of ring strain
D) It is more substituted

Answer: C

JEE/NEET Concept

Heat of hydrogenation order: $\text{Cyclopropene > Cyclobutene > Cyclopentene > Cyclohexene}$

Higher heat release means less stable alkene → faster hydrogenation.

MCQs

1. Which cyclic alkene hydrogenates fastest?

A) Cyclohexene
B) Cyclopentene
C) Cyclobutene
D) Cyclopropene

Answer: D

2. The fastest hydrogenation is shown by the alkene having:

A) Maximum stability
B) Maximum ring strain
C) Maximum substitution
D) Minimum heat of hydrogenation

Answer: B

3. Arrange in increasing rate of hydrogenation:

A) Cyclohexene < Cyclopentene < Cyclobutene < Cyclopropene
B) Cyclopropene < Cyclobutene < Cyclopentene < Cyclohexene
C) Cyclobutene < Cyclopropene < Cyclohexene < Cyclopentene
D) Cyclopentene < Cyclohexene < Cyclobutene < Cyclopropene

Answer: A

4. Cyclohexene hydrogenates slower than cyclobutene because:

A) It is more substituted
B) It has less ring strain
C) It has higher energy
D) It forms unstable product

Answer: B

5. Heat of hydrogenation is maximum for:

A) Cyclohexene
B) Cyclopentene
C) Cyclobutene
D) Cyclopropene

Answer: D

Numerical-Type Concept Question

The heats of hydrogenation are:

Cyclohexene → −119 kJ/mol
Cyclopentene → −120 kJ/mol
Cyclobutene → −127 kJ/mol
Cyclopropene → −151 kJ/mol

Question:

Which is most stable? Which reacts fastest?

Solution:

✔ Most stable: Cyclohexene (least heat released)
✔ Fastest reaction: Cyclopropene (highest heat released)

“Hydrogenation rate of cyclic alkenes depends on ring strain. Smaller rings possess greater angle strain and hence higher heat of hydrogenation. Therefore, they react faster.”

JEE/NEET Important Order

Rate of hydrogenation: $\textbf{Isolated diene > Conjugated diene > Aromatic compound}$

Comparison Example (Very Important for Exams)

Compare:

1,3-Butadiene (conjugated)
1,4-Pentadiene (isolated)

Stability Order:

$\textbf{Conjugated > Isolated}$

Rate of Hydrogenation:

$\textbf{Isolated > Conjugated}$

JEE/NEET Important Order

Rate of hydrogenation: $\textbf{Isolated diene > Conjugated diene > Aromatic compound}$

(Benzene hydrogenates very slowly due to aromatic stability)

Mechanism Of Wurtz reaction (Free Radical / SET Mechanism) JEE/NEET

Mechanism (Free Radical / SET Mechanism)

Intramolecular Wurtz Reaction (Cyclic Formation)

JEE / NEET Important Points

Works best with primary alkyl halides
Secondary & tertiary → elimination side reactions
Dry ether is essential
Same alkyl halide → symmetrical alkane
Different alkyl halides → mixture of products
Free radical intermediate
SET (Single Electron Transfer) mechanism
Even-numbered carbon chain formed
Not suitable for unsymmetrical alkane synthesis
Intramolecular reaction gives cyclic alkane

Important Concept

✔ Only alkali metals (Na, K, Li) give Wurtz-type coupling
✔ Sodium is preferred due to controlled reactivity
✔ Mg and Zn mainly form organometallic intermediates instead of direct alkane

Wurtz Reaction – Metals Other Than Sodium

Although sodium (Na) is the classical metal used in Wurtz reaction, some other metals can also promote coupling of alkyl halides

Potassium (K)

$2R–X + 2K \rightarrow R–R + 2KX$

✔ More reactive than sodium
✔ Reaction is more vigorous
❌ Harder to control
📌 Rarely used in practice (safety issue)

Lithium (Li)

$2R–X + 2Li \rightarrow R–R + 2LiX$

✔ Can participate in similar coupling
✔ Often forms organolithium intermediates
📌 Used more in organometallic synthesis than classical Wurtz

Silver (Ag)

$R–X + Ag \rightarrow R^\bullet$

✔ Used mainly for radical formation studies
❌ Not common for alkane synthesis
📌 Often used in rearrangement or carbocation studies

Zinc (Zn)

Zinc does not give classical Wurtz coupling easily but forms: $R–X + Zn \rightarrow R–ZnX$

✔ Forms organozinc compounds
✔ Used in coupling reactions (e.g., Reformatsky type)

Magnesium (Mg)

R–X+Mg→R–MgX

✔ Forms Grignard reagent
❌ Does NOT directly give Wurtz alkane
📌 Very important alternative pathway

Cyclic Wurtz Reaction (Intramolecular Wurtz)

Cyclic Wurtz reaction is an intramolecular version of the Wurtz reaction in which a dihaloalkane reacts with sodium metal in dry ether to form a cycloalkane.

JEE / NEET Important Points

Best for small rings (3–5 members).
Large rings are difficult due to entropy factor.
Competes with intermolecular Wurtz (polymerization possible).
Primary dihalides give better yield.
Dry ether is essential.
Mechanism involves free radicals.

Product of Tertiary Alkyl Halide in Wurtz Reaction

When a tertiary alkyl halide (3° RX) is treated with sodium (Na) in dry ether, it does not give Wurtz coupling product efficiently.

Major product = Alkene (Elimination product)
Coupling product is minor or negligible.

Example

Why Elimination Occurs?

✔ Tertiary halides easily form stable tertiary radicals / carbanions
✔ Strong base character of sodium promotes β-elimination
✔ Steric hindrance prevents effective coupling

Thus, E2-type elimination dominates over coupling.

JEE / NEET Important Points

Primary RX → best for Wurtz coupling
Secondary RX → mixture
Tertiary RX → alkene (major)
Elimination increases with substitution.
Reaction follows radical pathway but elimination competes strongly.

Reactivity Order (Bond Strength Basis)

$R–I > R–Br > R–Cl \gg R–F$

JEE / NEET Important Points

Alkyl bromides are preferred.
Primary halides give best results.
Aryl halides do NOT undergo normal Wurtz (need Wurtz–Fittig).
Fluorides are practically inactive.
Dry ether is essential for the reaction.

Reasons for Using Dry Ether

1. Provides Anhydrous Medium

✔ Sodium reacts violently with water: $2Na + 2H_2O \rightarrow 2NaOH + H_2\uparrow$

If moisture is present, sodium will react with water instead of alkyl halide.
Therefore, ether must be dry.

Why is solvent needed in Wurtz reaction?
Solvent used → Dry ether
1. Medium for Reaction
Sodium is solid
Alkyl halide is liquid
Reaction occurs at sodium surface
Ether provides: Proper contact between reactant.

2. Uniform reaction medium

Without solvent → reaction is uncontrolled / incomplete.
3. Stabilizes Reactive Intermediates
Wurtz reaction proceeds via:
Radical pathway
or Organosodium intermediate (R–Na)
Dry ether:

MCQ Questions for practice:

In the reaction of 1-bromo-3-chlorocyclobutane with two equivalents of sodium in ether, the major product is

Effect of Solvent on Nucleophilicity (Simple Point-wise) 🧪

Relative Nucleophilicity toward CH3l in Methanol

(a) Which is a stronger base?

(b) Better nucleophile in aqueous solution (polar protic solvent)

(c) Better nucleophile in Dimethyl sulfoxide (polar aprotic solvent)

Which is stronger nucleophile?

Nucleophilicity Is Affected by Steric Effects

1.Charge on Species (Most Important First)

2. Electronegativity (Across a Period)

3.Hybridization Effect

4.Resonance

5.Inductive Effect

6. Aromaticity

Quick NEET/JEE Order (Decision Flow)

20 NEET / JEE MCQs – Basic Strength

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

16

17

18

19

20

Answer Key

Types of Cracking

(A) Thermal Cracking

(B) Catalytic Cracking

Thermal Cracking Example (Free Radical)

Example 1:

Example 2:

Example 3:

Catalytic Cracking Example (Carbocation Mechanism)

Example 5:

Theoretical Point of View

🔹 Practical / Industrial Point of View

🔹 Key Concept

Conditions Required

Mechanism (Simplified for JEE/NEET)

Industrial Importance

The mechanism for butane:

Isomerisation of Pentane (C₅H₁₂)

Major Product → 2-Methylbutane (Isopentane)

Isomerisation of n-Hexane — Major Product (Isohexane)

Practically (Industrial Catalytic Isomerisation)

Mechanism:

Order of Reactivity of Halogens

Selectivity Rule (Very Important for JEE)

Allylic Halogenation (Very Important Exception)

Benzylic Halogenation

Energy Profile Concept (JEE Advanced Angle)

Iodination in Presence of HIO₃ (Why It Works?)

The Problem (Normal Iodination

⚡ Role of HIO₃ (Iodic Acid)

What happens?

Meaning of the Ratio

Why This Order?

Comparison with Bromination (Very Important for JEE)

Numerical 1

Calculate percentage of products formed when propane undergoes chlorination.

Step 1: Identify hydrogens

Step 2: Multiply by reactivity ratio

Step 3: Total

Preparation of Gilman Reagent

Important JEE/NEET Points

Mechanistic Reason

1. Based on Substitution of Double Bond

2.Cis vs Trans Alkenes

Reason:

Relative Nucleophilicity toward CH₃l in Methanol