Hydrocarbon JEE/NEET Test Paper SET-1

Hydrocarbons form the fundamental backbone of Organic Chemistry, consisting only of carbon and hydrogen atoms.
They are broadly classified into alkanes, alkenes, alkynes, and aromatic hydrocarbons based on bonding and structure.
Understanding their properties and reactions is crucial for mastering concepts in competitive exams like JEE and NEET.
This test paper is designed to evaluate conceptual clarity, problem-solving ability, and application skills.
It covers important topics such as nomenclature, isomerism, and reaction mechanisms of hydrocarbons.

Arrange the following alkenes in order of their stability

2. But-2-yne and hydrogen (one mole each) are separately treated with (i) P d / C and (ii) N a / liq. N H 3 to give the products X and Y respectively

Identify the incorrect statements.

A. X and Y are stereoisomers.

B. Dipole moment of X is zero

C. Boiling point of X is higher than Y .

D. X and Y react with O₃ / Z n + H ₂ O to give different products.

Choose the correct answer from the options given below :

(1) B and C only

(2) B and D only

(3) A and B only

(4) A and C only

The product Y formed is :

(1) 2-methylhex-2-yne

(2) 5-methylhex-2-yne

(3) 2-methylhex-3-yne

(4) Isopropylbut-1-yne

5. Which compound would give 3-methyl-6- oxoheptanal upon ozonolysis ?

6. Given below are two statements :

Statement (I) : Neopentane forms only one monosubstituted derivative.

Statement (II) : Melting point of neopentane is higher than n-pentane.

In the light of the above statements, choose the most appropriate answer from the options given below :

(1) Statement I is correct but Statement II is incorrect

(2) Both Statement I and Statement II are correct

(3) Both Statement I and Statement II are incorrect

7. Total number of sigma (σ) and pi(π) bonds respectively present in hex-1-en-4-yne are:

(1) 13, 3

(2) 14, 3

(3) 3, 14

(4) 14, 13

8. Identify product [A], [B] and [C] in the following reaction sequence.

9.Statement I : Nitration of benzene involves the following step –

Statement II: Use of Lewis base promotes the electrophilic substitution of benzene. In the light of the above statements, choose the most appropriate answer from the options given below :

(1) Both Statement I and Statement II are incorrect

(2) Statement I is correct but Statement II is incorrect (3) Both Statement I and Statement II are correct (4) Statement I is incorrect but Statement II is correct

10.Compound A formed in the following reaction reacts with B gives the product C. Find out A and B

11. Product ‘A’ of following sequence of reactions is

A) Cyclohexanol
B) Hexane-1,6-dioic acid (Adipic acid)
C) Cyclohexanone
D) 1,2-diol (Glycol)

13.Consider the following reaction :

Which of these reaction(s) will not produce Saytzeff product ?

Which of these reaction(s) will not produce Saytzeff product ? (

1) (c) only

(2) (a) (c) and (d)

(3) (d) only

(4) (b) and (d)

15. In the following skew conformation of ethane, H’–C–C–H” dihedral angle is :

In the following skew conformation of ethane, H’–C–C–H” dihedral angle is :

(1) 120º

(2) 58º

(3) 151º

(4) 149º

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

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

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).

Competition between E2 and E1 Reactions

In organic chemistry, competition between E1 Reaction and E2 Reaction occurs when a substrate and reaction conditions allow both elimination pathways. The dominant mechanism depends on several factors.

Primary and secondary alkyl halides undergo elimination mainly through the E2 mechanism because the carbocations that would form in an E1 Reaction are relatively unstable, and solvolysis reactions usually involve a high concentration of base that favors the E2 Reaction.

However, tertiary alkyl halides can undergo both E2 and E1 reactions because they are capable of forming relatively stable tertiary carbocations.

Because tertiary alkyl halides are able to undergo both E2 and E1 elimination reactions, the overall rate of the reaction depends on both pathways. Therefore, the rate law is the sum of the rate laws for the E2 and E1 reactions. $\text{Rate} = k_2[\text{alkyl halide}][\text{base}] + k_1[\text{alkyl halide}]$

Thus, An E2 reaction is favored by a high concentration of a strong base.
An E1 reaction is favored by a low concentration of a weak base

For each of the following reactions, (1) decide whether an E2 or an E1 occurs, and (2) draw the major
elimination product:

When 2‑Bromobutane reacts with Methoxide Ion, the reaction mainly proceeds through an E2 Reaction because:

2-Bromobutane is a secondary alkyl halide.
CH₃O⁻ (methoxide) is a strong base.
Strong bases favor E2 elimination.

The elimination forms bu-2-tene:

Major product: 2‑Butene (especially trans-2-butene, most stable)
Minor product: 1‑Butene

Reason (Zaitsev rule)

According to Zaitsev Rule, the more substituted alkene forms as the major product.

Order of products: $\text{trans-2-butene} > \text{cis-2-butene} > \text{1-butene}$

Substrate: secondary alkyl halide

Base: strong (CH₃O⁻)

Mechanism: E2 elimination

Major product: trans-2-butene

Problem: For each of the following reactions, (1) decide whether an E2 or an E1 occurs, and (2) draw the major
elimination product:

Strong base → E2
Weak base + tertiary substrate → E1
Zaitsev alkene = major product

let’s analyze each reaction using two steps:
Decide E1 or E2
Write the major elimination product

(a)

Substrate: 2-Bromobutane
Reagent: CH₃O⁻ (methoxide) → strong base

Reasoning

Secondary alkyl halide
Strong base present
E2 mechanism dominates

Elimination (Zaitsev rule)
The base removes a β-hydrogen giving the more substituted alkene.

Major product

2-Butene (mainly trans)

Reaction:

CH₃CH₂CH(Br)CH₃ + CH₃O⁻
→ CH₃CH=CHCH₃ + Br⁻ + CH₃OH

Major alkene:
trans-2-butene

(b)

Substrate: 2-Fluorobutane
Reagent: CH₃O⁻

Reasoning

Strong base → favors E2
Fluorine is a poor leaving group, so reaction is slower
But mechanism is still E2

Major product
Again Zaitsev alkene

Product
2-Butene (trans major)

CH₃CH₂CH(F)CH₃
→ CH₃CH=CHCH₃

(c)

Substrate: tert-Butyl chloride
Structure: (CH₃)₃C–Cl
Reagent: H₂O (weak base, polar protic)

Reasoning

Tertiary carbocation easily forms
Weak base
E1 mechanism

Steps:

Carbocation formation
Deprotonation

Product

2-Methylpropene (isobutene)

Reaction:

(CH₃)₃CCl → (CH₃)₂C=CH₂

(d)

Substrate: tert-Butyl chloride
Reagent: HO⁻ (strong base)

Reasoning

Strong base present
Tertiary substrate
E2 dominates

Product

2-Methylpropene

(CH₃)₃CCl + OH⁻
→ (CH₃)₂C=CH₂

E2 AND E1 REACTIONS ARE STEREOSELECTIVE

Yes, both E2 and E1 elimination reactions are stereoselective, but the degree and reason are different.

1. E2 Reactions – Highly Stereoselective

E2 reactions require a specific geometric arrangement.

Requirement

The β-hydrogen and leaving group must be anti-periplanar (180° apart).

Because of this requirement, the reaction selectively forms the more stable alkene stereoisomer.

Usually:

Trans (E) alkene is formed more than cis (Z).

Example
2-Bromobutane + strong base

Major product:

trans-2-butene

Reason:

Trans alkene is more stable due to less steric repulsion.

Key points:

Anti-periplanar geometry required
Reaction is stereospecific and stereoselective

2. E1 Reactions – Moderately Stereoselective

E1 occurs through a carbocation intermediate.

Steps:

Leaving group leaves → carbocation forms
Base removes β-hydrogen → alkene forms

Since the carbocation is planar, the base can remove hydrogen from either side.

Result:

Both cis and trans alkenes may form.

However:

Trans alkene is usually major because it is thermodynamically more stable.

Example
tert-Butyl chloride + H₂O

Product:

2-methylpropene

(Only one alkene possible here)