📌 SECTION D
Questions 34 to 36 · Detailed step-by-step solutions · Internal choice in Q34 & Q35
🧪 Carbon Compounds
(a) (i) Draw two isomeric structures of Butene (C₄H₈).
(ii) Name the following compounds:
CH₃-CH₂-CH₂-CH₂-Cl and
CH₃-CH₂-CH₂-CHO
(iii) Write chemical equations for: (I) Ethanol complete oxidation, (II) Propene hydrogenation, (III) Ethanoic acid + ethanol. Mention essential conditions.
🔍 VIEW DETAILED ANSWER (5 MARKS)
(a)(i) Isomers of Butene (C₄H₈):
- But-1-ene: CH₂=CH-CH₂-CH₃
- But-2-ene: CH₃-CH=CH-CH₃ (can be cis/trans)
(a)(ii) Naming:
– CH₃-CH₂-CH₂-CH₂-Cl : 1-Chlorobutane (or Butyl chloride)
– CH₃-CH₂-CH₂-CHO : Butanal (Butyraldehyde)
(a)(iii) Chemical equations with conditions:
(I) Ethanol complete oxidation: C₂H₅OH + 3O₂ → 2CO₂ + 3H₂O + heat+light (Condition: ignition/combustion)
(II) Propene hydrogenation: CH₃-CH=CH₂ + H₂ → CH₃-CH₂-CH₃ (Condition: Ni/Pd catalyst, heat)
(III) Ethanoic acid + ethanol (Esterification): CH₃COOH + C₂H₅OH ⇌ CH₃COOC₂H₅ + H₂O (Condition: conc. H₂SO₄, heat)
━━━━ OR ━━━━
(b) (i) A carbon compound X is a good solvent. On reaction with sodium, X forms two products Y and Z. Z is used to convert vegetable oil into vegetable ghee. Identify X, Y, Z. Write equation of X with sodium.
(ii) Write chemical equations when: (I) Ethanol burns in air, (II) Ethanol heated with excess conc. H₂SO₄ at 443 K, (III) Ethanol reacts with acidified K₂Cr₂O₇.
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(b)(i) Identification:
X = Ethanol (C₂H₅OH) – good solvent.
Reaction: 2C₂H₅OH + 2Na → 2C₂H₅ONa (Y = Sodium ethoxide) + H₂ (Z)
Z = Hydrogen gas – used in hydrogenation of vegetable oils to vanaspati ghee.
(b)(ii) Equations:
(I) C₂H₅OH + 3O₂ → 2CO₂ + 3H₂O (combustion)
(II) C₂H₅OH →(conc. H₂SO₄, 443 K) CH₂=CH₂ + H₂O (dehydration to ethene)
(III) 3C₂H₅OH + 2K₂Cr₂O₇ + 8H₂SO₄ → 3CH₃COOH + 2Cr₂(SO₄)₃ + 2K₂SO₄ + 11H₂O (oxidation to ethanoic acid)
👶 Reproduction
(a) (i) Write functions of: (I) Ovary, (II) Fallopian tube, (III) Uterus.
(ii) State briefly two contraceptive methods used by human males.
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(a)(i) Functions:
Ovary: Produces ova (eggs) and secretes female hormones (estrogen, progesterone).
Fallopian tube (Oviduct): Site of fertilization; transports ovum from ovary to uterus.
Uterus: Site of embryo implantation; nourishes and protects developing fetus.
(a)(ii) Male contraceptive methods:
1. Condom: Physical barrier preventing sperm entry.
2. Vasectomy: Surgical ligation of vas deferens to prevent sperm release.
━━━━ OR ━━━━
(b) (i) Differentiate between self-pollination and cross-pollination.
(ii) Identify A, B, C in the diagram and write one function of each.
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(b)(i) Difference:
Self-pollination: Transfer of pollen from anther to stigma of same flower or same plant. Occurs in bisexual flowers.
Cross-pollination: Transfer of pollen from anther of one flower to stigma of another flower of same species. Requires external agents (wind, insects).
(b)(ii) Diagram parts (typical flower):
A = Anther – produces pollen grains.
B = Stigma – receives pollen during pollination.
C = Ovary – contains ovules; develops into fruit.
🔍 Lenses
(a) (i) A lens X has power –2.5 D. Name the lens and find its focal length in cm. For which eye defect is it used?
(ii) For a lens, magnification m = –2, object distance = 20 cm. Using sign convention, state: nature of image, size comparison, position, sign of image height.
(iii) Lenses A (f=10 cm) and B (f=20 cm). Which shows higher convergence/divergence? Why?
🔍 VIEW DETAILED ANSWER (5 MARKS)
(a)(i) Lens X: Power negative → concave lens. f = 1/P = 1/(-2.5) = –0.4 m = –40 cm. Used for myopia (nearsightedness).
(a)(ii) m = –2, u = –20 cm (sign convention):
m = v/u ⇒ –2 = v/(-20) ⇒ v = +40 cm (real image, opposite side).
(I) Nature: Real and inverted
(II) Size: Enlarged (2×)
(III) Position: 40 cm from lens, on opposite side
(IV) Sign of image height: Negative (inverted)
(a)(iii) Higher convergence/divergence: Lens A (10 cm) shows higher power (P=1/f). Shorter focal length ⇒ greater bending ability.
━━━━ OR ━━━━
(b) (i) Draw ray diagram for refraction through rectangular glass slab (oblique incidence).
(ii) State Snell’s law.
(iii) Differentiate between virtual images formed by convex lens and concave lens based on object distance and magnification.
🔍 VIEW DETAILED ANSWER (OR)
(b)(i) Ray diagram: Incident ray bends towards normal at air-glass interface, away from normal at glass-air interface; emergent ray parallel but laterally shifted.
(b)(ii) Snell’s law: n₁ sin i = n₂ sin r, where i = angle of incidence, r = angle of refraction, n = refractive index.
(b)(iii) Difference in virtual images:
Convex lens virtual image: Object between optical centre and focus (u < f); image enlarged, erect, on same side; m > 1.
Concave lens virtual image: Always virtual for any object distance; image diminished, erect, on same side; m < 1.
✅ CBSE 2025 Science · Section D · Detailed 5-Mark Solutions