NCERT Exemplar Solutions Class 10 Science Chapter 4

Source-and-sink angle. The main solution listed the two gases; let me show why each one is there and in what amount.

Where CO₂ comes from. Every living cell respires: C₆H₁₂O₆ + 6O₂ → 6CO₂ + 6H₂O. Add to this the burning of coal, petrol, diesel and wood, plus rotting of dead matter. These sources keep CO₂ at roughly 0.04% by volume, and it is removed by green plants during photosynthesis. This balance is the carbon cycle.

Where CO comes from. When a fuel burns with too little oxygen, the carbon is only half-oxidised: 2C + O₂ → 2CO instead of C + O₂ → CO₂. Vehicle engines and coal stoves running rich do this, so a trace of CO always exists near roads and chimneys. CO is poisonous because it locks onto haemoglobin.

Why not the other options. ``Carbon monoxide only'' and ``carbon dioxide only'' both ignore one of the two real gases, and coal is a mined solid, not an air component. Only option (b) names both gases with the correct relative amounts.

CO₂ (major) plus CO (trace); option (b).

Property-from-bonding angle. Each of the four statements is a direct read-off from the nature of the covalent bond. Let me link bonding to property one by one.

Electrical conductivity. Conduction needs mobile charge. Metals have a sea of free electrons; ionic melts/solutions have free ions. Covalent carbon compounds have neither, because every valence electron is locked in a shared pair between two specific atoms. So a solid or molten covalent compound (sugar, naphthalene, kerosene) does not conduct. Statement (ii) is correct and (i) is wrong.

Intermolecular forces. A molecule of, say, methane is neutral overall. The only attractions to its neighbours are weak van der Waals (London) forces. Because these are weak, carbon compounds melt and boil at low temperatures and are often gases, liquids or soft, low-melting solids. Statement (iv) is correct and (iii) is wrong.

Putting it together. The two correct statements are (ii) and (iv), so the answer pair is option (d). A quick cross-check: high melting point and good conduction are signatures of ionic or metallic substances, not of covalent carbon compounds.

Carbon compounds are poor conductors with weak intermolecular forces; option (d).

Octet-budget angle. Instead of memorising the structure, let me build it from the electron count, which is the safe way to settle any ``how many bonds'' question.

Electron accounting. Nitrogen (atomic number 7) has the configuration 2, 5, so 5 valence electrons. To reach a stable octet of 8 it must gain a share in 3 more electrons. Three hydrogen atoms each bring 1 electron. Pairing one N electron with one H electron three times uses 3 of nitrogen's electrons and all 3 hydrogen electrons.

What is left. Nitrogen still has 5-3 = 2 electrons unused; they sit as one lone pair on the nitrogen. A lone pair is not a bond, so it does not change the bond count. Every bond formed is a single shared pair, hence a single bond.

Ruling out the others. A double bond would need nitrogen to share two pairs with one atom, and a triple bond three pairs; neither is possible here because each hydrogen can only ever share one pair. So options (b), (c) and (d) are impossible for NH₃.

Three N–H single bonds plus one lone pair; option (a).

Structure-decides-property angle. The main solution named the allotrope; let me explain the three carbon allotropes side by side so the ``football'' shape makes sense.

Diamond. Every carbon is bonded to four others in a rigid 3-D network. There are no free electrons, so diamond is the hardest natural substance and does not conduct electricity.

Graphite. Each carbon bonds to only three others, forming flat hexagonal sheets. The fourth electron is free to move, so graphite conducts electricity, and the sheets slide over one another, making it soft and slippery (used in pencils and as a lubricant).

Buckminsterfullerene (C₆₀). Here 60 carbon atoms close up into a hollow cage of 12 five-membered and 20 six-membered rings, exactly like the seams on a football. It was named after the architect Buckminster Fuller, whose geodesic domes have the same look.

Why the other options fail. Phosphorus (white/red), sulphur (rhombic/monoclinic) and tin (grey/white) all have their own allotropes, but none of them forms a C₆₀ cage. The football molecule is uniquely carbon.

It is a carbon allotrope (C₆₀ cage); option (c).

Skeleton-first angle. The reliable way to count isomers is to forget the hydrogens for a moment and just draw the carbon skeletons; then fill hydrogens to satisfy carbon's valency of four.

Step 1: carbon skeletons of C₄H₁₀. With four carbons there are only two skeletons. One is the unbranched line of four carbons. The other puts three carbons in a row and hangs the fourth carbon off the middle carbon as a branch. There is no third arrangement, because moving the branch to an end just regenerates the straight chain.

Step 2: name them. The straight chain is n-butane, CH₃-CH₂-CH₂-CH₃. The branched one is 2-methylpropane (common name isobutane), (CH₃)₂CH-CH₃. Both have formula C₄H₁₀, so they are true structural isomers.

Why (iii) and (iv) are decoys. The exam draws the same n-butane chain twice more, once bent into a zig-zag and once into a hook. Rotating or bending bonds does not change which atom is joined to which, so these are not new compounds. A common trick: count the longest continuous carbon chain and check the branch position; if those match, it is the same isomer.

Only two real isomers exist, drawn as (i) and (ii); option (c).

Electron-transfer angle. The main solution used the oxygen/hydrogen count. Let me check the same answer using oxidation numbers, which never lies.

Carbon oxidation state. In ethanol the carbon that carries the -OH is at oxidation state -1. In ethanoic acid that carbon becomes +3 (it is now bonded to two oxygens). The carbon's oxidation number has gone up by 4, which is the definition of oxidation of the organic substrate.

What happens to manganese. If ethanol is oxidised, something must be reduced to take the electrons. In alkaline KMnO₄ the manganese falls from +7 (in MnO₄^-, purple) to +4 (in MnO₂, brown). The purple colour fading is the visible sign that KMnO₄ has acted as the oxidising agent.

Why the wrong options fail. A reducing agent would lower the carbon oxidation state (wrong direction). A catalyst would speed the reaction without being consumed, but KMnO₄ is consumed (it changes to MnO₂). A dehydrating agent only removes water; it would give an alkene, not an acid.

Carbon is oxidised (-1 → +3) as Mn drops +7 → +4; KMnO₄ is the oxidising agent; option (b).

Everyday-chemistry angle. This is the reaction that turns liquid vegetable oil into vanaspati ghee, so it is worth understanding fully.

What unsaturation means. A vegetable oil molecule has long carbon chains with one or more C=C double bonds. These bends in the chain stop the molecules from packing tightly, so the oil stays liquid at room temperature.

The hydrogenation step. Passing H₂ over the oil with a finely divided nickel (or palladium) catalyst makes one H atom add to each carbon of a double bond, turning each C=C into a C–C single bond. Removing the double bonds straightens the chains, they now pack tightly, and the product is a semi-solid fat.

Why each other option is wrong. Substitution swaps one atom for another (as in chlorination of methane); nothing is swapped here. Displacement is a metal-replacing-metal idea from inorganic chemistry, not relevant. Oxidation would add oxygen, but here we add hydrogen. So the only correct label is addition.

Health note. Hydrogenation can also form harmful trans-fats if done partially, which is why this reaction matters beyond the exam.

Hydrogen adds across C=C, converting oil to fat; addition reaction; option (a).

Naming-system angle. This question rewards anyone who knows the suffix code. Let me lay out the code and then apply it to all four names.

The suffix code. In IUPAC naming, the part before the suffix tells the number of carbons (but- = 4 carbons here for all four), and the suffix names the functional group. ``-ol'' is reserved for the hydroxyl group -OH, the defining group of alcohols.

Applying it. Butanol = C₄H₉OH, an alcohol. The other three carry different groups: butanone is a ketone (CH₃COCH₂CH₃, C=O (carbonyl) in the middle); butanoic acid is a carboxylic acid (CH₃CH₂CH₂COOH, -COOH at the end); butanal is an aldehyde (CH₃CH₂CH₂CHO, -CHO at the end).

Why this matters. The functional group, not the carbon chain, decides chemical behaviour. So spotting -OH from the ``-ol'' ending immediately tells you the compound will react like ethanol (reacts with sodium, can be oxidised to an acid, forms esters).

Only ``butan-ol'' has the -OH group; option (b).

Polarity-matching angle. The main solution named the two ends. Let me explain the cleaning action so you see why this exact arrangement is needed.

Like dissolves like. Water is polar, so it dissolves polar or ionic things. Oil and grease are non-polar, so they dissolve non-polar things. A single molecule that has both a polar end and a non-polar end can therefore touch both worlds at once. The soap molecule is exactly that bridge.

How dirt is removed. When greasy cloth is washed, the hydrophobic tails dig into the oil drop while the hydrophilic heads stay outside in the water. Many soap molecules surround one oil drop, heads out, tails in, forming a tiny ball called a micelle. The outside of the micelle is charged, so the micelles repel each other and stay suspended; rinsing then carries the grease away.

Why the other options fail. If both ends were hydrophilic the soap could not grip grease; if both ends were hydrophobic it could not stay in water; swapping the head and tail roles is simply not how the carboxylate ion is built. Only ``hydrophilic head, hydrophobic tail'' allows cleaning.

Polar ionic head in water, non-polar tail in grease; option (a).

Octet-completion angle. The main solution built the triple bond from the electron need. Let me double-check by counting total electrons, which is the foolproof method for any dot structure.

Total valence electrons. Two nitrogen atoms give 5 + 5 = 10 valence electrons, i.e. 5 pairs to place. These must be split between bonding pairs (between the atoms) and lone pairs (on the atoms) so that each N ends up with 8 around it.

Placing the pairs. Put 3 pairs between the two atoms (the triple bond). That leaves 5 - 3 = 2 pairs, one to sit as a lone pair on each nitrogen. Now check each N: it sees 6 shared electrons (the triple bond) plus its own lone pair of 2, total 8. Octet satisfied for both.

Why fewer bonds fail. A single bond would leave each N with only 6 electrons; a double bond leaves 7; only the triple bond delivers the full octet. So options that show one or two shared pairs cannot be right, and four shared pairs would over-fill the shell.

Triple bond plus one lone pair on each N gives both atoms an octet; option (d).

Valency-check angle. The main solution gave the structure. Let me show how to eliminate every wrong option just by counting bonds on carbon, which must always total four.

Carbon's rule of four. Each carbon atom must form exactly four covalent bonds. In H-C#C-H each carbon uses three bonds for the triple bond to the other carbon and one bond to a hydrogen: 3 + 1 = 4. Perfect, and the formula is C₂H₂, matching ethyne.

Rejecting the decoys. Option (b) H₃-C#C-H asks one carbon to bond to 3 hydrogens and hold a triple bond, which needs 6 bonds: impossible. Option (c) H₂C=CH₂ is ethene (a double bond, formula C₂H₄), an alkene, not an alkyne. Option (d) H₃C-CH₃ is ethane (all single bonds, C₂H₆), an alkane.

Family check. Alkanes end in -ane (single bonds), alkenes in -ene (double bond), alkynes in -yne (triple bond). ``Ethyne'' ends in -yne, so it must have the triple bond, confirming option (a).

H-C#C-H satisfies carbon's valency and matches C₂H₂; option (a).

Formula-test angle. The main solution read the suffixes. Let me confirm using the general formulae, which is handy when the name is unfamiliar.

General formulae. A saturated alkane fits C_nH_2n+2. An alkene (one double bond) fits C_nH_2n, with two fewer hydrogens. An alkyne (one triple bond) fits C_nH_2n-2, with four fewer. Fewer hydrogens for the same carbons is the fingerprint of unsaturation.

Apply to the three carbons (n=3). Propane is C₃H₈ (2×3+2), fully saturated. Propene is C₃H₆ (2×3), short by 2 H, so one double bond, unsaturated. Propyne is C₃H₄ (2×3-2), short by 4 H, so one triple bond, unsaturated. Chloropropane C₃H₇Cl comes from propane with one H swapped for Cl, still only single bonds, so saturated.

Quick test reminder. Unsaturated compounds decolourise bromine water and alkaline KMnO₄; propene and propyne would, propane and chloropropane would not. So the two unsaturated members are (ii) and (iii).

Propene (C₃H₆) and propyne (C₃H₄) are unsaturated; option (d).

Energy-source angle. The main solution named sunlight. Let me explain why light, and not water or acid, is the thing that makes alkanes react.

Why alkanes are stubborn. An alkane has only strong C–H and C–C single bonds and no reactive site. To make it react you must first create a reactive species. Chlorine on its own at room temperature in the dark does nothing to methane.

What sunlight does. Ultraviolet light has enough energy to break the relatively weak Cl-Cl bond into two chlorine atoms (free radicals), each with an unpaired electron. These radicals are very reactive: one of them snatches a hydrogen from the alkane, and a chain of substitution steps follows, giving CH₃Cl, then CH₂Cl₂, and so on, releasing HCl.

Why the wrong options fail. The absence of sunlight means no radicals, so no reaction. Water and hydrochloric acid do not split the chlorine molecule, so they cannot start the substitution. Only sunlight provides the trigger.

Sunlight breaks Cl₂ into radicals that substitute alkane H; option (b).

Free-energy angle. The main solution gave the layout. Let me say why nature picks this exact arrangement and not the reverse.

The driving force. Water molecules hydrogen-bond strongly to each other. An oily tail cannot join this network, so when tails are exposed to water, the water has to cage them, which is unfavourable. The system lowers its energy by hiding all the tails together inside a ball, so the least water surface touches oil. This is the same hydrophobic effect that makes oil droplets merge in water.

Why heads face out. The ionic heads (-COO^-) are happy in water and even attract it. Placing them on the outside surface lets them hydrogen-bond and ion-attract the surrounding water, which is favourable. So heads out, tails in, is the lowest-energy structure.

Why this enables cleaning. The oily interior can trap a grease droplet, while the charged outer surface keeps micelles apart (like charges repel) and suspended in water, so rinsing carries the dirt away. The reversed arrangements in options (b), (c) and (d) would either expose the tails to water or bury the heads, both energetically impossible in water.

Heads outward (water side), tails inward (oily core); option (a).

Counting-by-formula angle. The main solution drew the chain. Let me give a formula method that works for any straight-chain alkane, so you never need the picture.

Set up the count. A straight-chain alkane C_nH_2n+2 has n carbons. The carbons link in a single line, so there are (n-1) C–C bonds. Every hydrogen makes exactly one C–H bond, and there are 2n+2 hydrogens, so (2n+2) C–H bonds.

Total bond formula. Total = (n-1) + (2n+2) = 3n + 1. For pentane n = 5, so total = 3(5) + 1 = 16. This matches the hand count of 4 C–C plus 12 C–H.

Why the decoys fail. ``5'' counts only the carbons, ``12'' counts only the hydrogens or only the C–H bonds, and ``17'' over-counts by one (a common slip if you write n C–C bonds instead of n-1). The careful total is 16.

3n+1 = 3(5)+1 = 16 covalent bonds; option (c).

Valency-on-a-ring angle. The main solution described the hexagon. Let me build it from scratch so the alternate double bonds are clearly necessary, not just memorised.

Start with the formula. Benzene is C₆H₆. With 6 carbons, each needing 4 bonds, the total bonding capacity is 6×4 = 24 bonds worth of electrons. The 6 hydrogens use up 6 of those (one C–H each), leaving 18 to be shared among the carbon ring.

Why double bonds appear. Six carbons joined in a plain ring with single bonds would use only 6 C–C bonds, leaving each carbon one bond short. To satisfy every carbon's valency of 4, three of the ring bonds must become double bonds, arranged alternately so each carbon has exactly one double and one single bond to its neighbours plus one C–H. That is the classic Kekul'e structure.

Rejecting the decoys. A five-membered ring or an open chain would not give the formula C₆H₆ with the right valencies; a ring of all single bonds (option d) leaves each carbon with only 3 bonds, which is impossible for carbon. Only the six-membered ring with alternate double bonds works.

Hexagonal C₆H₆ ring with alternating double bonds; option (c).

Acid-base angle. The main solution swapped Na for H. Let me frame this as a simple acid–base style reaction, which makes the products obvious.

Ethanol's weakly acidic H. The hydrogen of the -OH group is very slightly acidic. A reactive metal like sodium can displace it, just as sodium displaces hydrogen from water (2Na + 2H₂O → 2NaOH + H₂). The alcohol behaves like a very weak ``acid'' here.

Products by analogy. Replacing water's H by Na gives the hydroxide; replacing ethanol's O-H hydrogen by Na gives the ethoxide, CH₃CH₂O^-Na⁺. In both cases the hydrogen that leaves pairs up to form H₂ gas (which you see as bubbles). This effervescence is in fact a test for the -OH group.

Why the wrong options fail. ``Sodium ethanoate'' would need oxidation of ethanol to the acid first, which sodium does not do. Oxygen is never a product; the gas released is hydrogen. So only ``sodium ethoxide and hydrogen'' is correct.

Na displaces the O-H hydrogen → sodium ethoxide + H₂; option (c).

Name-to-structure angle. The main solution decoded the name. Let me also rule out the look-alikes, because the exam draws very similar four-carbon pictures.

Decode the name fully. The stem ``but'' fixes the carbon count at four. The suffix ``-oic acid'' fixes the functional group as carboxylic acid, -COOH, which must sit at the end of the chain (carbon number 1). So the skeleton is C-C-C-COOH, written CH₃CH₂CH₂COOH.

Why the decoys are wrong. A three-carbon acid CH₃CH₂COOH is propanoic acid, not butanoic, so it has the wrong carbon count. A ketone structure has C=O (carbonyl) in the middle, not -COOH at the end, so it is the wrong functional group. An ester structure has -COO- linking two carbon groups, again not a free acid. Only the straight four-carbon chain ending in -COOH fits.

Confirm by formula. CH₃CH₂CH₂COOH is C₄H₈O₂, the known formula of butanoic acid (the smell of rancid butter). This cross-check seals option (d).

CH₃CH₂CH₂COOH (C₄H₈O₂); option (d).

Everyday-product angle. The main solution gave the figures. Let me explain where vinegar comes from and why both the solvent and the concentration in option (c) are correct.

How vinegar is made. Vinegar is produced by the souring of dilute alcohol: bacteria oxidise the ethanol in fermented liquids to ethanoic acid in the presence of air. The starting liquid is already mostly water, and only part of the alcohol turns to acid, so the final product is a watery solution with a small acid content.

Why 5–8%. Food-grade vinegar is kept at about 5–8% acetic acid. This is acidic enough to give the sour taste and to preserve pickles by stopping microbial growth, but dilute enough to be safe to eat. A 50–60% solution would be corrosive and dangerous.

Why the solvent is water. Acetic acid is fully miscible with water and forms hydrogen bonds with it, so water is the natural solvent. Options that name alcohol as the solvent are simply wrong; alcohol is the starting material, not the solvent of the finished vinegar.

Dilute (5–8%) acetic acid in water; option (c).

Equilibrium angle. The main solution compared the extents of ionisation. Let me sharpen this with the idea of an ionisation equilibrium, which is what ``partly ionised'' really means.

Strong acid, no equilibrium. For HCl, ionisation goes almost fully to completion; we write a single arrow. Almost every molecule splits into H+ and Cl-, so a 0.1 M HCl solution has nearly 0.1 M H+. That high H+ concentration is what makes it strongly acidic (low pH).

Weak acid, true equilibrium. For CH₃COOH, ionisation sets up a balance, written with a double arrow: CH₃COOH ⇌ CH₃COO- + H+. At equilibrium only a small fraction (a few per cent) has ionised, because the acetate ion readily grabs H+ back. So a 0.1 M acetic acid solution has far less H+ and a higher pH.

Reading the options. Statement (i) (mineral acids completely ionised) and statement (iv) (carboxylic acids partially ionised) are both true and together explain the strength gap. Statements (ii) and (iii) reverse the facts, so any option containing them is wrong. The answer is (a).

Complete ionisation (mineral) vs partial ionisation (carboxylic): (i) and (iv); option (a).

Nearest-noble-gas angle. The main solution reached neon by adding to the octet. Let me explain why carbon picks neon and not helium.

Counting after bonding. Carbon starts with 4 outer electrons. Each of the four shared pairs in CH₄ contributes one extra electron to carbon's count, so it now has access to 4 + 4 = 8 electrons in its outer (second) shell. The full electron picture is 2 (inner) and 8 (outer), i.e. 2, 8.

Match to a noble gas. The noble gas with configuration 2, 8 is neon. Helium has only 2 electrons (a full first shell), which suits hydrogen, not carbon. Argon is 2, 8, 8 and krypton even larger, both far beyond carbon's two-shell size.

Why sharing, not transfer. Carbon would need to gain or lose 4 electrons to become an ion, which costs too much energy. Sharing four pairs is the easy route to the neon octet, and it is why carbon forms covalent, not ionic, compounds.

Carbon's 2, 8 outer count matches neon; option (b).

Electron-bookkeeping angle. The main solution placed the pairs. Let me confirm by a full electron count, which is the safest check for any dot structure.

Total valence electrons. Oxygen brings 6, and each hydrogen brings 1, so 6 + 1 + 1 = 8 valence electrons, i.e. 4 pairs to place. These must be arranged so oxygen ends with 8 around it and each hydrogen with 2.

Placing the pairs. Use one pair for each O–H bond, two bond pairs in total. That leaves 4 - 2 = 2 pairs, which sit as lone pairs on oxygen. Check oxygen: 2 bond pairs (4 electrons) plus 2 lone pairs (4 electrons) = 8, an octet. Check each hydrogen: it shares 1 pair, so it has 2 electrons, a complete first shell.

Why the decoys fail. A structure with no or one lone pair leaves oxygen short of its octet. A double bond to hydrogen is impossible, since hydrogen can hold only one shared pair (2 electrons). Only the two-bond-pair, two-lone-pair picture satisfies every atom.

O has 2 O–H bond pairs and 2 lone pairs (octet); option (c).

Skeleton-shape angle. The main solution spotted the branch. Let me make the straight-versus-branched test crisp, because the exam disguises it.

Definition first. ``Straight chain'' does not mean the drawing is a straight line on paper; it means there is one single, unbranched sequence of carbons. A zig-zag drawing of CH₃CH₂CH₂CH₃ is still straight chain because every carbon links to at most two other carbons.

The branch test. Count the bonds each carbon makes to other carbons. In a straight chain the two end carbons bond to one carbon each and every middle carbon bonds to two carbons. The moment a single carbon bonds to three (or four) carbons, a branch exists, and the molecule is branched.

Apply it. In option (d) one middle carbon bonds to three carbons (two along the chain plus the -CH₃ branch), so it fails the test and is the branched, non-straight-chain compound. Options (a), (b) and (c) pass the test (the double bond in (c) does not make a branch).

Option (d) has a carbon bonded to three carbons (a branch); option (d).

Bond-type angle. The main solution sorted by bonds. Let me add a test-tube check that distinguishes saturated from unsaturated, which is the practical side of this idea.

The structural test. Saturation is about whether carbon's bonds to its neighbours are all single. A structure with one or more double or triple carbon–carbon bonds is unsaturated; if every carbon–carbon link is single, it is saturated. So just scanning each structure for a = or ≡ sign between two carbons settles it: structures (ii) and (iv) carry them.

The chemical test. Unsaturated compounds add bromine across their multiple bond, so they decolourise bromine water quickly. They also decolourise dilute alkaline KMnO₄. Saturated alkanes like (i) and (iii) do not react this way at room temperature, so the bromine-water colour stays. This is how a chemist would confirm which samples are unsaturated.

Reading the options. The two unsaturated structures are (ii) and (iv), so the answer is option (c). Any option that includes (i) or (iii) is wrong because those are saturated alkanes.

Double-bond and triple-bond structures (ii) and (iv) are unsaturated; option (c).

Formula-pattern angle. The main solution tested each member. Let me stress what ``same series'' means so the choice is airtight.

What defines a series. Members of one homologous series share a single general formula and the same kind of bonding. The alkane series is C_nH_2n+2: methane, ethane, propane, butane (C₄H₁₀), and so on, each adding one CH₂ unit. They all have only single bonds.

Spot the intruder. The first three given formulae are exactly the first three alkanes. The fourth, C₄H₈, has two hydrogens fewer than butane (C₄H₁₀), which is the tell-tale of one double bond. So C₄H₈ is butene, a member of the alkene series (C_nH_2n), not the alkane series.

Why this matters. Because C₄H₈ belongs to a different series with a C=C double bond, it behaves differently (it gives addition reactions and decolourises bromine water). So it does not belong with the three alkanes.

C₄H₈ follows C_nH_2n (alkene), breaking the alkane series; option (d).

Naming-discipline angle. The main solution named it directly. Let me rule out each wrong option so the logic is complete.

Carbon count and group. The chain CH₃-CH₂-CHO has three carbons, fixing the stem as ``propan''. The terminal -CHO is an aldehyde group, fixing the suffix as ``-al''. Hence propanal. The aldehyde carbon is always carbon number 1 because -CHO must lie at a chain end.

Eliminate the decoys. Propanone (option b) is the three-carbon ketone CH₃COCH₃, with C=O (carbonyl) in the middle, not -CHO at the end. Ethanol (option c) is a two-carbon alcohol C₂H₅OH, wrong carbon count and wrong group. Ethanal (option d) is the two-carbon aldehyde CH₃CHO, the right group but one carbon short.

Final match. Only ``propanal'' has both the correct three-carbon count and the correct aldehyde group, so option (a) is the answer.

Three carbons + -CHO = propanal; option (a).

Definition-first angle. The main solution applied the rule. Let me make sure the meaning of ``heteroatom'' is rock solid, since options (a), (b) and (c) all try to slip carbon or hydrogen into the list.

The word itself. ``Hetero'' means different. In organic chemistry the backbone is built from carbon and hydrogen, so any different atom, anything other than C or H, is called a heteroatom. Common heteroatoms are O, N, S, and the halogens (F, Cl, Br, I).

Scan the molecule. Reading CH₃-CH₂-O-CH₂-CH₂-Cl left to right: the carbons and hydrogens form the chain, the oxygen sits in the middle (an ether linkage), and the chlorine sits at the right end (a halogen). Oxygen and chlorine are the only non-C, non-H atoms, so they are the two heteroatoms.

Rejecting the decoys. Any option that names carbon (ii) or hydrogen (iii) as a heteroatom is wrong by definition, which removes (a), (b) and (c). Only option (d), oxygen and chlorine, is correct.

Only O and Cl are non-C, non-H atoms; option (d).

Reaction-type sorting angle. The main solution matched the pattern. Let me name each of the four reactions so the difference is unmistakable.

What saponification looks like. The signature is: an ester (-COO- link) on the left reacting with a base (NaOH or KOH), giving a carboxylate salt and a free alcohol on the right. The word comes from ``sapo'' (soap), because boiling fats (which are esters) with alkali makes soap.

Naming the four reactions. Option (b) is the reverse process, esterification (acid plus alcohol making ester and water). Option (c) is the action of sodium metal on an acid, giving a salt and hydrogen gas. Option (a) is a salt heated with alkali (a decarboxylation to give methane). None of these starts from an ester plus a base.

Confirm option (d). Only option (d) has an ester (CH₃COOC₂H₅) meeting NaOH and yielding the sodium salt (CH₃COONa) plus ethanol. That is the exact definition of saponification, so option (d) is correct.

Only (d) is ester + alkali → salt + alcohol; option (d).

Series-start angle. The main solution found the smallest alkyne. Let me clear up why the other options cannot be the first alkyne.

Why two carbons is the floor. A triple bond is three shared pairs between two carbon atoms. You simply cannot have a carbon–carbon triple bond with only one carbon, so methane (CH₄, one carbon, an alkane) is ruled out at once. The series therefore must begin at two carbons, ethyne.

Eliminate the look-alikes. Ethene (C₂H₄) has a double bond, so it is the first alkene, not alkyne. Propyne (C₃H₄) is a genuine alkyne but has three carbons, so it is the second member, not the first. Only ethyne has the minimum two carbons with a triple bond.

Formula check. Ethyne fits C_nH_2n-2 at n=2: C₂H₂. This is the well-known gas used in oxy-acetylene welding torches, confirming it as the simplest alkyne.

Smallest alkyne is two-carbon ethyne (C₂H₂); option (a).

Build-from-electrons angle. The main solution drew both forms. Let me build the picture step by step from the electron count so the triple bond is clearly forced.

Count the electrons. Two carbons give 4 + 4 = 8 valence electrons, and two hydrogens give 2 more, total 10 electrons (5 pairs). Each hydrogen must end with 2 electrons and each carbon with 8.

Place the pairs. Use one pair for each C–H bond (2 pairs used). That leaves 3 pairs, all of which must go between the two carbons, because there is nowhere else for them and each carbon still needs to reach 8. Three shared pairs between the carbons is a triple bond. Now each carbon sees 6 electrons from the triple bond plus 2 from its C–H bond, total 8, an octet.

The two representations. In the electron dot (Lewis) structure we draw those three pairs as dots/crosses between the carbons and one pair for each C–H. In the line structural formula we replace each shared pair by a short line, giving three lines between the carbons: H-C#C-H. Both say the same thing.

A useful check. Ethyne is a linear molecule (bond angle 180^°) because the triple bond keeps the four atoms in a straight line. Drawing it straight is therefore also geometrically correct.

H-C#C-H: three shared pairs (triple bond) between the carbons, one C–H pair each, all octets satisfied.

Systematic-naming angle. The main solution gave the four names. Let me show the full two-step method on each so you can name any structure without guessing.

Step 1: longest carbon chain (stem). Count every carbon in the longest continuous chain. (a) has 5 carbons (pent), (b) has 4 (but), (c) has 7 (the CH₃, five CH₂ units, and the CHO carbon, so hept), (d) has 5 (pent). The stem is fixed by this count alone.

Step 2: functional group (suffix). Read the group and attach its suffix. (a) -COOH gives ``-oic acid'' → pentanoic acid. (b) a carbon–carbon triple bond gives ``-yne'' → butyne (strictly but-2-yne, as the triple bond is between carbons 2 and 3). (c) -CHO gives ``-al'' → heptanal. (d) -OH gives ``-ol'' → pentanol.

Why the order matters. The functional group always gets the lowest possible locant number, which is why the acid and aldehyde carbons are numbered 1. For the symmetrical butyne the triple bond is naturally at 2. Following this fixed order keeps everyone's names identical, which is the point of IUPAC naming.

(a) Pentanoic acid, (b) Butyne, (c) Heptanal, (d) Pentanol.

Spot-the-group angle. The main solution listed the four groups. Let me explain how to tell a ketone from an aldehyde and an alkene from an alkane, since these are the usual trip-ups.

Alcohol vs acid. An -OH joined to a plain carbon is a hydroxyl group (alcohol), as in ethanol (a). When that oxygen sits in a -C(=O)-OH arrangement, the whole thing is the carboxyl group -COOH of a carboxylic acid, as in ethanoic acid (b). The extra double-bonded oxygen is the difference.

Ketone vs aldehyde. Both contain the carbonyl C=O. In a ketone the carbonyl carbon is bonded to two other carbons, as in propanone CH₃COCH₃ (c). In an aldehyde the carbonyl carbon is at the chain end with an H, written -CHO. So (c) is a ketone, not an aldehyde, because the C=O is sandwiched between two methyls.

The alkene. A carbon–carbon double bond, as in propene CH₂=CHCH₃ (d), is itself the functional group; it makes the compound unsaturated and able to undergo addition reactions. No oxygen is involved.

(a) hydroxyl (-OH, alcohol), (b) carboxyl (-COOH, acid), (c) carbonyl (C=O, ketone), (d) C=C (alkene).

Clue-by-clue angle. The main solution decoded the puzzle. Let me walk through each clue so the identification is fully justified.

Clue 1: the acid formula. C₂H₄O₂ has 2 carbons, 4 hydrogens and 2 oxygens. The carboxylic acid with this formula is ethanoic acid CH₃COOH (write it out: CH₃ contributes CH₃, and COOH contributes CO₂H, total C₂H₄O₂). So acid = ethanoic acid.

Clue 2: the alcohol via oxidation. The alcohol, on oxidation with alkaline KMnO₄ and then acidification, gives back the same ethanoic acid. A primary alcohol oxidises first to an aldehyde and then to the acid with the same carbon count. The two-carbon alcohol that gives two-carbon ethanoic acid is ethanol CH₃CH₂OH. So alcohol = ethanol.

Clue 3: the product X. A carboxylic acid heated with an alcohol and a little concentrated H₂SO₄ undergoes esterification, losing water and forming an ester. Ethanoic acid + ethanol gives ethyl ethanoate CH₃COOC₂H₅, a sweet-smelling liquid. So X = ethyl ethanoate.

The balanced reaction. CH₃COOH + C₂H₅OH → CH₃COOC₂H₅ + H₂O (conc. H₂SO₄ acts as catalyst and dehydrating agent, pulling out the water and pushing the reaction forward).

Acid = ethanoic acid, alcohol = ethanol, X = ethyl ethanoate; CH₃COOH + C₂H₅OH → CH₃COOC₂H₅ + H₂O.

Ion-chemistry angle. The main solution named the scum problem. Let me show the exact chemistry so you see why detergents escape it.

What hard water contains. Hard water carries dissolved calcium and magnesium salts. The free Ca²⁺ and Mg²⁺ ions are the troublemakers; they react with the carboxylate head of soap.

Why soap fails. The head of soap is a carboxylate (-COO^-). With Ca²⁺ it forms calcium stearate, an insoluble solid (the grey scum on bath tubs). So part of the soap is used up making scum instead of cleaning, lather is poor, and the scum redeposits on cloth.

Why detergents work. The head of a detergent is a sulphonate (-OSO₃^- or -SO₃^-). The calcium and magnesium salts of sulphonates are soluble, so no scum forms. The detergent molecules stay free to surround grease and lift it away, giving good lather and cleaning even in hard water.

Extra advantage. Detergents can also be made from petroleum products and can be designed to work in cold water, which is why most modern washing powders are detergents, not soaps.

Detergent heads (sulphonate) give soluble Ca/Mg salts, so no scum forms; soap heads (carboxylate) give insoluble scum in hard water. Hence detergents clean better.

Position-of-the-oxygen angle. The main solution named the four groups. Let me make the distinctions sharp, because (a), (b) and (c) all contain a C=O and students often mix them.

Ketone (a). Here the carbonyl carbon (CO) is flanked by two carbon groups (CH₃ on one side and a butyl chain on the other), with no hydrogen on it. A carbonyl between two carbons is the defining mark of a ketone, so (a) is a ketone.

Aldehyde vs acid (b, c). In (c) the -CHO sits at the chain end with an H on the carbonyl carbon, the signature of an aldehyde. In (b) the end group is -COOH, which adds a hydroxyl on the carbonyl carbon, making it a carboxylic acid. So (b) is an acid and (c) is an aldehyde, even though both end in a carbonyl.

Alcohol (d). The simplest case: an -OH attached to a saturated carbon, with no neighbouring C=O, is a plain hydroxyl group, so (d) is an alcohol (ethanol).

(a) Ketone, (b) Carboxylic acid, (c) Aldehyde, (d) Alcohol.

Mechanism-lite angle. The main solution stated the dehydration. Let me explain how the acid does the job and why the conditions matter.

Role of conc. H₂SO₄. Concentrated sulphuric acid is a powerful dehydrating agent; it has a strong appetite for water. It pulls a hydrogen from one carbon of ethanol and the -OH from the next carbon, combining them as H₂O that the acid then holds. What remains on the two carbons is a double bond, giving ethene CH₂=CH₂.

Why 443 K and excess acid. The reaction needs heat (about 443 K, i.e. 170 C) to break the bonds and drive water off. Excess acid keeps the medium dehydrating and shifts the balance towards the alkene. If the temperature is lower, ethanol can instead form an ether, so the high temperature steers the product to ethene.

Conditions as text, not on the arrow. Note that the conditions (conc. H₂SO₄ and 443 K) are stated beside or below the equation, so that the long labels do not crowd onto the arrow and stay easy to read: CH₃CH₂OH → CH₂=CH₂ + H₂O (conc. H₂SO₄, 443 K).

Heat ethanol with excess conc. H₂SO₄ at 443 K (dehydration): CH₃CH₂OH → CH₂=CH₂ + H₂O.

Metabolism angle. The main solution traced the poisoning. Let me explain why the body's own enzymes turn a small dose into a deadly one.

The enzyme that does the harm. The liver enzyme that normally handles drinking alcohol (ethanol) also acts on methanol. It oxidises methanol to methanal (formaldehyde). With ethanol the product is acetaldehyde, which the body clears; with methanol the product is methanal, which is far more reactive and toxic, and it is then oxidised further to methanoic (formic) acid.

Why methanal is dangerous. Methanal cross-links proteins, so it makes cell protoplasm clot. The build-up of methanoic acid also upsets the body's acid balance. The tissues most quickly hit are those of the eye, so the optic nerve is damaged and the person may go blind.

Why a small amount is enough. Because the body keeps converting methanol into the toxic methanal, even a few millilitres can produce enough poison to cause permanent blindness or death. This is the danger behind illicit ``hooch'' tragedies, where drinks are contaminated with methanol.

The liver oxidises methanol to highly toxic methanal, which clots protoplasm and destroys the optic nerve, so even a small intake causes blindness and can kill.

Balancing-care angle. The main solution identified hydrogen. Let me show why the equation needs a 2 in front, which students often miss.

Why hydrogen is the gas. Sodium is a very reactive metal. It displaces the slightly acidic hydrogen of the alcohol's -OH group, just as it displaces hydrogen from water. The displaced hydrogen atoms cannot exist alone, so two of them join to make one molecule of H₂ gas.

Balancing the equation. One ethanol gives one hydrogen atom, but H₂ contains two atoms. To make a whole H₂ molecule we need two ethanol molecules and two sodium atoms. That gives 2CH₃CH₂OH + 2Na → 2CH₃CH₂ONa + H₂, where every atom now balances on both sides.

Cross-check the count. Left side: 2 Na, 2 ethanol units. Right side: 2 sodium ethoxide units and 1 H₂. The two hydrogens of H₂ came from the two O-H groups, so the books balance.

Hydrogen gas is evolved: 2CH₃CH₂OH + 2Na → 2CH₃CH₂ONa + H₂.

Role-clarification angle. The main solution named the role. Let me contrast it with the other things sulphuric acid can do, so the answer is precise.

What ``dehydrating agent'' means. A dehydrating agent removes the elements of water (H and OH) from a compound. Concentrated sulphuric acid is one of the best, because it bonds strongly to water. From ethanol it strips an -H from one carbon and an -OH from the neighbouring carbon, releasing them as H₂O and leaving a double bond.

Why not ``catalyst'' or ``oxidant''. It is sometimes loosely called a catalyst, but in dehydration it actively takes up the water, so ``dehydrating agent'' is the precise term the exam wants. It is not acting as an oxidising agent here; the carbon count and oxidation state of carbon do not increase, only water is removed.

The equation and conditions. The reaction is CH₃CH₂OH → CH₂=CH₂ + H₂O at 443 K with excess concentrated acid. Writing the conditions beside the arrow as plain text keeps the equation clean and avoids cramming labels onto the arrow.

H₂SO₄ is a dehydrating agent; CH₃CH₂OH → CH₂=CH₂ + H₂O (conc. H₂SO₄, 443 K).

Valency-balance angle. The main solution gave both compounds. Let me show how the valencies decide the formula, so you could predict any such compound.

Carbon with chlorine. Carbon has a valency of 4 and each chlorine a valency of 1. To satisfy carbon's four bonds, four chlorine atoms attach, one bond each. The result is CCl₄, carbon tetrachloride, a liquid once used in fire extinguishers and dry cleaning. Every atom is satisfied: carbon makes 4 single bonds, each chlorine makes 1.

Carbon with oxygen. Oxygen has a valency of 2. Carbon's four bonds are met by two oxygen atoms, each joined by a double bond, giving the linear molecule O=C=O, carbon dioxide. Two double bonds use up all four of carbon's bonds.

Why this matters. The same valency-matching idea explains why carbon forms CH₄ with hydrogen (4 H atoms) and CS₂ with sulphur (2 S atoms). Knowing carbon's valency of 4 lets you write the formula with any partner element.

(a) CCl₄ (carbon tetrachloride); (b) CO₂ (carbon dioxide, O=C=O).

Shell-filling angle. The main solution gave the configuration and the bond. Let me show how the configuration directly predicts the dot structure, tying the two parts together.

Filling the shells. Electrons fill K (max 2), then L (max 8), then M. With 17 electrons: K gets 2, L gets 8, and the remaining 7 go into M. That is the configuration 2, 8, 7. The outermost shell (M) has 7 electrons, so chlorine has 7 valence electrons, drawn as 7 dots around the symbol.

From valence count to bonding. Seven valence electrons means chlorine is just one short of the stable octet of 8. The cheapest way to get that one electron is to share with another chlorine atom. Each atom puts in one electron to make a shared pair, the single Cl–Cl bond. After bonding, each chlorine has its 6 unshared electrons (three lone pairs) plus the shared pair, totalling 8.

The completed picture. So the dot structure of Cl₂ shows two chlorine symbols joined by one shared pair, with three lone pairs drawn on each atom. This is why chlorine exists naturally as the diatomic molecule Cl₂, not as single atoms.

(a) 2, 8, 7; (b) a single shared pair (Cl–Cl bond) with three lone pairs on each chlorine.

Bond-strength angle. The main solution compared the two elements. Let me explain the size–strength link and what it means for the number of compounds each element forms.

Why size controls bond strength. When two atoms bond, their nuclei attract the shared electrons. In a small atom like carbon the shared electrons are close to both nuclei, so the bond is short and the pull is strong. In a larger atom like silicon the shared electrons are farther from the nuclei, so the bond is longer and weaker. The C–C bond energy is much higher than the Si–Si bond energy.

Effect on chain length. Because C–C bonds are strong, carbon can join into chains of thousands of atoms (as in plastics and biological molecules) and they stay intact. Silicon's weak Si–Si bonds break easily, especially when attacked by water or oxygen, so silicon chains rarely get beyond a few atoms.

The big consequence. This is why carbon forms millions of stable compounds and is the basis of all life, while silicon, despite being in the same group, forms only a limited number of catenated compounds. The single factor of atomic size makes the difference.

Carbon's smaller size gives strong C–C bonds and long stable chains; silicon's larger size gives weak Si–Si bonds, so carbon catenates far more.

Two-test angle. The main solution used bromine water. Let me give both the standard tests and explain the chemistry behind each.

Bromine water test (the main one). Bromine water is orange-brown. When ethene is bubbled through, the bromine adds across the C=C double bond: CH₂=CH₂ + Br₂ → CH₂Br-CH₂Br. Because bromine is used up, the orange-brown colour fades to colourless. Ethane, having only single bonds, cannot do addition, so the bromine water stays coloured. The fading colour cleanly marks the ethene.

The flame (combustion) test. Unsaturated hydrocarbons have a higher proportion of carbon, so they burn with a sooty, yellow, smoky flame. Saturated hydrocarbons burn with a clean, almost non-luminous blue flame. So ethene gives a yellow smoky flame and ethane a blue flame; this is a quick supporting test.

Which to prefer. The bromine water test is the safer and more reliable laboratory test because it is a clear colour change at room temperature and does not need burning. The flame test is a useful backup.

Ethene decolourises orange-brown bromine water (addition); ethane does not. Ethene also burns with a sooty yellow flame, ethane with a clean blue flame.

Pattern-recognition angle. The main solution matched all four. Let me give the one-line fingerprint of each reaction type so matching becomes automatic.

Esterification (a). The clue is an alcohol plus a carboxylic acid giving an ester (the -COO- link) and water. Here methanol and ethanoic acid give methyl ethanoate and water. Sweet smell plus water out equals esterification, so (a) matches (iv).

Addition (b). The clue is an unsaturated compound (a double or triple bond) taking on extra atoms with no atoms leaving. Ethene plus hydrogen becomes ethane; the double bond is gone and two H atoms are added. That is addition (hydrogenation), so (b) matches (i).

Substitution (c) and neutralisation (d). In (c) one hydrogen of methane is replaced by chlorine, with HCl leaving; an atom is swapped, the mark of substitution, so (c) matches (ii). In (d) an acid reacts with a base to give a salt and water, the textbook acid–base neutralisation, so (d) matches (iii).

(a)→(iv) esterification, (b)→(i) addition, (c)→(ii) substitution, (d)→(iii) neutralisation.

Systematic-search angle. The main solution listed five. Let me show the method that guarantees you find exactly five and no duplicates.

Start with the longest chain. The longest possible chain has all 6 carbons: that is n-hexane, isomer 1. This is the only 6-carbon-chain isomer.

Drop to a 5-carbon chain. Now make the main chain 5 carbons and hang the 6th carbon as a -CH₃ branch. The branch can go on carbon 2 or carbon 3 (putting it on 4 or 5 just repeats carbon 2 or 1 from the other end). That gives 2-methylpentane (isomer 2) and 3-methylpentane (isomer 3).

Drop to a 4-carbon chain. With a 4-carbon main chain, two extra carbons become branches. Either both methyls sit on the same carbon (2,2-dimethylbutane, isomer 4) or on adjacent carbons (2,3-dimethylbutane, isomer 5). No other distinct arrangement exists.

Why exactly five. Going to a 3-carbon main chain would force more than four bonds on a carbon, which is impossible. So the search stops, and hexane has precisely five structural isomers.

Five isomers: n-hexane, 2-methylpentane, 3-methylpentane, 2,2-dimethylbutane, 2,3-dimethylbutane.

Job-of-each-reagent angle. The main solution named the three roles. Let me explain how to tell a catalyst from an oxidising agent in each case.

(a) Nickel as catalyst. In the hydrogenation of ethene, nickel provides a surface on which hydrogen and ethene meet and react faster. The nickel is not consumed and appears unchanged at the end, which is the mark of a catalyst. The same reaction is what hardens vegetable oils into fats.

(b) Concentrated sulphuric acid as catalyst. In esterification the acid speeds up the joining of the carboxylic acid and the alcohol, and it also soaks up the water formed, pushing the balance towards the ester. So it acts as both a catalyst and a dehydrating agent, but it is not used up overall.

(c) Alkaline KMnO₄ as oxidising agent. Here the reagent actively donates oxygen to ethanol, turning it into ethanoic acid. The manganese is reduced from +7 (purple) to +4 (brown) as it gives up that oxygen, so KMnO₄ is changed by the reaction: that is the behaviour of an oxidising agent, not a catalyst.

(a) Ni catalyses hydrogenation; (b) conc. H₂SO₄ catalyses esterification and removes water; (c) alkaline KMnO₄ is the oxidising agent.

Identify-the-gas angle. The main solution set up the apparatus. Let me explain the chemistry of both the reaction and the confirming test.

The reaction. Ethanoic acid is a weak acid, but it is strong enough to react with the mild base sodium hydrogencarbonate. The acid gives a hydrogen ion to the hydrogencarbonate, which breaks down into water and carbon dioxide. The full equation is CH₃COOH + NaHCO₃ → CH₃COONa + H₂O + CO₂. The brisk fizzing you see is the carbon dioxide escaping.

Why the lime water test works. To prove the gas is CO₂ and not something else, pass it through lime water (calcium hydroxide solution). The carbon dioxide reacts to form insoluble calcium carbonate, which makes the clear lime water look milky: Ca(OH)₂ + CO₂ → CaCO₃ + H₂O. This milkiness is the recognised test for carbon dioxide.

A neat extra check. If you keep passing excess CO₂, the milky calcium carbonate dissolves again to form soluble calcium hydrogencarbonate and the solution clears: CaCO₃ + H₂O + CO₂ → Ca(HCO₃)₂. The appearance and then disappearance of milkiness is a strong confirmation.

X = sodium ethanoate, gas = CO₂; lime water turns milky (Ca(OH)₂ + CO₂ → CaCO₃ + H₂O), confirming carbon dioxide.

Definitions-with-purpose angle. The main solution answered all three parts. Let me deepen each definition so the differences are clear.

(a) Hydrocarbons. These are the simplest organic compounds, built from just carbon and hydrogen. They form the backbone of fuels: methane (natural gas), the main alkanes in petrol and diesel, and so on. Methane CH₄ and ethane C₂H₆ are the two smallest examples.

(b) Saturated vs unsaturated. In a saturated hydrocarbon every carbon is bonded to the maximum number of atoms using only single bonds, so no more atoms can be added; methane and ethane are saturated. In an unsaturated hydrocarbon at least one carbon–carbon bond is a double bond (ethene, CH₂=CH₂) or triple bond (ethyne, HC#CH); these can add more atoms across the multiple bond, which is why they give addition reactions and decolourise bromine water.

(c) Functional groups. A functional group is the specific atom or group of atoms responsible for the characteristic reactions of a compound. The same group behaves the same way no matter how long the carbon chain is. Four common ones are the hydroxyl group -OH (alcohols), the aldehyde group -CHO, the carboxyl group -COOH (carboxylic acids), and the carbonyl group C=O in ketones.

Hydrocarbons = C-and-H compounds; saturated have single bonds only, unsaturated have C=C/C≡C; functional groups include -OH, -CHO, -COOH, C=O.

Process-detail angle. The main solution named hydrogenation. Let me explain the reaction fully, including why a catalyst and why the product is solid.

Why oils are liquid. An oil molecule has long hydrocarbon chains with one or more C=C double bonds. Each double bond puts a fixed bend in the chain, so the molecules cannot stack neatly; the weak forces between them keep the substance liquid at room temperature.

The hydrogenation step. Passing hydrogen over the oil with a finely divided nickel catalyst makes hydrogen add across the double bonds: each C=C becomes a C–C single bond with two new C–H bonds. The nickel speeds this up by holding the hydrogen and oil molecules on its surface; it is not used up, so it is a catalyst.

Why the product is a fat. Once the double bonds are gone, the chains are straight and flexible, so they pack closely together. The stronger packing raises the melting point, and the once-liquid oil becomes a solid or semi-solid fat such as vanaspati ghee. This is the everyday use of this reaction in the food industry.

A caution. Partial hydrogenation can leave some bonds in the ``trans'' arrangement (trans-fats), which are linked to heart disease, so the reaction is controlled carefully in modern processing.

Hydrogenation: unsaturated oil + H₂ over a nickel catalyst becomes saturated fat; the loss of double bonds lets chains pack tightly into a solid.

Two-part angle. The main solution covered both parts. Let me add the electron count for CCl₄ and the real-world meaning of saponification.

(a) Electron count for CCl₄. Carbon brings 4 valence electrons and each chlorine brings 7, but for bonding we focus on the shared pairs. Carbon forms one single bond to each of the four chlorines, using all 4 of its valence electrons in four shared pairs. After bonding, carbon is surrounded by 8 electrons (octet) and each chlorine has its shared pair plus three lone pairs (also an octet). The molecule is symmetrical and tetrahedral.

(b) What saponification means. The word comes from soap-making. Fats and oils are esters of long-chain fatty acids with glycerol. Boiling them with sodium hydroxide breaks the ester links and gives the sodium salts of the fatty acids (which is soap) plus glycerol. Using a small ester as the classroom example: CH₃COOC₂H₅ + NaOH → CH₃COONa + C₂H₅OH, where CH₃COONa is the salt and C₂H₅OH the alcohol.

Why it is the reverse of esterification. Esterification joins an acid and an alcohol into an ester using an acid catalyst, releasing water. Saponification does the opposite: it splits an ester with an alkali into a salt and an alcohol. Remembering this pairing prevents the common mix-up.

(a) CCl₄: four C–Cl single bonds, octet on every atom. (b) Saponification = alkaline hydrolysis of an ester to soap + alcohol; CH₃COOC₂H₅ + NaOH → CH₃COONa + C₂H₅OH.

Lab-procedure angle. The main solution gave the activity. Let me explain why each step is done that way and what the chemistry is.

Choice of chemicals. ``Glacial'' acetic acid is pure, water-free ethanoic acid; ``absolute'' alcohol is water-free ethanol. Starting dry matters because esterification is reversible, and water would push the balance back towards the acid and alcohol. A few drops of concentrated sulphuric acid act as a catalyst and soak up the water formed, driving the reaction forward.

Why a water bath. The mixture needs gentle warmth (about 60 C) to react at a good rate, but ethanol is highly flammable. Heating in a water bath gives even, controlled warmth and keeps the flammable vapours away from a direct flame, which is the safe method.

Confirming the ester. When the warmed mixture is poured into water and smelt, the sweet, fruity odour confirms that the ester ethyl ethanoate has formed. The equation is CH₃COOH + C₂H₅OH → CH₃COOC₂H₅ + H₂O with concentrated H₂SO₄ as catalyst. Esters like this are the source of many fruit flavours and perfumes.

Warm ethanol + glacial acetic acid + conc. H₂SO₄ in a water bath; sweet smell confirms ethyl ethanoate; CH₃COOH + C₂H₅OH → CH₃COOC₂H₅ + H₂O.

Deduction-chain angle. The main solution identified all four compounds. Let me retrace the clues one at a time to show the logic is forced, not guessed.

Clue from the formula and the pop test. C₂H₄O₂ with two carbons and two oxygens, reacting with sodium to release a gas that pops, is the signature of a carboxylic acid. The pop gas is hydrogen, displaced from the -COOH group. So C is ethanoic acid CH₃COOH and R, the sodium salt left behind, is sodium ethanoate CH₃COONa.

Clue from the sweet-smelling ester. Compound C plus an alcohol A with an acid catalyst gives a sweet-smelling compound S of formula C₃H₆O₂. Sweet smell means an ester. Counting carbons: the acid part contributes two carbons (CH₃CO-), so the alcohol must contribute one carbon. The one-carbon alcohol is methanol CH₃OH (A). The ester S is therefore methyl ethanoate CH₃COOCH₃, which indeed is C₃H₆O₂.

Confirming with the last clues. Adding NaOH to C gives R and water, the neutralisation CH₃COOH + NaOH → CH₃COONa + H₂O, which fits. Treating the ester S with NaOH gives back R and A, the saponification CH₃COOCH₃ + NaOH → CH₃COONa + CH₃OH, which also fits. Every clue agrees, so the identifications are certain.

C = ethanoic acid, R = sodium ethanoate, A = methanol, S = methyl ethanoate; the four reactions follow as listed.

Apparatus-and-test angle. The main solution answered all four parts. Let me explain the chemistry that links the two test tubes shown in Figure 4.1.

What happens in tube A. Ethanoic acid is acidic enough to react with the carbonate, breaking it down into the salt sodium ethanoate, water, and carbon dioxide gas: 2CH₃COOH + Na₂CO₃ → 2CH₃COONa + H₂O + CO₂. The fizzing produces the CO₂ that travels along the delivery tube to tube B.

What happens in tube B. The carbon dioxide bubbles into the lime water (calcium hydroxide). It reacts to form insoluble calcium carbonate, which is what makes the clear solution go milky: Ca(OH)₂ + CO₂ → CaCO₃ + H₂O. This is the recognised test for carbon dioxide.

Why ethanol fails, and making lime water. Ethanol is neutral and does not react with sodium carbonate, so no carbon dioxide is produced and tube B stays clear; the change is therefore not the same. To prepare lime water, calcium oxide (quicklime) is added to water to give calcium hydroxide; the mixture is left to settle and the clear liquid above is decanted off for use.

Lime water turns milky from CaCO₃; A makes CO₂, B traps it as CaCO₃; ethanol gives no CO₂; lime water is made by dissolving CaO in water and decanting.

Right-tool angle. The main solution named both processes. Let me explain why each needs its own reagent, so you never swap them.

(a) Ethanol to ethene: dehydration. To turn an alcohol into an alkene you must remove the elements of water and create a C=C double bond. Concentrated sulphuric acid is the dehydrating agent; at about 443 K it strips an -H and an -OH from neighbouring carbons of ethanol, leaving ethene: CH₃CH₂OH → CH₂=CH₂ + H₂O. Heat and excess acid push the reaction to the alkene.

(b) Propanol to propanoic acid: oxidation. To turn a primary alcohol into a carboxylic acid you must add oxygen. Alkaline potassium permanganate (followed by acidification) is the oxidising agent. It first oxidises the -CH₂OH end to -CHO and then to -COOH: CH₃CH₂CH₂OH → CH₃CH₂COOH. The purple permanganate colour fading marks the oxidation.

Why not interchange them. Dehydration removes atoms (water) and makes a double bond; oxidation adds oxygen and makes an acid. Using H₂SO₄ on propanol would not give the acid, and using KMnO₄ on ethanol would give ethanoic acid, not ethene. So each conversion needs its matched reagent.

(a) Dehydration with conc. H₂SO₄ at 443 K gives ethene; (b) oxidation with alkaline KMnO₄ gives propanoic acid.

Functional-isomer angle. The main solution drew both isomers. Let me explain why these two, and not others, are the answer, and how to draw their dot structures.

Find the isomers. C₃H₆O has one oxygen and three carbons. With this formula the oxygen is most naturally a carbonyl (C=O). If the carbonyl carbon sits in the middle of the three carbons, you get propanone (a ketone). If it sits at the end (carrying a hydrogen as -CHO), you get propanal (an aldehyde). These two are called functional isomers because they differ in functional group.

Electron dot structures. In both, the carbon and oxygen of the carbonyl share two pairs of electrons (a double bond). The oxygen also keeps two lone pairs, so it has its octet (4 shared + 4 lone). Each carbon completes its four bonds with hydrogens or carbons as shown. Drawing the C=O as two shared pairs and adding the two lone pairs on oxygen gives the correct Lewis structure for each isomer.

A note on a third possibility. One could also write prop-2-en-1-ol (CH₂=CH-CH₂OH) as a formal C₃H₆O isomer, but at Class 10 the expected answers are the two carbonyl isomers, propanone and propanal, which the Exemplar key gives.

Two isomers of C₃H₆O: propanone (ketone) and propanal (aldehyde), each with a C=O double bond and two lone pairs on oxygen.

Five-types angle. The main solution gave one example each. Let me explain the change happening in each reaction so the examples make sense.

(a) and (b): adding atoms. Hydrogenation adds hydrogen across a multiple bond; ethene plus hydrogen over nickel gives ethane, CH₂=CH₂ + H₂ → CH₃CH₃. Oxidation adds oxygen (or removes hydrogen); ethanol with alkaline KMnO₄ on heating becomes ethanoic acid, CH₃CH₂OH → CH₃COOH.

(c): swapping an atom. Substitution replaces one atom with another without changing the carbon skeleton. In sunlight, methane reacts with chlorine and one hydrogen is replaced by chlorine, CH₄ + Cl₂ → CH₃Cl + HCl. This is typical of saturated hydrocarbons.

(d) and (e): breaking and burning. Saponification breaks an ester with an alkali into a soap and an alcohol, CH₃COOC₂H₅ + NaOH → CH₃COONa + C₂H₅OH, the reaction behind soap making. Combustion is the burning of a carbon compound in oxygen, releasing carbon dioxide, water, heat and light, as in CH₄ + 2O₂ → CO₂ + 2H₂O, the reaction that powers cooking gas.

Five reactions with examples: hydrogenation (CH₂=CH₂ + H₂ → CH₃CH₃), oxidation (CH₃CH₂OH → CH₃COOH), substitution (CH₄ + Cl₂ → CH₃Cl + HCl), saponification (CH₃COOC₂H₅ + NaOH → CH₃COONa + C₂H₅OH), combustion (CH₄ + 2O₂ → CO₂ + 2H₂O).

Work-backwards angle. The main solution identified all three. Let me retrace the logic from the last clue to the first, which is the natural way to crack this puzzle.

Start at the combustion of C. One mole of C burns to give 2 moles of carbon dioxide and 3 moles of water. Each CO₂ carries one carbon, so C has 2 carbons; each H₂O carries two hydrogens, so 3 waters mean 6 hydrogens. That makes C = C₂H₆, which is the saturated alkane ethane.

Step back to B. C was made by adding one mole of hydrogen to B over a nickel catalyst (hydrogenation). Removing that H₂ from ethane gives C₂H₄, ethene, an unsaturated alkene that can take up hydrogen. So B = ethene.

Step back to A. B was formed by heating A with concentrated H₂SO₄, which is dehydration. The compound that loses water to give ethene is ethanol, C₂H₅OH. So A = ethanol.

The three equations. Dehydration: C₂H₅OH → C₂H₄ + H₂O (conc. H₂SO₄). Hydrogenation: C₂H₄ + H₂ → C₂H₆ (Ni). Combustion: 2C₂H₆ + 7O₂ → 4CO₂ + 6H₂O (with heat and light). Each clue is satisfied, confirming the identities.

A = ethanol, B = ethene, C = ethane; the dehydration, hydrogenation and combustion equations follow as written.