Leaving Group Order

For common organic chemistry leaving groups, the order of leaving-group ability is:

I⁻ > Br⁻ > Cl⁻ > F⁻

So:

RI > RBr > RCl > RF
for SN1/SN2 reactions, generally.

Common leaving groups — good → poor

OTf⁻ > OTs⁻ ≈ OMs⁻ > I⁻ > Br⁻ > Cl⁻ > OAc> F⁻>OH^–> NH₂->H-

• OTf = triflate
• OTs = tosylate
• OMs = mesylate

Important trick

“The weaker the base, the better the leaving group.”

For halide ions, basicity increases:

F⁻ > Cl⁻ > Br⁻ > I⁻

Therefore leaving-group ability is the reverse:

I⁻ > Br⁻ > Cl⁻ > F⁻.

The key point is that –SO₃H acts as a temporary blocking group.

Step 1: Desulphonation

The starting compound is p-hydroxybenzenesulphonic acid:

On treatment with aqueous bromine, the strongly activating –OH group directs bromination to the ortho and para positions.

But the para position is occupied by –SO₃H.

Step 2: Bromination

Therefore, Br enters the two ortho positions:

The final product is:$$\boxed{\text{2,4,6-Tribromophenol (white precipitate)}}$$

Why does –SO₃H disappear?

Under the aqueous bromination conditions, desulphonation occurs, replacing SO₃H by H. This allows bromination at the para position as well.

So the overall idea is:$$\boxed{\text{Blocking group removed + bromination at all available o/p positions}}$$

Hence 3 Br atoms finally occupy the 2, 4 and 6 positions of phenol.

Benzophenone is even more hindered than acetophenone:$$(C_6H_5)_2C=O$$

Two effects make $HSO_3^-$ addition difficult:

1. Very high steric hindrance
   The carbonyl carbon is surrounded by two phenyl groups, so $HSO_3^-$ cannot easily attack it.
2. Resonance/conjugation
   The carbonyl group is conjugated with both benzene rings, reducing the electrophilic character of the carbonyl carbon.

Reactivity comparison

$$\boxed{\text{Aldehydes} > \text{acetophenone} > \text{benzophenone}}$$

So for the NaHSO₃ addition test:

• Aldehydes → generally react readily ✅
• Acetophenone → poor/slow reaction
• Benzophenone → generally does not react under usual conditions ❌

Key JEE/NEET fact: Benzophenone is a classic example of a ketone that does not form the usual sodium bisulfite addition compound readily because of steric hindrance + resonance stabilization.

Relation between Specific Conductivity (κ) and Solubility

For a sparingly soluble salt (e.g., AgCl), its solubility can be determined from its conductivity.

At infinite dilution:$$\boxed{\kappa = \frac{S}{1000}(\lambda_m^\circ)}$$

The answer is (B) benzyl chloride, C₆H₅–CH₂Cl.

The reagent is NaI in DMSO, which strongly favors an SN2 reaction.

Step 1: Compare the substrates

• A: tertiary alkyl chloride → highly hindered ❌
• B: benzyl chloride, C₆H₅–CH₂Cl → primary benzylic chloride ✅
• C: chlorobenzene, C₆H₅–Cl → aryl chloride, SN2 doesn’t occur ❌
• D: cyclohexylmethyl chloride → primary, but not benzylic

Step 2: Why is B exceptionally fast?

In benzyl chloride:

C₆H₅–CH₂–Cl

The carbon attacked by I⁻ is the CH₂ next to the benzene ring.

During SN2, the transition state has some electron deficiency at this carbon. The benzene ring can stabilize this transition state by resonance.

So:

benzyl chloride → very fast SN2

Also, DMSO is a polar aprotic solvent, which makes I⁻ a strong nucleophile and favors SN2.

Important order to remember

For SN2 reactivity:

benzyl/allyl halide > primary > secondary >> tertiary

And aryl/vinyl halides do not normally undergo SN2.

Answer: (A)

$$\boxed{\text{III} < \text{I} < \text{II} < \text{IV}}$$

Reason:

• III → more steric hindrance → slowest $S_N2$
• I → secondary alkyl chloride
• II → secondary alkyl bromide; Br⁻ is a better leaving group than Cl⁻
• IV → $\alpha$-bromoketone; carbonyl group strongly activates the $S_N2$ reaction → fastest

Correct option: (A) III < I < II < I

V is highest because it is an α-bromo ketone:$$\mathrm{CH_3CO-CH(Br)-CH_2CH_3}$$

The key effect is the carbonyl group (C=O)

Why does $C=O$ increase $S_N2$ rate?

The carbonyl group has a strong −I (electron-withdrawing) effect:$$\mathrm{C=O \rightarrow CH(Br)}$$

It pulls electron density away from the carbon bearing Br, making that carbon more electrophilic.

Iodoform test is used to identify compounds containing:$$\boxed{CH_3CO-}$$

or compounds that can be oxidised to a methyl ketone.

1. Reagent

$$\boxed{I_2 + NaOH}$$

Positive test: formation of a yellow precipitate of iodoform$$\boxed{CHI_3 = \text{yellow precipitate}}$$

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2. Main compounds giving the test

A. Methyl ketones

General structure:$$\boxed{R-CO-CH_3}$$

Example:$$CH_3COCH_3 \xrightarrow{I_2/NaOH} CHI_3\downarrow$$

So acetone gives iodoform test.

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B. Ethanol

Ethanol is oxidised to acetaldehyde, which gives the test:$$CH_3CH_2OH \rightarrow CH_3CHO \rightarrow CHI_3$$

Therefore:$$\boxed{CH_3CH_2OH \text{ gives iodoform test}}$$

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C. Secondary alcohols of type

$$\boxed{CH_3-CH(OH)-R}$$

These are oxidised to:$$CH_3-CO-R$$

which is a methyl ketone.

Example:$$CH_3CH(OH)CH_3$$

Propan-2-ol → positive.

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3. Important examples

Compound	Iodoform test
$CH_3COCH_3$	✅
$CH_3COCH_2CH_3$	✅
$CH_3CHO$	✅
$CH_3CH_2OH$	✅
$CH_3CH(OH)CH_3$	✅
$CH_3CH(OH)C_2H_5$	✅
$CH_3CH_2CH_2OH$	❌
$CH_3COOH$	❌

JEE/NEET shortcut

Look for:$$\boxed{CH_3CO-}$$

OR$$\boxed{CH_3CH(OH)-}$$

If either is present → Iodoform test is generally positive.

Special exception

Acetaldehyde (CH3CHO) gives the iodoform test even though it is not a methyl ketone.$$\boxed{CH_3CHO \rightarrow CHI_3\downarrow}$$

Fehling’s test is mainly used to detect aliphatic aldehydes.

1. Fehling’s reagent

It consists of two solutions:

• Fehling A: $CuSO_4$ solution
• Fehling B: alkaline sodium potassium tartrate (Rochelle salt) + NaOH

They are mixed freshly before the test.

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2. Reaction with aldehyde

For an aliphatic aldehyde:$$\boxed{RCHO + 2Cu^{2+} + 5OH^- \rightarrow RCOO^- + Cu_2O\downarrow + 3H_2O}$$

Observation:$$\boxed{\text{Blue solution} \rightarrow \text{brick-red }Cu_2O\text{ precipitate}}$$

Aldehyde is oxidised → carboxylate
$Cu^{2+}$ is reduced → $Cu_2O$

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3. Which compounds give Fehling’s test?

Compound	Fehling test
Aliphatic aldehyde	✅ Positive
Formaldehyde, HCHO	✅ Positive
Acetaldehyde, CH₃CHO	✅ Positive
Aromatic aldehyde, C₆H₅CHO	❌ Generally negative
Ketones	❌ Negative
Formic acid/formates	✅ Can reduce Cu²⁺
Reducing sugars	✅ Positive

Most important distinction

$$\boxed{\text{Aliphatic aldehyde → Fehling positive}}$$$$\boxed{\text{Aromatic aldehyde → Fehling generally negative}}$$

But both can give Tollens’ test:$$\boxed{\text{RCHO → Tollens positive}}$$$$\boxed{\text{RCHO (aliphatic) → Fehling positive}}$$$$\boxed{\text{ArCHO → Fehling generally negative}}$$

JEE/NEET trap

Benzaldehyde (C6H5CHO)

• Tollens → ✅ Silver mirror
• Fehling → ❌ No brick-red precipitate

Acetaldehyde (CH3CHO)

• Tollens → ✅
• Fehling → ✅

Memory trick:
“Fehling likes aliphatic aldehydes, not aromatic aldehydes.”

Tollens’ reagent = ammoniacal AgNO₃

It contains the complex ion:$$\boxed{[Ag(NH_3)_2]^+}$$

It is mainly used to identify aldehydes.

1. Basic reaction

For an aldehyde:$$RCHO + 2[Ag(NH_3)_2]^+ + 3OH^- \rightarrow RCOO^- + 2Ag\downarrow +4NH_3+2H_2O$$

Observation: Silver mirror is formed.

Aldehyde → oxidised to carboxylate
Ag⁺ → reduced to metallic Ag

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2. Which compounds give Tollens’ test?

Compound	Tollens’ test
Aldehyde, R–CHO	✅ Positive
Formic acid, HCOOH	✅ Positive
Formate ion, HCOO⁻	✅ Positive
α-hydroxy aldehyde	✅ Positive
Reducing sugars	✅ Positive
Ketones, R–CO–R	❌ Usually negative
Most carboxylic acids	❌ Negative
Esters	❌ Negative

3. Important JEE trick

Formic acid is an exception.$$\boxed{HCOOH \xrightarrow{\text{Tollens}} CO_2 + Ag}$$

Because formic acid can be oxidised to $CO_2$, it reduces Ag⁺.

So:$$\boxed{\text{Aldehydes + HCOOH + formates → Tollens positive}}$$

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4. Why ketones generally don’t react?

Ketones do not readily oxidise under the mild conditions of Tollens’ reagent.$$R-CO-R’ \quad \xrightarrow{\text{Tollens}} \quad \text{No reaction}$$

But beware: some compounds containing a ketone group can give a positive test if they can tautomerise/rearrange to an aldehyde-type reducing species.

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5. Glucose gives Tollens’ test

Glucose predominantly exists in cyclic form, but it can establish equilibrium with its open-chain aldehyde form:$$\text{Glucose} \rightleftharpoons \text{open-chain glucose}$$

The open-chain form reduces Tollens’ reagent.

Hence:$$\boxed{\text{Glucose → positive Tollens’ test}}$$

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6. Common JEE/NEET comparison

$$\boxed{\text{Tollens’ test → Aldehyde}}$$$$\boxed{\text{Fehling’s test → Aliphatic aldehydes mainly}}$$$$\boxed{\text{Iodoform test → } CH_3CO- \text{ or } CH_3CH(OH)-}$$

One-line memory

“Tollens = Silver mirror = Aldehyde oxidation.”

Alpha-Particle Scattering Experiment

Scientists: H. Geiger and E. Marsden, under Rutherford
Year: 1911
Source: Radioactive $^{214}_{83}\mathrm{Bi}$
Particles: $\alpha$-particles of energy 5.5 MeV
Target: Thin gold foil
Gold foil thickness: $2.1\times10^{-7}\,\text{m}$

Why gold foil was used?

Gold can be beaten into an extremely thin foil, so α-particles can pass through it.

Observation

1. Most α-particles passed straight through the foil.
   • Therefore, most of the atom is empty space.
2. Very few α-particles were deflected through small angles.
   • Only about 0.14% scattered by more than $1^\circ$.
3. About 1 in 8000 α-particles was deflected through an angle greater than $90^\circ$.
4. A very small number of α-particles were scattered backwards.

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Rutherford’s Conclusions

1. Most of the atom is empty space

Since most α-particles passed through without deflection, the atom must contain a large amount of empty space.

2. Positive charge is concentrated

Large deflections require a strong repulsive force.

Therefore, the positive charge of the atom must be concentrated in a very small region at the centre.

This region is called the nucleus.

3. Most of the mass is in the nucleus

The nucleus contains:

• Entire positive charge
• Almost entire mass of the atom

4. Nucleus is extremely small

Size of nucleus:$$\boxed{10^{-15}\text{ m to }10^{-14}\text{ m}}$$

Size of atom:$$\boxed{10^{-10}\text{ m}}$$

Therefore, the nucleus is extremely small compared with the atom.

5. Electrons revolve around nucleus

Electrons are present at a considerable distance from the nucleus and move around it, similar to planets revolving around the Sun.

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Rutherford Nuclear Model

According to Rutherford:$$\boxed{\text{Atom}=\text{tiny positive nucleus}+\text{electrons around it}}$$

Main points

• Nucleus is at the centre of the atom.
• Nucleus is positively charged.
• Almost the entire mass of the atom is concentrated in the nucleus.
• Electrons revolve around the nucleus.
• Most of the atom is empty space.
• Atomic electrons are very light and therefore do not appreciably affect α-particle scattering.

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Coulomb Force in α-Particle Scattering

An α-particle has charge:$$q_1=+2e$$

Gold nucleus has charge:$$q_2=+Ze$$

For gold:$$Z=79$$

Using Coulomb’s law:$$F=\frac{1}{4\pi\varepsilon_0} \frac{(2e)(Ze)}{r^2}$$

Therefore,$$\boxed{F=\frac{1}{4\pi\varepsilon_0}\frac{2Ze^2}{r^2}}$$

This force is repulsive because both α-particle and gold nucleus are positively charged.

Important assumption

The gold nucleus is about 50 times heavier than an α-particle, so the nucleus can be considered stationary during scattering.

Also, because the gold foil is very thin, an α-particle is assumed to undergo not more than one scattering.

Most important exam line

$$\boxed{\text{Most particles pass straight} \Rightarrow \text{atom mostly empty space}}$$$$\boxed{\text{Few particles deflect strongly} \Rightarrow \text{small, dense, positively charged nucleus}}$$