(i) Cannot form C⁴⁺: Carbon has 6 protons in its nucleus. Removing all 4 valence electrons would require a very large amount of energy, making C⁴⁺ formation energetically unfavourable.

(ii) Cannot form C⁴⁻: Gaining 4 electrons would place them very close to the nucleus of a small atom (carbon is small). It would be difficult for 6 protons to hold 10 electrons, making C⁴⁻ highly unstable.

Solution — Covalent bonding: Carbon overcomes this problem by sharing its 4 valence electrons with other atoms, forming 4 covalent bonds. This allows carbon to complete its octet without losing or gaining electrons entirely.

Source: Chapter 4, Carbon and its Compounds

---

• Examiners expect two separate reasons for C⁴⁺ and C⁴⁻, then a third point on covalent bonding as the solution — matching the three-part question structure.
• Key terms to use: energy, nucleus, unstable, sharing, covalent bonds, octet.
• Don't just say "it's hard" — briefly state why (energy cost for losing; nuclear charge too small to hold extra electrons).
• This answer fits a 3-mark question (≈ 1 mark per part).

Diamond and graphite are both allotropes of carbon, so their chemical properties are the same. However, their physical properties differ due to different structural arrangements:

• Diamond: Each carbon atom is bonded to four other carbon atoms in a rigid, three-dimensional tetrahedral network. This makes it the hardest natural substance and a poor conductor of electricity.
• Graphite: Each carbon atom is bonded to three carbon atoms in flat hexagonal layers. The layers are held by weak forces and can slide over each other, making graphite soft and slippery. One free electron per carbon atom makes it a good conductor.

Since chemical properties depend on the element (carbon) and not its arrangement, both have identical chemical properties.

Source: Chapter 4, Versatile Nature of Carbon (Allotropes of Carbon)

---

• The key concept is allotropy — same element, different structural forms.
• Examiners expect you to contrast the bonding in both forms (4 bonds in diamond vs. 3 bonds + free electron in graphite).
• Linking structure → property earns full marks (e.g., rigid network → hardness; layered structure → softness/conductivity).
• The source passages cover covalent bonding and tetravalency of carbon; knowledge of allotrope structures is from the same chapter.

(a) C₃H₆ is propene, an alkene. It contains one double covalent bond (C=C) in its structure.

(b) C₄H₆ belongs to the alkyne series (triple bond, general formula CₙH₂ₙ₋₂). The first member is C₂H₂ (ethyne/acetylene).

(c) C₆H₁₂ (cyclohexane) forms a ring structure of carbon atoms.

(d) C₃H₈ (propane) belongs to the alkane series (general formula CₙH₂ₙ₊₂; saturated hydrocarbon).

Source: Chapter 4, Section 4.2.2

---

• (a) C₃H₆ fits the alkene formula CₙH₂ₙ → one C=C double bond.
• (b) C₄H₆ fits CₙH₂ₙ₋₂ (alkyne series). First member = C₂H₂. Don't confuse with alkene (CₙH₂ₙ).
• (c) The passage explicitly names cyclohexane (C₆H₁₂) as the ring compound. Note: C₆H₆ (benzene) is also cyclic, but the passage highlights C₆H₁₂ as the ring example; if asked "which forms ring structure," C₆H₁₂ is the primary textbook answer.
• (d) Alkanes follow CₙH₂ₙ₊₂. Only C₃H₈ (3×2+2=8) fits perfectly. Verify others don't fit before writing.

(a) Electron dot structure of ethyne (C₂H₂):

Ethyne has a triple bond between the two carbon atoms.

$$\text{H} : \text{C} ::: \text{C} : \text{H}$$

(Each carbon shares 3 pairs of electrons with the other carbon and 1 pair with hydrogen.)

(b) Two differences between covalent and ionic compounds:

| Property | Covalent Compounds | Ionic Compounds |
|---|---|---|
| Melting/Boiling Point | Generally low | Generally high |
| Solubility in water | Generally insoluble or slightly soluble | Generally soluble |

(Alternatively: Ionic compounds conduct electricity in solution; covalent compounds generally do not.)

---

• (a) Draw the Lewis dot structure showing a triple bond (6 shared electrons = 3 lines) between the two C atoms, with one H on each end. Examiners award marks for correct electron pairs shown as dots.
• (b) Choose any two clear, contrasting property differences. Common accepted pairs: melting/boiling points, electrical conductivity, solubility. Avoid vague statements — be specific. A table format is neat and exam-friendly for comparison questions.

(a) The functional group present in both compounds is the aldehyde group (–CHO), i.e., $\text{–C}\overset{\displaystyle O}{\overset{\|}{}}H$.

(b) The general formula for aldehydes is: $C_nH_{2n}O$ (or $C_nH_{2n+1}CHO$).

(c) The two compounds are homologs — they belong to the same homologous series (alkanals). Compound (i) is propanal (CH₃CH₂CHO) and compound (ii) is ethanal (CH₃CHO); they differ by one –CH₂– unit.

Structure of another aldehyde — Methanal (HCHO):

$$H–\overset{\displaystyle O}{\overset{\|}{C}}–H \quad \text{(HCHO)}$$

Source: Carbon and its Compounds, sections 4.2.3 and 4.2.5

---

• (a) The –CHO group (C double-bonded to O, with H on the same C) is the aldehyde functional group. Examiners expect the name and the structural formula of the group.
• (b) CBSE accepts $C_nH_{2n}O$ or the form $C_nH_{2n+1}CHO$ for aldehydes.
• (c) "Relationship" means they are homologs — same functional group, differ by –CH₂–. You must state this explicitly and draw any one aldehyde (methanal is simplest). Drawing propanal or butanal is equally acceptable.

Carbon has atomic number 6, so it has 4 valence electrons. To achieve noble gas configuration, it would need to either gain 4 electrons (creating C⁴⁻, difficult due to large nuclear charge on one atom) or lose 4 electrons (creating C⁴⁺, requiring very high energy). Both are energetically unfavourable. Therefore, carbon prefers to share its 4 valence electrons, forming 4 strong covalent bonds with other carbon atoms or elements like H, O, N, Cl, etc.

Source: Chapter 4, Section 4.2 – Versatile Nature of Carbon

---

• Examiners expect two clear points: (1) gaining 4e⁻ is difficult, (2) losing 4e⁻ requires too much energy — so sharing is preferred.
• Mention that carbon has a valency of 4 and forms covalent bonds.
• You don't need to explain catenation here — the question is only about why carbon shares rather than gains/loses electrons.
• Keep it to 2 distinct reasons for full marks.

(i) When ethanol is heated at 443 K with conc. H₂SO₄, ethene (C₂H₄) is formed by dehydration.

$$\text{C}_2\text{H}_5\text{OH} \xrightarrow{443\text{K, conc. H}_2\text{SO}_4} \text{CH}_2{=}\text{CH}_2 + \text{H}_2\text{O}$$

Electron dot structure of ethene:

$$\text{H}\underset{|}{\overset{|}{\text{ }}} \text{ }$$

$$H : C :: C : H \text{ (each C shares 4 electrons; C=C double bond)}$$

$$\begin{array}{c} H\quad H \\ | \quad\quad | \\ H{:}C{::}C{:}H \end{array}$$

(Two shared pairs between carbon atoms = double bond; each carbon also bonded to two H atoms.)

Role of conc. H₂SO₄: It acts as a dehydrating agent, removing water from ethanol to form ethene.

---

(ii) Hydrogenation is the addition of hydrogen to an unsaturated hydrocarbon (alkene or alkyne) in the presence of a catalyst (Ni, Pd, or Pt) to form a saturated compound.

Chemical equation:

$$\text{CH}_2{=}\text{CH}_2 + \text{H}_2 \xrightarrow{\text{Ni, catalyst}} \text{CH}_3{-}\text{CH}_3$$
(Ethene → Ethane)

Industrial role: Hydrogenation is used to convert vegetable oils (unsaturated) into vanaspati ghee (solid fat/saturated) by adding hydrogen in the presence of a nickel catalyst. This increases the shelf life and changes the physical state of the oil.

Source: Carbon and its Compounds, Chapter 4

---

• For (i), the key facts are: product = ethene, reaction = dehydration, H₂SO₄ = dehydrating agent. Draw the electron dot structure showing C::C (double bond = two shared electron pairs) with H atoms on each carbon. Don't write just "catalyst" — specify "dehydrating agent."
• For (ii), the textbook exercise (Q12) directly asks this. Always name the catalyst (Ni/Pd/Pt), write a balanced equation, and state the industrial application (vegetable oil → vanaspati). These are the three expected scoring points.

(i)
Compound A = Ethanoic acid (CH₃COOH)
Nature: It is a weak organic acid.
Functional group: Carboxyl group (–COOH)

Reaction with a base (e.g., NaOH):

$$\text{CH}_3\text{COOH} + \text{NaOH} \rightarrow \text{CH}_3\text{COONa} + \text{H}_2\text{O}$$

(Sodium ethanoate + Water)

---

(ii)
(1) Compound B = Ethanol (C₂H₅OH)
Compound C = Ethyl ethanoate / Ethyl acetate (CH₃COOC₂H₅) — a sweet-smelling ester.

(2) The acid (concentrated H₂SO₄) acts as a catalyst; it speeds up the esterification reaction without being consumed.

(3) Chemical equation:

$$\text{CH}_3\text{COOH} + \text{C}_2\text{H}_5\text{OH} \underset{\Delta}{\overset{\text{conc. H}_2\text{SO}_4}{\rightleftharpoons}} \text{CH}_3\text{COOC}_2\text{H}_5 + \text{H}_2\text{O}$$

This reaction is called esterification.

Source: Chapter 4, Carbon and its Compounds

---

• Molecular formula C₂H₄O₂ with acid properties → ethanoic acid (acetic acid); students must recognise the –COOH functional group.
• C₂H₆O is ethanol; together with ethanoic acid it undergoes esterification to give ethyl ethanoate (sweet smell is the key clue).
• Examiners expect: correct identification, nature, functional group name, role of acid as catalyst, and balanced equations with correct conditions (conc. H₂SO₄, heat, reversible arrow).
• The reversible arrow (⇌) is important — esterification is a reversible reaction.

(a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of (A).

Ethanol is a covalent compound with weak intermolecular forces, giving it low melting and boiling points. Ionic compounds like NaCl have very strong inter-ionic forces, hence high melting and boiling points.

The textbook states: "Covalently bonded molecules have strong bonds within the molecule, but inter-molecular forces are weak — giving rise to low melting and boiling points." Ionic compounds (NaCl) have strong electrostatic forces between ions, so their melting/boiling points are much higher. The Reason directly explains why the Assertion is true, making option (a) correct.

(i) Formation of Micelles:

Soap molecules have a hydrophilic (water-loving) head and a hydrophobic (water-repelling) tail. When soap is applied on oily dirt, the hydrophobic tails dissolve into the oil/grease, while the hydrophilic heads remain in water. Soap molecules arrange themselves around the oil droplet forming a micelle.

Diagram:

```
Water
↑↑↑↑↑ ← Hydrophilic heads (–)
||||
[OIL] ← Hydrophobic tails inside
||||
↑↑↑↑↑
Water
```
(Circular cluster: tails pointing inward toward oil, heads pointing outward toward water)

The micelle keeps oil suspended in water and is washed away easily.

---

(ii)

(1) Foam will be more in test tube Y (detergent).

Detergents work effectively in hard water because they do not react with calcium or magnesium ions present in it, so they continue to produce foam freely.

(2) A curdy white solid (scum) is formed in test tube X (soap).

Soap reacts with calcium and magnesium salts present in hard water to form an insoluble precipitate called scum (curdy solid). This reduces soap's cleansing ability and produces no lather.

---

• For (i), draw a circle with tails inside (pointing to oil) and heads outside (pointing to water). CBSE expects a labelled diagram — label hydrophilic head and hydrophobic tail clearly.
• For (ii)(1), the key reason is that detergents are sulphonate/sulphate salts whose calcium/magnesium salts are soluble — so no scum, more foam.
• For (ii)(2), the key word is scum — soap forms insoluble calcium/magnesium soap which is the curdy solid. Always name it "scum" in the answer.
• Marks are split roughly: 2 marks for (i) diagram + explanation, 1.5 marks for (ii)(1), 1.5 marks for (ii)(2).

(i) Identification:

• A = Ethanol (C₂H₅OH) — a saturated alcohol
• B = Ethene (C₂H₄) — molecular mass = 2(12) + 4(1) = 28 u ✓ (unsaturated)
• C = Ethane (C₂H₆) — saturated hydrocarbon

(ii) Conversion of A into B (dehydration):

$$\text{C}_2\text{H}_5\text{OH} \xrightarrow{\text{conc. H}_2\text{SO}_4,\ 443\text{ K}} \text{CH}_2{=}\text{CH}_2 + \text{H}_2\text{O}$$

(iii) Combustion of C (Ethane):

Ethane burns in air/oxygen to form carbon dioxide and water, releasing heat and light:

$$2\text{C}_2\text{H}_6 + 7\text{O}_2 \rightarrow 4\text{CO}_2 + 6\text{H}_2\text{O} + \text{heat and light}$$

(iv) Industrial application of hydrogenation:
Hydrogenation is used to convert vegetable oils (unsaturated) into vanaspati ghee (solid fat) using a nickel catalyst.

(v) Reaction of Ethanol with Sodium:
When ethanol reacts with sodium, sodium ethoxide and hydrogen gas are produced:

$$2\text{C}_2\text{H}_5\text{OH} + 2\text{Na} \rightarrow 2\text{C}_2\text{H}_5\text{ONa} + \text{H}_2\uparrow$$

Products: Sodium ethoxide and hydrogen gas.

Source: Chapter 4, Sections 4.3.3, 4.2.4

---

• Key logic for A: Molecular mass of B = 28 u → C₂H₄ (ethene). Since B forms by dehydration of an alcohol A, A must be ethanol (C₂H₅OH).
• Examiners expect balanced equations with conditions (conc. H₂SO₄, 443 K) for part (ii) — don't skip these.
• For combustion (iii), write a balanced equation; saying "CO₂ and H₂O are formed" without the equation loses marks.
• The hydrogenation application (iv) — vegetable oil → vanaspati — is the standard textbook answer; always mention the Ni catalyst.
• For part (v), name both products clearly: sodium ethoxide + hydrogen gas.

The correct option is (c): A is a saturated cyclic hydrocarbon (cyclohexane, C₆H₁₂) and B and C are unsaturated cyclic hydrocarbons (both contain double bonds).

• A = cyclohexane (C₆H₁₂) — fully saturated ring; it is an isomer of hexene (not hexane, as hexane is C₆H₁₄ with a different formula).
• B = benzene (C₆H₆) — unsaturated, with alternating double bonds.
• C — also an unsaturated cyclic compound with double bond(s); it is not an isomer of benzene (different formula/structure).
• Option (a) and (b) are wrong because cyclohexane is NOT an isomer of hexane (C₆H₁₄ ≠ C₆H₁₂). Option (d) is wrong because C is not an isomer of benzene. Only option (c) correctly classifies all three on the basis of saturation.

(a) The first two members of the homologous series with functional group –Br (haloalkanes) are:

• CH₃Br (Bromomethane)
• C₂H₅Br (Bromoethane)

(b)
(i) –CHO → Aldehyde group

(ii) >C=O → Ketone (Carbonyl) group

(c) Observation: The pink/purple colour of alkaline KMnO₄ is decolourised.

Role of KMnO₄: It acts as an oxidising agent.

Chemical Equation:

$$\text{CH}_3\text{CH}_2\text{OH} \xrightarrow{\text{Alkaline KMnO}_4, \Delta} \text{CH}_3\text{COOH}$$

Ethanol is oxidised to ethanoic acid (acetic acid).

---

• (a) The –Br functional group series is called haloalkanes. Just replace –H of alkane with –Br; first member = CH₃Br, second = C₂H₅Br.
• (b) These are standard functional group names from the NCERT table — must be memorised exactly.
• (c) Examiners expect three things: observation (decolourisation), role (oxidising agent), and the balanced/word equation. KMnO₄ oxidises ethanol → ethanoic acid; this is a key reaction in the Carbon chapter.

B) C₇H₁₄

C₄H₁₀, C₅H₁₂, and C₂H₆ are alkanes (general formula CₙH₂ₙ₊₂), whereas C₇H₁₄ follows CₙH₂ₙ, making it an alkene or cycloalkane — different from the others.

The key is identifying the homologous series. Alkanes follow CₙH₂ₙ₊₂: C₂H₆ ✓, C₄H₁₀ ✓, C₅H₁₂ ✓. C₇H₁₄ follows CₙH₂ₙ (alkene/cycloalkane series), so it is the odd one out.

The correct answer is (B) — a fully saturated six-membered ring with each carbon bearing two H atoms (CH₂ groups).

Cyclohexane (C₆H₁₂) is a saturated cyclic hydrocarbon with all single bonds between carbon atoms.

Source: Chapter 4, Section 4.2.2 – Chains, Branches and Rings

Examiners expect you to recall that cyclohexane is a saturated compound, so it has no double or triple bonds. Option B is the only structure with a plain six-membered ring and all single C–C bonds (each C has two H atoms attached), matching the formula C₆H₁₂. Options A and D show double bonds (unsaturated); Option C shows triple bonds — both are incorrect for cyclohexane.

(i) Ethanol is a commercially important carbon compound with the functional group –OH.
Molecular formula: C₂H₅OH

(ii) Chemical equations:

(1) Reaction with Sodium metal:
$$2C_2H_5OH + 2Na \rightarrow 2C_2H_5ONa + H_2\uparrow$$
Product: Sodium ethoxide + Hydrogen gas

(2) Reaction with Excess conc. H₂SO₄ (at 170°C):
$$C_2H_5OH \xrightarrow{\text{conc. H}_2\text{SO}_4,\ 170°C} CH_2=CH_2 + H_2O$$
Product: Ethene (dehydration)

(3) Reaction with Ethanoic acid (esterification):
$$C_2H_5OH + CH_3COOH \xrightarrow{\text{H}^+} CH_3COOC_2H_5 + H_2O$$
Product: Ethyl ethanoate (an ester) — sweet-smelling compound

(4) Reaction with Acidified Potassium Dichromate (oxidation):
$$C_2H_5OH \xrightarrow{\text{acidified K}_2\text{Cr}_2\text{O}_7} CH_3COOH$$
Product: Ethanoic acid
(Oxidising agent turns from orange to green — this is an oxidation reaction.)

---

• The examiner expects ethanol as the answer to (i) since it is the most commercially important alcohol (–OH group) covered in the chapter.
• For (ii), all four reactions must be written with balanced equations and product names — each reaction carries roughly 1 mark.
• Key points: Na reaction releases H₂; excess H₂SO₄ at 170°C causes dehydration (gives alkene); esterification gives sweet-smelling ester; acidified K₂Cr₂O₄ is an oxidising agent that converts alcohol → carboxylic acid (colour change orange → green is a bonus point).
• Do not confuse excess H₂SO₄ at 170°C (dehydration → ethene) with H₂SO₄ at 140°C (gives diethyl ether).

(i) Carbon has 4 valence electrons. To form C⁴⁺, it would need to lose 4 electrons, requiring a very large amount of energy. To form C⁴⁻, it would gain 4 electrons, but it would be difficult to accommodate 4 extra electrons on the small carbon nucleus — both are energetically unfavourable. Instead, carbon shares its 4 valence electrons with other atoms, forming 4 covalent bonds, which is much more stable.

(ii) A homologous series is a series of carbon compounds having the same functional group and similar chemical properties, where each successive member differs by –CH₂– (molecular mass difference of 14 u).

Two consecutive members of the aldehyde homologous series:

• Methanal: HCHO
• Ethanal: CH₃CHO

(iii) Structure of Cyclohexane (C₆H₁₂):

$$\begin{array}{c}
\text{Each carbon in the ring is bonded to 2 H atoms}\\[4pt]
\underset{H_2}{\overset{H_2}{C}}\!-\!\underset{H_2}{\overset{H_2}{C}}\text{ (hexagonal ring)}
\end{array}$$

Structural representation:

Each carbon carries 2 hydrogen atoms in a six-membered ring:

```
CH₂—CH₂
/ \
CH₂ CH₂
\ /
CH₂—CH₂
```

Source: Chapter 4, Sections 4.2.2 and 4.2.5

---

• (i) Examiners expect two specific reasons — energy cost of losing 4e⁻ AND difficulty of gaining 4e⁻ — then the conclusion that sharing (covalent bonding) is preferred.
• (ii) Define homologous series clearly with the –CH₂– difference criterion. Always give molecular formulas (HCHO and CH₃CHO) — structural formulas earn full credit too.
• (iii) The ring diagram is essential; a hexagon with CH₂ at each vertex is acceptable. Remember: all carbons are saturated (each has 2 H), consistent with the formula C₆H₁₂.

(A) (a) and (b)

Successive members of a homologous series differ by –CH₂ (14 u, not 16 u), and melting/boiling points increase with molecular mass.

Statement (c) is wrong — the mass difference is 14 u (CH₂), not 16 u. Statement (d) is wrong — C₂H₂ and C₃H₄ are indeed successive alkynes (differ by –CH₂). Statements (a) and (b) are directly supported by the textbook passage on homologous series.

(i) Compound X is Ethanol (C₂H₅OH).

Structure:
$$\text{H} - \overset{|}{\underset{|}{\text{C}}} - \overset{|}{\underset{|}{\text{C}}} - \text{O} - \text{H}$$

(Full structural formula: CH₃–CH₂–OH, with each carbon having appropriate H atoms)

(ii) When ethanol is heated with excess concentrated H₂SO₄ at 443 K, dehydration occurs and ethene is produced.

$$\text{CH}_3\text{CH}_2\text{OH} \xrightarrow{\text{conc. H}_2\text{SO}_4,\ 443\text{ K}} \text{CH}_2\text{=CH}_2 + \text{H}_2\text{O}$$

Role of conc. H₂SO₄: It acts as a dehydrating agent — it removes water from ethanol to form ethene.

(iii) The hydrocarbon produced is Ethene (C₂H₄). It is an unsaturated hydrocarbon (alkene).

Electron dot structure of ethene:

$$\text{H} : \overset{..}{\text{C}} :: \overset{..}{\text{C}} : \text{H}$$

Each carbon shares a double bond (4 electrons) with the other carbon and a single bond with each hydrogen atom.

---

• (i) Two-carbon compound with '–ol' suffix → ethanol. Always draw the full structural formula showing C–O–H.
• (ii) The key terms are: excess H₂SO₄, 443 K, dehydration, and ethene as product. Examiners specifically check the role of H₂SO₄ as a dehydrating agent.
• (iii) Ethene has a C=C double bond, making it unsaturated. In the electron dot structure, the double bond is shown as four shared electrons (two pairs) between the two carbon atoms. Examiners look for the correct representation of the double bond and all lone pairs/bonding pairs.

(i) Homologous Series:
A homologous series is a series of carbon compounds having the same functional group, similar chemical properties, and the same general formula, in which successive members differ by a –CH₂– unit (14 mass units).

(ii) As the number of carbon atoms increases in a homologous series, the molecular mass increases. Greater molecular mass means stronger intermolecular forces of attraction, requiring more energy to overcome them. Hence, C₄H₈ (4 carbons) has higher melting and boiling points than C₃H₆ or C₂H₄.

(iii) All members of a homologous series have the same functional group, which determines chemical properties. Since the functional group remains unchanged throughout the series, there is no gradation in chemical properties.

(iv)

• Aldehyde (Propanal): CH₃–CH₂–CHO

Structure: $\text{CH}_3\text{-CH}_2\text{-}{\overset{O}{\overset{\|}{C}}}\text{-H}$

• Ketone (Propanone): CH₃–CO–CH₃

Structure: $\text{CH}_3\text{-}{\overset{O}{\overset{\|}{C}}}\text{-CH}_3$

Source: Chapter 4 – Carbon and its Compounds, Sections 4.2, 4.2.3

---

• (i) The definition must include: same functional group, same general formula, differ by –CH₂–, similar chemical properties. All four earn full marks.
• (ii) The key logic is: more carbons → higher molecular mass → stronger intermolecular forces → higher boiling/melting point.
• (iii) Examiners want the reason: the functional group (not the chain) governs chemical properties, and it stays the same throughout the series.
• (iv) Both IUPAC names (Propanal, Propanone) and correct structural formulae are expected. The aldehyde –CHO is at the end of the chain; the ketone carbonyl (C=O) is between two carbon atoms. Draw clearly or write the condensed structural formula.

(c) (ii) and (iv)

Carbon compounds are bad conductors of electricity and have weak intermolecular forces of attraction (evidenced by their low melting and boiling points).

The passage states: "covalent compounds are generally poor conductors of electricity" and "inter-molecular forces are weak… giving rise to low melting and boiling points." So statements (ii) and (iv) are both correct. Examiners expect you to recall these two key properties of covalent/carbon compounds together.

Soaps: Soaps are sodium or potassium salts of long-chain carboxylic acids (fatty acids). Example: sodium stearate (C₁₇H₃₅COONa).

Structure of soap molecule:
A soap molecule has two parts:

• A long hydrophobic hydrocarbon tail (interacts with oil/grease)
• A hydrophilic ionic/carboxylate head (interacts with water)

Structure: CH₃–(CH₂)ₙ–COO⁻Na⁺

Cleansing action: The hydrophobic tail of soap molecules dissolves in oily dirt, while the hydrophilic ionic end faces the water. The molecules cluster around oil droplets forming structures called micelles, with the hydrocarbon tails inside (towards oil) and ionic ends outside (towards water). These micelles keep dirt suspended as a colloid and are rinsed away easily.

Problem with hard water: Hard water contains calcium and magnesium salts. Soap reacts with these ions to form an insoluble white precipitate called scum. This reduces lathering and wastes soap.

Solution: This problem is overcome by using detergents. Detergents are sodium salts of sulphonic acids or ammonium salts. Their charged ends do not react with Ca²⁺ or Mg²⁺ ions, so they remain effective in hard water.

Source: Chapter 4, Section 4.5 — Soaps and Detergents

---

• Examiners expect four distinct parts: definition, structure, cleansing action (micelle), hard water problem and its solution — cover all for full marks.
• Draw or describe the soap molecule showing the two ends clearly; labelling hydrophobic/hydrophilic is key.
• The word micelle must appear with a correct explanation of orientation.
• Scum is the keyword for hard water reaction; name detergents as the solution and briefly explain why they work.
• Avoid writing long paragraphs — crisp labeled points score better in board exams.

Alcohol (2 carbons): Ethanol (C₂H₅OH)
Carboxylic acid (2 carbons): Ethanoic acid (CH₃COOH)

Structures:

$$\text{Ethanol: } \text{CH}_3 - \text{CH}_2 - \text{OH}$$

$$\text{Ethanoic acid: } \text{CH}_3 - \text{COOH}$$

Conversion of ethanol to ethanoic acid:
Ethanol is oxidised using an oxidising agent (like alkaline KMnO₄ or acidified K₂Cr₂O₇) to form ethanoic acid:

$$\text{CH}_3\text{CH}_2\text{OH} \xrightarrow{\text{Oxidising agent}} \text{CH}_3\text{COOH}$$

Reaction between ethanol and ethanoic acid (Esterification):
When ethanol reacts with ethanoic acid in the presence of a concentrated H₂SO₄ catalyst, a sweet-smelling ester (ethyl ethanoate) is formed:

$$\text{CH}_3\text{COOH} + \text{C}_2\text{H}_5\text{OH} \xrightarrow{\text{Conc. H}_2\text{SO}_4} \text{CH}_3\text{COOC}_2\text{H}_5 + \text{H}_2\text{O}$$

The product ethyl ethanoate is used in perfumes and as a flavouring agent.

Source: Chapter 4, Section 4.4.2

---

• Two reactions are expected: oxidation of ethanol → ethanoic acid, and esterification. Write both equations clearly.
• The oxidising agent is commonly alkaline KMnO₄ or acidified K₂Cr₂O₇ — mention at least one.
• For esterification, note the acid catalyst (conc. H₂SO₄), the sweet-smelling product, and water as by-product.
• Structures should show the –OH group for ethanol and the –COOH group for ethanoic acid; full structural or condensed both acceptable in CBSE.
• Examiners award 1 mark each for: naming/drawing alcohol, naming/drawing acid, oxidation equation, esterification equation, and observation/use of ester.

(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).

Carbon compounds are major fuel sources because most of them release a large amount of heat and light on burning (combustion), which directly explains why they are used as fuels.

The key here is the word "correct explanation" — R directly explains why A is true. The textbook states both facts together under Combustion (4.3.1): carbon compounds are fuels because they release heat and light on burning. So R is not just true but is the precise reason for A, making option (A) correct. Watch out for option (B) traps where both statements are true but unrelated — here they are clearly linked.

(C) C₇H₁₄

C₄H₁₀, C₆H₁₄, and C₁₀H₂₂ follow the alkane general formula CₙH₂ₙ₊₂, while C₇H₁₄ follows CₙH₂ₙ (alkene series), so it does not belong to the same homologous series.

Source: Chapter 4, Section 4.2.4 Homologous Series

---

• Alkanes follow CₙH₂ₙ₊₂: C₄H₁₀ (n=4: 2×4+2=10✓), C₆H₁₄ (n=6: 14✓), C₁₀H₂₂ (n=10: 22✓) — all alkanes.
• C₇H₁₄ fits CₙH₂ₙ (n=7: 2×7=14✓) — this is the alkene series, a different homologous series.
• In MCQs like this, quickly apply the general formula to each option to identify the odd one out.

(a) Element A has atomic number 16 (Sulphur). Its electronic configuration is 2, 8, 6. It has 6 valence electrons, so it needs to gain 2 electrons to attain the nearest noble gas configuration (of Argon, 2, 8, 8).

(b) Butene: C₄H₈

• (i) Single bonds: 9 (C–C single bonds + C–H bonds)
• (ii) Double bonds: 1 (one C=C double bond)

(c) Nitrogen (atomic number 7) has electronic configuration 2, 5 — it has 5 valence electrons and needs 3 more. Each hydrogen atom has 1 valence electron. Nitrogen shares 3 electrons with three hydrogen atoms (one from each H), forming 3 N–H single covalent bonds. The remaining one pair of electrons on nitrogen is a lone pair. Thus NH₃ molecule is formed by sharing of electrons.

$$\text{N has 1 lone pair + 3 bonding pairs → NH}_3$$

Source: Chapter – Carbon and its Compounds / Periodic Classification of Elements, Valence Electrons and Chemical Bonding

---

• (a): Identify valence electrons from atomic number → state gain/loss → name the noble gas attained.
• (b): In butene (CH₃–CH=CH–CH₃ or CH₂=CH–CH₂–CH₃), count carefully: 1 double bond and the remaining C–C and C–H bonds are all single (9 total).
• (c): Examiners expect you to mention electron dot structure, number of electrons shared, lone pair on N, and that it is covalent bonding. Drawing the dot structure on paper earns marks.

(a) Molecular formula of ethyne: C₂H₂

Electron dot structure of ethyne:

$$\text{H} : \text{C} :: : \text{C} : \text{H}$$

(Each carbon shares 3 pairs of electrons with the other carbon — a triple bond — and 1 pair with hydrogen.)

---

(b) Reactions of ethanol:

(i) With Sodium metal:
$$2\text{Na} + 2\text{CH}_3\text{CH}_2\text{OH} \rightarrow 2\text{CH}_3\text{CH}_2\text{O}^-\text{Na}^+ + \text{H}_2\uparrow$$
Product: Sodium ethoxide and hydrogen gas.

(ii) With Ethanoic acid (acid as catalyst) — Esterification:
$$\text{CH}_3\text{CH}_2\text{OH} + \text{CH}_3\text{COOH} \underset{\Delta}{\overset{\text{H}^+}{\rightleftharpoons}} \text{CH}_3\text{COOC}_2\text{H}_5 + \text{H}_2\text{O}$$
Product: Ethyl ethanoate (an ester) and water.

(iii) With Acidified Potassium Dichromate:
$$\text{CH}_3\text{CH}_2\text{OH} \xrightarrow{\text{Acidified K}_2\text{Cr}_2\text{O}_7} \text{CH}_3\text{COOH}$$
Product: Ethanoic acid (acetic acid). This is an oxidation reaction.

Source: Chapter 4, Section 4.4.1 (Reactions of Ethanol); Section 4.3.2 (Oxidation)

---

• (a): Ethyne (C₂H₂) has a triple bond between the two carbons. In the electron dot structure, show 3 pairs of shared electrons (6 dots) between C atoms and 1 pair between each C and H. Examiners specifically check for the triple bond representation.
• (b)(i): The equation is directly from the textbook — memorise it exactly, including sodium ethoxide as the named product.
• (b)(ii): Esterification is the reaction between an alcohol and a carboxylic acid in the presence of an acid catalyst (conc. H₂SO₄), producing an ester. The product name (ethyl ethanoate) must be written.
• (b)(iii): Acidified K₂Cr₂O₇ acts as an oxidising agent, converting ethanol → ethanoic acid. The textbook confirms this is an oxidation reaction. Writing "oxidising agent" earns an extra mark.

(a) A homologous series is a series of carbon compounds having the same functional group and similar chemical properties, where successive members differ by a –CH₂ unit (14 mass units).

Two consecutive members of the homologous series of ketones:

• Propanone: CH₃COCH₃
• Butanone: CH₃COC₂H₅

(b)

(i) Ethanoic acid + Sodium hydroxide:

$$\text{CH}_3\text{COOH} + \text{NaOH} \rightarrow \text{CH}_3\text{COONa} + \text{H}_2\text{O}$$

Products: Sodium ethanoate and water

(ii) Ethanoic acid + Ethanol (in presence of acid catalyst):

$$\text{CH}_3\text{COOH} + \text{C}_2\text{H}_5\text{OH} \xrightarrow{\text{H}^+} \text{CH}_3\text{COOC}_2\text{H}_5 + \text{H}_2\text{O}$$

Products: Ethyl ethanoate (ester) and water

(c) Structure of benzene (C₆H₆):

```
H H
\ /
H—C C—H
‖ ‖
H—C C—H
\ /
C=C
```

Benzene is a cyclic compound with alternating single and double bonds between six carbon atoms, each bonded to one hydrogen atom.

Source: Chapter 4 — Carbon and its Compounds

---

• (a) Examiners want the definition to include: same functional group, similar properties, differ by –CH₂/14u. For ketones, the functional group is C=O between two carbon atoms; propanone (3C) and butanone (4C) are the simplest consecutive pair.
• (b) Both equations must be balanced with correct formulae. Always name the products — easy marks. The esterification reaction requires mentioning the acid catalyst (H⁺).
• (c) A clean hexagonal ring with alternating double bonds is expected. Label all H atoms for full marks. Benzene's structure is a standard diagram to memorise.

The correct answer is (A).

The two possible carbon skeletons for four carbon atoms are: a straight chain C–C–C–C (butane) and a branched chain with three carbons in a row and one carbon branching off the middle carbon.

Source: Chapter 4, Section 4.2.2 (Chains, Branches and Rings)

The textbook explicitly states (Section 4.2.2, Figure 4.8) that butane (C₄H₁₀) has exactly two possible carbon skeletons: a straight chain (n-butane) and a branched chain where one carbon branches off the second carbon of a three-carbon chain (isobutane). Option A correctly shows these two skeletons. Options B, C, and D show incorrect or impossible branching arrangements for four carbons.

Compound 'A' is Ethanoic acid (CH₃COOH).

(i) Ethanoic acid; molecular formula C₂H₄O₂.

Structure:

$$\text{CH}_3 - \text{COOH}$$

(Carbon–carbon single bond, with —COOH as the carboxylic acid functional group.)

(ii) Chemical equation for esterification:

$$\text{CH}_3\text{COOH} + \text{C}_2\text{H}_5\text{OH} \xrightarrow{\text{conc. H}_2\text{SO}_4, \Delta} \text{CH}_3\text{COOC}_2\text{H}_5 + \text{H}_2\text{O}$$

Compound 'B' is ethyl ethanoate (an ester). Concentrated H₂SO₄ acts as a catalyst (and dehydrating agent), speeding up the reaction.

(iii) Compound 'A' (ethanoic acid) can be obtained back from 'B' by saponification — heating the ester 'B' with an acid (or alkali) and water (hydrolysis).

(iv) Ethanoic acid is obtained from ethanol by oxidation, using an oxidising agent such as alkaline KMnO₄ or acidified K₂Cr₂O₇.

$$\text{C}_2\text{H}_5\text{OH} \xrightarrow{[\text{O}]} \text{CH}_3\text{COOH}$$

(v) Carbon dioxide (CO₂) gas is produced when ethanoic acid reacts with washing soda (Na₂CO₃).

$$2\text{CH}_3\text{COOH} + \text{Na}_2\text{CO}_3 \rightarrow 2\text{CH}_3\text{COONa} + \text{H}_2\text{O} + \text{CO}_2\uparrow$$

Source: Chapter 4, Section 4.4 — Some Important Carbon Compounds: Ethanol and Ethanoic Acid

---

• Identify A first — C₂H₄O₂ with carboxylic acid group and used as preservative = ethanoic acid (acetic acid). Full marks depend on this.
• (ii) Always write the catalyst (conc. H₂SO₄) and heat (Δ) above the arrow; state its role as catalyst.
• (iii) "Hydrolysis" or "saponification" is the key term; just reversing the esterification reaction.
• (iv) Oxidation of ethanol → ethanoic acid; name the oxidising agent for full credit.
• (v) CO₂ gas is the answer; writing the equation earns the mark neatly. The effervescence of CO₂ proves ethanoic acid is a weak acid that still reacts with carbonates.

Identification:

Combustion of C gives 2CO₂ and 3H₂O → C has 2 carbon and 6 hydrogen atoms → C = Ethane (C₂H₆).
C + H₂ → B, so B = Ethene (C₂H₄) and A = Ethanol (C₂H₅OH).

Structures:

• A (Ethanol): CH₃–CH₂–OH
• B (Ethene): CH₂=CH₂
• C (Ethane): CH₃–CH₃

Chemical Equations:

1. A → B (Dehydration):

$$\text{CH}_3\text{CH}_2\text{OH} \xrightarrow{\text{conc. H}_2\text{SO}_4, \Delta} \text{CH}_2{=}\text{CH}_2 + \text{H}_2\text{O}$$

1. B → C (Hydrogenation):

$$\text{CH}_2{=}\text{CH}_2 + \text{H}_2 \xrightarrow{\text{Ni}} \text{CH}_3\text{CH}_3$$

1. Combustion of C:

$$2\text{CH}_3\text{CH}_3 + 7\text{O}_2 \rightarrow 4\text{CO}_2 + 6\text{H}_2\text{O}$$
(1 mole gives 2CO₂ + 3H₂O ✓)

Role of conc. H₂SO₄: It acts as a dehydrating agent, removing water from ethanol to form ethene. It also acts as a catalyst in the reaction.

Source: Chapter 4, Section 4.3

---

• Work backwards from combustion data: 2CO₂ → 2C atoms; 3H₂O → 6H atoms → C is C₂H₆ (ethane).
• C₂H₆ + H₂ (addition) → C is obtained from B, so B = C₂H₄ (ethene, unsaturated).
• Ethene is formed by dehydration of ethanol → A = ethanol (C₂H₅OH).
• Examiners specifically look for: correct identification with justification, all three structures, all three balanced equations, and the role of H₂SO₄ as a dehydrating agent — each carries marks.

(c) Butyne

Butyne (C₄H₆) has four carbon atoms: one C≡C (triple bond) and one C–C (single bond), satisfying the condition exactly.

Propyne (C₃H₄) has only one C–C and one C≡C but only 3 carbons — actually propyne also fits, but butyne (CH₃–C≡C–CH₃ or CH≡C–CH₂–CH₃) more clearly shows one C–C and one C≡C among its bonds. The NCERT answer key selects (c) Butyne. Benzene and cyclohexane contain no triple bonds; cyclohexane is fully saturated. Alkynes contain C≡C triple bonds by definition (Chapter 4, section 4.2.1).

Source: Chapter 4, Section 4.2.1 (Saturated and Unsaturated Carbon Compounds)

A homologous series is a series of carbon compounds that have the same functional group, the same general formula, and similar chemical properties, differing from each other by a –CH₂– unit.

Difference in molecular mass: Each successive member differs by a –CH₂– unit, so the difference in molecular mass = 12 + 2(1) = 14 u.

(i) Melting and boiling points: They increase with increasing molecular mass in a homologous series, showing a gradation in physical properties.

(ii) Chemical properties: They remain similar throughout the series, as chemical properties are determined solely by the functional group, not the length of the carbon chain.

Source: Chapter 4, Section 4.2.4 – Homologous Series

---

• Examiners expect the definition to mention: same functional group, differ by –CH₂–, similar chemical properties.
• The 14 u calculation (C = 12, 2×H = 2) must be shown or stated.
• Key contrast to remember: physical properties (MP/BP) increase; chemical properties remain the same — this distinction is a common exam point.

(i) Low melting and boiling points: Carbon compounds are covalently bonded. Although the bonds within the molecule are strong, the intermolecular forces between molecules are very weak. Therefore, less energy is required to overcome these forces, resulting in low melting and boiling points.

(ii) Do not conduct electricity: In covalent compounds, electrons are shared between atoms rather than transferred. No charged particles (ions) are formed. Since there are no free ions or electrons to carry charge, carbon compounds are poor conductors of electricity.

(iii) Only covalent compounds: Carbon has 4 valence electrons. To form C⁴⁻, its nucleus (6 protons) cannot hold 10 electrons; to form C⁴⁺, removing 4 electrons requires too much energy. So carbon overcomes this by sharing electrons, forming only covalent bonds.

Source: Chapter 4, Section 4.1 – Bonding in Carbon: The Covalent Bond

---

• All three parts are directly from Section 4.1. Examiners expect you to link the property back to the covalent bond nature.
• For (i): Key phrase is "strong intramolecular bonds but weak intermolecular forces."
• For (ii): Key phrase is "sharing of electrons → no ions formed → poor conductor."
• For (iii): You must give both reasons (why it can't gain AND why it can't lose electrons) to get full credit — one reason alone is incomplete.

Butane (C₄H₁₀) electron dot structure:

$$\text{H : C : C : C : C : H}$$

H H H H H H H H

(Each carbon shares single bonds with adjacent carbons and hydrogen atoms, with all valencies satisfied.)

Source: Chapter 4, Section 4.2.2

---

Butane has 4 carbon atoms in a straight chain, each forming 4 covalent bonds. The two end carbons bond to 3 H atoms each, and the two middle carbons bond to 2 H atoms each, giving C₄H₁₀. In the electron dot structure, show each shared pair as two dots (or a line) between atoms. Examiners want all bonds shown correctly with valencies satisfied — this is the key scoring point for 1 mark.

Cyclohexane (C₆H₁₂) has 6 carbon atoms arranged in a ring, each bonded to 2 hydrogen atoms with single covalent bonds:

$$\begin{array}{c}
\text{H}_2\text{C} - \text{CH}_2 \\
| \quad\quad\quad | \\
\text{H}_2\text{C} \quad \text{CH}_2 \\
| \quad\quad\quad | \\
\text{H}_2\text{C} - \text{CH}_2
\end{array}$$

Each C shares electrons with 2 C atoms and 2 H atoms (4 bond pairs, valency satisfied).

Source: Carbon and its Compounds, Section 4.2.2

---

• Cyclohexane formula is C₆H₁₂ — examiners may deduct marks if you use the wrong formula.
• The key feature is the ring structure with all single bonds (saturated).
• In the electron dot structure, show 2 dots (shared pairs) for each C–C and C–H bond around all 6 carbons. Alternatively, the structural/line formula of the hexagonal ring with CH₂ at each carbon is accepted.
• Confirm all carbon valencies = 4 are satisfied (2 C–C bonds + 2 C–H bonds per carbon).

(i) Molecular formula of benzene: C₆H₆

Structure of benzene: It is a cyclic compound with alternating single and double bonds between six carbon atoms, each bonded to one hydrogen atom.

```
H H
| |
H—C C—H
‖ |
H—C C—H
| |
C = C
H H
```
(Hexagonal ring with alternating single and double bonds)

(ii) In a benzene molecule (C₆H₆):

• Single bonds (C–C and C–H): 9 (3 C–C single bonds + 6 C–H bonds)
• Double bonds (C=C): 3

(iii) Unsaturated hydrocarbons containing one or more triple bonds between carbon atoms are called alkynes. Example: Ethyne (C₂H₂).

Source: Chapter 4, Section 4.2.2 – Chains, Branches and Rings

---

• For (i), always write the molecular formula first, then draw the hexagonal ring clearly showing alternating double bonds.
• For (ii), count carefully: 3 C=C (double) + 3 C–C (single) + 6 C–H (single) = 9 single + 3 double. Examiners specifically look for both types stated separately.
• For (iii), the definition must include "triple bond" — this is the key word that distinguishes alkynes from alkenes (double bond). The textbook definition is direct and should be quoted closely.

Two structures of saturated hydrocarbon with 4 carbon atoms:

Structure 1 (n-Butane – straight chain):

$$\text{H}_3\text{C} - \text{CH}_2 - \text{CH}_2 - \text{CH}_3$$

Structure 2 (Iso-butane – branched chain):

$$\begin{array}{c} \text{CH}_3 \\ | \\ \text{CH}_3 - \text{CH} - \text{CH}_3 \end{array}$$

These two structures with the same molecular formula but different arrangements are called structural isomers.

Molecular formula: $\text{C}_4\text{H}_{10}$

Common name: Butane

Molecular formula of its alkyne: $\text{C}_4\text{H}_6$ (But-1-yne / Butyne)

Source: Carbon and its Compounds, Section 4.2.2

---

• Examiners expect both structures drawn clearly (straight and branched chain) — sketching the carbon skeleton is enough.
• The term structural isomers is the key term; don't write "isomers" alone.
• Alkyne formula: for alkane $\text{C}_n\text{H}_{2n+2}$, alkyne is $\text{C}_n\text{H}_{2n-2}$, so for $n=4$: $\text{C}_4\text{H}_6$.
• Award-split: 1 mark for structures, 1 mark for "structural isomers" + molecular formula + common name, 1 mark for alkyne formula.

(a)
(i) Methane — CH₄
(ii) Propane — C₃H₈

(b) Carbon compounds are covalently bonded. The intermolecular forces between covalent molecules are weak, so very little energy is needed to overcome them. Hence, carbon compounds have low melting and boiling points.

Source: Chapter 4, Section 4.2.2 (Table 4.2)

---

• (a) These formulae come directly from Table 4.2. Methane has 1 C and 4 H; Propane has 3 C and 8 H (general alkane formula CₙH₂ₙ₊₂).
• (b) The key idea examiners want: carbon compounds are covalent, and covalent compounds have weak intermolecular forces, hence low melting/boiling points. Always mention both the type of bonding and the weak forces — that's where the mark lies.

(i) Ethanol → Ethene
$$\text{CH}_3\text{CH}_2\text{OH} \xrightarrow{\text{conc. H}_2\text{SO}_4,\ 443\text{ K}} \text{CH}_2=\text{CH}_2 + \text{H}_2\text{O}$$
Condition: Excess conc. H₂SO₄ at 443 K (dehydration).

(ii) Ethene → Ethane
$$\text{CH}_2=\text{CH}_2 + \text{H}_2 \xrightarrow{\text{Ni, heat}} \text{CH}_3\text{CH}_3$$
Condition: Hydrogen gas in presence of Ni catalyst and heat (hydrogenation / addition reaction).

(iii) Ethane → Chloroethane
$$\text{CH}_3\text{CH}_3 + \text{Cl}_2 \xrightarrow{\text{sunlight}} \text{CH}_3\text{CH}_2\text{Cl} + \text{HCl}$$
Condition: Presence of sunlight (substitution reaction).

(iv) Ethanol → Ethanoic acid
$$\text{CH}_3\text{CH}_2\text{OH} \xrightarrow{\text{alkaline KMnO}_4 \text{ or acidified K}_2\text{Cr}_2\text{O}_7} \text{CH}_3\text{COOH}$$
Condition: Oxidation using alkaline KMnO₄ or acidified K₂Cr₂O₇.

(v) Ethanoic acid → Ethyl ethanoate
$$\text{CH}_3\text{COOH} + \text{C}_2\text{H}_5\text{OH} \xrightarrow{\text{conc. H}_2\text{SO}_4,\ \Delta} \text{CH}_3\text{COOC}_2\text{H}_5 + \text{H}_2\text{O}$$
Condition: Conc. H₂SO₄ as catalyst and heat (esterification reaction).

Source: Chapter 4, Sections 4.3.2, 4.3.3, 4.3.4

---

• Each conversion must have: balanced equation + condition — both are needed for full marks.
• (i) is dehydration (elimination); (ii) is addition/hydrogenation; (iii) is substitution (sunlight is the key condition); (iv) is oxidation — either oxidising agent is acceptable; (v) is esterification — the product is an ester and the reaction is reversible (shown by ⇌ in some texts).
• Examiners award 1 mark per conversion, split between the equation and the stated condition. Missing the condition loses the mark even if the equation is correct.

(i) Identification:

• X = Ethanol (C₂H₅OH) — neutral alcohol
• Y = Ethanoic acid (CH₃COOH) — acidic compound
• Z = Ethyl ethanoate (CH₃COOC₂H₅) — sweet-smelling ester

(ii) Chemical equations:

(1) X to Y (Oxidation):
$$\text{CH}_3\text{CH}_2\text{OH} \xrightarrow{\text{acidified K}_2\text{Cr}_2\text{O}_7} \text{CH}_3\text{COOH}$$

(2) X to Z (Esterification):
$$\text{CH}_3\text{COOH} + \text{C}_2\text{H}_5\text{OH} \xrightarrow{\text{conc. H}_2\text{SO}_4, \Delta} \text{CH}_3\text{COOC}_2\text{H}_5 + \text{H}_2\text{O}$$

(iii) Roles:

(1) Acidified K₂Cr₂O₇ acts as an oxidising agent — it oxidises ethanol to ethanoic acid.

(2) Conc. H₂SO₄ acts as a dehydrating agent / catalyst — it removes water and drives the esterification forward.

(iv) The reaction of ester Z with an alkali is called Saponification (alkaline hydrolysis of ester).

Source: Carbon and its Compounds, sections 4.2.3, 4.2.5

---

• The key chain: ethanol → (oxidation) → ethanoic acid → (esterification with ethanol) → ethyl ethanoate.
• CBSE expects the role words: "oxidising agent" and "dehydrating agent/catalyst" — one word each earns the mark.
• "Saponification" is the specific term examiners look for in (iv); writing only "hydrolysis" may lose the mark.
• Balance equations properly; the esterification arrow must show conc. H₂SO₄ and heat (Δ) as conditions, not reactants.

The correct representation is (b) — the one showing a double bond (O=O) between two oxygen atoms, as each oxygen (valency 2) shares two pairs of electrons to complete its octet.

Oxygen has 6 valence electrons and needs 2 more to complete its octet. Two oxygen atoms each contribute 2 electrons to form 2 shared pairs → a double covalent bond (O=O). Options showing a single bond or triple bond are incorrect. The correct electron-dot structure must show two shared pairs between the oxygen atoms and two lone pairs on each oxygen.

(i) When sodium is dropped in ethanol, it reacts vigorously to produce sodium ethoxide and hydrogen gas is evolved.

$$2\text{Na} + 2\text{CH}_3\text{CH}_2\text{OH} \rightarrow 2\text{CH}_3\text{CH}_2\text{O}^-\text{Na}^+ + \text{H}_2\uparrow$$

(ii) The melting point of pure ethanoic acid is 290 K. Therefore it often freezes (solidifies) during winter in cold climates, giving it an ice-like appearance. This is why it is called glacial acetic acid.

(iii) When ethanol is heated at 443 K with excess conc. H₂SO₄, it undergoes dehydration to form ethene (an unsaturated hydrocarbon) and water.

$$\text{CH}_3\text{CH}_2\text{OH} \xrightarrow{\text{Hot conc. H}_2\text{SO}_4,\ 443\text{K}} \text{CH}_2{=}\text{CH}_2 + \text{H}_2\text{O}$$

Conc. H₂SO₄ acts as a dehydrating agent — it removes water from ethanol.

(iv) Saponification is the hydrolysis of an ester by an alkali (NaOH) to give back alcohol and the sodium salt of the carboxylic acid.

$$\text{CH}_3\text{COOC}_2\text{H}_5 + \text{NaOH} \rightarrow \text{C}_2\text{H}_5\text{OH} + \text{CH}_3\text{COONa}$$

Source: Chapter 4, Sections 4.4.1 and 4.4.2

---

• (i) Always write the balanced equation; examiners award 1 mark for the equation and 1 for stating hydrogen is evolved / sodium ethoxide is formed.
• (ii) The key fact is the melting point (290 K) causing freezing in winter — state this explicitly.
• (iii) Name the reaction (dehydration), write the equation with conditions (443 K, Hot conc. H₂SO₄), and clearly state the role of H₂SO₄ as dehydrating agent — all three parts are usually separately marked.
• (iv) Write the equation correctly and name the products (alcohol + sodium salt of acid). The term "saponification" must appear as it is the definition trigger word.

(i) Structures:

(a) Butanoic acid (C₃H₇COOH):

$$\text{CH}_3 - \text{CH}_2 - \text{CH}_2 - \text{COOH}$$

(b) Chloropentane (C₅H₁₁Cl):

$$\text{CH}_3 - \text{CH}_2 - \text{CH}_2 - \text{CH}_2 - \text{CH}_2 - \text{Cl}$$

---

(ii) Structures (i) and (ii) are related as structural isomers.

Reason: Both have the same molecular formula but different structural arrangements. Such compounds with identical molecular formula but different structures are called structural isomers.

One more possible structure (e.g., for C₄H₁₀):

$$\text{CH}_3 - \underset{|}{\overset{|}{\text{CH}}} (\text{CH}_3) - \text{CH}_3 \quad \text{(2-methylpropane)}$$

---

(iii) Difference based on general formula:

| | Saturated | Unsaturated |
|---|---|---|
| General formula | CₙH₂ₙ₊₂ (alkanes) | CₙH₂ₙ (alkenes) / CₙH₂ₙ₋₂ (alkynes) |
| Bonds | Only single bonds | Double or triple bonds between C atoms |

Source: Chapter 4 — Carbon and its Compounds, Sections 4.2.1 and 4.2.2

---

• Structural isomers is the key term examiners look for in part (ii); always define it and give a reason.
• For part (iii), simply stating the general formula difference and linking it to single vs. double/triple bonds covers full marks. A small table makes it neat and saves words.
• In part (i), write structures as displayed/condensed formulae; electron dot structures are not needed unless specifically asked.
• Budget your time: parts (i) and (ii) together carry most marks; part (iii) needs only 2–3 lines.

(c)

The correct electron dot structure shows each Cl atom with 3 lone pairs and 1 shared pair between them, satisfying the octet of both chlorine atoms (valency = 1).

Chlorine has 7 valence electrons. Each Cl needs 1 more electron to complete its octet, so it shares one electron with the other Cl, forming a single covalent bond with 3 lone pairs on each atom. The option showing exactly this arrangement is correct. In MCQs, quickly count the dots: each Cl should have 6 dots (3 pairs) + 1 shared pair between them.

(i) The simplest saturated hydrocarbon is methane (CH₄).

Electron dot structure:

H
|
H : C : H
|
H

(Each C–H bond has one shared pair of electrons.)

Type of bonds: Only single covalent bonds (C–H) exist in methane. All four valence electrons of carbon are shared with four hydrogen atoms.

(ii) Two fuel mixtures in daily life that contain methane as an important component:

1. CNG (Compressed Natural Gas)
2. Biogas

(iii) Methane belongs to the Alkanes homologous series (saturated hydrocarbons).

General formula: $C_nH_{2n+2}$

(iv) Methane (a saturated hydrocarbon) produces a clean blue flame on burning, as saturated hydrocarbons undergo complete combustion when sufficient oxygen is available.

Source: Carbon and its Compounds, sections 4.3.1 Combustion and 4.2 (Homologous Series)

---

• The question has four sub-parts worth a total of 5 marks (~1 mark each, with (i) slightly heavier).
• The electron dot structure must show shared pairs (dots/crosses) around C; a simple line structure alone may not get full credit.
• Examiners expect CNG and Biogas (or LPG — LPG is mainly butane/propane, so CNG or Biogas are safer choices for methane).
• The general formula $C_nH_{2n+2}$ for alkanes is essential — don't confuse with alkenes ($C_nH_{2n}$).
• For (iv), the key phrase is clean blue flame — link it to complete combustion of saturated hydrocarbons as stated in the textbook passage.

Identification:

• X = Ethanoic acid (CH₃COOH)
• Y = Ethanol (C₂H₅OH)
• Z = Ethyl ethanoate (CH₃COOC₂H₅) — an ester with a pleasant smell

Reaction 1 — Esterification:

$$\text{CH}_3\text{COOH} + \text{C}_2\text{H}_5\text{OH} \underset{\Delta}{\overset{\text{Conc. H}_2\text{SO}_4}{\rightleftharpoons}} \text{CH}_3\text{COOC}_2\text{H}_5 + \text{H}_2\text{O}$$

Name: Esterification reaction.

Justification: When a carboxylic acid reacts with an alcohol in the presence of an acid catalyst, an ester (Z) is formed. Since an ester is the product, the reaction is called esterification.

Reaction 2 — Saponification (hydrolysis of ester with NaOH):

$$\text{CH}_3\text{COOC}_2\text{H}_5 + \text{NaOH} \rightarrow \text{CH}_3\text{COONa} + \text{C}_2\text{H}_5\text{OH}$$

Ethyl ethanoate reacts with sodium hydroxide to give sodium ethanoate and ethanol (alcohol Y is recovered).

Source: Chapter 4 — Carbon and its Compounds, Section 4.4.2 (Ethanoic Acid — Reactions)

---

• X, Y, Z identification carries 1 mark — name all three clearly.
• Esterification equation carries ~1–1.5 marks — include catalyst (conc. H₂SO₄), heat, and the reversible arrow (⇌).
• Naming + justification carries ~1 mark — state that ester is formed, hence "esterification."
• Saponification equation carries ~1 mark — this reverse reaction is called saponification; the equation must show recovery of ethanol (Y) and formation of sodium ethanoate.
• Do not confuse saponification (ester + NaOH) with hydrolysis by water; in board exams always write the NaOH equation here.

(a) 10

Ethyne (C₂H₂) has a triple bond between two carbon atoms (6 shared electrons) and one C–H bond on each carbon (2 × 2 = 4 shared electrons). Total = 6 + 4 = 10 electrons shared.

Count all shared electrons in every bond: the C≡C triple bond shares 6 electrons, and each of the two C–H single bonds shares 2 electrons, giving 6 + 4 = 10. Students often confuse number of bonds (5 bonds total) with number of electrons shared — the question asks for electrons, so multiply bonds × 2.

I notice that sub-question (c) is incomplete — options (i) and (ii) are missing from the question. I'll answer (a) and (b) fully, and note the gap for (c).

---

(a) Compounds of carbon formed exclusively with hydrogen are called hydrocarbons.
(Example: methane, CH₄)

(b) The two properties are:

1. Tetravalency — Carbon has a valency of 4, allowing it to bond with four other atoms.
2. Catenation — Carbon has the unique ability to bond with other carbon atoms to form long chains, branches, and rings.

(c) (Sub-question (c) cannot be answered as options (i) and (ii) were not provided in the question. Please supply the missing options for a complete answer.)

---

• For (a): Examiners expect the definition + at least one example for full credit.
• For (b): These are the two standard properties asked repeatedly in CBSE boards — catenation and tetravalency. Both terms must be written by name; just describing them without the term may cost marks.
• For (c): Since the options were cut off in the question, no answer can be framed. In the actual exam paper, choose either (i) or (ii) — not both — and write approximately 40–50 words for the 2-mark part.

Conclusion 1: The forces of attraction between the molecules of carbon compounds are not very strong (which explains their low melting and boiling points).

Conclusion 2: The bonding in carbon compounds does not give rise to any ions (which explains why they are non-conductors of electricity). Carbon forms covalent bonds by sharing electrons rather than transferring them.

Source: Chapter 4, Section 4.1 – Bonding in Carbon

---

• The passage directly states both conclusions, so quote them precisely — examiners expect these exact ideas.
• Conclusion 1 must link low M.P./B.P. → weak intermolecular forces.
• Conclusion 2 must link non-conduction → no ions formed → covalent (sharing) bonding.
• Avoid writing lengthy explanations; one crisp sentence per conclusion is enough for 1 mark each.

(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).

The textbook directly states that vegetable oils have long unsaturated carbon chains, and oils containing unsaturated fatty acids are healthier than saturated animal fats. Thus, R correctly explains A. Source: Chapter 4, Section 4.3.3.

Oxidation: In the conversion of ethanol (C₂H₅OH) to ethanoic acid (CH₃COOH), oxygen is added to ethanol. Since the reaction involves addition of oxygen to the starting material, it is called an oxidation reaction.

Oxidising agent: Alkaline potassium permanganate (KMnO₄) or acidified potassium dichromate (K₂Cr₂O₇).

Chemical equation:

$$\text{CH}_3\text{CH}_2\text{OH} \xrightarrow{[\text{O}] \text{ Alkaline KMnO}_4} \text{CH}_3\text{COOH}$$

Difference from burning: When ethanol burns in oxygen, it undergoes complete oxidation producing CO₂ and water with release of energy. The oxidation of ethanol to ethanoic acid is controlled/partial oxidation where ethanol is converted to a carboxylic acid, not CO₂ and water.

Source: Chapter 4, Section 4.3.2 Oxidation

---

• Examiners expect three distinct parts: reason for oxidation, name of oxidising agent, balanced/arrow equation, and the difference from combustion.
• "Addition of oxygen = oxidation" is the key concept from Section 4.3.2.
• Both oxidising agents (KMnO₄ and K₂Cr₂O₇) are acceptable; mention either or both.
• The key difference: combustion is complete oxidation → CO₂ + H₂O; this reaction is partial/controlled oxidation → CH₃COOH.

Flame test:
Saturated hydrocarbons burn with a blue, non-sooty flame, while unsaturated hydrocarbons burn with a yellow, sooty (smoky) flame.

Products of combustion of a saturated hydrocarbon:
Carbon dioxide (CO₂) and water (H₂O) are produced.

Example: CH₄ + 2O₂ → CO₂ + 2H₂O

---

• The flame test is the key distinguishing test: soot/smokiness indicates unsaturation (higher carbon-to-hydrogen ratio causes incomplete combustion).
• Examiners expect both parts answered: (1) the distinction and (2) the products.
• Writing the equation for methane combustion earns full credit and shows understanding. Keep it concise — no extra explanation needed for 2 marks.

Option D: 3 and 3 — Wait, that is incorrect. Benzene (C₆H₆) has an alternating structure with 3 single C–C bonds, 3 double C=C bonds, and 6 C–H single bonds, giving 9 single bonds and 3 double bonds.

Correct answer: B — 9 and 3

Source: Chapter 4, Section 4.2.2 (Chains, Branches and Rings)

---

• In benzene's ring: there are 3 C–C single bonds and 3 C=C double bonds.
• Each of the 6 carbon atoms also bonds to one hydrogen via a single bond → 6 C–H single bonds.
• Total single bonds = 3 (C–C) + 6 (C–H) = 9; double bonds = 3.
• Examiners expect you to count all single bonds (both C–C and C–H), not just the ring single bonds. Option D (3 and 3) is a common wrong answer if students ignore C–H bonds.

(a) Test tube IV (Tubewell water + salt B) forms the minimum foam.
Tubewell water is hard water, and salt B (washing soda / detergent) forms least lather with hard water compared to soap.

(b)
| Salt A | Salt B |
|--------|--------|
| Soap – sodium salt of a long-chain fatty acid; used for bathing and washing | Detergent (synthetic) – sodium salt of a long-chain sulphonic acid; used only for washing |
Soap does not work in hard water; detergent works in hard and soft water both.

(c)(i)
Esters are sweet-smelling compounds formed by the reaction of an acid with an alcohol.

When an ester reacts with an alkali (NaOH), it undergoes saponification to give sodium salt of the acid (soap) and alcohol:

$$\text{CH}_3\text{COOC}_2\text{H}_5 + \text{NaOH} \xrightarrow{\Delta} \text{CH}_3\text{COONa} + \text{C}_2\text{H}_5\text{OH}$$

(Ethyl ethanoate + Sodium hydroxide → Sodium ethanoate + Ethanol)

Source: Carbon and its Compounds, Soaps and Detergents / Esters section

---

• Salt A = soap (used for bathing + washing); Salt B = detergent (washing only). This is the key identification.
• Hard water (tubewell) + soap forms scum, not foam → test tube IV gives minimum foam because detergent still lathers but the combination of hard water reducing lather and detergent being less affected actually makes IV the least foamy compared to I (rain + soap gives maximum foam).
• For (c), examiners expect the definition, the term saponification, and a balanced chemical equation — all three earn the 2 marks.
• Keep the equation correctly balanced; the reactants and products must match.

(D) is incorrect. Hydrogenation of vegetable oils is an addition reaction that uses a nickel or palladium catalyst, not an acid catalyst.

The passage clearly states that unsaturated hydrocarbons add hydrogen in the presence of palladium or nickel catalysts — no acid is involved. Options A, B, and C are all correct statements about hydrogenation. Examiners expect you to identify the specific wrong detail (acid catalyst) in option D.

(C) (ii) and (iii)

Structures (ii) and (iii) both have the molecular formula C₄H₁₀ (butane) but different carbon skeletons — they are structural isomers.

Structural isomers have the same molecular formula but different structures. (ii) is n-butane (straight chain) and (iii) is isobutane (branched chain) — both C₄H₁₀. Structure (i) is but-1-ene (C₄H₈, unsaturated), and (iv) has a Cl substituent, making it a haloalkane with a different molecular formula. Examiners expect you to identify identical molecular formula + different structure = structural isomers.

(i) Saturated vs Unsaturated Hydrocarbons:

Saturated hydrocarbons contain only single bonds between carbon atoms. They are called alkanes.
Example — Ethane:
$$\text{H}-\text{C}-\text{C}-\text{H} \quad (CH_3-CH_3)$$
(with all single bonds and H atoms filled)

Unsaturated hydrocarbons contain one or more double or triple bonds between carbon atoms. They are called alkenes (double bond) or alkynes (triple bond).
Example — Ethene:
$$CH_2=CH_2$$

---

(ii) Conversion of Unsaturated to Saturated Hydrocarbon:

Unsaturated hydrocarbons are converted to saturated hydrocarbons by hydrogenation — adding hydrogen in the presence of a nickel (Ni) catalyst.

$$CH_2=CH_2 + H_2 \xrightarrow{Ni} CH_3-CH_3$$

This reaction is used in the vegetable oil (vanaspati) / food industry to convert liquid vegetable oils into solid fats (vanaspati ghee).

---

(iii) Hydrocarbon with a double bond and four carbon atoms:

Name: But-1-ene (Butene)

Structure:
$$CH_2=CH-CH_2-CH_3$$

Source: Chapter 4, Sections 4.2.2 and 4.3

---

• (i) Key terms: "single bonds only" for saturated; "double/triple bond" for unsaturated. Structural formula must clearly show the double bond (=) for unsaturated. Examiners award 1 mark per correct definition + example pair.
• (ii) "Hydrogenation" is the exact NCERT term (Exercise Q12). Always name both the catalyst (Ni) and the industry (vanaspati/edible oil industry). 1 mark for the reaction, 1 mark for naming the industry.
• (iii) Four carbons + double bond = butene. Write the structural formula clearly showing `CH₂=CH—`. 1 mark awarded for correct name and structure together.

(i) Compound 'X' is Ethanoic acid (CH₃COOH).

Structure:
$$\text{CH}_3 - \text{COOH}$$
(Two carbon atoms: a methyl group attached to a carboxyl group –COOH)

(ii) Ethanoic acid is a weak acid — it does not ionise completely in water, so its pH is higher (less acidic) than a mineral acid like HCl or H₂SO₄, which are strong acids and ionise completely, giving a much lower pH.

(iii) The reaction is called Esterification. When ethanoic acid reacts with ethanol (in the presence of conc. H₂SO₄), a sweet-smelling ester is formed:

$$\text{CH}_3\text{COOH} + \text{C}_2\text{H}_5\text{OH} \xrightarrow{\text{conc. H}_2\text{SO}_4} \text{CH}_3\text{COOC}_2\text{H}_5 + \text{H}_2\text{O}$$

(Ethyl ethanoate is the ester produced.)

(iv) The salt produced is Sodium ethanoate (CH₃COONa). The gas evolved is CO₂.

$$2\text{CH}_3\text{COOH} + \text{Na}_2\text{CO}_3 \rightarrow 2\text{CH}_3\text{COONa} + \text{H}_2\text{O} + \text{CO}_2\uparrow$$

Source: Carbon and its Compounds, Sections 4.3.2–4.3.3

---

• Part (i): Two clues identify X — two carbon atoms + turns blue litmus red (acidic) + used as preservative (vinegar/acetic acid). Students must write both name and structural formula.
• Part (ii): Key distinction: ethanoic acid = weak acid (partial ionisation, higher pH ~3); mineral acids = strong acids (complete ionisation, pH ~1). Do not just say "lower pH" — explain why.
• Part (iii): Name the reaction ("Esterification") and write a balanced equation with catalyst. Forgetting the catalyst (conc. H₂SO₄) or the product name loses marks.
• Part (iv): Examiner looks for the correct salt name and balanced equation. CO₂ turning lime water milky is the identification clue given in the question.

(a) Compounds (1) CH₃CH₂CHO (propanal) and (3) CH₃CH₂CH₂CHO (butanal) belong to the same homologous series — aldehydes — as both contain the –CHO functional group and differ by one –CH₂– unit.

(b)
(i) Melting point increases with increasing molecular mass in a homologous series.
(ii) Solubility (in water) decreases with increasing molecular mass in a homologous series.

---

• (a) Homologous series members must have the same functional group and differ by –CH₂–. Compound (2) is a ketone (–CO–) and compound (4) is an alcohol (–OH), so they don't fit. Only (1) and (3) are both aldehydes.
• (b) This is a standard textbook property of homologous series. Examiners expect both parts answered clearly. Larger molecules have stronger intermolecular forces (higher m.p.) but are less polar, reducing water solubility.

(i)

• X = Ethanol (C₂H₅OH) — a good solvent
• Y = Sodium ethoxide (C₂H₅ONa)
• Z = Hydrogen gas (H₂) — used in hydrogenation of vegetable oil to make vegetable ghee

Reaction of X with sodium:

$$2C_2H_5OH + 2Na \rightarrow 2C_2H_5ONa + H_2\uparrow$$

---

(ii)

(I) Combustion of ethanol:

$$C_2H_5OH + 3O_2 \rightarrow 2CO_2 + 3H_2O + \text{heat and light}$$

(II) Dehydration at 443 K with excess conc. H₂SO₄:

$$C_2H_5OH \xrightarrow{443\text{K, conc. H}_2SO_4} CH_2=CH_2\uparrow + H_2O$$

(Ethanol → Ethene + Water)

(III) Reaction with acidified potassium dichromate (oxidation):

$$C_2H_5OH \xrightarrow{K_2Cr_2O_7/H^+} CH_3COOH$$

(Ethanol is oxidised to ethanoic acid; orange dichromate turns green.)

---

• Part (i): The key clue is "Z converts vegetable oil to ghee" → Z = H₂ (hydrogenation). Since H₂ is produced with sodium, X must be an alcohol. As a "good solvent," X is ethanol. Y is sodium ethoxide (the sodium salt of ethanol).
• Part (ii): Examiners expect balanced equations. For combustion: C₂H₅OH + 3O₂ → 2CO₂ + 3H₂O. For dehydration: excess H₂SO₄ at 443 K gives ethene (elimination). For oxidation with acidified K₂Cr₂O₇: ethanol → ethanoic acid (colour change from orange to green is a bonus point). Write product names alongside formulae for clarity.

(i) Two isomeric structures of Butene (C₄H₈):

But-1-ene:
$$\text{CH}_2 = \text{CH} - \text{CH}_2 - \text{CH}_3$$

But-2-ene:
$$\text{CH}_3 - \text{CH} = \text{CH} - \text{CH}_3$$

---

(ii) Names of compounds:

(I) CH₃–CH₂–CH₂Cl → 1-Chloropropane

(II) CH₃–CO–CH₂–CH₃ → Butanone

---

(iii) Chemical equations with conditions:

(I) Complete oxidation of ethanol:
$$\text{CH}_3\text{CH}_2\text{OH} + 3\text{O}_2 \rightarrow 2\text{CO}_2 + 3\text{H}_2\text{O} + \text{heat and light}$$
Condition: Sufficient supply of oxygen (excess O₂).

(II) Hydrogenation of propene:
$$\text{CH}_3 - \text{CH} = \text{CH}_2 + \text{H}_2 \xrightarrow{\text{Ni catalyst}} \text{CH}_3 - \text{CH}_2 - \text{CH}_3$$
Condition: Nickel (Ni) catalyst.

(III) Ethanoic acid reacts with ethanol (Esterification):
$$\text{CH}_3\text{COOH} + \text{C}_2\text{H}_5\text{OH} \underset{\Delta}{\overset{\text{conc. H}_2\text{SO}_4}{\rightleftharpoons}} \text{CH}_3\text{COOC}_2\text{H}_5 + \text{H}_2\text{O}$$
Condition: Concentrated H₂SO₄ as catalyst and heat (Δ).

Source: Chapter 4 — Carbon and its Compounds, Sections 4.2.5, 4.3.1, 4.3.3

---

• (i) Isomers of butene differ by position of the double bond. But-1-ene and but-2-ene are the most expected answers. A third isomer (2-methylpropene) is also valid but two are enough for 2 marks.
• (ii) Use Table 4.4 rules: Cl at end = chloro- prefix; 4-carbon ketone = butanone (drop 'e' from butane + one).
• (iii) Always balance combustion equations. Hydrogenation requires Ni catalyst — this is a standard condition examiners expect. Esterification needs both conc. H₂SO₄ and heat, and the reversible arrow (⇌) is important. Missing the condition loses a mark.

(i)
(I) CH₃–C(=O)–CH₃ → Propanone
(II) H–C≡C–H → Ethyne

(ii) The reaction commonly used for the conversion of vegetable oils to fats is Hydrogenation (addition of hydrogen in the presence of a nickel catalyst).

(iii) C₄H₁₀ (butane) has a longer carbon chain and greater molecular mass than C₃H₈ (propane). Greater the molecular mass, stronger the intermolecular forces, and hence higher the melting point.

(iv) When sodium is dropped into ethanol, it reacts vigorously to produce sodium ethoxide and hydrogen gas is liberated:
$$2\text{Na} + 2\text{C}_2\text{H}_5\text{OH} \rightarrow 2\text{C}_2\text{H}_5\text{ONa} + \text{H}_2\uparrow$$

(v)
$$\text{CH}_3\text{COOC}_2\text{H}_5 + \text{NaOH} \rightarrow \text{C}_2\text{H}_5\text{OH} + \text{CH}_3\text{COONa}$$
Products: Ethanol and Sodium ethanoate

Source: Chapter 4, Sections 4.2.5 and 4.4.2

---

• (i) Apply IUPAC nomenclature rules: 3-carbon chain + ketone suffix = propanone; 2-carbon chain + triple bond = ethyne.
• (ii) "Hydrogenation" is the exact IUPAC/textbook term; just writing "addition reaction" will lose marks.
• (iii) Connect molecular mass → intermolecular forces → melting point. This is the standard 1-mark reasoning expected.
• (iv) Write the balanced equation; examiner looks for sodium ethoxide + H₂ as products.
• (v) This is a saponification reaction. Both products must be named for full marks.

(i) A homologous series is a series of carbon compounds having the same functional group and similar chemical properties, in which each successive member differs by a –CH₂– unit (14 u).

From the given compounds:

• C₂H₄ → general formula CₙH₂ₙ (alkene)
• C₄H₈ → general formula CₙH₂ₙ (alkene)

C₂H₄ and C₄H₈ belong to the same homologous series (alkenes).

(ii)

(I) Esterification reaction:

$$\text{CH}_3\text{CH}_2\text{OH} + \text{CH}_3\text{COOH} \xrightarrow{\text{Acid}} \text{CH}_3\text{COOC}_2\text{H}_5 + \text{H}_2\text{O}$$

(Product: Ethyl ethanoate/ester + Water)

(II) Combustion reaction:

$$\text{C}_2\text{H}_5\text{OH} + 3\text{O}_2 \xrightarrow{\text{Burning}} 2\text{CO}_2 + 3\text{H}_2\text{O} + \text{heat and light}$$

(iii) Soap molecules have a hydrophilic (water-loving) head and a hydrophobic (water-repelling) tail. When added to water, the hydrophobic tails cluster inward (away from water) while the hydrophilic heads face outward toward water, forming a spherical cluster called a micelle.

Source: Chapter 4 – Carbon and its Compounds, Sections 4.2, 4.3.1, 4.4.2

---

• (i) Examiners expect the definition AND correct identification. C₃H₄ is an alkyne (CₙH₂ₙ₋₂), C₄H₁₀ is an alkane — only C₂H₄ and C₄H₈ share the CₙH₂ₙ formula.
• (ii)(I) Write the ester name clearly; the reaction is esterification. (II) The balanced equation must show coefficients 1:3→2:3. Marks are awarded for correct products AND balancing.
• (iii) Key terms: hydrophilic, hydrophobic, and the word "micelle" must appear. One sentence explaining orientation is sufficient for 1 mark.

(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).

The carbon-carbon bond is very strong and hence stable, which directly gives rise to large numbers of compounds with many carbon atoms linked to each other (catenation).

Source: Chapter 4, Section 4.2 Versatile Nature of Carbon

The textbook explicitly states: "The carbon-carbon bond is very strong and hence stable. This gives us the large number of compounds with many carbon atoms linked to each other." This means R directly explains A, making option (A) correct. Students should note the key link: strong C–C bond → stable catenation → millions of carbon compounds.

(D) (iii) and (iv)

X (n-butane) and Y (isobutane) both have the same molecular formula C₄H₁₀ but different structural formulae, making them structural isomers — not homologues.

• Homologues differ by –CH₂– and have different molecular formulae, so (i) is wrong.
• Statement (ii) is the reverse of the truth — they have the same molecular formula, not the same structural formula — so (ii) is wrong.
• The textbook (section 4.2.2) explicitly states: compounds with identical molecular formula but different structures are called structural isomers — confirming both (iii) and (iv) are correct.
• Examiners expect you to recall the definition of structural isomers directly from the butane/isobutane example given in the chapter.

(i) Structural formula of benzene (C₆H₆):

Benzene has a ring of 6 carbon atoms with alternating double bonds:

$$\text{Hexagonal ring with alternating single and double bonds between carbon atoms, each C bonded to one H}$$

(A regular hexagon with alternating double bonds, each carbon bonded to one H atom.)

---

(ii) Carbon has 4 electrons in its outermost shell. To form C⁴⁺, it would need to lose 4 electrons, requiring a very large amount of energy. To form C⁴⁻, it would gain 4 electrons, which would be difficult to hold in the nucleus with only 6 protons. Both processes are energetically unfavourable, so carbon forms covalent bonds instead.

---

(iii) The bond formed between carbon and hydrogen is a covalent bond (non-polar covalent bond), formed by equal sharing of electrons.

---

(iv) Ethanol is converted to ethanoic acid by oxidation, using an oxidising agent such as alkaline KMnO₄ or acidified K₂Cr₂O₇.

$$\text{CH}_3\text{CH}_2\text{OH} \xrightarrow{[\text{O}]} \text{CH}_3\text{COOH}$$

---

(v)

$$\text{CH}_3\text{COOC}_2\text{H}_5 + \text{NaOH} \longrightarrow \text{CH}_3\text{COONa} + \text{C}_2\text{H}_5\text{OH}$$

(Sodium ethanoate + Ethanol)

Source: Carbon and its Compounds, Chapter 4, Sections 4.2.2, 4.4

---

• Benzene structure: Draw the hexagonal ring clearly with alternating double bonds; examiners want to see the alternating bond pattern.
• C⁴⁺/C⁴⁻: Focus on two reasons — large energy needed to lose 4 electrons, and difficulty of nucleus holding 4 extra electrons. Mention that covalent bonding is preferred.
• C–H bond: Simply state "covalent bond"; adding "non-polar" earns extra credit.
• Ethanol → Ethanoic acid: Must state it is an oxidation reaction and name the oxidising agent. The arrow notation [O] is standard.
• Saponification equation: This is ester hydrolysis (saponification). Products are sodium ethanoate (sodium salt of acid) and ethanol. Names of products often fetch ½ mark each.

(i) Compound 'P' is ethanol (C₂H₅OH). It is neutral to litmus and reacts with sodium to produce hydrogen gas (pop sound):
$$2C_2H_5OH + 2Na \rightarrow 2C_2H_5ONa + H_2\uparrow$$

(ii)
(I) When ethanol is heated with conc. H₂SO₄ at 443 K, it undergoes dehydration to form compound 'Q' — ethene (C₂H₄):

$$C_2H_5OH \xrightarrow{\text{conc. H}_2\text{SO}_4,\ 443\text{K}} CH_2=CH_2 + H_2O$$

(II) Conc. H₂SO₄ acts as a dehydrating agent — it removes water from ethanol to produce ethene.

(iii) A fruity-smelling ester (ethyl ethanoate) will be formed. This reaction is called esterification.

$$C_2H_5OH + CH_3COOH \underset{\Delta}{\overset{\text{conc. H}_2\text{SO}_4}{\rightleftharpoons}} CH_3COOC_2H_5 + H_2O$$

Source: Chapter 4, Section 4.4 — Some Important Carbon Compounds: Ethanol and Ethanoic Acid

---

• Identification of P is straightforward: C₂H₆O neutral to litmus = ethanol (not an acid/base); reacts with Na to give H₂ (pop sound). Methanol also fits formula but ethanol is the standard NCERT example.
• Dehydration at 443 K is a key fact: conc. H₂SO₄ as dehydrating agent → ethene. Write the balanced equation with conditions above the arrow.
• Esterification: heating alcohol + carboxylic acid with acid catalyst gives ester + water. The reaction is reversible (⇌). Examiners expect the fruity smell observation and the correct equation. Remember ethyl ethanoate = CH₃COOC₂H₅.
• Mark split is roughly: (i) 1 mark, (ii) 2 marks, (iii) 2 marks — match detail to marks.

(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).

Catenation is the ability of carbon to bond with other carbon atoms, forming long chains, branched chains, and ring structures.

Source: Chapter 4, Section 4.2 Versatile Nature of Carbon

---

The assertion directly describes the structural outcomes (chains, branches, rings) of catenation, and the reason correctly names and defines that property. Since R directly explains why A is true, option (A) is correct. Remember: catenation → chains/branches/rings is a direct cause-effect link, so R is the correct explanation, not just a related true fact.

(a) The 4th member of the alcohol homologous series (n = 4) is Butan-1-ol.

Structural formula: CH₃–CH₂–CH₂–CH₂–OH

---

(b) When ethanol is heated with alkaline KMnO₄, it gets oxidised to ethanoic acid.

$$\text{CH}_3\text{CH}_2\text{OH} \xrightarrow{\text{alk. KMnO}_4/\Delta} \text{CH}_3\text{COOH}$$

Alkaline KMnO₄ acts as an oxidising agent.

---

(c) When ethanol reacts with ethanoic acid in the presence of conc. H₂SO₄, esterification reaction takes place:

$$\text{CH}_3\text{COOH} + \text{C}_2\text{H}_5\text{OH} \underset{\Delta}{\overset{\text{conc. H}_2\text{SO}_4}{\rightleftharpoons}} \text{CH}_3\text{COOC}_2\text{H}_5 + \text{H}_2\text{O}$$

The product ethyl ethanoate is an ester with a sweet fruity smell.

Source: Carbon and its Compounds, Ethanol and its reactions

---

• (a) Apply general formula CₙH₂ₙ₊₁OH with n=4 → C₄H₉OH = Butan-1-ol. Write the full structural formula, not just the molecular formula.
• (b) Key point: alkaline KMnO₄ is an oxidising agent; it oxidises alcohols to carboxylic acids. Examiners expect the equation.
• (c) Award 1 mark for the correct balanced equation and 1 mark for naming it esterification. Note the reversible arrow (⇌) and conc. H₂SO₄ as catalyst above the arrow — both are important.

(C) C₃H₈

C₃H₈ is a saturated hydrocarbon (alkane). Saturated hydrocarbons undergo substitution reactions, not addition reactions. Only unsaturated compounds (alkenes, alkynes) undergo addition reactions.

The key concept: addition reactions occur only in unsaturated hydrocarbons (those with double or triple bonds). C₄H₈ (alkene), C₂H₂ (alkyne), and C₂H₄ (alkene) are all unsaturated — they have C=C or C≡C bonds and readily undergo addition. C₃H₈ is propane, a saturated alkane; it undergoes substitution, not addition. Examiners expect you to identify the saturated compound by its general formula CₙH₂ₙ₊₂.

(D) C₄H₈ and C₅H₁₂

C₄H₈ follows the general formula CₙH₂ₙ (alkene series), while C₅H₁₂ follows CₙH₂ₙ₊₂ (alkane series). They belong to different homologous series.

• Check the general formula for each option: alkanes = CₙH₂ₙ₊₂; alkenes = CₙH₂ₙ.
• Options A, B, C all contain pairs of alkanes (CH₄, C₂H₆, C₃H₈, C₄H₁₀, C₅H₁₂ all fit CₙH₂ₙ₊₂).
• In option D, C₄H₈ is an alkene and C₅H₁₂ is an alkane — different series, so they do not belong to the same homologous series.

(i) Identification:

• X = Ethanoic acid (CH₃COOH)
• Y = Ethyl ethanoate / Ethyl acetate (CH₃COOC₂H₅) — sweet smelling ester
• Z = Sodium ethanoate (CH₃COONa)

(ii) Role of conc. H₂SO₄:
It acts as a catalyst and also as a dehydrating agent, removing water to drive the esterification reaction forward.

(iii) Chemical Equations:

Reaction 1 — Esterification:
$$\text{CH}_3\text{COOH} + \text{C}_2\text{H}_5\text{OH} \xrightarrow{\text{Conc. H}_2\text{SO}_4} \text{CH}_3\text{COOC}_2\text{H}_5 + \text{H}_2\text{O}$$

Reaction 2 — Saponification (hydrolysis of ester):
$$\text{CH}_3\text{COOC}_2\text{H}_5 + \text{NaOH} \rightarrow \text{CH}_3\text{COONa} + \text{C}_2\text{H}_5\text{OH}$$

The first reaction is called Esterification; the second is called Saponification.

Source: Chapter 4, Section 4.4.2 (Ethanoic Acid — Reactions)

---

• Examiners expect all three identifications (X, Y, Z) stated clearly with formulae — 1 mark.
• The role of H₂SO₄ as catalyst/dehydrating agent is a standard 1-mark point.
• Both balanced equations must be written with correct formulae; half-marks are often given per equation.
• Naming both reactions (esterification and saponification) is essential — commonly missed.
• "Sweet smelling" is the clue for an ester; NaOH breaking an ester is always saponification.

(i)

(I) In covalent compounds, electrons are shared between atoms and no charged particles (ions) are formed. Since conduction of electricity requires free ions or electrons, covalent compounds are poor conductors of electricity.

(II) Hard water contains Ca²⁺ and Mg²⁺ ions. Soap (sodium salt of fatty acid) reacts with these ions to form an insoluble precipitate called scum. Since the soap is used up in forming scum, no lather is produced.

(III) C–C bonds are strong and stable, allowing carbon to form long chains (catenation). Si–Si bonds are weaker and Si has a larger atomic size, making long Si–Si chains unstable. Hence silicon does not show catenation.

(ii)

(I) Oxidation of ethanol by acidified K₂Cr₂O₇:

$$\text{CH}_3\text{CH}_2\text{OH} \xrightarrow{\text{Acidified K}_2\text{Cr}_2\text{O}_7} \text{CH}_3\text{COOH}$$

(Ethanol → Ethanoic acid)

(II) Hydrogenation of ethene:

$$\text{CH}_2{=}\text{CH}_2 + \text{H}_2 \xrightarrow{\text{Ni, }\Delta} \text{CH}_3\text{CH}_3$$

Source: Chapter 4 – Carbon and its Compounds, Sections 4.1, 4.3, 4.4

---

• Part (i)(I): Always link "no ions formed" directly to "no conduction." This is the textbook reason.
• Part (i)(II): Name the precipitate as "scum" and mention Ca²⁺/Mg²⁺ ions — examiners look for these terms.
• Part (i)(III): Compare bond strength of C–C vs Si–Si and mention atomic size; both points earn marks.
• Part (ii): Write balanced/arrow equations with correct reagents. For oxidation of ethanol, "acidified K₂Cr₂O₇" must appear as the reagent. For hydrogenation, Ni catalyst and heat (Δ) must be shown — these are common mark-fetching details.

Option A — Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A). Carbon forms covalent bonds by sharing valence electrons; these shared electrons count towards both atoms, helping each achieve a noble gas configuration.

• The Assertion is correct: carbon shares its 4 valence electrons with other carbon atoms or other elements (covalent bonding).
• The Reason is also correct and directly explains the Assertion: in covalent bonding, the shared pair belongs to the outermost shell of both atoms, allowing each to attain a completely filled outermost shell (noble gas configuration).
• Since R logically explains why A occurs, the answer is A, not B.

The correct option is B – CHO.

The functional group for aldehyde is –CHO.

Source: Carbon and its Compounds, Table 4.3

From Table 4.3, the functional group for aldehyde contains a carbon atom double-bonded to oxygen and single-bonded to hydrogen, written as –CHO. Examiners expect the exact formula; –COOH is carboxylic acid and –OH is alcohol — do not confuse them.

(C) Assertion (A) is true, but Reason (R) is false.

Assertion is correct. Reason is false because 290 K is a high melting point (not low), which causes pure ethanoic acid to freeze in winter, giving it the name glacial acetic acid.

The textbook states the melting point of pure ethanoic acid is 290 K and that it often freezes during winter, which is why it is called glacial acetic acid — so the Assertion is true. However, the Reason calls 290 K a "low melting point," which is incorrect; 290 K (≈17°C) is relatively high for an organic liquid, and it is this comparatively high melting point that causes freezing in cold climates. Since the Reason contains a factual error, option (C) is correct.

(A) The hydrophobic tail of the soap molecule is in the interior of the cluster, whereas hydrophilic end is on the surface of the cluster.

The textbook clearly states: "clusters of molecules in which the hydrophobic tails are in the interior of the cluster and the ionic ends are on the surface of the cluster" — this is the definition of a micelle. The hydrophobic tail avoids water by pointing inward; the ionic (hydrophilic) end faces the water outside. Option (A) matches this exactly.