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

Degree of Unsaturation (DU) formula for organic compound

Degree of Unsaturation (DU) / Index of Hydrogen Deficiency (IHD) Formula

The Degree of Unsaturation (DU) tells you the total number of rings and π (pi) bonds present in an organic compound.

General Formula

Where:

C = Number of carbon atoms
H = Number of hydrogen atoms
N = Number of nitrogen atoms
X = Number of halogen atoms (F, Cl, Br, I)
O, S, Se = Ignored (they do not affect DU)

Shortcut Formula

Effect of Different Atoms

Atom	Effect on DU Calculation
C	Count normally
H	Subtract
N	Add 1 H equivalent (+1)
F, Cl, Br, I	Treat as H
O, S	Ignore

Meaning of DU

DU	Possible Structure
0	Saturated compound (no ring, no π bond)
1	One ring or one double bond
2	Two double bonds, or one triple bond, or one ring + one double bond
3	Three π bonds/rings in any combination
4	Benzene ring (1 ring + 3 double bonds = 4 DU)

Which is stronger acid in Picric acid and formic acid?

Picric acid has three –NO₂ groups which are strong electron-withdrawing groups.
These groups stabilize the conjugate base (picrate ion) by resonance and –I effect.
Formic acid has only one –COOH group, so its conjugate base is less stabilized.
Greater stabilization of conjugate base ⇒ higher acidity, so picric acid is stronger.

HCOOH (Picric Acid)

Picric acid is soluble in NaHCO₃ solution?

Picric acid (2,4,6-trinitrophenol) is a strong acid due to the presence of three –NO₂ groups, which strongly withdraw electrons and stabilize the phenoxide ion. Because of this high acidity, it can react even with a weak base like NaHCO₃.

Reaction:
Picric acid + NaHCO₃ → Sodium picrate (soluble) + CO₂ + H₂O

So, it dissolves in NaHCO₃ with effervescence of CO₂ gas.

Compare polarity between CH₃CN (acetonitrile) and CH₃OH (methanol):

Key idea: Dipole moment + bond polarity

CH₃CN (acetonitrile)
Structure: CH₃–C≡N
Strongly polar due to C≡N (nitrile group), which has a large electronegativity difference and linear structure → strong net dipole.
CH₃OH (methanol)
Structure: CH₃–OH
Polar due to O–H and C–O bonds, but overall dipole is slightly less compared to nitrile.

Final order of polarity:

CH₃CN > CH₃OH

Reason:

The C≡N group in CH₃CN creates a stronger dipole moment (~3.9 D)
Methanol has ~1.7 D, so comparatively less polar.

Extra insight (important for exams):

CH₃OH shows hydrogen bonding → higher boiling point
CH₃CN is more polar (dipole-wise) but no H-bonding like alcohols

Polarity (dipole moment):
CH₃CN > CH₃OH
Boiling point:
CH₃CN (≈82°C) > CH₃OH (≈65°C)

Mechanism of Kolbe’s reaction

The Kolbe’s reaction (Kolbe–Schmitt reaction) is used to introduce a –COOH group into phenol, mainly at the ortho position.

Overall reaction

Phenol + NaOH + CO₂ (pressure, heat) → o-hydroxybenzoic acid (salicylic acid)

Key Points

Reagents: NaOH + CO₂ (high pressure, ~125°C)
Electrophile: CO₂
Major product: ortho (salicylic acid)
Para product is minor
Ortho favored due to intramolecular H-bonding

Reimer Tiemann reaction mechanism

The Reimer–Tiemann reaction is used to introduce a –CHO (aldehyde) group into a phenol ring, usually at the ortho position (forming salicylaldehyde).

Overall reaction

Phenol + CHCl₃ + NaOH → o-hydroxybenzaldehyde (major) + p-hydroxybenzaldehyde (minor)

Key Points to Remember

Reagent: CHCl₃ + NaOH
Reactive intermediate: dichlorocarbene (:CCl₂)
Directing group: –O⁻ (phenoxide) → ortho/para directing
Major product: ortho-hydroxybenzaldehyde
Para product is minor due to intramolecular H-bonding stabilizing ortho product

Oxymercuration–Demercuration Reaction

Reagents:

Step 1: Hg(OAc)₂ / H₂O
Step 2: NaBH₄

Overall Reaction:

Alkene → Alcohol (Markovnikov addition, no rearrangement)

Example:

CH₃–CH=CH₂ → CH₃–CHOH–CH₃

(Propene → 2-propanol)

Key Features (VERY IMPORTANT)

1. Markovnikov Addition

–OH goes to more substituted carbon

2. NO Carbocation Rearrangement ❌

This is the most asked point in JEE

Unlike acid-catalyzed hydration:

No hydride shift
No methyl shift

3. Anti Addition (mostly)

OH and HgOAc add from opposite sides initially

4. Intermediate Formation

Forms mercurinium ion (cyclic intermediate)
More stable than carbocation → prevents rearrangement

Mechanism (Simplified)

Step 1: Formation of mercurinium ion

Alkene + Hg(OAc)₂ → cyclic intermediate

Step 2: Nucleophilic attack

H₂O attacks more substituted carbon

Step 3: Demercuration

NaBH₄ replaces Hg with H

Important Comparison (JEE trap)

Reaction	Rearrangement	Markovnikov	Intermediate
Oxymercuration	❌ No	✔ Yes	Mercurinium ion
Acid hydration	✔ Yes	✔ Yes	Carbocation

Special Case (VERY IMPORTANT)

With Alcohol instead of Water:

Hg(OAc)₂ / ROH → Ether formation

👉 Alkene → Ether (Markovnikov)

Tough Concept Question

Q: Why no rearrangement?

Answer:

Because intermediate is mercurinium ion (bridged), not free carbocation

Memory Trick:

Mercury locks structure”
→ No rearrangement possible

Exam Trap

Given:

3° carbocation possible after rearrangement

Leaving Group Ability Order

Leaving Group Ability (Best → Worst)

$\boxed{ \text{Sulfonium} > \text{Quaternary ammonium} > \text{Sulfonate (OTs/OMs)} > I^- > Br^- > Cl^- > F^- > H_2O > ROH > OH^- > OR^- > NH_2^- }$

Sulfonate ester

Sulfonium ion structure

Why HO⁻ and RO⁻ are Poor Leaving Groups

HO⁻ (hydroxide ion) and RO⁻ (alkoxide ion) are:
- Strong bases
- High-energy species
A good leaving group should be:
- Stable after leaving
- Weak base

Since HO⁻ and RO⁻ are strong bases, they are unstable when free, so they do not want to leave.

2. Comparison with Alkyl Halides

In alkyl halides (R–X), the leaving group is X⁻ (Cl⁻, Br⁻, I⁻):
- These are weak bases
- Highly stable ions

Consequence in Reactions

Alcohols (R–OH) and ethers (R–O–R’):
- Do not undergo SN1/SN2/E1/E2 easily
- Because –OH or –OR cannot leave directly

So they are less reactive than alkyl halides.

4. Activation of Alcohols & Ethers

To make them reactive, we convert the poor leaving group into a good one.

Common activation methods:

(A) Protonation (Acidic Medium)

In presence of acid (like H₂SO₄ or HCl):

$R-OH + H^+ \rightarrow R-OH_2^+$

Now leaving group = H₂O (water)
Water is neutral and stable → good leaving group

Conversion to Better Leaving Groups

Convert –OH into:
- Tosylate (–OTs)
- Mesylate (–OMs)

These are excellent leaving groups.

$R-OH + HCl \rightarrow R-Cl + H_2O$

Summary (Exam Ready)

HO⁻ and RO⁻ = strong bases → poor leaving groups
Alkyl halides = weak base leaving groups → high reactivity
Alcohols/ethers = less reactive
Need activation (protonation or conversion) before reaction

How Niels Bohr found radius of hydrogen-like atom

According to Bohr, electrons move in stationary orbits where they do not radiate energy, hence energy of an orbit remains constant with time.

Bohr assumed stationary orbits (not proved mathematically) This explains atomic stability Energy is quantized and constant in each orbit.

Electrostatic force = Centripetal force

Step 1: Electron is attracted to nucleus, so: $\frac{mv^2}{r} = \frac{1}{4\pi \varepsilon_0} \cdot \frac{Z e^2}{r^2}$

Step 2 : Quantization of angular momentum

$mvr = \frac{nh}{2\pi}$

Step 3: Substitute v into Step 1

$m \left( \frac{nh}{2\pi mr} \right)^2 = \frac{1}{4\pi \varepsilon_0} \cdot \frac{Z e^2}{r}$

Step 4: Simplify

$\frac{n^2 h^2}{4\pi^2 m r^2} = \frac{1}{4\pi \varepsilon_0} \cdot \frac{Z e^2}{r}$

Step 5: Solve for radius r

$r = \frac{4\pi \varepsilon_0 n^2 h^2}{4\pi^2 m Z e^2}$

Simplify: $r = \frac{\varepsilon_0 h^2}{\pi m e^2} \cdot \frac{n^2}{Z}$

Final Result

$r_n = \frac{n^2 a_0}{Z}$

where, $a_0 = \frac{\varepsilon_0 h^2}{\pi m e^2} = 0.529 \, \text{Å}$

Key Results

$r_n \propto n^2$
$r_n \propto \frac{1}{Z}$
Valid for hydrogen-like species (H, He⁺, Li²⁺)

Bohr model, hydrogen-like atom, radius derivation, angular momentum quantization, electrostatic force

The limiting line (or limit) of a spectral series (like Lyman, Balmer..)

The line corresponding to the transition where the electron comes from an infinitely large orbit (n = ∞) to a fixed lower energy level.

🔹 What it means simply

When an electron falls from very high energy levels (n = very large), the spectral lines get closer and closer.
Finally, at n = ∞, they merge into one line → this is the series limit (limiting line).

🔹 Formula (Rydberg equation)

$\frac{1}{\lambda} = R \left( \frac{1}{n_1^2} – \frac{1}{n_2^2} \right)$

For limiting line $\frac{1}{\lambda} = \frac{R}{n_1^2}$

Examples of limits

Lyman series (n₁ = 1)
- Limit: electron falls from ∞ → 1
- Region: Ultraviolet (UV)
Balmer series (n₁ = 2)
- Limit: ∞ → 2
- Region: Visible
Paschen series (n₁ = 3)
- Limit: ∞ → 3
- Region: Infrared

Limiting line = shortest wavelength (maximum energy) line of a series.

Lyman Alpha (α) Line

The Lyman α (alpha) line is the first spectral line of the Lyman series.

Period 2 elements — Anomalous Behaviour

Elements: Li, Be, B, C, N, O, F

These show properties different from the rest of their groups.

Main Reasons

Very small atomic size
High ionization enthalpy
High electronegativity
No d-orbitals available (cannot expand octet)
High charge density / polarising power

Important Examples (Exam Focus)

Li (vs other Group 1)

Forms Li₃N (others don’t easily)
LiCl shows covalent character
Li₂CO₃ decomposes on heating

Be (vs other Group 2)

Compounds are covalent
BeO, Be(OH)₂ are amphoteric
Strong complex formation

B (vs other Group 13)

Always covalent
Does not form B³⁺ ion easily
Forms electron-deficient compounds (e.g., BF₃)

🔹 C, N, O, F

Show strong pπ–pπ bonding
Form multiple bonds (C=C, C≡C, N≡N, etc.)
Higher electronegativity than heavier elements