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

R and S Configuration (JEE/NEET Concept – Easy Explanation)

R/S configuration is used to describe the absolute configuration of a chiral carbon (asymmetric carbon with 4 different groups)

Step-by-Step Rules (Cahn–Ingold–Prelog Rules)

Step 1: Assign Priority (1 → 4)

Higher atomic number = higher priority
Example:
I > Br > Cl > S > O > N > C > H

If same atom, move to next atoms (tie-breaker rule)

Step 2: Orient the Molecule

Keep the lowest priority group (4) away from you (behind the plane)

Step 3: Trace Path (1 → 2 → 3)

Clockwise direction → R (Rectus)
Anticlockwise direction → S (Sinister)

Important Shortcut

If the lowest priority (4) is towards you (front):

Reverse the result
- Clockwise → S
- Anticlockwise → R

Key Points for Exams

R/S is about 3D arrangement, not optical rotation (+/−)
Always check lowest priority position
Double/triple bonds are treated as multiple single bonds

Quick Trick (Exam Speed)

“4 behind → direct result“4 in front → reverse result”

The R and S configuration for each stereogenic centre in this from top to bottom is:

A.R, R, R
B.R, S, S
C.R, S, R
D.S, S, R

The correct option is (C)

The R and S Configuration Practice Problems

(i)

Groups: OH, CHO, CH₂OH, H
Priority: OH (1) > CHO (2) > CH₂OH (3) > H (4)

H is horizontal (front) → ❗reverse result
1 → 2 → 3 = anticlockwise → reversed → R

✔️ Answer: (i) → R

(ii)

Groups: OH, COOH, CH₃, H
Priority: OH (1) > COOH (2) > CH₃ (3) > H (4)

H is horizontal (front) → ❗reverse
1 → 2 → 3 = clockwise → reversed → S

✔️ Answer: (ii) → S

(iii)

Groups: NH₂, COOH, CH₃, H
Priority: NH₂ (1) > COOH (2) > CH₃ (3) > H (4)

H is horizontal (front) → ❗reverse
1 → 2 → 3 = anticlockwise → reversed → R

✔️ Answer: (iii) → R

(iv) (Two chiral centers)

🔹 Upper carbon

Priority: NH₂ (1) > COOH (2) > lower C (3) > H (4)

H is horizontal → reverse
1 → 2 → 3 = anticlockwise → reversed → R

🔹 Lower carbon

Priority: OH (1) > upper C (2) > CH₃ (3) > H (4)

H is horizontal → reverse
1 → 2 → 3 = clockwise → reversed → S

✔️ Answer: (iv) → (R, S)

Final Answers:

(i) → R
(ii) → S
(iii) → R
(iv) → (R, S)

Position of Elements on Lothar Meyer Curve (JEE/NEET)

Lothar Meyer’s Contribution to Periodic Table (JEE/NEET Concepts)

Julius Lothar Meyer (1869) was a German chemist who independently worked on classification of elements, around the same time as Dmitri Mendeleev.

Key Idea of Lothar Meyer

He arranged elements based on:

Atomic mass
Valency (combining capacity)
Physical properties

Most Important Contribution

Atomic Volume vs Atomic Mass Graph

He plotted a graph between:

Atomic Volume (volume occupied by 1 mole of element)
Atomic Mass

This graph showed a periodic pattern, proving that properties repeat regularly.

On the atomic volume vs atomic mass curve given by
Julius Lothar Meyer, elements occupy characteristic positions depending on their properties.

1. Peaks (Maxima) → Highly Reactive Metals

Elements at the top of the curve

Alkali metals:
- Li, Na, K, Rb, Cs

✔ Features:

Maximum atomic volume
Highly reactive
Valency = +1

2. Just After Peaks → Alkaline Earth Metals

Slightly lower than peaks

Be, Mg, Ca, Sr, Ba

✔ Features:

Less reactive than alkali metals
Valency = +2

3. Descending Region → Transition Elements

Moving downward from peak

Fe, Co, Ni, etc.

✔ Features:

Moderate atomic volume
Variable valency
Metallic character

4. Troughs (Minima) → Least Reactive Elements

Bottom of the curve

Elements like:
- Cu, Zn (near minima region)

✔ Features:

Minimum atomic volume
Stable nature

5. Rising Region → Non-metals

Moving upward after minima

C, N, O, F

✔ Features:

Increasing electronegativity
Non-metallic character increases

Newlands’ Law of Octaves (JEE/NEET Concepts)

Proposed by John Alexander Reina Newlands (1864), this was another early attempt to classify elements.

🔹 What is Newlands’ Law of Octaves?

Statement:
When elements are arranged in increasing order of atomic mass,
every 8th element has properties similar to the first, just like musical octaves 🎵

Why “Octaves”?

Inspired by music:

In music, the 8th note repeats
Similarly, in elements:
- 1st ≈ 8th
- 2nd ≈ 9th
- and so on

🔹 Example (Important for Exams)

Element No.	Element	Similar To
1	Li	Na (8th)
2	Be	Mg
3	B	Al
4	C	Si
5	N	P
6	O	S
7	F	Cl

Example:

Li and Na → similar properties
Be and Mg → similar properties

Key Features

First to show periodicity in a systematic way
Applicable mainly to lighter elements (up to Ca)
Based on atomic mass

🔹 Limitations (VERY IMPORTANT )

Worked only up to Calcium (Ca)
No space for new elements
Dissimilar elements grouped together:
- Co and Ni placed with F, Cl
Ignored transition elements
Assumed only 56 elements existed

🔹 Why Important?

First clear attempt to show periodic repetition
Helped in development of:
- Dmitri Mendeleev Periodic Table

Dobereiner’s Triads (JEE/NEET Concepts)

Johann Wolfgang Döbereiner (1829) proposed one of the earliest attempts to classify elements based on their properties.

What are Dobereiner’s Triads?

A triad is a group of three elements having:

Similar chemical properties
Gradual change in physical properties
A special relationship in atomic masses

Key Rule (Important for exams):
The atomic mass of the middle element ≈ average of the other two

Examples of Dobereiner’s Triads

1. Alkali Metal Triad

Lithium (Li) = 7
Sodium (Na) = 23
Potassium (K) = 39

✔ Check: $\frac{7 + 39}{2} = 23 \approx \text{Na}$ 27+39=23≈Na

2. Alkaline Earth Metal Triad

Calcium (Ca) = 40
Strontium (Sr) = 88
Barium (Ba) = 137

✔ Check: $\frac{40 + 137}{2} = 88.5 \approx \text{Sr}$

3. Halogen Triad

Chlorine (Cl) = 35.5
Bromine (Br) = 80
Iodine (I) = 127

✔ Check: $\frac{35.5 + 127}{2} = 81.25 \approx \text{Br}$

🔹 Important Characteristics

Properties change gradually across the triad
Middle element shows intermediate behavior
First hint toward periodicity in elements

🔹 Limitations (VERY IMPORTANT for NEET/JEE)

Only a few elements could be grouped
Many known elements did not fit into triads
Did not explain why periodicity occurs

Conformations of Propane (JEE/NEET Concepts)

Propane (C₃H₈) shows conformational isomerism due to free rotation around the C–C single bond (σ-bond).

Key Idea

Rotation is considered around the C₁–C₂ bond (or C₂–C₃; both are equivalent).

Use Newman projection to understand conformations.

Types of Conformations in Propane

1. Staggered Conformation (Most Stable)

Hydrogen atoms on front and back carbon are as far apart as possible
Dihedral angle = 60°
Minimum torsional strain

This is the lowest energy conformation

2. Eclipsed Conformation (Least Stable)

Hydrogen atoms on front and back carbon overlap each other
Dihedral angle = 0°
Maximum torsional strain

This is the highest energy conformation

Important Difference from Ethane

In propane, one carbon has CH₃ group, so:
- There are H–H eclipsing interactions
- And also H–CH₃ eclipsing interactions (more repulsion)

Hence, eclipsed propane is slightly more unstable than ethane eclipsed

Energy Profile of Propane Rotation

Energy varies as bond rotates:
- Maxima → Eclipsed
- Minima → Staggered

There are:

3 staggered conformations (all equivalent)
3 eclipsed conformations (all equivalent)

JEE/NEET Important Points

Only two types → staggered & eclipsed
Staggered = most stable
Eclipsed = least stable
Rotation barrier ≈ 14 kJ/mol (slightly higher than ethane)
Stability depends on torsional strain

Quick Memory Trick

“Spread = Stable, Clash = Unstable”

Why is propane eclipsed more unstable than ethane?”
Answer: Because of H–CH₃ repulsion > H–H repulsion

Hybridisation in Borazine (Inorganic Benzene)

Structure Overview

Ring: B–N–B–N–B–N (6-membered ring)
Similar to Benzene
Planar structure

Hybridisation of Atoms

1. Boron (B)

Hybridisation = sp²
Forms:
- 2 σ bonds with N
- 1 σ bond with H
Has empty p-orbital

2. Nitrogen (N)

Hybridisation = sp²
Forms:
- 2 σ bonds with B
- 1 σ bond with H
Has lone pair in p-orbital (participates in π bonding)

π-Bonding (Important Concept)

p-orbitals of B and N overlap → π system
But unlike benzene:
- π electrons are not equally distributed
- Due to electronegativity difference (B < N)

So, borazine is:

Partially aromatic
Less stable than benzene

Key Differences from Benzene

Property	Borazine	Benzene
Hybridisation	All atoms sp²	All C sp²
Bond nature	Polar (B–N)	Nonpolar
Aromaticity	Weak	Strong

Exam Points (JEE/NEET)

✔ Both B and N are sp² hybridised
✔ Structure is planar hexagon
✔ Contains delocalised π electrons
✔ Called “inorganic benzene” due to similarity

Conformations of Butane (JEE/NEET Concepts)

Types of Conformations of Butane

1. Anti Conformation (Most Stable)

Dihedral angle = 180°
Two CH₃ groups are opposite
Minimum steric hindrance

Stability: Highest

✔ No torsional strain
✔ No steric crowding

2. Gauche Conformation

Dihedral angle = 60°
CH₃ groups are close but not eclipsed

Stability: Moderate

✔ Some steric hindrance
✔ Still relatively stable

3. Eclipsed Conformations (Unstable)

(a) CH₃–H Eclipsed

Dihedral angle = 120° or 240°
CH₃ eclipses H

Stability: Low

✔ Torsional strain present

(b) Fully Eclipsed (CH₃–CH₃)

Dihedral angle = 0°
CH₃ groups directly overlap

Stability: Lowest

❌ Maximum torsional strain
❌ Maximum steric repulsion

Energy Order (Important for Exams)

$\text{Anti} < \text{Gauche} < \text{Eclipsed (CH₃–H)} < \text{Fully Eclipsed (CH₃–CH₃)}$

Stereoisomerism organic chemistry JEE/NEET Concepts

What is Stereoisomerism?

Compounds with:

Same molecular formula
Same connectivity
Different 3D arrangement

Types of Stereoisomerism

(A) Geometrical Isomerism

(B) Optical Isomerism

Geometrical Isomerism (cis–trans / E–Z)

Condition:

Restricted rotation (C=C, ring)
Each carbon must have two different groups

Types:

cis–trans (simple case)

Same groups → cis
Opposite → trans

Example:

cis-2-butene vs trans-2-butene

E–Z system (priority based)

Use Cahn–Ingold–Prelog priority rules

Higher priority same side → Z
Opposite side → E

Priority Rules

Based on Cahn–Ingold–Prelog priority rules

When to use E–Z?

Use E–Z when:

Alkene has 4 different groups
cis–trans is not applicable

Priority Rules (STEP-BY-STEP)

Rule 1: Atomic Number

Higher atomic number = higher priority

Example:

I > Br > Cl > F
O > N > C > H

Rule 2: If same atom → go to next atom

Compare atoms attached to that atom

Example:

–CH₃ vs –CH₂CH₃
\ Compare next atoms → ethyl gets higher priority

Rule 3: Multiple Bonds

Treat as if atom is bonded to duplicate atoms

Example:

C=O → C attached to O, O
C≡N → C attached to N, N, N

Rule 4: Isotopes

Higher mass number = higher priority
Example:

D > H

MCQs (E–Z Configuration – Advanced)

Q1.

The correct configuration of the alkene is:

CH₃–CH = C(Cl)–CH₂CH₃

(A) E
(B) Z
(C) Cannot be determined
(D) No geometrical isomerism

Q2.

Which of the following has Z-configuration?

(A) CH₃–CH=CH–Br
(B) CH₃–CH=CH–CH₃
(C) CH₃–CH=CH–Cl
(D) CH₃–CH=CH–F

Q3.

The compound having E-configuration is:

(A) CH₃–CH=CH–NO₂
(B) CH₃–CH=CH–CN
(C) CH₃–CH=CH–CHO
(D) CH₃–CH=CH–CH₃

Q4.

Consider:

CH₃–CH = C(CH₃)–CHO

Its configuration is:

(A) E
(B) Z
(C) Both possible
(D) No GI

Q5.

Which has highest priority group correctly identified?

(A) –OH > –NH₂
(B) –CH₃ > –Cl
(C) –Br > –I
(D) –H > –D

Q6.

In the compound:

CH₃–C(Br)=C(Cl)–CH₂OH

The configuration is:

(A) E
(B) Z
(C) No GI
(D) Cannot decide

Q7.

Which shows no geometrical isomerism?

(A) CH₃–CH=CH–CH₃
(B) CH₂=CH–CH₃
(C) CHCl=CHCl
(D) CH₃–CH=CH–Cl

Q8.

In:

CH₃–CH = C(CN)–COOH

Configuration is:

(A) E
(B) Z
(C) Both
(D) None

Q9.

Priority order is correct for:

(A) –COOH > –CHO
(B) –CN > –COOH
(C) –CH₂OH > –CH₃
(D) –NH₂ > –OH

Q10.

In:

CH₃–CH = C(Br)–C≡N

Configuration is:

(A) E
(B) Z
(C) No GI
(D) Cannot be determined

Here are the correct options only

Optical Isomerism

Based on Chirality

A molecule is chiral if:

It is non-superimposable mirror image

Chiral Carbon (asymmetric carbon)

Attached to 4 different groups

Types:

(1) Enantiomers

Mirror images
Rotate plane polarized light in opposite directions

(2) Diastereomers

Not mirror images
Different physical properties

Meso Compounds

Have chiral centers but optically inactive
Due to internal plane of symmetry

Optical Activity

d(+) → dextrorotatory

l(–) → levorotatory

Optical Activity of Diastereomers

Case 1: Optically Active

If molecule:

Has chiral center(s)
No plane of symmetry

✔ Then diastereomer rotates plane polarized light

Case 2: Optically Inactive

If molecule:

Has internal plane of symmetry (meso form)

✔ Then diastereomer is optically inactive

Example

Tartaric acid:

(R,R) and (S,S) → Enantiomers → optically active
(R,S) → Meso form → optically inactive

here:

(R,R) vs (R,S) = Diastereomers
But:
- (R,R) → active
- (R,S) → inactive

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Final Answer

Enantiomers:

(i) & (ii)

Meso:

(iii)

Diastereomers:

(i) & (iii)
(ii) & (iii)
(i) & (iv)
(ii) & (iv)
(iii) & (iv)

Key Recall

(i) & (ii) → enantiomers → both optically active
(iii) → meso → optically inactive
(iv) → optically active

Diastereomer Pairs Analysis

(i) & (iii)

(i) → active
(iii) → inactive (meso)
Pair: one active + one inactive

(ii) & (iii)

(ii) → active
(iii) → inactive
Pair: one active + one inactive

(iii) & (iv)

(iii) → inactive
(iv) → active
Pair: one active + one inactive

(i) & (iv)

(i) → active
(iv) → active
Pair: both optically active

(ii) & (iv)

(ii) → active
(iv) → active
Pair: both optically active

3. Effect on number of stereoisomers

General:

$Maximum = 2^{n}$

With plane of symmetry:

Some structures become identical (meso) $Actual = 2^{n} - (meso forms)$

How to find the plane of symmetry in biphenyls and substituted cyclohexane?

Optical activity is shown by compounds in which plane of symmetry is not present, if plane of symmetry is present in any compound then they will never show optical activity.

Optical activity is a kind of property of any compound which rotates the plane polarized light to the right or left side from the direction of upcoming light.
In biphenyls two benzene rings are present and if a plane of symmetry is present within the molecule then they don’t show optical activity.

This molecule does not have a plane of symmetry

Biphenyl rotation:

Structure: Biphenyl has two benzene rings connected by a single C–C bond.
- So, in principle, it’s a single bond, which allows rotation.
Rotation limitation:
- Each benzene ring is planar.
- If there are no substituents at ortho positions, the two rings can rotate freely at room temperature.
- If there are bulky groups at the ortho positions (next to the connecting bond), steric hindrance prevents free rotation, and the molecule may adopt a twisted conformation.
- Extremely hindered biphenyls can even be atropisomers (stable enantiomers due to restricted rotation)

. It is chiral if the biphenyl can’t rotate freely (restricted rotation around the central C–C bond) and may exist as atropisomers.

Position matters most

Only ortho substituents (2,2’ positions on biphenyl) cause steric hindrance.
Meta and para positions do not block rotation.

Organic Reagents MCQs (NEET/JEE Practice) set 1/level NCERT

This question set is designed to strengthen key concepts of Organic Chemistry, especially reactions involving amines, diazonium salts, and important named reactions frequently asked in NEET and JEE. It covers a wide range of reagent-based transformations to help students improve accuracy and reaction recognition skills. Practice these questions regularly to build a strong conceptual foundation and boost exam confidence

1.

Nitrile on reduction gives:
A) Alcohol
B) Aldehyde
C) Primary amine
D) Secondary amine

2.

Amide + Br₂/NaOH gives:
A) Same carbon amine
B) One carbon higher amine
C) One carbon lower amine
D) Alcohol

3.

Which reagent converts alkyl halide to nitrile?
A) KOH
B) AgCN
C) KCN
D) NH₃

4.

Reduction of isocyanide gives:
A) Primary amine
B) Secondary amine
C) Tertiary amine
D) Alcohol

5.

Nitrile hydrolysis produces:
A) Alcohol
B) Aldehyde
C) Carboxylic acid
D) Amine

6.

Carboxylic acid reacts with NH₃ to give:
A) Ester
B) Amide
C) Amine
D) Nitrile

7.

Primary amine + HNO₂ gives:
A) Alkene
B) Alcohol
C) Nitrile
D) Amide

8.

Aniline + HNO₂ (0–5°C) forms:
A) Nitrobenzene
B) Diazonium salt
C) Phenol
D) Benzene

9.

C₆H₅N₂⁺Cl⁻ + KI gives:
A) Chlorobenzene
B) Bromobenzene
C) Iodobenzene
D) Fluorobenzene

10.

Sandmeyer reaction involves:
A) HNO₃
B) Cu salts
C) KMnO₄
D) NaBH₄

11.

C₆H₅N₂Cl + H₂O (warm) gives:
A) Benzene
B) Phenol
C) Aniline
D) Nitrobenzene

12.

C₆H₅N₂Cl + H₃PO₂ gives:
A) Phenol
B) Benzene
C) Aniline
D) Chlorobenzene

13.

C₆H₅N₂Cl + HBF₄ / heat gives:
A) Chlorobenzene
B) Bromobenzene
C) Fluorobenzene
D) Iodobenzene

14.

LiAlH₄ reduces CH₃COOH to:
A) Aldehyde
B) Primary alcohol
C) Ketone
D) Secondary alcohol

15.

Aldehyde reduction gives:
A) Secondary alcohol
B) Primary alcohol
C) Ketone
D) Acid

16.

Ketone reduction gives:
A) Primary alcohol
B) Secondary alcohol
C) Tertiary alcohol
D) Acid

17.

Carbylamine test is given by:
A) Primary amine
B) Secondary amine
C) Tertiary amine
D) Amide

18.

Aniline + Br₂/H₂O gives:
A) Monobromo
B) Dibromo
C) 2,4,6-tribromoaniline
D) No reaction

19.

Aniline nitration gives mainly:
A) Meta product
B) Ortho + para + some aniline
C) Only para
D) Only ortho

20.

Aniline + acetic anhydride forms:
A) Benzamide
B) Acetanilide
C) Nitrobenzene
D) Phenol

21.

Acetanilide + Br₂/CH₃COOH gives:
A) Ortho product
B) Para product
C) Meta product
D) Mixture

22.

Gabriel phthalimide synthesis gives:
A) Primary amine
B) Secondary amine
C) Tertiary amine
D) Amide

23.

Nitrile partial hydrolysis gives:
A) Acid
B) Amide
C) Amine
D) Alcohol

24.

Aniline + conc. H₂SO₄ gives:
A) Nitrobenzene
B) Sulphanilic acid
C) Phenol
D) Benzene

25.

Coupling reaction of diazonium salt gives:
A) Alcohol
B) Azo dye
C) Acid
D) Alkene

Answers:

1-C, 2-C, 3-C, 4-B, 5-C, 6-B, 7-B, 8-B, 9-C, 10-B, 11-B, 12-B, 13-C, 14-B, 15-B, 16-B, 17-A, 18-C, 19-B, 20-B, 21-B, 22-A, 23-B, 24-B, 25-B

Predicting the Products of the Reaction of an Alkyl Halide with a Nucleophile/Base

Primary and secondary alkyl halides can undergo SN2 and E2 reactions.

Tertiary alkyl halides generally undergo E2 reactions or SN1 and E1 reactions.

Tertiary alkyl halides do not undergo SN2 reactions because of steric hindrance.

If the reagent is a good nucleophile and a strong base, a tertiary alkyl halide undergoes an E2 reaction.

If the weakly basic solvent is the only nucleophile/base, the tertiary alkyl halide undergoes SN1 and E1 reactions.

Because a primary alkyl halide is the most reactive in an SN2 reaction, substitution will
win the competition.

Case 1 : SN2/E2 Reactions of Primary Alkyl Halides

Problem: Bromopropane + CH₃O⁻ (methoxide ion)

First identify the substrate. Bromopropane (1-bromopropane) is a primary alkyl halide, and CH₃O⁻ is a strong nucleophile and strong base.

Because the substrate is primary, SN2 reaction dominates over E2.

Reaction

CH₃CH₂CH₂Br + CH₃O⁻ → CH₃CH₂CH₂OCH₃ + Br⁻

Major Product

Methoxypropane (methyl propyl ether)

Structure:
CH₃CH₂CH₂OCH₃ (90%)

Minor Product (possible but small amount)

E2 elimination → Propene (CH₂=CHCH₃) (10%)

Steric Hindrance Favors the Elimination Product (E2.)

Case II : SN2/E2 Reactions of Primary Alkyl Halides

The stronger and bulkier the nucleophile/base, the greater the percentage of the elimination product.

The weaker and less bulky the nucleophile/base, the greater the percentage of the substitution product

eaction:
2-Chloropropane + Acetate ion (CH₃COO⁻)

Substrate: Secondary alkyl halide
Reagent: Acetate ion → good nucleophile but weak base

Reaction Type

Because acetate ion is a good nucleophile and weak base, the reaction mainly proceeds through substitution (SN2) rather than elimination.

Reaction

(CH₃)₂CHCl + CH₃COO⁻ → (CH₃)₂CH–OCOCH₃ + Cl⁻

Major Product

Isopropyl acetate (propan-2-yl acetate)

Structure:
CH₃–CH(OCOCH₃)–CH₃

Reason

Secondary alkyl halide

Weak base → elimination not favored
Good nucleophile → substitution product dominates

Major product: Isopropyl acetate (100%)

2-Chloropropane + Ethoxide ion (C₂H₅O⁻)

Substrate

2-Chloropropane → secondary alkyl halide

Reagent

Ethoxide ion (C₂H₅O⁻) → strong base and good nucleophile

Reaction Type

With a secondary alkyl halide and a strong base, E2 elimination is the major reaction.

Reaction

(CH₃)₂CHCl + C₂H₅O⁻ → CH₃–CH=CH₂ + C₂H₅OH + Cl⁻

Major Product

Propene

Structure:
CH₃–CH=CH₂

Reason

Secondary substrate
Strong base present
Elimination favored over substitution

Minor Product (possible)

Isopropyl ethyl ether via SN2.

Major product: Propene (E2 elimination) (75%)

Minor Product

Isopropyl ethyl ether
(IUPAC: 2-ethoxypropane) (25%)

A Bulky Base Favors the Elimination Product

Bulky bases such as DBN and DBU are strong bases but poor nucleophiles because their bulky structure makes it difficult to attack the carbon atom.

When a secondary (2°) alkyl halide reacts with these bulky bases:

SN2 substitution is hindered due to steric bulk around the base.
The base instead abstracts a β-hydrogen from the alkyl halide.
This leads to E2 elimination rather than substitution.
Therefore, alkene formation becomes the major product.
In many cases, bulky bases may favor the Hofmann (less substituted) alkene.

Conclusion:
With 2° alkyl halides, bulky bases like DBN and DBU mainly give E2 elimination products (alkenes) rather than substitution products.

The full forms are:

DBN
DBN = 1,5-Diazabicyclo[4.3.0]non-5-ene
DBU
DBU = 1,8-Diazabicyclo[5.4.0]undec-7-ene

A High Temperature Favors the Elimination Product

2-Bromopropane + OH⁻

Substrate: Secondary alkyl halide

Both substitution (SN2) and elimination (E2) can occur.
Temperature decides the major product.

At 45 °C (lower temperature)

Lower temperature favors substitution.

Reaction (SN2):
(CH₃)₂CHBr + OH⁻ → (CH₃)₂CHOH + Br⁻

Minor product:
2-Propanol (Isopropyl alcohol) (47%)

Major product: Propene (53%)

At 100 °C (higher temperature)

Higher temperature favors elimination.

Reaction (E2):
(CH₃)₂CHBr + OH⁻ → CH₃–CH=CH₂ + H₂O + Br⁻

Major product:
Propene (71%)

Minor product: 2-Propanol (29%)

CASE III: E2 Reaction of a Tertiary Alkyl Halide

Because a tertiary alkyl halide cannot undergo an SN2 reaction, only an elimination product is
formed when a tertiary alkyl halide reacts with a strong base.

Tert-butyl bromide + Ethoxide ion (C₂H₅O⁻)

Substrate: Tertiary alkyl halide
Reagent: Strong base and good nucleophile

Key Concept

Tertiary alkyl halides do not undergo SN2 reactions due to strong steric hindrance.
When a strong base like ethoxide ion is present, the reaction proceeds mainly by E2 elimination.

Reaction (E2)

$(CH_3)_3CBr + C_2H_5O^- \rightarrow (CH_3)_2C=CH_2 + C_2H_5OH + Br^-$

Major Product

2-Methylpropene (Isobutene)

Structure:
(CH₃)₂C=CH₂

Reason

Substrate is tertiary
SN2 is blocked due to steric hindrance
Strong base present
Therefore E2 elimination dominates

Major product: 2-Methylpropene (alkene) (100%)

CASE IV: SN1/E1 Reactions of Tertiary Alkyl Halides

Tert-butyl bromide + Ethanol

Substrate: tertiary alkyl halide
Reagent/Solvent: ethanol (weak nucleophile and weak base, polar protic solvent)

Reaction Type

Tertiary alkyl halides in a weak nucleophile solvent undergo SN1 and E1 reactions.

Step 1: Formation of tert-butyl carbocation
Step 2: Nucleophile or base reacts with the carbocation.

Substitution Reaction (SN1) – Major Product

$(CH_3)_3CBr + C_2H_5OH \rightarrow (CH_3)_3COC_2H_5 + HBr$

Major product:
tert-Butyl ethyl ether (81%)

Structure: $(CH_3)_3C-O-CH_2CH_3$

Elimination Reaction (E1) – Minor Product

$(CH_3)_3CBr \rightarrow (CH_3)_2C=CH_2$

Minor product:
2-Methylpropene (isobutene) (19 %)

tert-Butyl bromide + Ethoxide ion (C₂H₅O⁻)

Substrate: tertiary alkyl halide
Reagent: strong base (ethoxide ion)

Reaction Type

SN2 cannot occur because the tertiary carbon is highly sterically hindered.
With a strong base, the reaction proceeds mainly by E2 elimination.

Reaction (E2)

$(CH_3)_3CBr + C_2H_5O^- \rightarrow (CH_3)_2C=CH_2 + C_2H_5OH + Br^-$

Major Product

2-Methylpropene (Isobutene) (100%)

Structure:
(CH₃)₂C=CH₂

Reason

Substrate is tertiary.
Strong base present.
SN2 blocked due to steric hindrance.
Therefore E2 elimination dominates.

Major product: 2-Methylpropene (alkene).