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

General Reaction

$R–NC \xrightarrow[\text{reduction}]{H_2 / Ni \; \text{or} \; LiAlH_4} R–NH–CH_3$

Product: Secondary amine

What Happens?

Isocyanide (R–NC) gets reduced
The carbon of –NC becomes –CH₃
Nitrogen forms secondary amine (R–NH–CH₃)

Example

$C_2H_5–NC \xrightarrow{H_2/Ni} C_2H_5–NH–CH_3$

Hofmann Degradation (Hofmann Bromamide Reaction)

It is the conversion of an amide (R–CONH₂) into a primary amine (R–NH₂) with one carbon less using Br₂ + KOH/NaOH.

Overall Reaction

$RCONH_2 + Br_2 + 4KOH \rightarrow RNH_2 + K_2CO_3 + 2KBr + 2H_2O$

Key point: Chain length decreases by one carbon

Stepwise Mechanism

Step 1: Formation of N-bromoamide

Amide reacts with bromine in basic medium: $RCONH_2 + Br_2 + OH^- \rightarrow RCONHBr + H_2O$

Step 2: Deprotonation

$RCONHBr + OH^- \rightarrow RCONBr^- + H_2O$

Step 3: Rearrangement (Key Step )

Alkyl group (R) migrates from carbonyl carbon → nitrogen
Br⁻ leaves

Forms isocyanate (R–N=C=O) $RCONBr^- \rightarrow R–N=C=O + Br^-$

Step 4: Hydrolysis of Isocyanate

$R–N=C=O + H_2O \rightarrow RNHCOOH$

Step 5: Decarboxylation

$RNHCOOH \rightarrow RNH_2 + CO_2$

Final Product

✔ Primary amine (R–NH₂)
✔ CO₂ released

Important Features

✔ One carbon less than starting amide
✔ Always gives 1° amine
✔ Rearrangement step is key (migration)

Quick Trick“CONH₂ → NH₂ (minus one carbon)”

Exam Points

Migration order:

$3^\circ > 2^\circ > 1^\circ > CH_3$

Works only for primary amides
Intermediate: Isocyanate (very important)

Reaction of amines with nitrous acid (HNO₂)

1°, 2°, 3° Amines with HNO₂

Primary (1°) Amine

(Just for comparison)

Aliphatic:

$RNH_2 + HNO_2 \rightarrow ROH + N_2 \uparrow + H_2O$

Brisk effervescence (N₂ gas)

Aromatic: forms diazonium salt

2° (Secondary) Amine

Reaction:

$R_2NH + HNO_2 \rightarrow R_2N–N=O + H_2O$ Product: N-nitrosamine

Observation:

Yellow oily liquid
No gas evolution

✔ Used as a test for 2° amines

3° (Tertiary) Amine

(A) Aliphatic 3° Amine:

$R_3N + HNO_2 \rightarrow R_3NH^+ NO_2^-$

Forms ammonium nitrite salt

Observation:

No visible reaction
No gas

(B) Aromatic 3° Amine:

Undergoes electrophilic substitution (ring reaction)

Example: aniline derivatives like dimethylaniline $\text{Ar–N(CH₃)₂} + HNO_2 \rightarrow p\text{-nitroso compound}$

✔ Green-colored solution (often)

Summary Table

Amine Type	Product	Observation
1°	Alcohol / Diazonium	N₂ gas
2°	N-nitrosamine	Yellow oily liquid
3° (aliphatic)	Salt	No reaction
3° (aromatic)	p-nitroso compound	Colored solution

Quick Trick

1° → Gas (N₂)
2° → Nitroso (yellow oil)
3° → No gas

Correct Reagent

We use NaNO₂ + HCl, but the actual reacting species is HNO₂ (nitrous acid)

Why?

HNO₂ is unstable, so it is not stored directly
It is generated in situ:

$NaNO_2 + HCl \rightarrow HNO_2 + NaCl$

Conclusion

✔ In reaction/mechanism → write HNO₂
✔ In practical/lab conditions → write NaNO₂ + HCl

emperature is very important in reactions with nitrous acid (HNO₂ / NaNO₂ + HCl).

Required Temperature

✔ Primary (1°) Aromatic Amines

0–5°C (ice-cold conditions)

Needed to form stable diazonium salt
Above 5°C → diazonium salt decomposes

Primary (1°) Aliphatic Amines

Also done at 0–5°C

But diazonium salt is unstable → immediately decomposes to alcohol + N₂

Secondary (2°) Amines

0–5°C preferred

Forms N-nitrosamine
Low temperature prevents side reactions

Tertiary (3°) Amines

Aliphatic: 0–5°C (no major reaction, just salt formation)
Aromatic: slightly higher temp possible, but usually cold conditions used

Summary Table

Amine Type	Temperature
1° Aromatic	0–5°C (very important)
1° Aliphatic	0–5°C
2°	0–5°C
3°	0–5°C (generally)

Basic strength of amines is a very important concept for NEET/JEE

What is basic strength?

Amines act as bases because of the lone pair on nitrogen (N) that can accept a proton (H⁺)

Key Concepts Affecting Basic Strength

1. Availability of Lone Pair

More available lone pair → stronger base
If lone pair is involved in resonance or bonding → weaker base

2. Inductive Effect (+I and –I)

Alkyl groups donate electrons (+I effect) → increase electron density on N → increase basicity

Order in gas phase: $3^\circ > 2^\circ > 1^\circ > NH_3$

But this changes in aqueous solution (see point 4).

3. Resonance Effect

If lone pair participates in resonance → less available → weaker base

Example:

Aniline is less basic because lone pair is delocalized in benzene ring

4. Solvation Effect (VERY IMPORTANT for NEET)

In aqueous solution, stability of protonated amine matters.

More H-bonding → more stable → stronger base
1° amines are better solvated than 3° amines

Order in aqueous solution: $2^\circ > 1^\circ > 3^\circ > NH_3$

5. Steric Hindrance

Bulky groups around N hinder protonation
More crowding → weaker base

This is why 3° amines are weaker in water

6. Hybridization of Nitrogen

$sp^3 > sp^2 > sp$

More s-character → electrons closer to nucleus → less available → weaker base

7. Aromatic vs Aliphatic Amines

Aliphatic amines → stronger base
Aromatic amines (like aniline) → weaker due to resonance

Important Orders to Remember

Gas Phase:

$3^\circ > 2^\circ > 1^\circ > NH_3$

Aqueous Solution:

$2^\circ > 1^\circ > 3^\circ > NH_3$

Aromatic vs Aliphatic:

$\text{Aliphatic amine} > \text{Aromatic amine}$

Basic strength in heterocyclic compounds

Core Idea (MOST IMPORTANT)

Is the nitrogen lone pair part of the aromatic system or not?

Case 1: Lone pair NOT involved in aromaticity → Basic

Example: Pyridine-type

Nitrogen is sp² hybridized
Lone pair lies in an sp² orbital (outside π-system)
Available for protonation → basic

Example:

Pyridine

Case 2: Lone pair involved in aromaticity → Very weak base

Example: Pyrrole-type

Lone pair is part of aromatic sextet (6π electrons)
Not available for H⁺ → very weak base

Example:

Pyrrole

Case 3: Multiple heteroatoms (competition effect)

Example: Imidazole, Pyrazole

One N behaves like pyridine (basic)
Other N behaves like pyrrole (non-basic)

So overall: moderately basic

key Rule

The nitrogen whose lone pair is NOT involved in aromaticity is basic.

Imidazole

N–H nitrogen (shown with H)
❌ Lone pair is part of aromatic sextet → NOT basic
Other nitrogen (without H)
✅ Lone pair is free → BASIC

Basic position = N without H (pyridine-like N)

Pyrazole

N–H nitrogen
❌ Lone pair involved in aromaticity → NOT basic
Adjacent nitrogen (without H)
✅ Lone pair free → BASIC

Basic position = N without H

Important Examples & Order

1. Pyridine vs Pyrrole

$\text{Pyridine} > \text{Pyrrole}$

✔ Pyridine → basic
✔ Pyrrole → almost non-basic

2. Aliphatic vs Aromatic Heterocycles

$\text{Aliphatic amine} > \text{Pyridine} > \text{Pyrrole}$

3. Imidazole (VERY IMPORTANT)

$\text{Imidazole} > \text{Pyridine}$

Reason: Extra stabilization of conjugate acid

Summary Table

Compound Type	Lone Pair	Basic Strength
Pyridine	Free	Moderate
Pyrrole	In aromaticity	Very weak
Imidazole	One free	Moderate–strong

Standard Basic Strength Order (Important)

$\text{Piperidine} > \text{Pyrrolidine} > \text{Imidazole} > \text{Pyridine} > \text{Pyrazole} > \text{Pyrrole}$

Piperidine>Pyrrolidine>Imidazole>Pyridine>Aniline>Pyrazole>Pyrrole

Answers:

45 NEET-level MCQs covering Solutions, Classification, d-block elements, and Coordination Compounds

SECTION 1: SOLUTIONS (10 MCQs)

Which of the following is an ideal solution?
A) Acetone + Chloroform
B) Benzene + Toluene
C) Ethanol + Water
D) Water + HCl
Raoult’s law is applicable to:
A) Non-volatile solutes
B) Electrolytes
C) Ideal solutions
D) Strong acids
Mass of glucose $(C_{6} H_{12} O_{6})$ required to be dissolved to prepare one litre of its solution which is isotonic with 15 g $L^{- 1}$ solution of urea $({NH}_{2} {CONH}_{2})$ is(Given: Molar mass in g ${mol}^{- 1}$ C : 12, H : 1, O : 16, N : 14)
A) 55g
B) 30g
C) 15g
D) 45g
The Henry’s law constant (KH) values of three gases (A, B, C) in water are 145, 2 × 10^-5 and 35 kbar, respectively. The solubility of these gases in water follow the order:
A) B > A > C (B) B > C > A (C) A > C > B (D) A > B > C
The plot of osmotic pressure ( ∏ ) vs concentration ( mol^{− 1} ) for a solution gives a straight line with slope 25.73 L bar mol⁻¹ . The temperature at which the osmotic pressure measurement is done is (Use R = 0.083 L bar mol⁻¹ K ⁻¹ )
A) 1
B) 2
C) 3
D) 0
Osmotic pressure is given by:

$\pi = iCRT$

A) Boyle’s law
B) Ideal gas equation
C) van’t Hoff equation
D) Dalton’s law

Maximum boiling azeotrope shows:
A) Positive deviation
B) Negative deviation
C) No deviation
D) Infinite deviation
Which increases boiling point?
A) Adding volatile solute
B) Adding non-volatile solute
C) Decreasing pressure
D) Increasing solvent
Henry’s law relates:
A) Pressure and solubility
B) Volume and temperature
C) Mass and density
D) Temperature and pressure
Which amongst the following aqueous solution of electrolytes will have minimum elevation in boiling point? Choose the correct option :-
A) 0.05 M NaCl
B) 0.1 M KCl
C) 0.1MgSO₄
D) 1M NaCl

SECTION 2: CLASSIFICATION OF ELEMENTS (10 MCQs)

Which of the following statements are true?
A) Unlike Ga that has a very high melting point, Cs has a very low melting point.
B) On Pauling scale, the electronegativity values of N and Cl are not the same
C) A r , K⁺ , C l⁻ , C a^{2 +}, and S ^{2 −}are all isoelectronic species.
D) The correct order of the first ionization enthalpies of N a , M g , A l and Si is S i > A l > M g > Na

12. Which among the following electronic configurations belong to main group elements?

Choose the correct answer from the option given below :

(1) B and E only

(2) A and C only

(3) D and E only

(4) A, C and D

13.The correct decreasing order of atomic radii (pm) of Li, Be, B and C is

(1) Be > Li > B > C

(2) Li > Be > B > C

(3) C > B > Be > Li

(4) Li > C > Be > B

14.Match List-I with List-II :

Identify the correct answer from the options given below :

(1) A-III, B-I, C-IV, D-II

(2) A-I, B-IV, C-III, D-II

(3) A-II, B-IV, C-I, D-III

(4) A-III, B-IV, C-I, D-II

15.Arrange the following elements in increasing order of first ionization enthalpy: Li, Be, B, C, N

Choose the correct answer from the options given below:

(1) Li < Be < B < C < N

(2) Li < B < Be < C < N

(3) Li < Be < C < B < N
(4) Li < Be < N < B < CR

16. Arrange the following elements in increasing order of electronegativity:

N, O, F, C, Si

Choose the correct answer from the options given below :

Choose the correct answer from the options given below:

(1) Si < C < N < O < F

(2) Si < C < O < N < F

(3) O < F < N < C < Si

(4) F < O < N < C < Si

17 Which of the following is correctly matched?

18. The correct sequence given below containing neutral, acidic, basic and amphoteric oxide each, respectively, is

Options: A. NO, ZnO, CO₂, CaO

B. ZnO, NO, CaO, CO₂

C. NO, CO₂, ZnO, CaO

D. NO, CO₂, CaO, ZnO

19. For elements B, C, N, Li, Be, O and F the correct order of first ionization enthalpy is

(1) Li < Be < B < C < N < O < F

(2) B > Li > Be > C > N > O > F

(3) Li < B < Be < C < O < N < F

(4) Li < Be < B < C < O < N < F

20. Decrease in size from left to right in actinoid series is greater and gradual than that in lanthanoid series due to

1.4f orbitals are second to last

2. 4f orbitals have greater shielding effect

3. 5f orbitals have poor shielding effect

4.5f orbitals have greater shielding effect

SECTION 3: d-BLOCK ELEMENTS (10 MCQs)

Transition elements have:
A) Completely filled d-orbitals
B) Partially filled d-orbitals
C) No d-orbitals
D) Only s-orbitals
General configuration:
A) (n-1)d¹–¹⁰ ns¹–²
B) ns² np⁶
C) ns¹
D) np⁵
Highest oxidation state of Mn:
A) +2
B) +4
C) +7
D) +6
Color in transition metals due to:
A) s-electrons
B) d-d transition
C) p-electrons
D) nuclear charge
Which is not transition element?
A) Zn
B) Fe
C) Cu
D) Cr
Catalytic property due to:
A) Low density
B) Variable oxidation states
C) High melting point
D) Atomic size
Magnetic behavior depends on:
A) paired electrons
B) unpaired electrons
C) protons
D) neutrons
Which is diamagnetic?
A) Fe²⁺
B) Mn²⁺
C) Zn²⁺
D) Cu²⁺
Alloy of Cu and Zn:
A) Bronze
B) Brass
C) Steel
D) Solder
Highest melting point:
A) Hg
B) Fe
C) W
D) Cu

SECTION 4: COORDINATION COMPOUNDS (15 MCQs)

Coordination number of central atom is:
A) number of ligands
B) number of bonds
C) number of donor atoms
D) oxidation state
Ligand donating one pair:
A) Bidentate
B) Monodentate
C) Tridentate
D) Polydentate
Example of bidentate ligand:
A) NH₃
B) H₂O
C) en
D) Cl⁻
IUPAC name of [Cu(NH₃)₄]²⁺:
A) Copper tetraammine
B) Tetraammine copper(II)
C) Ammine copper
D) Copper ammonia
Oxidation state of Fe in [Fe(CN)₆]³⁻:
A) +2
B) +3
C) +6
D) −3
Coordination compounds show isomerism:
A) Yes
B) No
C) Rarely
D) Only optical
Geometrical isomerism in:
A) tetrahedral
B) square planar
C) linear
D) trigonal
Strong field ligand:
A) Cl⁻
B) H₂O
C) CN⁻
D) Br⁻
Crystal field splitting occurs in:
A) s-orbitals
B) p-orbitals
C) d-orbitals
D) f-orbitals
High spin complex has:
A) pairing
B) maximum unpaired electrons
C) no electrons
D) stable structure
Hybridization of [Ni(CN)₄]²⁻:
A) sp³
B) dsp²
C) sp²
D) d²sp³
Werner theory explains:
A) valency
B) bonding
C) structure of complexes
D) reactivity
Chelate effect increases:
A) instability
B) stability
C) reactivity
D) acidity
[Co(NH₃)₆]³⁺ is:
A) paramagnetic
B) diamagnetic
C) ferromagnetic
D) antiferromagnetic
Which shows optical isomerism?
A) [Co(en)₃]³⁺
B) [NiCl₄]²⁻
C) [PtCl₄]²⁻
D) [Zn(NH₃)₄]²⁺

Organic chemistry – Some Basic Principles and techniques JEE/NEET Concepts

Methods of Purification of Organic Compounds

Basic Idea of Wedge–Dash Notation

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

Conformations of Propane (JEE/NEET Concepts)

Stereoisomerism organic chemistry JEE/NEET Concepts

Key Concepts to Arrange Nucleophile Strength (NEET/JEE)

Key Concepts to Arrange Basic Strength (NEET/JEE)

Degree of Unsaturation (DU) formula for organic compound

Methods of Purification of Organic Compounds

Sublimation

What is Sublimation?

Sublimation is the process in which a solid directly changes into vapour without becoming liquid.

Example:

Ammonium chloride
Iodine
Camphor
Naphthalene

Why does it happen?

Some solids have:

Weak intermolecular forces
High vapour pressure even at lower temperatures

So, when heated, they directly escape into vapour form.

Sublimation as a Separation Technique

It is used to purify substances:

✔ Sublimable substance → turns into vapour
✔ Non-sublimable impurities → remain as solid

Then vapour is cooled → forms pure solid again.

Simple Setup

Heat the mixture
Sublimable substance vaporizes
Vapour condenses on a cold surface
Pure solid is collected

Example Use

Separating:

Ammonium chloride + sand

Ammonium chloride sublimates, sand stays behind.

Crystallisation

What is Crystallisation?

A purification technique used to purify solid organic compounds based on difference in solubility.

Principle

Substance is less soluble at low temperature
Substance is more soluble at high temperature

So:

Dissolve at high temp
Cool → pure crystals form

Steps

Dissolve impure solid in hot solvent
Make a saturated solution
Filter (if needed, remove insoluble impurities)
Cool the solution
Pure crystals separate out
Filter and dry crystals

Important Terms

Mother liquor → remaining liquid after crystallisation
Activated charcoal → removes colored impurities

🔹 Examples

Example 1: Benzoic Acid purification

Impure benzoic acid dissolved in hot water
On cooling → pure crystals of benzoic acid form

When repeated crystallisation is needed?

When

Simple Distillation (large boiling point difference)

When used:

Difference in b.p. > 25–30°C
Or liquid + non-volatile impurity

Examples:

Example 1:

Water + salt
Water distils, salt remains

Example 2:

Chloroform (334 K) + Aniline (457 K)
Easily separated

Example 3:

Alcohol + sugar solution
Alcohol vaporises, sugar stays

Example 4:

Acetone + water
Acetone (low b.p.) comes first

Fractional Distillation (close boiling points)

When used:

Difference in b.p. < 25°C

Examples:

Example 1:

Ethanol (78°C) + Water (100°C)
Cannot separate by simple distillation

Example 2:

Benzene (80°C) + Toluene (110°C)
Need fractionating column

Example 3:

Hexane + Heptane
Very close boiling points

Example 4 (Industrial):

Crude oil refining
Petrol, diesel, kerosene separated

3. Steam Distillation (heat-sensitive substances)

When used:

Substance:
- Immiscible with water
- Volatile with steam
- Decomposes at high temperature

Examples:

Example 1:

Extraction of essential oils
From plants (e.g., clove oil, eucalyptus oil)

Example 2:

Aniline purification
High b.p. but steam volatile

Example 3:

Nitrobenzene
Separated using steam

Example 4:

Rose oil extraction (attar)
Used in perfumes

Quick Comparison (Very Important)

Method	Condition	Example
Simple Distillation	Large b.p. difference	Water + salt
Fractional Distillation	Small b.p. difference	Ethanol + water
Steam Distillation	Heat-sensitive, steam volatile	Essential oils

Distillation Under Reduced Pressure (Vacuum Distillation)

Statement:

Distillation under reduced pressure → Glycerol + spent-lye

Why this method is used?

Some liquids:

Have very high boiling points
Decompose before boiling

Example: Glycerol

Principle

When pressure is reduced:
➡ Boiling point decreases

So, substance can distil at lower temperature without decomposition

Application in given case

Mixture:

Glycerol (high b.p., heat sensitive)
Spent-lye (impurities)

What happens?

Pressure ↓
Glycerol boils at lower temp
Vapours collected → pure glycerol
Impurities remain

Why not simple distillation?

❌ Glycerol decomposes at high temperature
✔ So vacuum distillation is required

🔹 More Examples

Example 1:

Glycerol purification (most common)

Example 2:

High boiling oils
Example 3:
Fatty acids

Example 4:

Petroleum residues

Quick Summary

Method	Used for
Simple distillation	Low b.p. liquids
Fractional distillation	Close b.p. liquids
Steam distillation	Heat-sensitive, steam volatile
Reduced pressure distillation	High b.p., decomposing liquids

Differential Extraction

When an organic compound is present in an aqueous medium, it is separated by shaking it with an organic solvent in which it is more soluble than in water. The organic solvent and aqueous solution should be immiscible so that they form two distinct layers which can be separated using a separating funnel. The organic solvent is later removed by distillation or evaporation to obtain the compound.

Concept: Differential (Liquid–Liquid) Extraction

Principle

Based on difference in solubility of a compound in two immiscible liquids

Usually:

Water (aqueous layer)
Organic solvent (like ether, benzene)

Key Conditions

✔ Two liquids must be immiscible
✔ Compound should be more soluble in organic solvent

Process (Separating Funnel)

Take mixture in separating funnel
Add organic solvent
Shake well
Allow layers to separate

Two layers form:

Upper layer (usually organic solvent)
Lower layer (water)

Separate layers
Evaporate solvent → get pure compound

Example

Example 1:

Iodine in water + carbon tetrachloride (CCl₄)
→ Iodine moves to organic layer

Example 2:

Benzoic acid from water using ether

Continuous Extraction

When used?

When compound is slightly soluble in solvent

Why?

Single extraction inefficient
Repeated extraction increases yield

Same solvent reused again and again

Advantages

Efficient separation
Better recovery of compound
Widely used in organic chemistry labs

Chromatography

What is Chromatography?

A separation technique where components of a mixture are separated based on their different affinities between two phases.

Principle

Based on distribution of components between two phases:

Stationary phase → fixed (solid or liquid)
Mobile phase → moving (liquid or gas)

Different substances move at different speeds → separation occurs

Types of Chromatography

1. Paper Chromatography

Stationary phase → paper
Mobile phase → solvent

Used for:

Ink separation
Plant pigments

2. Thin Layer Chromatography (TLC)

Stationary phase → silica gel layer
Faster and more accurate than paper chromatography

3. Column Chromatography

Column filled with adsorbent (silica/alumina)
Used for larger scale separation

4. Gas Chromatography (GC)

Mobile phase → gas
Used for volatile compounds

How Separation Happens?

Component with more attraction to stationary phase → moves slowly
Component with more attraction to mobile phase → moves faster

Example

Ink Separation

Black ink → separates into different colors on paper

Plant Pigments

Chlorophyll, carotene separated using chromatography

Important Term: Rf Value

$Rf = \frac{\text{Distance travelled by substance}}{\text{Distance travelled by solvent}}$

Uses

✔ Separation of mixtures
✔ Purification
✔ Checking purity
✔ Drug testing
✔ Food analysis

Stephen Reaction (Stephen Aldehyde Synthesis)

tephen Reaction (Stephen Aldehyde Synthesis) – Mechanism (JEE/NEET Concept)

Definition:
Conversion of nitriles (R–C≡N) into aldehydes (R–CHO) using SnCl₂/HCl, followed by hydrolysis.

Overall Reaction

$R-C \equiv N \xrightarrow[\text{HCl}]{\text{SnCl}_2} R-CH=NH \cdot HCl \xrightarrow{\text{H}_2O} R-CHO$

Stepwise Mechanism

Step 1: Formation of Iminium Salt (Reduction Step)

SnCl₂ (mild reducing agent) donates electrons.
Nitrile gets partially reduced.
Formation of iminium chloride salt:

$R-C \equiv N \xrightarrow{\text{SnCl}_2/HCl} R-CH=NH \cdot HCl$

Key point:

Triple bond (C≡N) → double bond (C=NH)
Controlled reduction (not all the way to amine)

Step 2: Hydrolysis of Iminium Salt

Iminium salt reacts with water.
Produces aldehyde + ammonium salt

$R-CH=NH \cdot HCl + H_2O \rightarrow R-CHO + NH_4Cl$ R−CH=NH⋅HCl+H2O→R−CHO+NH4Cl

Important Points (Exam-Oriented)

Works best for alkyl nitriles (aryl nitriles give poor yield).
Stops at aldehyde stage (unlike strong reduction → amine).
SnCl₂/HCl = selective mild reducing system
Intermediate = iminium salt (not imine directly)

Shortcut Trick (Memory Tip)

“Nitrile + Mild Reduction = Aldehyde”
Think: Stephen = Stops Early (Aldehyde, not amine)

Example

$CH_3CN \xrightarrow{\text{SnCl}_2/HCl} CH_3CHO$