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 🎵
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
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:
With plane of symmetry:
Some structures become identical (meso)
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.
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
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)
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
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.
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:
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.
Ring expansion is a type of carbocation rearrangement in which a smaller ring increases its size (usually by one carbon) to reduce ring strain.
It commonly occurs in SN1 or E1 reactions when a carbocation is adjacent to a small ring.
Why Ring Expansion Happens
Small rings such as:
Cyclopropane
Cyclobutane
have high angle strain.
When a carbocation forms next to these rings, a C–C bond shifts (1,2-shift) and the ring expands to:
Cyclobutane
Cyclopentane
which are more stable
PROBLEM : Propose a mechanism for the following reaction:
In this reaction the substrate is a tertiary alkyl chloride attached to a cyclobutane ring and the reagent is methanol (CH₃OH), which is a weak nucleophile and polar protic solvent.
So the reaction proceeds through an SN1 mechanism.
Step 1: Carbocation formation
The Cl⁻ leaves, forming a tertiary carbocation.
Step 2: Ring expansion (rearrangement)
Cyclobutane rings have high ring strain. To reduce strain, a 1,2-shift of a ring bond occurs, converting:
cyclobutane → cyclopentane carbocation
This produces a more stable carbocation and also reduces ring strain.
Step 3: Deprotonation
Methanol removes a proton from the carbocation, giving an alkene.
Final Product
The rearranged product is:
1,2-dimethylcyclopentene
In this reaction substitution does NOT take place as the final product
Reason
Leaving group departure Cl⁻ leaves first → tertiary carbocation forms (SN1 condition, because solvent is CH₃OH, weak nucleophile).
Carbocation rearrangement The carbocation is next to a cyclobutane ring, which has high ring strain. So a 1,2-C–C shift occurs, causing ring expansion:
Cyclobutane → Cyclopentyl carbocation
This rearrangement is very favorable because it relieves ring strain.
Elimination occurs Instead of nucleophilic attack by methanol, a proton is removed, forming a stable alkene.
Why elimination dominates
Rearranged carbocation becomes very stable
Formation of cyclopentene double bond is thermodynamically favorable
Relief of ring strain drives the reaction
Final product
1,2-dimethylcyclopentene
✔ Therefore substitution product (ether) is not observed as the major product. The reaction proceeds via SN1 → rearrangement → E1 elimination.
Important teaching point (useful for students)
Whenever you see:
Carbocation next to small ring (cyclobutane / cyclopropane)
Ring expansion possible
➡ Rearranged alkene products are usually formed.
If you want, I can also show the step-by-step rearrangement mechanism diagram (very useful for explaining this in class).
Can cyclopentane undergo ring expansion?
Yes, but it is much less common.
Reason
Ring expansion mainly occurs to reduce ring strain.
A base removes a proton (H⁺) from a carbon atom that is adjacent to the carbon bonded to the halogen. As the proton is removed, the electrons from the C–H bond move toward the neighboring carbon that is attached to the halogen. These electrons then form a double bond between the two carbons. At the same time, the halogen leaves with its bonding electron pair, because carbon cannot form more than four bonds. This process results in the formation of an alkene.
Why E2 Reaction is Regioselective
An E2 reaction is called regioselective because the elimination of hydrogen and the leaving group can produce more than one possible alkene, but one alkene is formed in greater amount than the others.
Reason
In E2 elimination, the base removes a β-hydrogen (hydrogen on the carbon adjacent to the carbon bearing the leaving group). If there are different β-carbons, the base can remove hydrogen from different positions, giving different alkenes.
However, according to Zaitsev’s rule, the more substituted alkene (the alkene with more alkyl groups attached to the double-bonded carbons) is usually more stable and therefore forms as the major product.
Zaitsev’s Rule
Alexander M. Zaitsev, a nineteenth-century Russian chemist, proposed a rule to predict the major alkene formed in elimination reactions.
According to Zaitsev’s rule, the major product is the more substituted alkene. This occurs when the base removes a hydrogen from the β-carbon that has fewer hydrogen atoms.
As a result, the double bond forms between the more substituted carbon atoms, producing the more stable alkene.
Limitations of Zaitsev’s Rule (JEE / NEET level)
Although Zaitsev’s rule usually predicts the major alkene in elimination reactions, there are several situations where it does not apply.
1. Bulky (Sterically Hindered) Bases
When a bulky base is used, it removes the most accessible hydrogen instead of the one predicted by Zaitsev’s rule.
Examples of bulky bases:
tert-butoxide (t-BuO⁻)
LDA
Result → Less substituted alkene (Hofmann product) becomes the major product.
If the alkyl halide is not sterically hindered and the base is only moderately hindered, the more stable alkene will be the major product, as expected. In other words, it takes a lot of steric hindrance for the less stable product to be the major product. Thus, the major product of the following reaction is 2-butene.
2. Poor Leaving Groups (e.g., Fluoride)
When the leaving group is F⁻, elimination often gives the less substituted alkene instead of the Zaitsev product.
Reason → Strong C–F bond and different transition state stability.
When a hydrogen and a chlorine, bromine, or iodine are eliminated from an alkyl halide, the halogen starts to leave as soon as the base begins to remove the proton. Consequently, the transition state resembles an alkene (see page 414). The fluoride ion, however, is the strongest base of the halide ions and, therefore, the poorest leaving group.
Elimination of quaternary ammonium hydroxides produces mainly the less substituted alkene.
The Hofmann elimination is the preparation of alkenes from the treatment of quaternary ammonium salts with silver oxide, water, and heat.
General features: The Hofmann elimination is a β-elimination. Thus, the β-hydrogen is abstracted by the base (hydroxide ion) from the β-carbon atom. It is also an anti elimination (the leaving group has to be antiperiplanar). Besides, some side reactions occur when the base acts as a nucleophile, delivering then alcohol byproducts.
4. In each of the following reactions, the major product is the alkene with conjugated double bonds because it is the more stable alkene, even though it is not the more substituted alkene.
Relative Reactivities in an E2 Reaction
In E2 elimination, the rate depends on how easily the base can remove a β-hydrogen and how stable the resulting alkene will be.
Elimination from more substituted alkyl halides usually forms more substituted and more stable alkenes. Therefore, the reaction occurs more easily.
Reactivity Order of Alkyl Halides in E2
Explanation
Tertiary alkyl halides (3°) eliminate fastest because they form highly substituted stable alkenes.
Secondary alkyl halides (2°) react at a moderate rate.
Primary alkyl halides (1°) react slowly because they form less substituted alkenes.
Methyl halides (CH₃X)do not undergo E2 because there is no β-hydrogen available for elimination.
Features of E2 Reaction (JEE / NEET level)
Bimolecular Reaction
Two species participate: alkyl halide and base.
Second-Order Kinetics
One-Step (Concerted) Mechanism
Proton removal, double bond formation, and leaving group departure occur simultaneously.
Strong Base Required
Examples: OH⁻, RO⁻, t-BuO⁻, NH₂⁻
β-Elimination Reaction
Hydrogen is removed from the β-carbon (adjacent to the carbon bearing the halogen).
No Carbocation Intermediate
Reaction occurs in one step, so rearrangement does not occur.
Anti-Periplanar Requirement
The β-hydrogen and leaving group must be in opposite (anti) orientation.
Regioselective Reaction
Usually follows Zaitsev’s rule → more substituted alkene forms as major product.
Effect of Substrate Structure Reactivity order:
3∘>2∘>1∘
Competes with SN2 Reaction
Strong bases with secondary or tertiary halides often favor E2 elimination.
Stereochemistry of E2 Mechanism
The E2 (Elimination Bimolecular) reaction has very important stereochemical requirements. For elimination to occur, the β-hydrogen and leaving group must have a specific spatial arrangement.
1. Anti-Periplanar Requirement
The β-hydrogen (H) and leaving group (X) must lie in the same plane but opposite directions (180° apart).
This arrangement is called anti-periplanar geometry.
Reason: This alignment allows proper overlap of orbitals so the π bond of the alkene can form easily.
2. Concerted Mechanism
E2 occurs in one single step:
Base removes β-H
C–H bond breaks
C–X bond breaks
C=C double bond forms simultaneously
There is no intermediate (no carbocation).
3. Stereospecific Reaction
Because the geometry must be anti-periplanar, the reaction is stereospecific.
Example:
Different stereoisomers of reactants can give different alkene stereoisomers (E or Z).
4. Anti Elimination
Most E2 reactions proceed by anti elimination because:
It is energetically more stable
Less steric repulsion between groups.
5. Syn Elimination (Rare)
If anti arrangement is not possible, syn-periplanar elimination may occur.
However, it is rare and higher in energy.
6. Important in Cyclic Compounds
In cyclohexane systems:
Leaving group must be axial
β-Hydrogen must also be axial This creates the required anti-periplanar arrangement.
7. Product Stereochemistry
E2 generally favors formation of the more stable alkene (Zaitsev product), though bulky bases may give Hofmann product.
THE E1 REACTION
E1 Mechanism (Elimination Unimolecular) is a two-step elimination reaction commonly seen with tertiary alkyl halides in polar protic solvents. These are the main features important for JEE/NEET:
1. First-Order Kinetics
Rate depends only on the concentration of alkyl halide.
Rate law:
Rate=k[R–X]
2. Two-Step Mechanism
Slow step: Leaving group leaves → carbocation formation
Fast step: Base removes β-hydrogen → alkene formation
3. Carbocation Intermediate
A planar carbocation is formed.
Because of this, rearrangement (hydride shift or methyl shift) can occur.
4. Favored with Tertiary Alkyl Halides
Carbocation stability determines reactivity:
3° > 2° >> 1°
Primary carbocations are unstable, so E1 rarely occurs with primary halides.
5. Polar Protic Solvent Favours E1
Solvents like:
H₂O
Alcohols (ROH)
These stabilize the carbocation and leaving group.
6. Weak Bases Are Sufficient
Strong base is not required because the slow step is carbocation formation.
7. Zaitsev Alkene Major Product
The more substituted (more stable) alkene is usually the major product.
8. Competes with SN1 Reaction
Because both involve carbocation intermediate, E1 and SN1 often occur together.
9. Rearrangement Possible
Carbocation may rearrange to form a more stable carbocation, giving unexpected products.
10. Usually Occurs at Higher Temperature
Higher temperature favors elimination over substitution.
both E1 and E2 reactions can give cis and trans alkenes, but the reason and control are different.
In E1 Reaction
Carbocation intermediate forms (planar).
Base can remove β-H from either side.
Therefore both cis (Z) and trans (E) alkenes form.
Trans usually major because it is more stable.
Not stereospecific.
In E2 Reaction
Reaction occurs in one step.
Requires anti-periplanar geometry between β-H and leaving group.
The geometry of the reactant determines the alkene formed.
So cis or trans may form depending on the starting stereochemistry.
Stereospecific reaction.
Stereochemistry of E1 Mechanism
The E1 (Elimination Unimolecular) reaction has different stereochemistry from E2 because it proceeds through a carbocation intermediate.
1. Planar Carbocation Intermediate
In the first step, the leaving group (X) leaves and forms a carbocation.
The carbocation is sp² hybridized and planar.
Because it is planar, the β-hydrogen can be removed from either side.
2. Not Stereospecific
Unlike E2, E1 is not stereospecific.
Reason: Since the intermediate carbocation is planar, the base can remove the β-hydrogen from different orientations, giving different alkene stereoisomers.
3. Formation of E and Z Alkenes
Both E (trans) and Z (cis) alkenes can form.
However:
Trans (E) alkene is usually the major product
because it is more stable (less steric repulsion).
4. No Anti-Periplanar Requirement
In E1, the β-hydrogen and leaving group do not need anti-periplanar geometry.
This is because:
the leaving group already left in the first step.
5. Rearrangement Possible
Since a carbocation intermediate exists, rearrangements such as:
