📖 Chapter CN07 · NDA Class 11–12 Level🎯 NDA Level : High Priority
Carbon and Its Compounds (Organic Chemistry) is a high-yield chapter in NDA — questions appear on allotropes, homologous series formulas, IUPAC nomenclature, functional group identification, and reaction types. The chapter is largely factual and visual, making it very accessible for average students who invest time memorising the general formulas, prefix table, and functional group structures.
📌 What to expect in NDA (based on 2022–2025 pattern): (1) Carbon allotropes — diamond vs graphite (structure, properties, uses); fullerene (C₆₀); (2) General formulas: Alkane CₙH₂ₙ₊₂; Alkene CₙH₂ₙ; Alkyne CₙH₂ₙ₋₂; (3) IUPAC naming — chain length prefix (meth/eth/prop/but…) + suffix (-ane/-ene/-yne); (4) Functional groups — –OH, –CHO, –CO–, –COOH, –O–, –COO– and their class names; (5) Combustion, substitution (alkanes), and addition reactions (alkenes/alkynes); (6) Isomerism — structural isomers from same molecular formula.
No other element forms as many compounds as carbon — over 10 million known organic compounds
Carbon (Z = 6) has electronic configuration 2, 4 — four valence electrons. This gives it two unique properties that explain the enormous diversity of organic chemistry.
① Catenation
Carbon can bond to other carbon atoms to form long chains, branches, and rings
C–C bond is strong and stable (347 kJ/mol)
Chains can be straight, branched, or cyclic
No other element does this as extensively
Basis of all organic molecules (petroleum, proteins, DNA)
② Tetravalency
Carbon forms 4 bonds (single, double, or triple) with other atoms
Can bond with: H, O, N, S, halogens, and other C atoms
Single bond (C–C): alkanes; Double (C=C): alkenes; Triple (C≡C): alkynes
Q1. Graphite is a good conductor of electricity whereas diamond is not, because:
(a) Diamond has a higher melting point (b) Graphite has delocalised electrons free to move between layers (c) Diamond is harder (d) Graphite has a layered structure
Answer: (b) Graphite has delocalised electrons free to move
In graphite, each C is sp² hybridised — bonded to 3 neighbours, leaving one electron per C atom delocalised across the layers. These π-electrons can move freely and carry electric current. In diamond, each C is sp³ hybridised — all 4 electrons are in strong σ-bonds with no free electrons. No free charge carriers = insulator. This is why graphite is used as electrodes and diamond is not.
Q2. The allotrope of carbon used as a lubricant and in pencil lead is:
(a) Diamond (b) Fullerene (c) Graphite (d) Carbon black
Answer: (c) Graphite
Graphite's layered structure has weak van der Waals forces between layers — layers slide over each other easily, making it an excellent solid lubricant (used where oils fail, e.g. high-temperature environments). Pencil "lead" contains graphite (not lead metal) mixed with clay. Diamond is used for cutting; fullerene for nanotechnology and research.
🧠 TRICKY QUESTIONS
Allotropes — Conceptual Traps
Q. Diamond and graphite are both pure carbon. Why do they have such different physical properties?
Answer: Different bonding arrangements (allotropy) — not different atoms, but different structures.
Diamond: every C bonded to 4 other C atoms in a rigid 3D network (sp³). Every electron is locked in a C–C bond. Result: extreme hardness, no conductivity.
Graphite: every C bonded to only 3 other C (sp²). The 4th electron is delocalised across the layer. Layers held by weak vdW forces. Result: soft (layers slide), good conductor.
This is the essence of allotropy: same element, completely different physical properties due to different structural arrangement. NDA sometimes asks: "Diamond and graphite are allotropes of carbon" — True. "They have similar properties" — False.
2. Hydrocarbons — Alkanes, Alkenes, Alkynes
2.1
General Formulas, Homologous Series & IUPAC Naming
The three series form the backbone of all organic chemistry
Homologous series: each successive member differs by –CH₂– (14 mass units). All members in a series have the same functional group, similar chemical properties, and gradually changing physical properties (MP, BP increase with chain length).
2.2
IUPAC Nomenclature — Systematic Naming
Chain length prefix + bond type suffix = IUPAC name
IUPAC naming gives every organic compound a unique, unambiguous name. For NDA, master the 10 chain-length prefixes and the three bond suffixes.
C₁
Meth-
Methane, Methanol
C₂
Eth-
Ethane, Ethanol
C₃
Prop-
Propane, Propene
C₄
But-
Butane, Butyne
C₅
Pent-
Pentane, Pentanol
C₆
Hex-
Hexane, Hexene
C₇
Hept-
Heptane
C₈
Oct-
Octane (petrol)
🔧 IUPAC Naming Rules — Step by Step for NDA: Step 1: Find the longest carbon chain — this gives the parent name (meth/eth/prop…). Step 2: Identify the functional group or multiple bond → suffix: –ane (single), –ene (double), –yne (triple), –ol (alcohol), –al (aldehyde), –one (ketone), –oic acid (carboxylic acid). Step 3: Number the chain from the end that gives the functional group the lowest number. Step 4: Name substituents with their position numbers (e.g. 2-methylpropane). Step 5: Alphabetical order for substituents if more than one.
Compound
Molecular Formula
IUPAC Name
Common Name
Key Feature
CH₄
CH₄
Methane
Marsh gas, Natural gas
Simplest alkane; greenhouse gas
C₂H₄
C₂H₄
Ethene
Ethylene
Simplest alkene; ripens fruits (plant hormone)
C₂H₂
C₂H₂
Ethyne
Acetylene
Simplest alkyne; oxyacetylene welding
C₆H₆
C₆H₆
Benzene
Benzene
Aromatic; DoU = 4; used in manufacture of dyes, plastics
C₈H₁₈
C₈H₁₈
Octane
Octane
Main component of petrol; octane rating
CH₃OH
CH₄O
Methanol
Wood alcohol, Methyl alcohol
Simplest alcohol; toxic; fuel, solvent
C₂H₅OH
C₂H₆O
Ethanol
Alcohol (drinking), Ethyl alcohol
In beverages, disinfectant, fuel
📝 TOPIC-WISE PYQ
Hydrocarbons & IUPAC — NDA Pattern Questions
Q1. The general formula of alkenes is:
(a) CₙH₂ₙ₊₂ (b) CₙH₂ₙ (c) CₙH₂ₙ₋₂ (d) CₙH₂ₙ₋₆
Answer: (b) CₙH₂ₙ
Alkanes = CₙH₂ₙ₊₂ (saturated, all single bonds). Alkenes = CₙH₂ₙ (one double bond, 2 fewer H than alkane). Alkynes = CₙH₂ₙ₋₂ (one triple bond, 4 fewer H than alkane). Each double bond or ring reduces H count by 2 from the saturated formula.
Q2. How many carbon atoms are present in a molecule of pentane?
(a) 3 (b) 4 (c) 5 (d) 6
Answer: (c) 5
"Pent-" = 5 carbon atoms. Pentane molecular formula: C₅H₁₂ (alkane, CₙH₂ₙ₊₂ → 5×2+2 = 12 H). Remember: meth(1), eth(2), prop(3), but(4), pent(5), hex(6), hept(7), oct(8), non(9), dec(10). This prefix table is the single most important memorisation task in organic chemistry nomenclature.
Q3. Which of the following is the IUPAC name for CH₃–CH₂–OH?
Answer: (b) Ethanol
CH₃–CH₂–OH: 2 carbon chain (eth-) + –OH functional group (–ol suffix) = Ethanol. Full IUPAC: Ethan-1-ol. This is ordinary drinking alcohol, also used as a disinfectant (70% solution) and biofuel. Methanol (CH₃OH) has only 1 carbon and is toxic. Propan-1-ol has 3 carbons.
🧠 TRICKY QUESTIONS
Hydrocarbons — Formula & Structure Traps
Q. A hydrocarbon has the molecular formula C₄H₈. Is it definitely an alkene?
Answer: Not necessarily — C₄H₈ fits both alkene (CₙH₂ₙ) AND cycloalkane (also CₙH₂ₙ).
The general formula CₙH₂ₙ is shared by: (1) alkenes (one C=C double bond) and (2) cycloalkanes (one ring — also reduces H by 2). C₄H₈ could be: But-1-ene (CH₂=CH–CH₂–CH₃), But-2-ene (CH₃–CH=CH–CH₃), 2-methylpropene, or cyclobutane (a 4-membered ring). The molecular formula alone is not enough to identify — you need the structural formula. NDA tests this with "which molecular formula can represent both an alkene and a cycloalkane?"
Q. Ethyne (C₂H₂) requires 2 moles of H₂ to become fully saturated. How many moles of Cl₂ does it require?
Answer: 2 moles of Cl₂ — same as H₂.
Ethyne has a triple bond (C≡C). Each addition reaction breaks one π bond and adds one molecule across it.
Step 1: C₂H₂ + Cl₂ → CHCl=CHCl (1,2-dichloroethene) — adds 1 Cl₂
Step 2: CHCl=CHCl + Cl₂ → CHCl₂–CHCl₂ (1,1,2,2-tetrachloroethane) — adds 2nd Cl₂
Total: 2 moles Cl₂ per mole of ethyne. Compare ethene (one double bond) — only 1 mole of Cl₂ needed. Alkynes need 2 moles of addend; alkenes need 1.
3. Functional Groups
3.1
Major Functional Groups — Structure, Class Name & Examples
The functional group determines the chemical behaviour of the molecule
A functional group is an atom or group of atoms that gives an organic compound its characteristic chemical properties. All members of the same class (same functional group) undergo similar reactions.
Fig. 2 — Six major functional groups with R-group notation. Key distinctions (aldehyde vs ketone; ether vs ester) shown in the lower panel — both are the most commonly confused pairs in NDA.
Esters have characteristic fruity odours — used in artificial flavours and perfumes. Ethyl acetate (nail polish remover), isoamyl acetate (banana flavour), benzyl acetate (jasmine). Natural fats and oils are triglycerides (esters of glycerol with three fatty acids).
📝 TOPIC-WISE PYQ
Functional Groups — NDA Pattern Questions
Q1. Which functional group is present in carboxylic acids?
(a) –OH (b) –CHO (c) –COOH (d) –CO–
Answer: (c) –COOH
The carboxyl group –COOH consists of a carbonyl (C=O) and a hydroxyl (–OH) bonded to the same carbon. It is the defining functional group of carboxylic acids. –OH alone = alcohol; –CHO = aldehyde; –CO– (carbonyl in middle) = ketone. Acetic acid (CH₃COOH) is the simplest well-known carboxylic acid — present in vinegar.
Q2. The reaction between ethanol and acetic acid in the presence of H₂SO₄ produces:
Answer: (c) Ethyl acetate
CH₃COOH + C₂H₅OH → CH₃COOC₂H₅ + H₂O (esterification). This is a classic NDA question. Ethyl acetate (ethyl ethanoate) has a fruity smell and is used as nail polish remover and a solvent. H₂SO₄ acts as a catalyst (concentrated acid) and is also a dehydrating agent. Heating is required.
Q3. Acetone (CH₃COCH₃) belongs to which class of organic compounds?
(a) Alcohol (b) Aldehyde (c) Ketone (d) Ester
Answer: (c) Ketone
Acetone (propanone) contains the C=O group in the middle of the carbon chain (flanked by two methyl groups: CH₃–CO–CH₃). This defines a ketone. Acetone is used as a solvent and nail polish remover. The difference from an aldehyde: in aldehydes the C=O is at the terminal carbon (–CHO), while in ketones it is internal. Acetone cannot be oxidised further (unlike aldehydes which give carboxylic acids).
🧠 TRICKY QUESTIONS
Functional Groups — Structure Confusion Traps
Q. Formic acid (HCOOH) has both –COOH and –CHO characteristics. Is it an acid, an aldehyde, or both?
Answer: It is classified as a carboxylic acid, but it also shows aldehyde-like reducing properties.
HCOOH structure: H–COOH. The –COOH makes it an acid (first and foremost). However, the "H" on the carboxyl carbon means it has a –CHO-like hydrogen — so formic acid can be oxidised and acts as a reducing agent (gives positive Tollens test / silver mirror reaction), unlike other carboxylic acids. NDA tests: "Which carboxylic acid can reduce Fehling's solution?" → Formic acid (HCOOH). It is unique because the carbonyl C in –COOH still has an H attached.
Q. CH₃OCH₃ and C₂H₅OH have the same molecular formula C₂H₆O. Are they the same compound?
Answer: No — they are structural isomers (same formula, different structure and different functional group).
CH₃OCH₃ = Dimethyl ether (functional group: ether –O–): gas at room temperature, used as propellant.
C₂H₅OH = Ethanol (functional group: alcohol –OH): liquid at room temperature, in beverages and as disinfectant.
They share the molecular formula C₂H₆O but have completely different properties. This is structural isomerism — same atoms connected differently. NDA frequently tests this as: "Which two compounds are isomers?" or "Give an isomer of ethanol."
4. Reactions of Organic Compounds
4.1
Combustion, Substitution & Addition Reactions
Three fundamental organic reaction types — each tied to a different class of compound
COMBUSTION
Burning in Oxygen
All hydrocarbons burn in O₂
Complete combustion: → CO₂ + H₂O (blue flame)
Incomplete combustion: → CO + H₂O + C (soot) (yellow flame)
Addition test: Br₂ water decolourised → unsaturated compound
⚛ Combustion — Balanced Equations for Common Compounds
Complete combustion (blue flame, enough O₂):
CH₄ + 2O₂ → CO₂ + 2H₂O (natural gas)
C₂H₆ + 7/2O₂ → 2CO₂ + 3H₂O (ethane)
C₂H₄ + 3O₂ → 2CO₂ + 2H₂O (ethylene)
C₂H₂ + 5/2O₂→ 2CO₂ + H₂O (acetylene — used in welding)
C₃H₈ + 5O₂ → 3CO₂ + 4H₂O (propane, LPG)
C₈H₁₈ + 25/2O₂→ 8CO₂+ 9H₂O (octane — petrol)
Incomplete combustion (yellow/red flame, limited O₂):
CH₄ + O₂ → CO + 2H₂ (or CO + H₂O)
CO is highly toxic — binds to haemoglobin (200× more than O₂)
Carbon (soot) deposited → black smoke
Key: Number of CO₂ = number of C atoms; moles of H₂O = H atoms ÷ 2
Balancing combustion equations: balance C first (as CO₂), then H (as H₂O), then O (as O₂ — may be fractional). Alkyne C₂H₂ burns with very bright luminous flame — used in oxyacetylene torch (3500°C). LPG = mainly propane (C₃H₈) + butane (C₄H₁₀).
🚨 Bromine Water Test — Unsaturation Test
Add Br₂ water (orange/brown) to compound
If colour disappears → unsaturated compound (alkene/alkyne) — addition occurred
If colour persists → saturated (alkane) or aromatic ring
Answer: (c) Addition reaction
C₂H₄ + Br₂ → C₂H₄Br₂ (1,2-dibromoethane, colourless). The π bond of ethene breaks; each Br adds to one carbon. Br₂ is consumed → orange/brown colour disappears. This is the standard test for unsaturation (C=C or C≡C). Alkanes do not decolourise Br₂ water at room temperature without UV light.
Q2. The complete combustion of methane (CH₄) produces:
(a) CO + H₂ (b) CO₂ + H₂O (c) CO₂ + H₂ (d) CO + H₂O
Answer: (b) CO₂ + H₂O
CH₄ + 2O₂ → CO₂ + 2H₂O (complete combustion — sufficient O₂). Products are always CO₂ and H₂O for complete combustion of any hydrocarbon. With insufficient O₂: incomplete combustion gives CO + H₂O (toxic CO) or C (soot). Natural gas appliances must be properly ventilated — incomplete combustion produces lethal CO.
Q3. Chlorination of methane requires which condition?
(a) Darkness and room temperature (b) UV light or high temperature (c) KOH catalyst (d) Acidic medium
Answer: (b) UV light or high temperature
CH₄ + Cl₂ → CH₃Cl + HCl (substitution). This is a free radical reaction. Cl₂ molecules are split into Cl· radicals by UV light (or heat): Cl₂ → 2Cl·. Cl· then attacks CH₄: Cl· + CH₄ → HCl + CH₃·, then CH₃· + Cl₂ → CH₃Cl + Cl·. In darkness, the reaction does not proceed — Cl₂ cannot form radicals without energy input.
🧠 TRICKY QUESTIONS
Reactions — Mechanism & Product Traps
Q. Why do alkanes undergo substitution but alkenes undergo addition — even though both react with Cl₂?
Answer: Alkanes have no π bond; alkenes have a reactive π bond.
Alkane (e.g. CH₄): All bonds are strong C–H and C–C σ bonds. Cl₂ cannot add across σ bonds — instead it substitutes (replaces) a H with Cl, requiring UV light to generate Cl· radicals. Product: CH₃Cl + HCl.
Alkene (e.g. C₂H₄): Has a weak π bond (part of C=C). Cl₂ can break the π bond and both Cl atoms add to the two carbons directly: C₂H₄ + Cl₂ → CH₂Cl–CH₂Cl. No UV needed — reaction happens at room temperature in the dark. The π bond is the reactive site; once it's consumed, the product (now a saturated compound) does not react further with Cl₂ easily.
Q. Incomplete combustion of hydrocarbons produces CO. Why is CO dangerous while CO₂ is not (at normal concentrations)?
Answer: CO binds to haemoglobin ~200× more strongly than O₂ — blocking oxygen transport.
CO (carbon monoxide) binds to haemoglobin (Hb) to form carboxyhaemoglobin (COHb), which is 200–250 times more stable than oxyhaemoglobin (HbO₂). Once CO occupies the Hb binding site, O₂ cannot bind → tissues are starved of oxygen (carbon monoxide poisoning). CO₂ does not bind to haemoglobin in the same way — it's transported mainly dissolved in plasma and as bicarbonate. Even small concentrations of CO (200–400 ppm) cause headache/dizziness; higher concentrations are fatal. This is why gas heaters need ventilation — incomplete combustion produces CO.
📄 CN07 Formula & Fact Sheet — Quick Reference
① Allotropes of Carbon
Diamond: sp³, 4 bonds, 3D network; hardest, insulator
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