माध्यमिक शिक्षा बोर्ड राजस्थान, अजमेर
उत्तर कुंजी — रसायन विज्ञान मॉडल पेपर
Answer Key — Chemistry Model Paper 2025-26
कक्षा : 11
Subject Code : 41
परीक्षा : 2026
Total : 56 Marks
For Teacher / Self-Study Use
| खण्ड | Section | प्रश्न | अंक | इस Key में |
|---|---|---|---|---|
| अ | A — MCQ | Q.1–15 | 15 | ✅ All answers + reason |
| ब | B — Very Short | Q.16–23 | 16 | ✅ Model answers |
| स | C — Short | Q.24–28 | 15 | ✅ Point-wise answers |
| द | D — Long (Internal choice) | Q.29–30 | 10 | ✅ Both options solved |
खण्ड — अ | Section A — MCQ (1 mark each)
Q.1
Standard enthalpy of combustion of carbon = −393.5 kJ mol⁻¹. Reaction is —
1 mark
✅
(B) Exothermic
— Negative ΔH always indicates heat is released → Exothermic reaction.
💡 याद रखें: ΔH < 0 = Exothermic | ΔH > 0 = Endothermic
Q.2
For spontaneous process at constant T & P, ΔG must be —
1 mark
✅
(C) ΔG < 0
— Gibbs criterion: spontaneous if ΔG negative; equilibrium if ΔG = 0; non-spontaneous if ΔG > 0.
ΔG = ΔH − TΔS | Spontaneous: ΔG < 0
Q.3
Relationship between Kp and Kc —
1 mark
✅
(B) Kp = Kc(RT)Δn
— Derived from ideal gas law PV = nRT. Here Δn = moles of gaseous products − moles of gaseous reactants.
Kp = Kc(RT)Δn where R = 0.0821 L·atm mol⁻¹K⁻¹
Q.4
pH of 0.01 M HCl solution —
1 mark
✅
(B) 2
— HCl is strong acid, fully dissociates. [H⁺] = 0.01 = 10⁻². pH = −log(10⁻²) = 2.
pH = −log[H⁺] = −log(0.01) = 2
Q.5
Zn + CuSO₄ → ZnSO₄ + Cu — which species is oxidised?
1 mark
✅
(C) Zn
— Zn goes from 0 → +2 (loses electrons = oxidised). Cu²⁺ goes from +2 → 0 (gains electrons = reduced).
OIL RIG: Oxidation Is Loss (of electrons) | Reduction Is Gain
Q.6
IUPAC name of CH₃–CH₂–CH₂–OH —
1 mark
✅
(C) 1-Propanol
— 3 carbons (propan-), –OH at C1 (-1-ol). Not 2-propanol which has OH at C2.
Q.7
Which is an electrophile? NH₃ / H₂O / OH⁻ / BF₃
1 mark
✅
(D) BF₃
— BF₃ has empty orbital on B (electron-deficient) → electron-seeker = electrophile. NH₃, H₂O, OH⁻ are nucleophiles (have lone pairs to donate).
Rule: Electrophile = electron-deficient / positive centre. Nucleophile = electron-rich / lone pair donor.
Q.8
Which hydrocarbon decolourises purple KMnO₄ (Baeyer's test)?
1 mark
✅
(C) Ethene (CH₂=CH₂)
— Alkenes & alkynes have π bonds that reduce KMnO₄ (purple → colourless). Alkanes and benzene don't react under mild conditions.
Q.9
Relationship between Cp and Cv for ideal gas —
1 mark
✅
(C) Cp − Cv = R
— Mayer's relation. Extra work done by gas against atmosphere when heated at constant pressure = R per mole.
Cp − Cv = R (Mayer's Relation)
Q.10
Buffer solution resists change in —
1 mark
✅
(C) pH
— Buffer is a solution that resists significant changes in pH on adding small amounts of acid or base.
Q.11
Oxidation number of Mn in KMnO₄ —
1 mark
✅
(C) +7
— K = +1, O = −2 × 4 = −8. Total neutral: +1 + x − 8 = 0 → x = +7.
+1 + x + 4(−2) = 0 → x = +7
Q.12
General formula of alkenes —
1 mark
✅
(B) CnH2n
— One C=C double bond causes 2H deficiency from alkane (CnH2n+2). Alkynes = CnH2n−2.
Series: Alkane CnH2n+2 | Alkene CnH2n | Alkyne CnH2n−2 | Benzene CnH2n−6
Q.13
Entropy of a perfect crystal at 0 K —
1 mark
✅
(C) Zero
— Third Law of Thermodynamics: perfect crystal at 0 K has only one microstate (W=1), so S = k lnW = k ln1 = 0.
S = kB ln W → W=1 at 0 K → S = 0
Q.14
cis-but-2-ene & trans-but-2-ene — type of isomerism?
1 mark
✅
(C) Geometrical Isomerism
— Same molecular formula, same connectivity, but different spatial arrangement around C=C (restricted rotation). Two different groups on each sp² carbon.
Condition: Both carbons of C=C must have two different groups attached for geometrical isomerism.
Q.15
Le Chatelier's principle applies to —
1 mark
✅
(C) Any system in equilibrium
— Le Chatelier's principle states: if a system at equilibrium is disturbed, it shifts to counteract the disturbance and restore a new equilibrium.
खण्ड — ब | Section B — Very Short Answer (2 marks each)
Q.16
Define Hess's Law with one example.
2 marks
Hess's Law of Constant Heat Summation:
The total enthalpy change for a reaction is the same regardless of whether the reaction takes place in one step or in multiple steps. It depends only on the initial and final states.
हेस का नियम: किसी रासायनिक अभिक्रिया में कुल एन्थैल्पी परिवर्तन, अभिक्रिया चाहे एक पद में हो या अनेक पदों में, सदा स्थिर रहती है।
Example / उदाहरण:
C(s) + O₂(g) → CO₂(g) ΔH = −393.5 kJ ...(directly)
OR via two steps:
C(s) + ½O₂(g) → CO(g) ΔH₁ = −110.5 kJ
CO(g) + ½O₂(g) → CO₂(g) ΔH₂ = −283 kJ
Total: ΔH = ΔH₁ + ΔH₂ = −110.5 + (−283) = −393.5 kJ ✓
The total enthalpy change for a reaction is the same regardless of whether the reaction takes place in one step or in multiple steps. It depends only on the initial and final states.
हेस का नियम: किसी रासायनिक अभिक्रिया में कुल एन्थैल्पी परिवर्तन, अभिक्रिया चाहे एक पद में हो या अनेक पदों में, सदा स्थिर रहती है।
Example / उदाहरण:
C(s) + O₂(g) → CO₂(g) ΔH = −393.5 kJ ...(directly)
OR via two steps:
C(s) + ½O₂(g) → CO(g) ΔH₁ = −110.5 kJ
CO(g) + ½O₂(g) → CO₂(g) ΔH₂ = −283 kJ
Total: ΔH = ΔH₁ + ΔH₂ = −110.5 + (−283) = −393.5 kJ ✓
1 mark — Correct definition
1 mark — Valid example with values
Q.17
Henderson–Hasselbalch equation for buffer. Components?
2 marks
pH = pKa + log([A⁻] / [HA])
Components (घटक):
• pKa — negative log of acid dissociation constant of the weak acid
• [A⁻] — molar concentration of conjugate base (e.g., CH₃COO⁻ from salt)
• [HA] — molar concentration of weak acid (e.g., CH₃COOH)
Example: Acetic acid + Sodium acetate buffer
pH = pKa(CH₃COOH) + log([CH₃COO⁻] / [CH₃COOH]) = 4.74 + log(ratio)
1 mark — Correct equation
1 mark — Components explained
Q.18
Balance Fe + H₂SO₄ → FeSO₄ + H₂ by oxidation number method.
2 marks
Step 1 — Assign oxidation numbers:
Fe: 0 → +2 (increase by 2, oxidised)
H: +1 → 0 (decrease by 1, reduced; 2 H atoms so total decrease = 2)
Step 2 — Equalise change:
Change in Fe = 2 ↑ | Change in 2H = 2 ↓ → already equal ✓
Step 3 — Balanced equation:
Verification: Fe(1), H(2), S(1), O(4) — both sides equal.
Fe: 0 → +2 (increase by 2, oxidised)
H: +1 → 0 (decrease by 1, reduced; 2 H atoms so total decrease = 2)
Step 2 — Equalise change:
Change in Fe = 2 ↑ | Change in 2H = 2 ↓ → already equal ✓
Step 3 — Balanced equation:
Fe + H₂SO₄ → FeSO₄ + H₂ ↑
Already balanced (1:1:1:1) ✓Verification: Fe(1), H(2), S(1), O(4) — both sides equal.
1 mark — Correct oxidation number assignment
1 mark — Balanced equation
Q.19
Markovnikov's Rule — definition + example.
2 marks
Markovnikov's Rule:
In the addition of an unsymmetrical reagent (like HBr) to an unsymmetrical alkene, the hydrogen atom adds to the carbon that already bears more hydrogen atoms (more substituted H goes to carbon with more H — "rich gets richer").
नियम: असममित अभिकर्मक का H परमाणु उस कार्बन से जुड़ता है जिस पर पहले से अधिक H परमाणु हों।
Example:
CH₃–CH=CH₂ + HBr →
CH₃–CH(Br)–CH₃ (2-Bromopropane — MAJOR ✓)
CH₃–CH₂–CH₂Br (1-Bromopropane — minor)
Reason: H⁺ attacks C1 (more H), forming more stable 2° carbocation at C2; then Br⁻ attacks C2.
In the addition of an unsymmetrical reagent (like HBr) to an unsymmetrical alkene, the hydrogen atom adds to the carbon that already bears more hydrogen atoms (more substituted H goes to carbon with more H — "rich gets richer").
नियम: असममित अभिकर्मक का H परमाणु उस कार्बन से जुड़ता है जिस पर पहले से अधिक H परमाणु हों।
Example:
CH₃–CH=CH₂ + HBr →
CH₃–CH(Br)–CH₃ (2-Bromopropane — MAJOR ✓)
CH₃–CH₂–CH₂Br (1-Bromopropane — minor)
Reason: H⁺ attacks C1 (more H), forming more stable 2° carbocation at C2; then Br⁻ attacks C2.
1 mark — Correct definition
1 mark — Example with major product identified
Q.20
Distinguish homogeneous and heterogeneous equilibrium.
2 marks
| Feature | Homogeneous Equilibrium सजातीय साम्य |
Heterogeneous Equilibrium विषमांगी साम्य |
|---|---|---|
| Phase | All species in same phase | Species in different phases |
| Example | N₂(g) + 3H₂(g) ⇌ 2NH₃(g) | CaCO₃(s) ⇌ CaO(s) + CO₂(g) |
| Kc expression | Includes all species | Pure solids/liquids excluded (activity = 1) |
1 mark — Homogeneous: definition + example
1 mark — Heterogeneous: definition + example
Q.21
IUPAC names of: (i) CH₃–CH(CH₃)–CH₂–CH₃ (ii) CH₂=CH–CH₃
2 marks
(i) CH₃–CH(CH₃)–CH₂–CH₃
Longest chain = 4 carbons (butane backbone). CH₃ branch at C2.
2-Methylbutane
(ii) CH₂=CH–CH₃
3 carbons with double bond starting at C1.
Prop-1-ene (Propene)
Longest chain = 4 carbons (butane backbone). CH₃ branch at C2.
2-Methylbutane
(ii) CH₂=CH–CH₃
3 carbons with double bond starting at C1.
Prop-1-ene (Propene)
1 mark — (i) 2-Methylbutane
1 mark — (ii) Prop-1-ene / Propene
Q.22
Differentiate inductive effect and resonance effect.
2 marks
| Feature | Inductive Effect (I) प्रेरण प्रभाव |
Resonance Effect (M/R) अनुनाद प्रभाव |
|---|---|---|
| Basis | Permanent displacement of σ-bond electrons due to electronegativity difference | Delocalisation of π-electrons or lone pairs in conjugated systems |
| Bond involved | σ-bond (single bond) | π-bond or lone pairs conjugated with π |
| Range | Decreases along chain (short range) | Transmitted through conjugated system |
| Example | –Cl (−I effect); –CH₃ (+I effect) | –OH in phenol (+M); –NO₂ (−M) |
1 mark — Inductive effect correct
1 mark — Resonance effect correct
Q.23
State the First and Second Laws of Thermodynamics.
2 marks
First Law (Energy Conservation):
Energy can neither be created nor destroyed; it can only be converted from one form to another. The total energy of the universe remains constant.
Second Law (Entropy / Spontaneity):
All spontaneous processes proceed in the direction of increasing entropy of the universe. The entropy of the universe always increases for irreversible processes.
Energy can neither be created nor destroyed; it can only be converted from one form to another. The total energy of the universe remains constant.
ΔU = q + w (q = heat absorbed, w = work done on system)
प्रथम नियम: ऊर्जा न तो उत्पन्न होती है और न ही नष्ट होती है, केवल रूपान्तरित होती है।Second Law (Entropy / Spontaneity):
All spontaneous processes proceed in the direction of increasing entropy of the universe. The entropy of the universe always increases for irreversible processes.
ΔSuniverse = ΔSsystem + ΔSsurroundings > 0 (spontaneous)
द्वितीय नियम: ब्रह्मांड की कुल एन्ट्रॉपी सदा बढ़ती है।
1 mark — First Law (statement + equation)
1 mark — Second Law (statement)
खण्ड — स | Section C — Short Answer (3 marks each)
Q.24
Entropy concept + Third Law of Thermodynamics + significance.
3 marks
Entropy (S): A thermodynamic state function that measures the degree of randomness or disorder in a system. Higher disorder = higher entropy.
एन्ट्रॉपी: तंत्र में अव्यवस्था की मात्रा का माप। S = kB ln W (Boltzmann equation)
Examples: Ice(s) < Water(l) < Steam(g) in entropy.
एन्ट्रॉपी: तंत्र में अव्यवस्था की मात्रा का माप। S = kB ln W (Boltzmann equation)
Examples: Ice(s) < Water(l) < Steam(g) in entropy.
Third Law of Thermodynamics (Nernst's Heat Theorem):
"The entropy of a perfectly crystalline substance is zero at absolute zero (0 K)."
"The entropy of a perfectly crystalline substance is zero at absolute zero (0 K)."
S = 0 at T = 0 K (perfect crystal)
ऊष्मागतिकी का तृतीय नियम: 0 K पर पूर्ण क्रिस्टलीय पदार्थ की एन्ट्रॉपी शून्य होती है।
Significance:
(a) Enables calculation of absolute entropies of substances.
(b) Explains why absolute zero cannot be attained experimentally.
(c) Provides a reference point for entropy calculations (S° at 298 K).
(a) Enables calculation of absolute entropies of substances.
(b) Explains why absolute zero cannot be attained experimentally.
(c) Provides a reference point for entropy calculations (S° at 298 K).
1 mark — Entropy definition with example
1 mark — Third Law statement
1 mark — Significance (any 2 points)
Q.25
Ionisation of water, ionic product Kw, effect of temperature.
3 marks
Ionisation of Water:
Water undergoes self-ionisation (autoprotolysis):
जल स्वयं आयनित होता है — यह अम्ल व क्षार दोनों की भूमिका निभाता है (उभयधर्मी)।
Water undergoes self-ionisation (autoprotolysis):
H₂O + H₂O ⇌ H₃O⁺ + OH⁻
or simply: H₂O ⇌ H⁺ + OH⁻जल स्वयं आयनित होता है — यह अम्ल व क्षार दोनों की भूमिका निभाता है (उभयधर्मी)।
Ionic Product (Kw):
Pure water: [H⁺] = [OH⁻] = 10⁻⁷ mol L⁻¹ → pH = 7 (neutral)
Kw = [H⁺][OH⁻]
At 25°C: Kw = 1.0 × 10⁻¹⁴ mol² L⁻²Pure water: [H⁺] = [OH⁻] = 10⁻⁷ mol L⁻¹ → pH = 7 (neutral)
Effect of Temperature:
Ionisation of water is endothermic. Increasing temperature shifts equilibrium forward → Kw increases.
At 60°C: Kw ≈ 10⁻¹³ (greater than 10⁻¹⁴). Water remains neutral (pH decreases but [H⁺] = [OH⁻] always).
ताप बढ़ने पर Kw बढ़ता है, परन्तु जल सदा उदासीन रहता है।
Ionisation of water is endothermic. Increasing temperature shifts equilibrium forward → Kw increases.
At 60°C: Kw ≈ 10⁻¹³ (greater than 10⁻¹⁴). Water remains neutral (pH decreases but [H⁺] = [OH⁻] always).
ताप बढ़ने पर Kw बढ़ता है, परन्तु जल सदा उदासीन रहता है।
1 mark — Ionisation equation
1 mark — Kw expression + value
1 mark — Temperature effect correctly explained
Q.26
Electrophilic substitution in benzene + mechanism of nitration.
3 marks
Electrophilic Aromatic Substitution (EAS):
In benzene, the π-electron cloud acts as an electron donor attacking an electrophile. The H is replaced (substituted), not the π-system. Aromaticity is maintained in the product.
In benzene, the π-electron cloud acts as an electron donor attacking an electrophile. The H is replaced (substituted), not the π-system. Aromaticity is maintained in the product.
Step 1 — Generation of electrophile (NO₂⁺):
HNO₃ + H₂SO₄ → NO₂⁺ + HSO₄⁻ + H₂O
NO₂⁺ (nitronium ion) is the actual electrophile.
Step 2 — Attack of NO₂⁺ on benzene π-cloud:
C₆H₆ + NO₂⁺ → [C₆H₆·NO₂]⁺ (σ-complex / Wheland intermediate / Arenium ion)
Aromaticity is temporarily disrupted. This is the rate-determining step.
C₆H₆ + NO₂⁺ → [C₆H₆·NO₂]⁺ (σ-complex / Wheland intermediate / Arenium ion)
Aromaticity is temporarily disrupted. This is the rate-determining step.
Step 3 — Loss of H⁺ (Aromaticity restored):
[C₆H₆·NO₂]⁺ → C₆H₅NO₂ + H⁺
[C₆H₆·NO₂]⁺ → C₆H₅NO₂ + H⁺
C₆H₆ + HNO₃ —(H₂SO₄, 55°C)→ C₆H₅NO₂ + H₂O
उत्पाद: नाइट्रोबेंजीन (हल्की पीली तेलीय गंध वाला द्रव)
1 mark — EAS concept + Step 1 electrophile generation
1 mark — Step 2 σ-complex formation
1 mark — Step 3 + overall equation
Q.27
Differences between SN1 and SN2 mechanisms.
3 marks
| Feature | SN1 (Unimolecular) | SN2 (Bimolecular) |
|---|---|---|
| Steps | 2 steps (ionisation then attack) | 1 step (concerted) |
| Intermediate | Carbocation (planar) | No stable intermediate (transition state) |
| Rate | Rate = k[RX] (only substrate) | Rate = k[RX][Nu] (both) |
| Stereochemistry | Racemisation (both inversion + retention) | Walden inversion (100% inversion) |
| Favoured by | 3° substrate, polar protic solvent | 1° substrate, polar aprotic solvent, strong nucleophile |
| Example | (CH₃)₃C–Br + H₂O | CH₃Br + OH⁻ |
3 marks — Any 3 well-explained differences (1 mark each)
Q.28
N₂ + 3H₂ ⇌ 2NH₃ at 500°C, Kc = 0.061 → Calculate Kp
3 marks
Find Δn:
Δn = moles of gaseous products − moles of gaseous reactants
= 2 − (1 + 3) = 2 − 4 = −2
Δn = moles of gaseous products − moles of gaseous reactants
= 2 − (1 + 3) = 2 − 4 = −2
Convert Temperature to Kelvin:
T = 500 + 273 = 773 K
⚠️ Always use Kelvin — never Celsius in Kp/Kc calculations!
Apply formula:
RT = 0.0821 × 773 = 63.46
(RT)−2 = 1 / (63.46)² = 1 / 4027.2 = 2.483 × 10⁻⁴
Kp = 0.061 × 2.483 × 10⁻⁴ = 1.51 × 10⁻⁵
Kp = Kc × (RT)Δn
Using R = 0.0821 L·atm mol⁻¹K⁻¹:RT = 0.0821 × 773 = 63.46
(RT)−2 = 1 / (63.46)² = 1 / 4027.2 = 2.483 × 10⁻⁴
Kp = 0.061 × 2.483 × 10⁻⁴ = 1.51 × 10⁻⁵
Final Answer: Kp ≈ 1.51 × 10⁻⁵ atm⁻² | Kp < Kc because Δn < 0 (fewer moles of gas in products)
1 mark — Δn correctly calculated (−2)
1 mark — T converted to 773 K
1 mark — Final Kp value correct
खण्ड — द | Section D — Long Answer (5 marks each, internal choice)
Q.29(a)
First Law of Thermodynamics + Derive ΔH = ΔU + ΔnRT + Calculate ΔH
5 marks
1. First Law of Thermodynamics:
"Energy can neither be created nor destroyed. The total energy of an isolated system remains constant."
Mathematically: ΔU = q + w
where q = heat absorbed by system, w = work done on system (w = −PΔV for expansion).
2. Derivation of ΔH = ΔU + ΔnRT:
At constant pressure: H = U + PV
ΔH = ΔU + Δ(PV) = ΔU + Δ(nRT) [using ideal gas: PV = nRT]
= ΔU + (Δn)RT
where Δn = nproducts(g) − nreactants(g)
3. Numerical Calculation:
Given: ΔU = −742 kJ, Δn = −2, T = 300 K
R = 8.314 J mol⁻¹K⁻¹ = 8.314 × 10⁻³ kJ mol⁻¹K⁻¹
ΔnRT = (−2) × 8.314 × 10⁻³ × 300 = −4.988 kJ
"Energy can neither be created nor destroyed. The total energy of an isolated system remains constant."
Mathematically: ΔU = q + w
where q = heat absorbed by system, w = work done on system (w = −PΔV for expansion).
2. Derivation of ΔH = ΔU + ΔnRT:
At constant pressure: H = U + PV
ΔH = ΔU + Δ(PV) = ΔU + Δ(nRT) [using ideal gas: PV = nRT]
= ΔU + (Δn)RT
where Δn = nproducts(g) − nreactants(g)
ΔH = ΔU + ΔnRT
3. Numerical Calculation:
Given: ΔU = −742 kJ, Δn = −2, T = 300 K
R = 8.314 J mol⁻¹K⁻¹ = 8.314 × 10⁻³ kJ mol⁻¹K⁻¹
ΔnRT = (−2) × 8.314 × 10⁻³ × 300 = −4.988 kJ
ΔH = −742 + (−4.988) = −746.988 ≈ −747 kJ
⚠️ Unit Tip: R given in J — always divide by 1000 to convert to kJ before adding to ΔU (kJ). Many students lose marks by adding 4988 J to −742 kJ directly!
1 mark — First Law statement + equation
2 marks — Derivation (logical steps)
2 marks — Correct numerical (Δn, T in K, unit conversion, final answer)
Q.29(b)
Kc & Kp definitions + Le Chatelier on N₂+3H₂⇌2NH₃ (ΔH=−92 kJ)
5 marks
1. Equilibrium Constants:
Kc (concentration equilibrium constant):
For aA + bB ⇌ cC + dD:
2. Le Chatelier's Analysis for N₂ + 3H₂ ⇌ 2NH₃, ΔH = −92 kJ:
Industrial significance (Haber Process): High pressure (200 atm), moderate temp (450°C), Fe catalyst — compromise between yield and rate.
Kc (concentration equilibrium constant):
For aA + bB ⇌ cC + dD:
Kc = [C]c[D]d / [A]a[B]b
Kp (pressure equilibrium constant):Kp = (PC)c(PD)d / (PA)a(PB)b
Relation: Kp = Kc(RT)Δn2. Le Chatelier's Analysis for N₂ + 3H₂ ⇌ 2NH₃, ΔH = −92 kJ:
| Disturbance (परिवर्तन) | Shift Direction | Effect on NH₃ yield |
|---|---|---|
| ↑ [N₂] or [H₂] (increase conc.) | → Forward | ↑ NH₃ increases |
| ↑ Pressure (compress) | → Forward (4 mol → 2 mol) | ↑ NH₃ increases (high pressure favoured) |
| ↓ Temperature (cool) | → Forward (exothermic) | ↑ NH₃ increases (but rate slows) |
| ↑ Temperature (heat) | ← Backward | ↓ NH₃ decreases |
| Add catalyst (Fe) | No shift (equilibrium unchanged) | Reaches equilibrium faster, K unchanged |
Industrial significance (Haber Process): High pressure (200 atm), moderate temp (450°C), Fe catalyst — compromise between yield and rate.
1 mark — Kc definition + expression
1 mark — Kp definition + relation
3 marks — Le Chatelier table (1 each: concentration, pressure, temperature)
Q.30(a)
Classification of organic compounds + Functional groups + Homologous series
5 marks
1. Classification of Organic Compounds:
Based on carbon skeleton:
(i) Acyclic (Open chain) — straight or branched: CH₃CH₃, (CH₃)₄C
(ii) Cyclic (Closed chain):
• Homocyclic: only C in ring (cyclohexane, benzene)
• Heterocyclic: other atoms in ring (pyridine, furan — N, O, S)
(iii) Aromatic — benzene and derivatives (Hückel's 4n+2 π electrons rule)
2. Functional Groups (क्रियात्मक समूह):
3. Homologous Series (समजातीय श्रेणी):
A series of compounds having the same functional group, same general formula, and differing by –CH₂– (14 mass units) from adjacent members.
Properties: (i) Same functional group (ii) Same general formula (iii) Physical properties change gradually (iv) Can be prepared by same general methods.
Example: Alkane series: CH₄, C₂H₆, C₃H₈, C₄H₁₀ ... (CnH2n+2)
Based on carbon skeleton:
(i) Acyclic (Open chain) — straight or branched: CH₃CH₃, (CH₃)₄C
(ii) Cyclic (Closed chain):
• Homocyclic: only C in ring (cyclohexane, benzene)
• Heterocyclic: other atoms in ring (pyridine, furan — N, O, S)
(iii) Aromatic — benzene and derivatives (Hückel's 4n+2 π electrons rule)
2. Functional Groups (क्रियात्मक समूह):
| Group | Structure | Example | Suffix |
|---|---|---|---|
| Alcohol (हाइड्रॉक्सिल) | –OH | C₂H₅OH (Ethanol) | -ol |
| Aldehyde (एल्डिहाइड) | –CHO | HCHO (Methanal) | -al |
| Ketone (कीटोन) | –CO– | CH₃COCH₃ (Propanone) | -one |
| Carboxylic acid | –COOH | CH₃COOH (Ethanoic acid) | -oic acid |
| Amine (एमीन) | –NH₂ | CH₃NH₂ (Methanamine) | -amine |
3. Homologous Series (समजातीय श्रेणी):
A series of compounds having the same functional group, same general formula, and differing by –CH₂– (14 mass units) from adjacent members.
Properties: (i) Same functional group (ii) Same general formula (iii) Physical properties change gradually (iv) Can be prepared by same general methods.
Example: Alkane series: CH₄, C₂H₆, C₃H₈, C₄H₁₀ ... (CnH2n+2)
2 marks — Classification (acyclic, cyclic, aromatic with examples)
2 marks — Functional groups table (any 5 correct)
1 mark — Homologous series definition + example
Q.30(b)
Hydrocarbons — Alkynes: nomenclature, preparation, reactions (focus on Ethyne)
5 marks
1. Hydrocarbons (हाइड्रोकार्बन):
Compounds containing only carbon and hydrogen. Classified as: Alkanes (saturated), Alkenes (one C=C), Alkynes (one C≡C), Aromatic (benzene ring).
2. Alkynes — General formula: CnH2n−2
Nomenclature: Parent chain → longest containing C≡C → suffix "-yne". Number from end closest to triple bond.
Examples: HC≡CH = Ethyne; HC≡C–CH₃ = Prop-1-yne; CH₃–C≡C–CH₃ = But-2-yne
3. Methods of Preparation of Alkynes:
(i) From calcium carbide + water:
CH₂Br–CH₂Br + 2KOH (alc.) → HC≡CH + 2KBr + 2H₂O
4. Reactions of Ethyne (HC≡CH):
(i) Addition of H₂ (Hydrogenation):
HC≡CH + H₂ → CH₂=CH₂ → CH₃–CH₃ (Lindlar's catalyst for partial)
(ii) Addition of Cl₂/Br₂:
HC≡CH + Cl₂ → CHCl=CHCl → CHCl₂–CHCl₂
(iii) Addition of HCl (Markovnikov):
HC≡CH + HCl → CH₂=CHCl → CH₃–CHCl₂
(iv) Acidic character (unique to terminal alkynes):
HC≡CH + CuCl/NH₃ → CuC≡CCu↓ (copper acetylide — red ppt.)
(v) Polymerisation:
3HC≡CH → Benzene (C₆H₆) at 300°C / red hot iron tube (Trimerisation)
Compounds containing only carbon and hydrogen. Classified as: Alkanes (saturated), Alkenes (one C=C), Alkynes (one C≡C), Aromatic (benzene ring).
2. Alkynes — General formula: CnH2n−2
Nomenclature: Parent chain → longest containing C≡C → suffix "-yne". Number from end closest to triple bond.
Examples: HC≡CH = Ethyne; HC≡C–CH₃ = Prop-1-yne; CH₃–C≡C–CH₃ = But-2-yne
3. Methods of Preparation of Alkynes:
(i) From calcium carbide + water:
CaC₂ + 2H₂O → Ca(OH)₂ + HC≡CH ↑ (Ethyne)
(ii) Dehydrohalogenation of vicinal dihalides:CH₂Br–CH₂Br + 2KOH (alc.) → HC≡CH + 2KBr + 2H₂O
4. Reactions of Ethyne (HC≡CH):
(i) Addition of H₂ (Hydrogenation):
HC≡CH + H₂ → CH₂=CH₂ → CH₃–CH₃ (Lindlar's catalyst for partial)
(ii) Addition of Cl₂/Br₂:
HC≡CH + Cl₂ → CHCl=CHCl → CHCl₂–CHCl₂
(iii) Addition of HCl (Markovnikov):
HC≡CH + HCl → CH₂=CHCl → CH₃–CHCl₂
(iv) Acidic character (unique to terminal alkynes):
HC≡CH + Na → HC≡C⁻Na⁺ + ½H₂ (pKa ≈ 25)
HC≡CH + AgNO₃/NH₃ → AgC≡CAg↓ (silver acetylide — white ppt.)HC≡CH + CuCl/NH₃ → CuC≡CCu↓ (copper acetylide — red ppt.)
(v) Polymerisation:
3HC≡CH → Benzene (C₆H₆) at 300°C / red hot iron tube (Trimerisation)
1 mark — Definition + nomenclature rules
1 mark — Preparation methods (any 2)
3 marks — Reactions of ethyne (1 mark each: addition, acidic character, polymerisation)
📊 अंक सारांश — Marking Summary
| Section | Q.Nos | Each | Max | मुख्य विषय |
|---|---|---|---|---|
| A | 1–15 | 1 | 15 | MCQ — concept check |
| B | 16–23 | 2 | 16 | परिभाषाएँ, नियम, सूत्र |
| C | 24–28 | 3 | 15 | विस्तृत उत्तर + संख्यात्मक |
| D | 29–30 | 5 | 10 | निबंधात्मक (दोनों विकल्प) |
| कुल / Total | 56 | — | ||
⭐ परीक्षक को यह ध्यान रखना चाहिए कि वैकल्पिक सही उत्तरों को भी पूर्ण अंक दिए जाएँ।
Examiners should award full marks for any scientifically correct alternative answer not listed here.
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