Organic Chemistry Some Basic Principles and Techni — Class 11 Chemistry NCERT Solutions (Free)
Free step-by-step NCERT solutions for Class 11 Chemistry chapter "Organic Chemistry Some Basic Principles and Techni" — 8 important questions with detailed answers for CBSE board exam preparation.
TL;DR: Free step-by-step NCERT solutions for Class 11 Chemistry chapter "Organic Chemistry Some Basic Principles and Techni" — 8 important questions with det…
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NCERT Solutions for Class 11 Chemistry — Organic Chemistry Some Basic Principles and Techni. Step-by-step answers to all 8 textbook questions from this chapter, written for CBSE board preparation and free to use.
Organic Chemistry Some Basic Principles and Techni — All 8 Questions Solved
Q1. What are homologous series? Give characteristics and an example.
Homologous Series: A series of organic compounds with:
1. Same general formula
2. Successive members differing by one CH2 unit
3. Similar chemical properties
4. Gradual change in physical properties
5. Prepared by similar methods
Characteristics:
1. Differ by CH2 (mass difference = 14 amu)
2. Each differs in one methylene group
3. Chemical properties similar but show gradual trends
4. Can show isomerism
5. Increase in molecular weight is regular
Example: Alkane Series (Alkanes)
General formula: CnH(2n+2)
Methane: CH4
Ethane: C2H6 (CH4 + CH2)
Propane: C3H8 (C2H6 + CH2)
Butane: C4H10 (C3H8 + CH2)
Pentane: C5H12 (C4H10 + CH2)
Trend in boiling points:
CH4: -161°C
C2H6: -89°C
C3H8: -42°C
C4H10: -0.5°C
C5H12: 36°C
Boiling points increase due to increasing intermolecular van der Waals forces.
Q2. Draw and name the structural isomers of C4H10 (butane)
Structural isomers of C4H10 have same molecular formula but different structural arrangements:
Isomer 1: n-Butane (Butane)
Structure: CH3-CH2-CH2-CH3
Straight chain of 4 carbons
Boiling point: -0.5°C
Isomer 2: 2-Methylpropane (Isobutane)
Structure: CH3
|
CH3-CH-CH3
Central carbon bonded to three methyl groups and one H
Boiling point: -11.6°C
Differences:
1. Different structural arrangements (skeletal isomerism)
2. Different boiling points (isobutane lower)
3. Different densities
4. Different reactivity toward halogenation
- n-Butane gives monohalogenation at either primary or secondary position
- Isobutane gives monohalogenation at either primary or secondary position (but limited by symmetry)
No other isomers possible for C4H10. These are the only two structural isomers.
Q3. What is covalent bond? Explain sigma (σ) and pi (π) bonds with examples.
Covalent Bond: Chemical bond formed by sharing of electrons between two atoms.
Sigma (σ) Bond:
- Strongest type of covalent bond
- Head-on overlap of atomic orbitals (or orbital lobes)
- Electron density concentrated along the internuclear axis
- Free rotation possible around σ bond
- All single bonds are σ bonds
Examples of σ bonds:
- H-H in H2
- C-C in CH3-CH3 (ethane)
- C-H in all organic compounds
- All bonds in saturated compounds
Pi (π) Bond:
- Weaker than σ bond
- Lateral (side-by-side) overlap of p-orbitals
- Electron density in two regions above and below internuclear axis
- Cannot rotate; restricts rotation
- Always exists in addition to σ bond (double bonds = σ + π)
Examples of π bonds:
- C=C in ethene: CH2=CH2
Contains 1 σ bond and 1 π bond
- C≡C in ethyne: HC≡CH
Contains 1 σ bond and 2 π bonds
- C=O in aldehydes and ketones
Comparison:
σ Bond π Bond
Strength: Strong Weak
Overlap: Head-on Lateral
Rotation: Free Restricted
Position: Along axis Above/below axis
Q4. Explain sp, sp2, and sp3 hybridization with examples.
Hybridization: Intermixing of atomic orbitals to form new equivalent hybrid orbitals.
sp Hybridization (Linear):
- 1 s orbital + 1 p orbital mix to form 2 sp hybrid orbitals
- Bond angle: 180°
- Geometry: Linear
- Example: Ethyne (HC≡CH)
Each C atom: sp hybridized
Bonds: C-C is σ bond + 2 π bonds (triple bond)
H-C bonds: σ bonds
Geometry around C: Linear
- Other examples: CO2, HCN, AlCl3
sp2 Hybridization (Trigonal Planar):
- 1 s orbital + 2 p orbitals mix to form 3 sp2 hybrid orbitals
- Bond angle: 120°
- Geometry: Trigonal planar
- Example: Ethene (CH2=CH2)
Each C atom: sp2 hybridized
Bonds: C=C is σ bond + 1 π bond
H-C bonds: σ bonds
Geometry around C: Planar
- Other examples: BF3, C2H4 (ethylene)
sp3 Hybridization (Tetrahedral):
- 1 s orbital + 3 p orbitals mix to form 4 sp3 hybrid orbitals
- Bond angle: 109.5°
- Geometry: Tetrahedral
- Example: Ethane (CH3-CH3)
Each C atom: sp3 hybridized
Bonds: C-C is σ bond, C-H are σ bonds
Geometry around C: Tetrahedral
- Other examples: CH4, NH3, H2O, CCl4
Summary Table:
Hybridization | Orbitals mixed | Hybrids | Angle | Shape | Examples
sp | s + p | 2 | 180° | Linear | HC≡CH
sp2 | s + 2p | 3 | 120° | Planar | CH2=CH2
sp3 | s + 3p | 4 | 109.5° | Tetrahedral | CH3CH3
Q5. What is IUPAC nomenclature? Give rules and name: CH3-CHBr-CH2-CH3
IUPAC Nomenclature: Systematic method to name organic compounds.
Steps:
1. Identify the longest continuous carbon chain
2. Identify the principal functional group
3. Number the chain to give the functional group lowest number
4. Name substituents and their positions
5. Arrange substituents alphabetically
Rules for Alkyl Halides:
- Functional group = halogen (F, Cl, Br, I)
- Use prefix: fluoro-, chloro-, bromo-, iodo-
- Number carbon chain to give halogen lowest position
- Parent name follows alkane series
Example: CH3-CHBr-CH2-CH3
Step 1: Longest chain = 4 carbons (butane)
Step 2: Principal group = Br (bromo)
Step 3: Number chain
From left: C1-C2(Br)-C3-C4
From right: C1-C2-C3(Br)-C4
Left numbering gives Br position 2 (lower)
Right numbering gives Br position 3
Choose left numbering
Step 4: Substituent = Br at position 2
Step 5: Alphabetical order = bromo comes before other substituents
IUPAC Name: 2-Bromobutane
Alternative structure example: (CH3)3CBr
Longest chain = 3 carbons (propane)
Number: C1(Br)-C2-C3
Br at position 1
IUPAC Name: 1-Bromo-2,2-dimethylpropane (common name: neopentyl bromide)
Q6. Classify organic compounds by functional groups. Give examples and properties.
Functional Groups: Specific atoms or groups of atoms in organic molecules responsible for characteristic chemical properties.
1. Alkanes (C-C, C-H single bonds)
Example: CH3CH2CH3 (Propane)
Properties: Non-polar, low reactivity, hydrophobic, combustible
2. Alkenes (C=C double bond)
Example: CH2=CHCH3 (Propene)
Properties: Polar bonds, reactive (addition reactions), unsaturated
3. Alkynes (C≡C triple bond)
Example: HC≡CCH3 (Propyne)
Properties: Reactive, high unsaturation, acidic H in terminal alkynes
4. Alcohols (-OH, hydroxyl)
Example: CH3CH2OH (Ethanol)
Properties: Polar, H-bonding, water-soluble, can be oxidized
5. Ethers (C-O-C)
Example: CH3OCH3 (Dimethyl ether)
Properties: Slightly polar, relatively inert, volatile
6. Aldehydes (CHO, carbonyl + H)
Example: CH3CHO (Acetaldehyde)
Properties: Polar, oxidizable, reactive in nucleophilic addition
7. Ketones (C=O, carbonyl)
Example: CH3COCH3 (Acetone)
Properties: Polar, less oxidizable than aldehydes, good solvents
8. Carboxylic Acids (-COOH)
Example: CH3COOH (Acetic acid)
Properties: Polar, acidic (Ka ~ 10^-5), form dimers
9. Esters (-COO-)
Example: CH3COOCH3 (Methyl acetate)
Properties: Polar, aromatic odor, hydrolyzable
10. Amines (-NH2, -NHR, -NR2)
Example: CH3NH2 (Methylamine)
Properties: Polar, basic, nucleophilic
Q7. What is oxidation-reduction in organic chemistry? Give example of alcohol oxidation.
Oxidation in Organic Chemistry:
- Removal of hydrogen atoms
- Addition of oxygen atoms
- Loss of electrons
- Increase in oxidation state of carbon
Reduction in Organic Chemistry:
- Addition of hydrogen atoms
- Removal of oxygen atoms
- Gain of electrons
- Decrease in oxidation state of carbon
Example: Oxidation of Primary Alcohol to Aldehyde to Carboxylic Acid
Step 1: Primary alcohol to aldehyde (mild oxidation)
R-CH2-OH -> R-CHO + [H]
Example: Ethanol to Acetaldehyde
CH3CH2OH --[O]--> CH3CHO + H2O
Oxidation state of C: -1 to +1 (increase by 2)
Oxidizing agent: K2Cr2O7/H2SO4 (mild) or PCC (precise control)
Step 2: Aldehyde to carboxylic acid (strong oxidation)
R-CHO -> R-COOH
Example: Acetaldehyde to Acetic Acid
CH3CHO --[O]--> CH3COOH
Oxidation state of C: +1 to +3 (increase by 2)
Oxidizing agent: KMnO4/NaOH or K2Cr2O7/H2SO4 (excess)
Overall: Ethanol to Acetic Acid
CH3CH2OH --2[O]--> CH3COOH + H2O
Mechanism: Primary alcohol -> aldehyde -> carboxylic acid
Oxidation state of carbon: -1 -> +1 -> +3
Electrons transferred = 4 per alcohol molecule (2 × 2e- for each oxidation step)
Secondary alcohol oxidizes to ketone (stops here, ketone resistant to further oxidation):
R2CHOH --[O]--> R2C=O
Oxidation state change: 0 to +2
Q8. Explain the mechanism of electrophilic addition to alkenes using ethene + HBr as example
Electrophilic Addition: Reaction where an electrophile attacks the π bond of an alkene.
Example: Ethene + HBr -> Ethyl Bromide
CH2=CH2 + HBr -> CH3-CH2Br
Mechanism (Two-step process):
Step 1: Formation of carbocation (rate-determining step)
- π electrons of C=C attack H+ from HBr
- This forms a C-H σ bond
- One carbon becomes electron-deficient (carbocation)
- One carbon retains the positive charge (secondary carbocation is unstable, but no rearrangement here)
H H
| | H+
C=C -> C----C+ + Br-
| | |
H H H
Carbocation intermediate
Step 2: Nucleophilic attack (fast)
- Br- (nucleophile) attacks the carbocation
- Forms C-Br σ bond
- Regenerates neutral product
H Br- H Br
| ↓ | |
C----C+ -> C--C
| | | |
H H H H
Ethyl Bromide (1-Bromoethane)
Regioselectivity (Markovnikov's Rule):
In unsymmetrical alkenes, H adds to C with more H atoms
Br adds to C with fewer H atoms
This gives more stable carbocation intermediate.
Example: Propene + HBr
CH3-CH=CH2 + HBr -> CH3-CHBr-CH3 (2-Bromopropane, major)
Not: CH3-CH2-CH2Br (1-Bromopropane, minor)
Reason: 2° carbocation (CH3-CHBr+) more stable than 1° carbocation (CH3-CH2-CH2Br+)
Stereochemistry: Addition is not stereospecific (carbocation is planar, Br can attack from either face).
How to Use These Solutions
Attempt each question yourself first and only then compare with the worked answer. Marks in the CBSE board exam are awarded for the METHOD as much as the final result, so reproduce the steps rather than memorising the last line. Where a solution states a law, a formula or a definition, learn that wording — examiners look for it.
Related practice for this chapter: chapter MCQs, previous-year questions, revision notes and sample papers.