Class 11 Chemistry

Chapter 8: Organic Chemistry – Some Basic Principles and TechniquesComplete Question Bank (MCQ + Fill in the Blanks + True/False + Assertion Reason + Short Answer)

PART A: Multiple Choice Questions (MCQs)

1. Organic chemistry is mainly the study of compounds containing:

A) Oxygen
B) Carbon
C) Nitrogen
D) Hydrogen

Answer: B) Carbon

2. The property of carbon by which it forms long chains is called:

A) Hybridisation
B) Catenation
C) Isomerism
D) Resonance

Answer: B) Catenation

3. Carbon is tetravalent because it has:

A) 2 valence electrons
B) 3 valence electrons
C) 4 valence electrons
D) 8 valence electrons

Answer: C) 4 valence electrons

4. Urea was first prepared artificially by:

A) Berzelius
B) Wöhler
C) Kekulé
D) Dalton

Answer: B) Wöhler

5. Wöhler prepared urea from:

A) Methane
B) Ammonium cyanate
C) Carbon dioxide
D) Ethane

Answer: B) Ammonium cyanate

6. The hybridisation of carbon in methane is:

A) sp
B) sp²
C) sp³
D) dsp²

Answer: C) sp³

7. The shape of an sp³ hybridised carbon atom is:

A) Linear
B) Planar
C) Tetrahedral
D) Angular

Answer: C) Tetrahedral

8. Bond angle in methane is:

A) 90°
B) 109.5°
C) 120°
D) 180°

Answer: B) 109.5°

9. Ethene contains carbon atoms with:

A) sp hybridisation
B) sp² hybridisation
C) sp³ hybridisation
D) d²sp³ hybridisation

Answer: B) sp² hybridisation

10. Ethyne contains carbon atoms with:

A) sp
B) sp²
C) sp³
D) sp³d

Answer: A) sp

11. A double bond consists of:

A) Two σ bonds
B) Two π bonds
C) One σ and one π bond
D) One π bond only

Answer: C) One σ and one π bond

12. A triple bond contains:

A) 3σ
B) 2σ + 1π
C) 1σ + 2π
D) 3π

Answer: C) 1σ + 2π

13. Sigma bond is formed by:

A) Sidewise overlap
B) Head-on overlap
C) Electron transfer
D) Hydrogen bonding

Answer: B) Head-on overlap

14. Pi bond is weaker than sigma bond because:

A) It has more electrons
B) It has less overlap
C) It is ionic
D) It is non-polar

Answer: B) It has less overlap

15. The general formula of alkanes is:

A) CₙH₂ₙ
B) CₙH₂ₙ₋₂
C) CₙH₂ₙ₊₂
D) CₙHₙ

Answer: C) CₙH₂ₙ₊₂

16. The general formula of alkenes is:

A) CₙH₂ₙ
B) CₙH₂ₙ₊₂
C) CₙH₂ₙ₋₂
D) CₙHₙ

Answer: A) CₙH₂ₙ

17. The functional group of alcohol is:

A) –CHO
B) –COOH
C) –OH
D) –NH₂

Answer: C) –OH

18. The functional group of carboxylic acid is:

A) –COOH
B) –OH
C) –CHO
D) –CO–

Answer: A) –COOH

19. Successive members of a homologous series differ by:

A) CH₃
B) CH₂
C) H₂
D) C₂H₂

Answer: B) CH₂

20. The IUPAC name of CH₃CH₂OH is:

A) Methanol
B) Ethanol
C) Propanol
D) Ethanal

Answer: B) Ethanol

PART B: Fill in the Blanks

Carbon shows the property of forming chains called ________.

Answer: Catenation

The modern organic chemistry started after the synthesis of ________ by Wöhler.

Answer: Urea

Carbon has ______ valency.

Answer: Four

sp³ hybridisation has ______ geometry.

Answer: Tetrahedral

The bond angle in sp² hybridisation is ______.

Answer: 120°

The bond angle in sp hybridisation is ______.

Answer: 180°

A double bond contains one ______ and one ______ bond.

Answer: Sigma, Pi

The functional group of aldehyde is ______.

Answer: –CHO

The functional group of ketone is ______.

Answer: >C=O

The suffix used for alcohol is ______.

Answer: –ol

The suffix used for aldehyde is ______.

Answer: –al

The suffix used for ketone is ______.

Answer: –one

Compounds having same molecular formula but different structures are called ______.

Answer: Isomers

Benzene is an example of ______ compound.

Answer: Aromatic

The formula used for calculating TLC movement is called ______ value.

Answer: Rf

Carbocation contains a ______ charge.

Answer: Positive

Carbanion contains a ______ charge.

Answer: Negative

Free radicals contain an ______ electron.

Answer: Unpaired

The strongest electron withdrawing effect is called ______ effect.

Answer: –I

Nitrogen detection test gives ______ blue colour.

Answer: Prussian

PART C: True or False

Carbon is tetravalent.
✅ True
sp hybridisation gives tetrahedral geometry.
❌ False
π bonds are stronger than σ bonds.
❌ False
Alkenes contain double bonds.
✅ True
Functional groups determine chemical properties.
✅ True
Homologous series members differ by CH₂.
✅ True
Ethyne has sp² hybridisation.
❌ False
Benzene is an aromatic compound.
✅ True
Carbocations are negatively charged species.
❌ False
Carbanions are electron-rich species.
✅ True
Chromatography is used for purification.
✅ True
Sublimation is used for liquids.
❌ False
Nitrogen gives Prussian blue colour test.
✅ True
Aldehydes contain –COOH group.
❌ False
Triple bond contains two π bonds.
✅ True

PART D: Assertion–Reason Questions

1. Assertion:

Carbon forms millions of compounds.

Reason:
Carbon shows catenation.

Answer: Both Assertion and Reason are true, and Reason correctly explains Assertion.

2. Assertion:

sp hybridised carbon is linear.

Reason:
It has bond angle of 180°.

Answer: Both are true and Reason explains Assertion.

3. Assertion:

Carbocations are stabilised by alkyl groups.

Reason:
Alkyl groups show +I effect.

Answer: Both are true and Reason explains Assertion.

4. Assertion:

Alkenes undergo addition reactions.

Reason:
They contain π bonds.

Answer: Both are true and Reason explains Assertion.

PART E: One Word Answer Questions

Father of modern organic chemistry?
Wöhler
Number of bonds formed by carbon?
Four
Shape of sp carbon?
Linear
Shape of sp² carbon?
Trigonal planar
Functional group of amines?
–NH₂
Purification method based on boiling point?
Distillation
Purification method using adsorption?
Chromatography
Positive carbon ion?
Carbocation
Negative carbon ion?
Carbanion
Unpaired electron species?
Free radical

Part 2: Higher-Level MCQs + IUPAC Practice + Numerical + Case-Based Questions

SECTION A: Advanced MCQs (NEET/JEE Level)

1. The hybridisation of carbon in CO₂ is:

A) sp³
B) sp²
C) sp
D) dsp²

Answer: C) sp

2. The increasing order of s-character is:

A) sp > sp² > sp³
B) sp³ > sp² > sp
C) sp² > sp > sp³
D) sp³ > sp > sp²

Answer: B) sp³ > sp² > sp

3. Which bond allows free rotation?

A) π bond
B) σ bond
C) Double bond
D) Triple bond

Answer: B) σ bond

4. The number of σ and π bonds in ethene (C₂H₄) are:

A) 4σ, 2π
B) 5σ, 1π
C) 6σ, 1π
D) 5σ, 2π

Answer: B) 5σ, 1π

5. The number of σ and π bonds in ethyne (C₂H₂) are:

A) 3σ, 2π
B) 2σ, 2π
C) 5σ, 1π
D) 1σ, 3π

Answer: A) 3σ, 2π

6. Which compound shows chain isomerism?

A) Ethane
B) Propane
C) Butane
D) Methane

Answer: C) Butane

7. Which compound shows functional group isomerism?

A) Ethanol and dimethyl ether
B) Butane and isobutane
C) But-1-ene and But-2-ene
D) Propane and propene

Answer: A) Ethanol and dimethyl ether

8. The IUPAC name of:

CH₃–CH(CH₃)–CH₂–CH₃

is:

A) Pentane
B) 2-Methylbutane
C) 3-Methylbutane
D) Isopentane

Answer: B) 2-Methylbutane

9. The IUPAC name of:

CH₃–CH₂–CH₂–OH

is:

A) Propan-1-ol
B) Propanol
C) Ethanol
D) Propanal

Answer: A) Propan-1-ol

10. The IUPAC name of:

CH₃COCH₂CH₃

is:

A) Butanal
B) Butan-1-one
C) Butan-2-one
D) Butanoic acid

Answer: C) Butan-2-one

SECTION B: IUPAC Naming Practice

Give IUPAC names:

1.

CH₃–CH₂–CH₂–CH₃

Answer: Butane

2.

CH₃–CH(CH₃)–CH₃

Answer: 2-Methylpropane

3.

CH₃–CH₂–OH

Answer: Ethanol

4.

CH₃–CHO

Answer: Ethanal

5.

CH₃–CO–CH₃

Answer: Propan-2-one

6.

CH₂=CH–CH₃

Answer: Prop-1-ene

7.

CH≡C–CH₃

Answer: Prop-1-yne

8.

CH₃CH₂COOH

Answer: Propanoic acid

SECTION C: Reaction-Based Questions

1. The reaction:

CH₄ + Cl₂ → CH₃Cl + HCl

is an example of:

A) Addition
B) Elimination
C) Substitution
D) Rearrangement

Answer: C) Substitution

2. Ethene reacts with hydrogen to form ethane. This is:

A) Addition reaction
B) Substitution reaction
C) Elimination reaction
D) Rearrangement reaction

Answer: A) Addition reaction

3. Ethanol on heating with conc. H₂SO₄ gives ethene. This reaction is:

A) Substitution
B) Elimination
C) Addition
D) Oxidation

Answer: B) Elimination

SECTION D: Stability-Based Questions

1. Correct order of carbocation stability is:

A) 1° > 2° > 3°
B) 3° > 2° > 1°
C) 2° > 1° > 3°
D) 1° > 3° > 2°

Answer: B) 3° > 2° > 1°

2. The most stable carbocation is:

A) CH₃⁺
B) CH₃CH₂⁺
C) (CH₃)₂CH⁺
D) (CH₃)₃C⁺

Answer: D) (CH₃)₃C⁺

3. The stability of carbanions follows:

A) 3° > 2° > 1°
B) 1° > 2° > 3°
C) 2° > 3° > 1°
D) 3° > 1° > 2°

Answer: B) 1° > 2° > 3°

SECTION E: Electronic Effect Questions

1. Electron withdrawing groups show:

A) +I effect
B) –I effect
C) +R effect
D) Hyperconjugation

Answer: B) –I effect

2. Alkyl groups show:

A) –I effect
B) +I effect
C) –R effect
D) Electromeric effect

Answer: B) +I effect

3. Resonance involves:

A) Movement of σ electrons
B) Delocalisation of π electrons
C) Transfer of protons
D) Breaking of ionic bonds

Answer: B) Delocalisation of π electrons

SECTION F: Numerical Questions

1. Calculate percentage of carbon if 4.4 g CO₂ is obtained from 2.0 g organic compound.

Formula: $\%C=\frac{12}{44}\times\frac{4.4}{2}\times100$ %C=4412×24.4×100

Answer: 60%

2. Calculate percentage of hydrogen if 1.8 g water is formed from 2 g compound.

$\%H=\frac{2}{18}\times\frac{1.8}{2}\times100$ %H=182×21.8×100

Answer: 10%

SECTION G: Case-Based Questions

Case 1:

A student analyses an organic compound. It contains carbon, hydrogen and nitrogen. On sodium fusion, the extract gives Prussian blue colour.

Questions:

1. Which element is present?

Answer: Nitrogen

2. Which test is performed?

Answer: Lassaigne’s test

3. Which compound causes the blue colour?

Answer: Prussian blue complex

Case 2:

An unknown liquid mixture contains two liquids having different boiling points.

Questions:

1. Which purification method is used?

Answer: Distillation

2. If boiling points are close, which method is preferred?

Answer: Fractional distillation

3. Principle of this method?

Answer: Difference in boiling points

SECTION H: Match the Following

Column A	Column B
Alcohol	–OH
Aldehyde	–CHO
Ketone	>C=O
Acid	–COOH
Amine	–NH₂

Answer:

Alcohol → –OH
Aldehyde → –CHO
Ketone → >C=O
Acid → –COOH
Amine → –NH₂

SECTION I: Important NEET/JEE One-Liners

Strongest bond in organic compounds → σ bond
Carbon with four different groups → Chiral carbon
Mirror image isomers → Enantiomers
Electron donor → Nucleophile
Electron acceptor → Electrophile
Purification by adsorption → Chromatography
TLC is based on → Adsorption
Nitrogen estimation method → Dumas/Kjeldahl
Halogen estimation method → Carius method
Sulphur is estimated as → BaSO₄

Part 3: Most Expected Board Exam Questions (2 Marks, 3 Marks, 5 Marks)

SECTION A: Very Short Answer Questions (1–2 Marks)

1. Define organic chemistry.

Answer:
Organic chemistry is the branch of chemistry that deals with the study of carbon compounds, especially hydrocarbons and their derivatives.

2. What is catenation?

Answer:
Catenation is the ability of carbon atoms to form covalent bonds with other carbon atoms, producing chains and rings.

3. Why does carbon form a large number of compounds?

Answer:

Carbon forms a large number of compounds because of:

Tetravalency
Catenation
Ability to form single, double and triple bonds

4. What is a functional group?

Answer:
A functional group is an atom or group of atoms responsible for the characteristic chemical properties of an organic compound.

Example:

–OH in alcohols
–COOH in acids

5. Define homologous series.

Answer:
A homologous series is a group of organic compounds having the same functional group and general formula, where consecutive members differ by –CH₂–.

6. What is hybridisation?

Answer:
Hybridisation is the mixing of atomic orbitals of similar energy to form new hybrid orbitals.

7. What is the hybridisation of carbon in methane?

Answer:
Carbon in methane is sp³ hybridised.

8. Give the shape of sp² hybridised carbon.

Answer:
Trigonal planar shape with bond angle 120°.

9. What is isomerism?

Answer:
The phenomenon in which compounds have the same molecular formula but different structures or arrangements is called isomerism.

10. What is a chiral carbon?

Answer:
A carbon atom attached to four different groups is called a chiral carbon.

SECTION B: Short Answer Questions (2–3 Marks)

1. Explain the types of hybridisation of carbon.

Answer:

sp³ Hybridisation:

Four orbitals mix.
Shape: Tetrahedral
Bond angle: 109.5°
Example: CH₄

sp² Hybridisation:

Three orbitals mix.
Shape: Trigonal planar
Bond angle: 120°
Example: C₂H₄

sp Hybridisation:

Two orbitals mix.
Shape: Linear
Bond angle: 180°
Example: C₂H₂

2. Differentiate between sigma and pi bonds.

Sigma Bond	Pi Bond
Formed by head-on overlap	Formed by sideways overlap
Stronger bond	Weaker bond
Allows free rotation	Restricts rotation
Present in all single bonds	Present in multiple bonds

3. Explain the classification of organic compounds.

Answer:

Organic compounds are classified as:

Acyclic compounds:

Open-chain compounds.

Example:
Ethane

Cyclic compounds:

Ring compounds.

Types:

Alicyclic compounds
Aromatic compounds
Heterocyclic compounds

4. Explain chain isomerism with example.

Answer:

Chain isomerism occurs when compounds have the same molecular formula but different carbon skeletons.

Example:

C₄H₁₀:

n-Butane
Isobutane

5. Explain position isomerism.

Answer:

Position isomerism occurs when the position of a functional group or multiple bond changes.

Example:

C₃H₈O:

Propan-1-ol
Propan-2-ol

6. Explain inductive effect.

Answer:

The permanent displacement of σ electrons due to electronegativity difference is called inductive effect.

Types:

+I Effect:

Electron releasing groups.

Example:
Alkyl groups

–I Effect:

Electron withdrawing groups.

Example:
–NO₂, –Cl

7. Explain resonance effect.

Answer:

Resonance is the delocalisation of π electrons or lone pair electrons in a molecule.

Effects:

Increases stability
Spreads electron density

Example:
Benzene

8. Explain hyperconjugation.

Answer:

Hyperconjugation is the delocalisation of electrons of a C–H bond adjacent to a double bond or positive charge.

It increases stability of:

Carbocations
Alkenes

9. Explain carbocation stability.

Answer:

Carbocations are stabilised by alkyl groups due to +I effect and hyperconjugation.

Stability order: $3^\circ > 2^\circ > 1^\circ > CH_3^+$ 3∘>2∘>1∘>CH3+

10. Explain purification by crystallisation.

Answer:

Crystallisation is used to purify solids based on difference in solubility.

Steps:

Dissolve compound in hot solvent.
Filter impurities.
Cool solution.
Collect crystals.

SECTION C: Long Answer Questions (5 Marks)

1. Explain IUPAC nomenclature rules.

Answer:

Steps:

Select the longest carbon chain.
Number the chain from the nearest functional group.
Identify substituents.
Write substituent names alphabetically.
Add suffix according to functional group.

Example:

CH₃–CH(CH₃)–CH₃

Name:

2-Methylpropane

2. Explain different types of organic reactions.

Answer:

1. Substitution Reaction

Replacement of an atom/group by another atom/group.

Example:

CH₄ + Cl₂ → CH₃Cl + HCl

2. Addition Reaction

Addition of atoms across double/triple bonds.

Example:

C₂H₄ + H₂ → C₂H₆

3. Elimination Reaction

Removal of atoms/groups forming multiple bonds.

Example:

Ethanol → Ethene + Water

4. Rearrangement Reaction

Change in arrangement of atoms forming a new structure.

3. Explain qualitative analysis of organic compounds.

Answer:

Qualitative analysis identifies elements present.

Carbon and Hydrogen:

Carbon → CO₂
Hydrogen → H₂O

Nitrogen:

Sodium fusion test
Gives Prussian blue colour

Sulphur:

Gives violet colour or black precipitate

Halogens:

Silver nitrate test
Gives AgCl, AgBr, AgI precipitates

4. Explain quantitative analysis methods.

Answer:

Carbon and Hydrogen:

Estimated by combustion method.

Nitrogen:

Methods:

Dumas method
Kjeldahl method

Halogens:

Estimated by Carius method.

Sulphur:

Estimated by precipitation as BaSO₄.

5. Explain stereoisomerism.

Answer:

Stereoisomerism occurs when compounds have the same connectivity but different spatial arrangements.

Types:

Geometrical Isomerism:

Cis form
Trans form

Example:
But-2-ene

Optical Isomerism:

Occurs due to chiral carbon.

Produces:

Enantiomers

SECTION D: Most Expected Board Questions ⭐

Explain catenation and tetravalency of carbon.
Write IUPAC names of given organic compounds.
Explain types of structural isomerism with examples.
Explain electronic effects in organic molecules.
Compare carbocations, carbanions and free radicals.
Explain methods of purification of organic compounds.
Explain Lassaigne’s test.
Explain chromatography and Rf value.
Explain preparation and importance of homologous series.
Explain different types of organic reactions.

Part 4: Full Chapter Practice Test (100 MCQs with Answer Key)

Section A: Basic Concepts (1–25)

1. The branch of chemistry dealing with carbon compounds is:

A) Physical chemistry
B) Organic chemistry
C) Inorganic chemistry
D) Analytical chemistry

Ans: B

2. Carbon belongs to which group of the periodic table?

A) Group 1
B) Group 2
C) Group 14
D) Group 18

Ans: C

3. Carbon has atomic number:

A) 4
B) 6
C) 12
D) 14

Ans: B

4. Carbon forms stable compounds because of:

A) Low electronegativity
B) Tetravalency and catenation
C) High atomic mass
D) Radioactivity

Ans: B

5. The first organic compound prepared artificially was:

A) Methane
B) Benzene
C) Urea
D) Ethanol

Ans: C

6. The scientist who prepared urea artificially was:

A) Kekulé
B) Wöhler
C) Dalton
D) Arrhenius

Ans: B

7. The ability of carbon to form long chains is:

A) Resonance
B) Isomerism
C) Catenation
D) Hybridisation

Ans: C

8. The valence electrons in carbon are:

A) 2
B) 3
C) 4
D) 6

Ans: C

9. Organic compounds mainly contain:

A) Carbon and hydrogen
B) Metals only
C) Noble gases
D) Radioactive elements

Ans: A

10. Methane contains:

A) sp hybridised carbon
B) sp² hybridised carbon
C) sp³ hybridised carbon
D) Unhybridised carbon

Ans: C

11. Bond angle in ethyne is:

A) 90°
B) 109.5°
C) 120°
D) 180°

Ans: D

12. Bond angle in ethene is:

A) 180°
B) 120°
C) 109.5°
D) 90°

Ans: B

13. The number of σ bonds in methane is:

A) 1
B) 2
C) 3
D) 4

Ans: D

14. Rotation around which bond is restricted?

A) σ bond
B) π bond
C) Single bond
D) Ionic bond

Ans: B

15. The strongest covalent bond is:

A) π bond
B) σ bond
C) Hydrogen bond
D) Coordinate bond

Ans: B

16. The formula of alkane is:

A) CₙH₂ₙ
B) CₙH₂ₙ₊₂
C) CₙH₂ₙ₋₂
D) CₙHₙ

Ans: B

17. The formula of alkyne is:

A) CₙH₂ₙ₊₂
B) CₙH₂ₙ
C) CₙH₂ₙ₋₂
D) CₙHₙ₊₂

Ans: C

18. Benzene is:

A) Alicyclic
B) Aromatic
C) Saturated
D) Heterocyclic

Ans: B

19. The functional group of alcohol is:

A) –COOH
B) –CHO
C) –OH
D) –NH₂

Ans: C

20. The functional group of aldehyde is:

A) –CHO
B) –COOH
C) –OH
D) –CO–

Ans: A

21. Ketones contain:

A) –NH₂
B) >C=O
C) –OH
D) –COOH

Ans: B

22. Amines contain:

A) –NH₂
B) –CHO
C) –OH
D) –COOH

Ans: A

23. Members of a homologous series differ by:

A) CH₃
B) CH₂
C) OH
D) COOH

Ans: B

24. Same molecular formula but different structures are:

A) Isotopes
B) Isomers
C) Homologues
D) Radicals

Ans: B

25. Chain isomerism is due to difference in:

A) Functional group
B) Carbon skeleton
C) Molecular mass
D) Element composition

Ans: B

Section B: Intermediate Level (26–50)

26. Ethanol and dimethyl ether show:

A) Chain isomerism
B) Position isomerism
C) Functional isomerism
D) Optical isomerism

Ans: C

27. Cis-trans isomerism is a type of:

A) Structural isomerism
B) Geometrical isomerism
C) Functional isomerism
D) Chain isomerism

Ans: B

28. A carbon attached to four different groups is:

A) Primary carbon
B) Chiral carbon
C) Carbonyl carbon
D) Radical carbon

Ans: B

29. Mirror image isomers are:

A) Homologues
B) Enantiomers
C) Metamers
D) Tautomers

Ans: B

30. The +I effect is shown by:

A) NO₂
B) Cl
C) Alkyl groups
D) COOH

Ans: C

31. The –I effect is shown by:

A) CH₃
B) C₂H₅
C) NO₂
D) Alkyl groups

Ans: C

32. Resonance involves movement of:

A) Protons
B) π electrons
C) Neutrons
D) Atoms only

Ans: B

33. Hyperconjugation is also called:

A) Hydrogen bonding
B) No bond resonance
C) Ionic effect
D) Electromeric effect

Ans: B

34. Most stable carbocation is:

A) CH₃⁺
B) Primary
C) Secondary
D) Tertiary

Ans: D

35. Carbocations are:

A) Electron rich
B) Electron deficient
C) Neutral
D) Stable always

Ans: B

36. Carbanions contain:

A) Positive charge
B) Negative charge
C) No electron
D) No charge

Ans: B

37. Free radicals contain:

A) Lone pair
B) Positive charge
C) Unpaired electron
D) Negative charge

Ans: C

38. Correct carbocation stability order:

A) 1° > 2° > 3°
B) 3° > 2° > 1°
C) 2° > 3° > 1°
D) 1° > 3° > 2°

Ans: B

39. Reaction involving replacement is:

A) Addition
B) Substitution
C) Elimination
D) Rearrangement

Ans: B

40. Ethene + H₂ → Ethane is:

A) Addition
B) Elimination
C) Substitution
D) Rearrangement

Ans: A

41. Removal of atoms producing double bond is:

A) Addition
B) Substitution
C) Elimination
D) Reduction

Ans: C

42. Electron-rich species are:

A) Electrophiles
B) Nucleophiles
C) Radicals
D) Carbocations

Ans: B

43. Electron-deficient species are:

A) Nucleophiles
B) Electrophiles
C) Carbanions
D) Alkyl groups

Ans: B

44. Purification method based on boiling point:

A) Chromatography
B) Distillation
C) Sublimation
D) Crystallisation

Ans: B

45. TLC is based on:

A) Boiling point
B) Adsorption
C) Melting point
D) Density

Ans: B

46. Rf value is used in:

A) Distillation
B) Chromatography
C) Crystallisation
D) Extraction

Ans: B

47. Nitrogen detection gives:

A) Green colour
B) Blue colour
C) Red colour
D) Yellow colour

Ans: B

48. Halogens are detected using:

A) NaOH
B) AgNO₃
C) H₂SO₄
D) CuSO₄

Ans: B

49. Halogen estimation method is:

A) Dumas
B) Kjeldahl
C) Carius
D) Lassaigne

Ans: C

50. Sulphur is estimated as:

A) AgCl
B) BaSO₄
C) CO₂
D) NH₃

Ans: B

Section C: Advanced MCQs (51–75)

51. The IUPAC name of CH₃–CH₂–CH₂–CH₃ is:

A) Propane
B) Butane
C) Pentane
D) Ethane

Ans: B) Butane

52. The IUPAC name of CH₃–CH(CH₃)–CH₃ is:

A) Butane
B) 2-Methylpropane
C) Propane
D) 3-Methylpropane

Ans: B) 2-Methylpropane

53. The IUPAC name of CH₂=CH₂ is:

A) Ethane
B) Ethene
C) Ethyne
D) Methene

Ans: B) Ethene

54. The IUPAC name of CH≡CH is:

A) Ethane
B) Ethene
C) Ethyne
D) Methyne

Ans: C) Ethyne

55. CH₃CH₂OH belongs to:

A) Aldehydes
B) Ketones
C) Alcohols
D) Acids

Ans: C) Alcohols

56. The suffix used for carboxylic acids is:

A) –ol
B) –al
C) –one
D) –oic acid

Ans: D) –oic acid

57. The IUPAC name of CH₃COOH is:

A) Methanoic acid
B) Ethanoic acid
C) Propanoic acid
D) Acetic acid

Ans: B) Ethanoic acid

58. The IUPAC name of CH₃CHO is:

A) Methanal
B) Ethanal
C) Propanal
D) Ethanol

Ans: B) Ethanal

59. The IUPAC name of CH₃COCH₃ is:

A) Propanal
B) Propanone
C) Propan-2-one
D) Ethanal

Ans: C) Propan-2-one

60. The functional group with highest priority is:

A) –OH
B) –NH₂
C) –COOH
D) –CHO

Ans: C) –COOH

61. A good solvent for crystallisation should:

A) React with solute
B) Dissolve impurities only
C) Dissolve compound at high temperature and poorly at low temperature
D) Have high boiling point always

Ans: C

62. Sublimation is used for purification of:

A) Ethanol
B) Benzene
C) Camphor
D) Acetic acid

Ans: C

63. Separation of liquids with close boiling points is done by:

A) Simple distillation
B) Fractional distillation
C) Sublimation
D) Filtration

Ans: B

64. Chromatography is based on differences in:

A) Molecular weight only
B) Adsorption or distribution
C) Colour only
D) Density only

Ans: B

65. Lassaigne’s extract is prepared using:

A) Potassium
B) Sodium
C) Calcium
D) Magnesium

Ans: B

66. Nitrogen in sodium fusion forms:

A) NaNO₃
B) NaCN
C) NH₃
D) N₂

Ans: B

67. Prussian blue colour confirms the presence of:

A) Sulphur
B) Halogen
C) Nitrogen
D) Carbon

Ans: C

68. White precipitate in AgNO₃ test indicates:

A) Bromine
B) Iodine
C) Chlorine
D) Sulphur

Ans: C

69. Yellow precipitate in AgNO₃ test indicates:

A) Chlorine
B) Iodine
C) Bromine
D) Nitrogen

Ans: B

70. Carbon and hydrogen are estimated by:

A) Carius method
B) Combustion method
C) Kjeldahl method
D) Dumas method

Ans: B

71. Kjeldahl method is used for estimation of:

A) Carbon
B) Hydrogen
C) Nitrogen
D) Halogens

Ans: C

72. Dumas method estimates:

A) Nitrogen
B) Sulphur
C) Carbon
D) Oxygen

Ans: A

73. Carius method is used for:

A) Nitrogen
B) Halogens
C) Carbon
D) Hydrogen

Ans: B

74. The formula of Rf value is:

A) Distance travelled by solvent / distance travelled by solute
B) Distance travelled by solute / distance travelled by solvent
C) Mass / volume
D) Volume / mass

Ans: B

75. The process of removing impurities from organic compounds is called:

A) Oxidation
B) Purification
C) Reduction
D) Hydrolysis

Ans: B

Section D: Numerical & Concept-Based MCQs (76–90)

76. If 44 g CO₂ is formed, the mass of carbon present is:

A) 12 g
B) 22 g
C) 44 g
D) 6 g

Ans: A

77. In 18 g of water, the mass of hydrogen is:

A) 1 g
B) 2 g
C) 8 g
D) 16 g

Ans: B

78. One mole of CO₂ contains carbon mass:

A) 6 g
B) 12 g
C) 16 g
D) 44 g

Ans: B

79. The empirical formula represents:

A) Actual molecule size
B) Simplest whole number ratio of atoms
C) Molecular weight
D) Bond angle

Ans: B

80. Molecular formula is obtained from:

A) Empirical formula only
B) Molecular mass and empirical formula
C) Density only
D) Boiling point

Ans: B

81. More substituted carbocations are stable due to:

A) –I effect
B) Hyperconjugation
C) Hydrogen bonding
D) Resonance only

Ans: B

82. Alkyl groups are:

A) Electron withdrawing
B) Electron donating
C) Neutral always
D) Oxidising agents

Ans: B

83. The strongest –I group among these is:

A) CH₃
B) NO₂
C) C₂H₅
D) H

Ans: B

84. A nucleophile attacks:

A) Electron-rich centre
B) Electron-deficient centre
C) Positive ion only
D) Negative ion only

Ans: B

85. Electrophiles are:

A) Electron pair acceptors
B) Electron pair donors
C) Neutral molecules only
D) Radicals only

Ans: A

86. Benzene stability is due to:

A) Ionic bonds
B) Resonance
C) Hydrogen bonds
D) Free radicals

Ans: B

87. Hyperconjugation involves:

A) C–H σ electrons
B) π electrons only
C) Lone pairs only
D) Protons

Ans: A

88. The least stable carbocation is:

A) Tertiary
B) Secondary
C) Primary
D) Methyl

Ans: D

89. The most stable carbanion is:

A) Tertiary
B) Secondary
C) Primary
D) Methyl

Ans: D

90. Free radicals contain:

A) Positive charge
B) Negative charge
C) Unpaired electron
D) Lone pair

Ans: C

Section E: Assertion–Reason MCQs (91–100)

Choose:

A) Both A and R are true and R explains A
B) Both A and R are true but R does not explain A
C) A is true but R is false
D) A is false but R is true

91.

Assertion: Carbon forms a large number of compounds.
Reason: Carbon shows catenation.

Ans: A

92.

Assertion: sp carbon is linear.
Reason: It has 180° bond angle.

Ans: A

93.

Assertion: Alkyl groups stabilise carbocations.
Reason: Alkyl groups show +I effect.

Ans: A

94.

Assertion: Benzene is stable.
Reason: It has resonance.

Ans: A

95.

Assertion: Alkenes undergo addition reactions.
Reason: They contain π bonds.

Ans: A

96.

Assertion: Carbanions are electron-rich species.
Reason: They carry negative charge.

Ans: A

97.

Assertion: Distillation separates liquids.
Reason: Liquids have different boiling points.

Ans: A

98.

Assertion: Chromatography separates mixtures.
Reason: Components have different adsorption abilities.

Ans: A

99.

Assertion: Nitrogen gives Prussian blue colour.
Reason: Sodium fusion forms sodium cyanide.

Ans: A

100.

Assertion: Optical isomers rotate plane-polarised light.
Reason: They contain chiral centres.

Ans: A