NCERT Exemplar Solutions Class 10 Science Carbon and its Compounds — Short and Long Answer Questions Q30–57

NCERT Exemplar Problems, Class 10 Science — Carbon and its Compounds, Short Answer Questions 30 to 46 and Long Answer Questions 47 to 57. All 28 questions solved.

This band asks you to produce organic chemistry rather than recognise it: draw the structure, name the compound, write the reaction, describe the experiment. Almost all of it runs on two small tables, and it is worth having both in front of you before starting.

The functional groups, because Questions 31, 32, 35 and 48 are nothing else:

Group	Formula	Class	Suffix
Hydroxyl	$\mathrm{-OH}$	alcohol	-ol
Aldehyde	$\mathrm{-CHO}$	aldehyde	-al
Ketone	$\mathrm{>C{=}O}$	ketone	-one
Carboxyl	$\mathrm{-COOH}$	carboxylic acid	-oic acid

The named reactions, because Questions 44, 46, 49, 54 and 56 are nothing else:

Reaction	What happens	Reagent
Addition (hydrogenation)	$\mathrm{H_2}$ adds across a double bond	Ni or Pd catalyst
Substitution	an atom replaces a hydrogen	$\mathrm{Cl_2}$ in sunlight
Oxidation	alcohol → carboxylic acid	alkaline $\mathrm{KMnO_4}$, heat
Dehydration	alcohol → alkene	conc. $\mathrm{H_2SO_4}$, 443 K
Esterification	acid + alcohol → ester	conc. $\mathrm{H_2SO_4}$
Saponification	ester + alkali → soap + alcohol	NaOH

Key insight. Ethanol and ethanoic acid between them account for most of this band, and the four things that can be done to ethanol are worth learning as a set: sodium takes its hydrogen and gives hydrogen gas; alkaline permanganate oxidises it to the acid; concentrated sulphuric acid at 443 K dehydrates it to ethene; and an acid with sulphuric acid as catalyst esterifies it. Each has a different reagent, and the reagent is what the question is really naming.

Short Answer Questions

Question 30

Draw the electron dot structure of ethyne and also draw its structural formula

Solution. Ethyne is $\mathrm{C_2H_2}$. Work out the bonding before drawing anything.

Each carbon has four valence electrons and each hydrogen one, so the molecule has $4 + 4 + 1 + 1 = 10$ valence electrons. Each carbon uses one of its electrons to bond to a hydrogen, leaving three each to share with the other carbon — a triple bond, three shared pairs.

Check the octets: each carbon is surrounded by the one pair it shares with hydrogen plus the three it shares with the other carbon, which is $2 + 6 = 8$. Each hydrogen has its single shared pair, which fills its only shell. Nothing is left over, so there are no lone pairs in ethyne — which is what distinguishes its dot structure from nitrogen’s in Question 10.

The structural formula replaces each shared pair with a line:

$$\mathrm{H} – \mathrm{C} \equiv \mathrm{C} – \mathrm{H}$$

Question 31

Write the names of the following compounds

(a) $\begin{array}{ccccccccccc} & \mathrm{H} & & \mathrm{H} & & \mathrm{H} & & \mathrm{H} & & \mathrm{O} & \\ & | & & | & & | & & | & & \| & \\ \mathrm{H}- & \mathrm{C} & – & \mathrm{C} & – & \mathrm{C} & – & \mathrm{C} & – & \mathrm{C} & -\mathrm{OH} \\ & | & & | & & | & & | & & & \\ & \mathrm{H} & & \mathrm{H} & & \mathrm{H} & & \mathrm{H} & & & \end{array}$

(b) $\begin{array}{ccccccc} & \mathrm{H} & & \mathrm{H} & & & \\ & | & & | & & & \\ \mathrm{H}- & \mathrm{C} & – & \mathrm{C} & – & \mathrm{C} \equiv \mathrm{C}-\mathrm{H} \\ & | & & | & & & \\ & \mathrm{H} & & \mathrm{H} & & & \end{array}$

(c) $\begin{array}{ccccccccccccccc} & \mathrm{H} & & \mathrm{H} & & \mathrm{H} & & \mathrm{H} & & \mathrm{H} & & \mathrm{H} & & \mathrm{H} & \\ & | & & | & & | & & | & & | & & | & & | & \\ \mathrm{H}- & \mathrm{C} & – & \mathrm{C} & – & \mathrm{C} & – & \mathrm{C} & – & \mathrm{C} & – & \mathrm{C} & – & \mathrm{C} & = \mathrm{O} \\ & | & & | & & | & & | & & | & & | & & & \\ & \mathrm{H} & & \mathrm{H} & & \mathrm{H} & & \mathrm{H} & & \mathrm{H} & & \mathrm{H} & & & \end{array}$

(d) $\begin{array}{ccccccccccc} & \mathrm{H} & & \mathrm{H} & & \mathrm{H} & & \mathrm{H} & & \mathrm{H} & \\ & | & & | & & | & & | & & | & \\ \mathrm{H}- & \mathrm{C} & – & \mathrm{C} & – & \mathrm{C} & – & \mathrm{C} & – & \mathrm{C} & -\mathrm{OH} \\ & | & & | & & | & & | & & | & \\ & \mathrm{H} & & \mathrm{H} & & \mathrm{H} & & \mathrm{H} & & \mathrm{H} & \end{array}$

Solution. Name each in two steps: count the carbons for the stem, then find the functional group for the suffix. Remember that a carbon in a $\mathrm{-COOH}$ or $\mathrm{-CHO}$ group counts towards the chain.

(a) Five carbons in the chain, and the last one carries a doubly bonded oxygen and a hydroxyl — that is the carboxyl group. Stem pent-, suffix -oic acid: pentanoic acid, $\mathrm{CH_3CH_2CH_2CH_2COOH}$.

(b) Four carbons, with a triple bond between the third and the fourth. Stem but-, suffix -yne: butyne, more precisely but-1-yne, $\mathrm{CH_3CH_2C \equiv CH}$. The triple bond is at the end of the chain, which is what the “1” records.

(c) Count carefully — there are seven carbons, the last carrying a doubly bonded oxygen and a hydrogen. A carbonyl at the end of a chain is an aldehyde. Stem hept-, suffix -al: heptanal, $\mathrm{CH_3(CH_2)_5CHO}$.

(d) Five carbons, all single bonds, ending in $\mathrm{-OH}$. Stem pent-, suffix -ol: pentanol, or pentan-1-ol, $\mathrm{CH_3CH_2CH_2CH_2CH_2OH}$.

Compare (a) and (d): both have five carbons and both end in an $\mathrm{-OH}$, and the only difference is the doubly bonded oxygen in (a). That single oxygen turns an alcohol into an acid, which is the whole of Question 54(b).

Question 32

Identify and name the functional groups present in the following compounds.

(a) $\begin{array}{ccccccc} & \mathrm{H} & & \mathrm{H} & & \mathrm{H} & \\ & | & & | & & | & \\ \mathrm{H}- & \mathrm{C} & – & \mathrm{C} & – & \mathrm{C} & -\mathrm{OH} \\ & | & & | & & | & \\ & \mathrm{H} & & \mathrm{H} & & \mathrm{H} & \end{array}$     (b) $\begin{array}{ccccccc} & \mathrm{H} & & \mathrm{H} & & \mathrm{O} & \\ & | & & | & & \| & \\ \mathrm{H}- & \mathrm{C} & – & \mathrm{C} & – & \mathrm{C} & -\mathrm{OH} \\ & | & & | & & & \\ & \mathrm{H} & & \mathrm{H} & & & \end{array}$

(c) $\begin{array}{ccccccccccc} & \mathrm{H} & & \mathrm{H} & & \mathrm{O} & & \mathrm{H} & & \mathrm{H} & \\ & | & & | & & \| & & | & & | & \\ \mathrm{H}- & \mathrm{C} & – & \mathrm{C} & – & \mathrm{C} & – & \mathrm{C} & – & \mathrm{C} & -\mathrm{H} \\ & | & & | & & & & | & & | & \\ & \mathrm{H} & & \mathrm{H} & & & & \mathrm{H} & & \mathrm{H} & \end{array}$     (d) $\begin{array}{ccccccccc} & \mathrm{H} & & \mathrm{H} & & \mathrm{H} & & \mathrm{H} & \\ & | & & | & & | & & | & \\ \mathrm{H}- & \mathrm{C} & – & \mathrm{C} & – & \mathrm{C} & = & \mathrm{C} & -\mathrm{H} \\ & | & & | & & & & & \\ & \mathrm{H} & & \mathrm{H} & & & & & \end{array}$

Solution. Look for whatever is not a plain carbon–hydrogen single bond. That is the functional group, and there is exactly one in each.

(a) An $\mathrm{-OH}$ on the end carbon, with everything else single-bonded. This is the hydroxyl group, and the compound is an alcohol — propan-1-ol.

(b) The end carbon carries both a doubly bonded oxygen and a hydroxyl. That combination is the carboxyl group, $\mathrm{-COOH}$, and the compound is a carboxylic acid — propanoic acid.

(c) A doubly bonded oxygen on the third carbon of a five-carbon chain, with carbons on both sides of it. A carbonyl with a carbon on each side is a ketone — pentan-3-one. Note that this carbon carries no hydrogen, which is why the row below it is blank.

(d) No oxygen anywhere, but a carbon–carbon double bond between the third and fourth carbons. The double bond is itself the functional group, and the compound is an alkene — but-1-ene.

Comparing (b) and (c) is the point of the question. Both contain $\mathrm{C{=}O}$; what makes one an acid and the other a ketone is where in the chain it sits and what else is attached. At the end with an $\mathrm{-OH}$ it is a carboxylic acid; at the end with an $\mathrm{-H}$ it is an aldehyde; in the middle it is a ketone.

Question 33

A compound X is formed by the reaction of a carboxylic acid $\mathrm{C_2H_4O_2}$ and an alcohol in presence of a few drops of $\mathrm{H_2SO_4}$. The alcohol on oxidation with alkaline $\mathrm{KMnO_4}$ followed by acidification gives the same carboxylic acid as used in this reaction. Give the names and structures of (a) carboxylic acid, (b) alcohol and (c) the compound X. Also write the reaction.

Solution.

(a) The formula is given: $\mathrm{C_2H_4O_2}$. A two-carbon carboxylic acid is ethanoic acid, $\mathrm{CH_3COOH}$ — check the atoms: two carbons, four hydrogens (three on the methyl, one on the $\mathrm{-OH}$), two oxygens. ✓

(b) The second sentence identifies the alcohol without naming it, and the argument is worth following. Oxidising an alcohol with alkaline permanganate gives the carboxylic acid with the same number of carbon atoms. Since that acid is ethanoic acid, with two carbons, the alcohol must have two carbons as well: ethanol, $\mathrm{C_2H_5OH}$.

$$\mathrm{CH_3CH_2OH} \xrightarrow[\text{Heat}]{\text{Alkaline } \mathrm{KMnO_4}} \mathrm{CH_3COOH}$$

(c) An acid plus an alcohol with sulphuric acid is esterification, so X is the ester ethyl ethanoate, $\mathrm{CH_3COOC_2H_5}$:

$$\mathrm{CH_3COOH} + \mathrm{C_2H_5OH} \xrightarrow{\text{conc. } \mathrm{H_2SO_4}} \mathrm{CH_3COOC_2H_5} + \mathrm{H_2O}$$

The ester is sweet-smelling, which is how it is recognised in the laboratory — Question 51 is that experiment.

Question 34

Why detergents are better cleansing agents than soaps? Explain.

Solution. The difference shows up only in hard water, which contains dissolved calcium and magnesium ions.

Soap is the sodium salt of a long-chain carboxylic acid. Its $\mathrm{-COO^-}$ head reacts with those $\mathrm{Ca^{2+}}$ and $\mathrm{Mg^{2+}}$ ions to form insoluble calcium and magnesium salts — the grey curdy scum that sticks to clothes and to the bath. Every soap molecule lost to scum is one that is not cleaning, so a great deal of soap is wasted before any lather appears at all.

Detergent is the sodium salt of a long-chain sulphonic acid or sulphate. Its charged head is a $\mathrm{-SO_3^-}$ or $\mathrm{-OSO_3^-}$ group, and the calcium and magnesium salts of those are soluble. No precipitate forms, so the detergent lathers and cleans just as well in hard water as in soft.

So the advantage is specific rather than general: a detergent works in both hard and soft water, while soap works properly only in soft water.

Two qualifications worth making, since “better” is doing a lot of work in the question. Detergents also work in acidic water, where soap is converted back to the insoluble free acid. But many detergents are not biodegradable, and those accumulate in rivers and cause foaming — so they are better cleansing agents and worse for the environment.

Question 35

Name the functional groups present in the following compounds

(a) $\mathrm{CH_3\,CO\,CH_2\,CH_2\,CH_2\,CH_3}$ (b) $\mathrm{CH_3\,CH_2\,CH_2\,COOH}$ (c) $\mathrm{CH_3\,CH_2\,CH_2\,CH_2\,CHO}$ (d) $\mathrm{CH_3\,CH_2\,OH}$

Solution. These are condensed formulae, so the functional group can be read straight off the letters — provided the four abbreviations are known exactly.

(a) The $\mathrm{CO}$ sits between two carbon groups: $\mathrm{CH_3}$ on one side, $\mathrm{CH_2}$ on the other. A carbonyl flanked by carbons is a ketone. (The compound is hexan-2-one.)

(b) $\mathrm{-COOH}$ at the end is the carboxyl group — a carboxylic acid, butanoic acid.

(c) $\mathrm{-CHO}$ at the end is the aldehyde group — pentanal.

(d) $\mathrm{-OH}$ at the end is the hydroxyl group — an alcohol, ethanol.

The pair to be careful with is (a) and (c). Both contain a $\mathrm{C{=}O}$, and in a condensed formula the difference is one letter: $\mathrm{CHO}$ carries a hydrogen and must therefore be at the end of the chain, making it an aldehyde, while $\mathrm{CO}$ carries no hydrogen and sits inside the chain, making it a ketone. Writing $\mathrm{CHO}$ in the middle of a formula is always an error.

Question 36

How is ethene prepared from ethanol? Give the reaction involved in it.

Solution. Ethene is made from ethanol by dehydration — removing a molecule of water. Heating ethanol with excess concentrated sulphuric acid at 443 K does it:

$$\mathrm{CH_3CH_2OH} \xrightarrow[\text{443 K}]{\text{conc. } \mathrm{H_2SO_4}} \mathrm{CH_2{=}CH_2} + \mathrm{H_2O}$$

The water removed is made of the $\mathrm{-OH}$ from one carbon and an $\mathrm{-H}$ from the next; the two carbons left short of a bond form a double bond between them, and a saturated alcohol becomes an unsaturated alkene.

Three conditions in that sentence each matter, which is why the question asks for them.

• Concentrated sulphuric acid, because dilute acid would supply the water the reaction is trying to remove.
• Excess, so that the acid is not diluted by the water produced as the reaction proceeds.
• 443 K — about $170^\circ\mathrm{C}$. At a lower temperature, around 413 K, the same reagents give ethoxyethane instead.

The sulphuric acid’s role here is as a dehydrating agent, which is Question 39.

Question 37

Intake of small quantity of methanol can be lethal. Comment.

Solution. Methanol is dangerous not in itself but because of what the body does to it.

In the liver, methanol is oxidised to methanal (formaldehyde). Methanal reacts very rapidly with the components of living cells — it makes the protoplasm coagulate, in much the way heat coagulates the white of an egg. Cells whose protoplasm has set cannot function.

It also attacks the optic nerve, and damage there is what causes the blindness associated with drinking methylated spirit. Even a small quantity can be enough, because the damage is done by a product that accumulates faster than it is cleared.

The contrast with ethanol is the point worth taking away. Ethanol is oxidised by the same liver enzymes but to ethanal, which is much less reactive and is broken down further to harmless products. The two alcohols differ by a single $\mathrm{CH_2}$ group, and one is an ingredient of drinks while the other is lethal — a reminder that in organic chemistry a small structural difference can make a total difference in behaviour.

This is also why industrial alcohol is denatured: methanol is deliberately added to ethanol to make it unfit to drink.

Question 38

A gas is evolved when ethanol reacts with sodium. Name the gas evolved and also write the balanced chemical equation of the reaction involved.

Solution. The gas is hydrogen.

Sodium attacks the hydrogen of the $\mathrm{-OH}$ group, exactly as it attacks the hydrogen of water. The ethanol loses that hydrogen and becomes the ethoxide ion:

$$2\mathrm{Na} + 2\mathrm{CH_3CH_2OH} \rightarrow 2\mathrm{CH_3CH_2ONa} + \mathrm{H_2}$$

Check the balance: two sodiums each side; two ethyl groups each side; on the left $2 \times 6 = 12$ hydrogens in the ethanol, on the right $2 \times 5 = 10$ in the ethoxide plus 2 in the hydrogen gas. ✓

The coefficient of 2 is where marks are lost. Hydrogen gas is diatomic, so two molecules of ethanol must each give up one hydrogen atom to make one molecule of $\mathrm{H_2}$ — and that forces the 2 on the sodium as well.

The product, sodium ethoxide, is the alkoxide of Question 17 in the multiple-choice band, and it is not sodium ethanoate.

Question 39

Ethene is formed when ethanol at 443 K is heated with excess of concentrated sulphuric acid. What is the role of sulphuric acid in this reaction? Write the balanced chemical equation of this reaction.

Solution. The sulphuric acid acts as a dehydrating agent — it removes a molecule of water from each molecule of ethanol.

$$\mathrm{CH_3CH_2OH} \xrightarrow[\text{443 K}]{\text{conc. } \mathrm{H_2SO_4}} \mathrm{CH_2{=}CH_2} + \mathrm{H_2O}$$

Concentrated sulphuric acid has a strong affinity for water, which is why it is used to dry gases and why it must never be diluted carelessly. Here that affinity is strong enough to pull an $\mathrm{-OH}$ and an $\mathrm{-H}$ out of the alcohol as water, leaving a double bond behind.

It is worth being precise about what “dehydrating agent” claims. The acid is not consumed in the overall equation and does not appear among the products, so in that sense it behaves like a catalyst — which is why it is written over the arrow. But it is called a dehydrating agent rather than a catalyst because it works by taking the water, and it has to be present in excess for exactly that reason: as water accumulates the acid is diluted and its dehydrating power falls.

Compare Question 46(b), where the same reagent in a different reaction is called a catalyst. The name given to $\mathrm{H_2SO_4}$ depends on the job it is doing.

Question 40

Carbon, Group (14) element in the Periodic Table, is known to form compounds with many elements. Write an example of a compound formed with

(a) chlorine (Group 17 of Periodic Table) (b) oxygen (Group 16 of Periodic Table)

Solution. Use the group numbers, which is why the question supplies them. The group fixes the number of outer electrons, and that fixes how many bonds each atom forms.

Carbon, Group 14, has four outer electrons and forms four bonds.

(a) Chlorine, Group 17, has seven outer electrons and needs one more, so it forms one bond. One carbon therefore holds four chlorine atoms: $\mathrm{CCl_4}$, carbon tetrachloride.

$$\mathrm{C} + 2\mathrm{Cl_2} \rightarrow \mathrm{CCl_4}$$

(b) Oxygen, Group 16, has six outer electrons and needs two, so it forms two bonds. Carbon’s four bonds are satisfied by two oxygen atoms with a double bond to each: $\mathrm{CO_2}$, carbon dioxide, $\mathrm{O{=}C{=}O}$.

$$\mathrm{C} + \mathrm{O_2} \rightarrow \mathrm{CO_2}$$

The general rule the question is teaching is worth stating: for the main-group elements, the number of bonds an atom forms is $18$ minus the group number for groups 14 to 17 — four for carbon, three for nitrogen, two for oxygen, one for the halogens.

Question 41

In electron dot structure, the valence shell electrons are represented by crosses or dots.

(a) The atomic number of chlorine is 17. Write its electronic configuration (b) Draw the electron dot structure of chlorine molecule.

Solution.

(a) Seventeen electrons are filled shell by shell — K takes 2, L takes 8, and the remaining 7 go into M:

$$\mathrm{K}\ 2, \quad \mathrm{L}\ 8, \quad \mathrm{M}\ 7 \qquad \text{that is, } 2, 8, 7$$

So chlorine has seven valence electrons, one short of an octet.

(b) Two chlorine atoms each need one more electron, and neither can take one from the other, since they are identical. So they share one pair — a single covalent bond — and each is left with three lone pairs of its own.

Count to check: each atom is surrounded by six electrons of its own plus the two it shares, which is eight — a complete octet. The shared pair is the single bond, so the structural formula is $\mathrm{Cl-Cl}$.

Question 42

Catenation is the ability of an atom to form bonds with other atoms of the same element. It is exhibited by both carbon and silicon. Compare the ability of catenation of the two elements. Give reasons.

Solution. Carbon catenates far better than silicon, and the difference is not small: carbon chains can run to thousands of atoms, while silicon chains rarely exceed seven or eight before becoming unstable.

The reason is atomic size. Carbon is a second-period element and is small, so the two nuclei in a $\mathrm{C-C}$ bond are close together and the shared electron pair is held tightly by both. Silicon sits one period lower and is appreciably larger, so in a $\mathrm{Si-Si}$ bond the shared pair is further from each nucleus and less strongly held.

The bond energies show the size of the effect:

Bond	Approximate bond energy
$\mathrm{C-C}$	$348\ \mathrm{kJ\,mol^{-1}}$
$\mathrm{Si-Si}$	$226\ \mathrm{kJ\,mol^{-1}}$

A $\mathrm{Si-Si}$ bond is barely two-thirds as strong as a $\mathrm{C-C}$ bond, so long silicon chains break apart readily — particularly in the presence of air or water.

This is ultimately why there are millions of carbon compounds and comparatively few of silicon, even though silicon is far more abundant in the earth’s crust. Catenation is what makes the enormous variety of organic chemistry possible, and only carbon does it well enough.

Question 43

Unsaturated hydrocarbons contain multiple bonds between the two C-atoms and show addition reactions. Give the test to distinguish ethane from ethene.

Solution. Two tests work, and the question’s own wording points at the second.

Burn them. Ethane is saturated and burns with a clean blue flame. Ethene is unsaturated, has a higher proportion of carbon, and does not burn completely in air — it gives a yellow, sooty flame with a good deal of black smoke. This is the test the printed answer gives, and it works for any saturated-versus-unsaturated pair.

Add bromine water, which is the sharper test and the one the question’s first sentence sets up. Bromine water is orange-brown. Ethene, being unsaturated, decolourises it immediately by an addition reaction across the double bond:

$$\mathrm{CH_2{=}CH_2} + \mathrm{Br_2} \rightarrow \mathrm{CH_2Br{-}CH_2Br}$$

Ethane has no double bond to add across, so the orange-brown colour persists.

The second test is the better one in practice because the result is unambiguous — a colour either goes or stays — whereas how sooty a flame looks is a matter of judgement. It also demonstrates the very property the question names: unsaturated compounds show addition reactions and saturated ones do not.

Question 44

Match the reactions given in Column (A) with the names given in column (B).

Column (A)	Column (B)
(a) $\mathrm{CH_3OH} + \mathrm{CH_3COOH} \xrightarrow{\mathrm{H^+}} \mathrm{CH_3COOCH_3} + \mathrm{H_2O}$	(i) Addition reaction
(b) $\mathrm{CH_2{=}CH_2} + \mathrm{H_2} \xrightarrow{\mathrm{Ni}} \mathrm{CH_3-CH_3}$	(ii) Substitution reaction
(c) $\mathrm{CH_4} + \mathrm{Cl_2} \xrightarrow{\text{Sunlight}} \mathrm{CH_3Cl} + \mathrm{HCl}$	(iii) Neutralisation reaction
(d) $\mathrm{CH_3COOH} + \mathrm{NaOH} \rightarrow \mathrm{CH_3COONa} + \mathrm{H_2O}$	(iv) Esterification reaction

Solution. Each reaction carries a signature that names it, and none needs to be worked out from first principles.

(a) An alcohol and a carboxylic acid give an ester and water, with an acid catalyst. That is esterification — (iv). The product formula $\mathrm{CH_3COOCH_3}$ is the giveaway.

(b) Two reactants become one product, with hydrogen adding across a double bond over a nickel catalyst and nothing leaving. That is an addition reaction — (i), specifically hydrogenation. Compare Question 49.

(c) A hydrogen of methane is replaced by chlorine, and the displaced hydrogen leaves as HCl. That is substitution — (ii). The sunlight over the arrow confirms it, from Question 13.

(d) An acid and a base give a salt and water. That is neutralisation — (iii).

Two quick structural tests separate (b) and (c), which are the pair most often confused. Count the products: an addition has one product, because nothing leaves; a substitution has two, because whatever was displaced has to go somewhere. And check the reagent: a metal catalyst means addition, sunlight means substitution.

Question 45

Write the structural formulae of all the isomers of hexane.

Solution. Hexane is $\mathrm{C_6H_{14}}$, and the question is to find every distinct way of joining six carbons. Work down from the longest chain, which guarantees none is missed and none is counted twice.

Six carbons in a row — one structure, n-hexane:

$\begin{array}{ccccccccccc} & \mathrm{H} & & \mathrm{H} & & \mathrm{H} & & \mathrm{H} & & \mathrm{H} & \\ & | & & | & & | & & | & & | & \\ \mathrm{H}- & \mathrm{C} & – & \mathrm{C} & – & \mathrm{C} & – & \mathrm{C} & – & \mathrm{C} & -\mathrm{CH_3} \\ & | & & | & & | & & | & & | & \\ & \mathrm{H} & & \mathrm{H} & & \mathrm{H} & & \mathrm{H} & & \mathrm{H} & \end{array}$

Five in a row with one methyl branch. The branch cannot go on carbon 1 or 5 — that would just extend the chain to six — so it goes on carbon 2 or carbon 3, giving two structures, 2-methylpentane and 3-methylpentane:

$\begin{array}{ccccccccc} \mathrm{CH_3} & – & \mathrm{CH} & – & \mathrm{CH_2} & – & \mathrm{CH_2} & – & \mathrm{CH_3} \\ & & | & & & & & & \\ & & \mathrm{CH_3} & & & & & & \end{array}$      $\begin{array}{ccccccccc} \mathrm{CH_3} & – & \mathrm{CH_2} & – & \mathrm{CH} & – & \mathrm{CH_2} & – & \mathrm{CH_3} \\ & & & & | & & & & \\ & & & & \mathrm{CH_3} & & & & \end{array}$

Four in a row with two methyl branches. Both on carbon 2, or one each on carbons 2 and 3 — two structures, 2,2-dimethylbutane and 2,3-dimethylbutane:

$\begin{array}{ccccccc} & & \mathrm{CH_3} & & & & \\ & & | & & & & \\ \mathrm{CH_3} & – & \mathrm{C} & – & \mathrm{CH_2} & – & \mathrm{CH_3} \\ & & | & & & & \\ & & \mathrm{CH_3} & & & & \end{array}$      $\begin{array}{ccccccc} \mathrm{CH_3} & – & \mathrm{CH} & – & \mathrm{CH} & – & \mathrm{CH_3} \\ & & | & & | & & \\ & & \mathrm{CH_3} & & \mathrm{CH_3} & & \end{array}$

Three in a row is not possible: three carbons offer only one middle position, and putting three methyl groups there would give a carbon with five bonds.

So hexane has exactly five structural isomers. Every one is $\mathrm{C_6H_{14}}$ — worth checking on at least one, say 2,2-dimethylbutane: three $\mathrm{CH_3}$ groups give 9 hydrogens, the quaternary carbon none, the $\mathrm{CH_2}$ two, and the last $\mathrm{CH_3}$ three, totalling $\mathrm{C_6H_{14}}$. ✓

Question 46

What is the role of metal or reagents written on arrows in the given chemical reactions?

(a) $\mathrm{(CH_3)_2C{=}C(CH_3)_2} + \mathrm{H_2} \xrightarrow{\mathrm{Ni}} \mathrm{(CH_3)_2CH{-}CH(CH_3)_2}$ (b) $\mathrm{CH_3COOH} + \mathrm{CH_3CH_2OH} \xrightarrow{\text{Conc. } \mathrm{H_2SO_4}} \mathrm{CH_3COOC_2H_5} + \mathrm{H_2O}$ (c) $\mathrm{CH_3CH_2OH} \xrightarrow[\text{Heat}]{\text{Alk. } \mathrm{KMnO_4}} \mathrm{CH_3COOH}$

Solution. Whatever is written over an arrow is a condition, not a reactant — and the question is what kind of condition each one is.

(a) Nickel is a catalyst. It speeds the addition of hydrogen across the double bond and is recovered unchanged at the end. The reaction is hydrogenation, and without the catalyst it is impractically slow.

(b) Concentrated sulphuric acid is a catalyst in this esterification. It supplies the $\mathrm{H^+}$ that speeds the reaction and is regenerated. (It also helps by absorbing the water formed, which pulls the equilibrium towards the ester — but its role as named here is catalytic.)

(c) Alkaline potassium permanganate is an oxidising agent. It is not a catalyst: it is consumed, being reduced as it oxidises the ethanol to ethanoic acid, and the purple colour disappears as the reaction proceeds.

The contrast is the point of the question. Two of these three reagents come out unchanged and one does not, and you can tell which by asking what happens to the reagent itself. A catalyst is recoverable; an oxidising agent has been reduced.

Note also that $\mathrm{H_2SO_4}$ appears here as a catalyst and in Question 39 as a dehydrating agent. The same substance gets a different name according to the job it is doing in that reaction.

Long Answer Questions

Question 47

A salt X is formed and a gas is evolved when ethanoic acid reacts with sodium hydrogencarbonate. Name the salt X and the gas evolved. Describe an activity and draw the diagram of the apparatus to prove that the evolved gas is the one which you have named. Also, write chemical equation of the reaction involved.

Solution. An acid on a hydrogencarbonate always gives a salt, water and carbon dioxide:

$$\mathrm{CH_3COOH} + \mathrm{NaHCO_3} \rightarrow \mathrm{CH_3COONa} + \mathrm{H_2O} + \mathrm{CO_2}$$

So X is sodium ethanoate (sodium acetate), $\mathrm{CH_3COONa}$, and the gas is carbon dioxide.

The activity. Put a spatula of sodium hydrogencarbonate in a test tube and add a few millilitres of dilute ethanoic acid. Brisk effervescence begins at once. Close the tube with a cork carrying a delivery tube, and lead the gas into a second test tube containing lime water.

What you see, and why it proves the gas. The lime water turns milky, because carbon dioxide reacts with the calcium hydroxide in it to give a fine suspension of insoluble calcium carbonate:

$$\mathrm{Ca(OH)_2} + \mathrm{CO_2} \rightarrow \mathrm{CaCO_3}\downarrow + \mathrm{H_2O}$$

Turning lime water milky is the characteristic test for carbon dioxide and no other common gas does it, so the observation identifies the gas. Passing excess gas turns the milkiness clear again as the carbonate dissolves to the soluble hydrogencarbonate, which is a useful confirmation.

Question 48

(a) What are hydrocarbons? Give examples. (b) Give the structural differences between saturated and unsaturated hydrocarbons with two examples each. (c) What is a functional group? Give examples of four different functional groups.

Solution.

(a) Hydrocarbons are compounds made of carbon and hydrogen only — no other element. Examples: methane $\mathrm{CH_4}$, ethane $\mathrm{C_2H_6}$, ethene $\mathrm{C_2H_4}$, ethyne $\mathrm{C_2H_2}$, benzene $\mathrm{C_6H_6}$.

The word only is what makes the definition work. Ethanol contains carbon and hydrogen but also oxygen, so it is not a hydrocarbon; chloromethane is not one either.

(b) The difference is in the carbon–carbon bonds.

Saturated hydrocarbons have only single bonds between carbon atoms. Each carbon holds as many hydrogens as it possibly can — it is “saturated” with hydrogen. Examples: methane, $\mathrm{CH_4}$, and ethane, $\mathrm{CH_3{-}CH_3}$.

Unsaturated hydrocarbons have at least one double or triple bond between carbons, and therefore carry fewer hydrogens than the same carbon skeleton could hold. Examples: ethene, $\mathrm{CH_2{=}CH_2}$, and ethyne, $\mathrm{CH \equiv CH}$.

The consequence worth adding, because it is what the distinction is for: saturated hydrocarbons are relatively unreactive and undergo substitution; unsaturated ones are more reactive and undergo addition.

(c) A functional group is an atom or group of atoms joined to a carbon chain in a particular way, which is responsible for the characteristic chemical properties of the compound. The hydrocarbon part is comparatively inert; the functional group is where the chemistry happens, and every member of a homologous series shares one.

Four examples: the hydroxyl group $\mathrm{-OH}$ (alcohols), the aldehyde group $\mathrm{-CHO}$, the ketone group $\mathrm{>C{=}O}$ and the carboxyl group $\mathrm{-COOH}$ (carboxylic acids).

Question 49

Name the reaction which is commonly used in the conversion of vegetable oils to fats. Explain the reaction involved in detail.

Solution. The reaction is hydrogenation, and it is a particular case of an addition reaction.

Why it works. Vegetable oils are unsaturated — their long carbon chains contain one or more carbon–carbon double bonds, which put kinks in the chains and stop the molecules packing closely, so the substance is a liquid at room temperature. Animal fats are saturated, their chains straight and closely packed, so they are solids.

The reaction. Hydrogen is added across each double bond in the presence of a nickel catalyst at about $473\ \mathrm{K}$:

$$\mathrm{-CH{=}CH-} + \mathrm{H_2} \xrightarrow{\text{Ni}} \mathrm{-CH_2{-}CH_2-}$$

or, written generally as the printed answer does,

$$\mathrm{R_2C{=}CR_2} + \mathrm{H_2} \xrightarrow{\text{Ni}} \mathrm{R_2CH{-}CHR_2}$$

Each double bond that takes up hydrogen becomes a single bond, the kink disappears, and the chain straightens. Once enough of them have been converted the oil sets to a solid — vanaspati, or vegetable ghee.

Nothing leaves the molecule, which is what makes this an addition rather than a substitution, and the nickel is a true catalyst: it speeds the reaction and is recovered unchanged, as Question 46(a) says.

A health note the textbook makes: because the product is saturated, hydrogenated vegetable oil is less healthy than the oil it came from, and saturated fats are implicated in heart disease.

Question 50

(a) Write the formula and draw electron dot structure of carbon tetrachloride. (b) What is saponification? Write the reaction involved in this process.

Solution.

(a) The formula is $\mathbf{CCl_4}$, as Question 40(a) established: carbon forms four bonds and each chlorine one.

For the electron dot structure, count first. Carbon brings 4 valence electrons and each chlorine 7, so there are $4 + 4 \times 7 = 32$ in all. Carbon shares one pair with each chlorine — four bonding pairs, 8 electrons — and the remaining 24 sit as three lone pairs on each chlorine.

Every atom now has an octet: carbon from its four shared pairs, each chlorine from three lone pairs plus one shared.

(b) Saponification is the hydrolysis of an ester by an alkali, giving the sodium salt of the carboxylic acid — which is the soap — and the alcohol the ester was made from:

$$\mathrm{CH_3COOC_2H_5} + \mathrm{NaOH} \rightarrow \mathrm{CH_3COONa} + \mathrm{C_2H_5OH}$$

It is the reverse of esterification, and the name means “soap-making”: in industry the ester is a natural fat or oil and the salt produced is soap.

A correction to the printed key. It defines saponification as converting esters into “salts of carboxylic acids and ethanol“. The alcohol released is whichever one the ester was built from, and it is ethanol only for an ethyl ester. The key’s own answer to Question 52(d) proves the point: saponifying methyl acetate there gives methanol, not ethanol. The definition should say an alcohol.

Question 51

Esters are sweet-smelling substances and are used in making perfumes. Suggest some activity and the reaction involved for the preparation of an ester with well labeled diagram.

Solution. The ester is made by esterification — warming a carboxylic acid with an alcohol in the presence of a little concentrated sulphuric acid.

The activity.

1. Take about 1 mL of ethanol (absolute alcohol) and 1 mL of glacial acetic acid in a test tube, and add a few drops of concentrated sulphuric acid.
2. Warm the tube in a water bath at about $60^\circ\mathrm{C}$ for at least five minutes.
3. Pour the mixture into a beaker containing 20–50 mL of water and smell it. The sweet, fruity smell is the ester.

Why a water bath rather than a flame. Ethanol is highly volatile and its vapours catch fire readily, so the tube must never be heated directly. A water bath cannot go above $100^\circ\mathrm{C}$ and heats the tube evenly, which is the whole reason for the apparatus.

The reaction:

$$\mathrm{CH_3COOH} + \mathrm{CH_3CH_2OH} \xrightarrow{\text{conc. } \mathrm{H_2SO_4}} \mathrm{CH_3COOCH_2CH_3} + \mathrm{H_2O}$$

$$\text{ethanoic acid} \qquad \text{ethanol} \qquad\qquad \text{ethyl ethanoate (ester)}$$

Pouring into water at the end serves two purposes: it dilutes the acid so the mixture is safe to smell, and the ester is only slightly soluble so it separates as a distinct sweet-smelling layer.

Question 52

A compound C (molecular formula, $\mathrm{C_2H_4O_2}$) reacts with Na-metal to form a compound R and evolves a gas which burns with a pop sound. Compound C on treatment with an alcohol A in presence of an acid forms a sweet smelling compound S (molecular formula, $\mathrm{C_3H_6O_2}$). On addition of NaOH to C, it also gives R and water. S on treatment with NaOH solution gives back R and A. Identify C, R, A, S and write down the reactions involved.

Solution. Take the clues in the order that fixes each compound.

C has the formula $\mathrm{C_2H_4O_2}$ and reacts with sodium to give a gas that burns with a pop — hydrogen. Both point to a carboxylic acid, and the two-carbon one is ethanoic acid, $\mathrm{CH_3COOH}$. ✓

R is what C gives with sodium, and also what C gives with sodium hydroxide: the sodium salt, sodium ethanoate, $\mathrm{CH_3COONa}$.

S is the sweet-smelling ester, $\mathrm{C_3H_6O_2}$. Here is the step that identifies A. The ester has three carbons and the acid supplied two, so the alcohol supplied one — and the one-carbon alcohol is methanol, $\mathrm{CH_3OH}$. So $\mathrm{A} = \mathrm{CH_3OH}$ and $\mathrm{S} = \mathrm{CH_3COOCH_3}$, methyl ethanoate.

Counting carbons in the ester is the whole trick of this question; without it there is no way to tell which alcohol was used.

The four reactions:

$$\text{(a)}\quad 2\mathrm{CH_3COOH} + 2\mathrm{Na} \rightarrow 2\mathrm{CH_3COONa} + \mathrm{H_2}$$

$$\text{(b)}\quad \mathrm{CH_3COOH} + \mathrm{CH_3OH} \xrightarrow{\text{conc. } \mathrm{H_2SO_4}} \mathrm{CH_3COOCH_3} + \mathrm{H_2O}$$

$$\text{(c)}\quad \mathrm{CH_3COOH} + \mathrm{NaOH} \rightarrow \mathrm{CH_3COONa} + \mathrm{H_2O}$$

$$\text{(d)}\quad \mathrm{CH_3COOCH_3} + \mathrm{NaOH} \rightarrow \mathrm{CH_3COONa} + \mathrm{CH_3OH}$$

Reaction (d) is the saponification of Question 50(b) — and note that it releases methanol, which is exactly why that question’s definition should say “an alcohol” rather than “ethanol”.

Question 53

Look at Figure 4.1 and answer the following questions

(a) What change would you observe in the calcium hydroxide solution taken in tube B? (b) Write the reaction involved in test tubes A and B respectively. (c) If ethanol is given instead of ethanoic acid, would you expect the same change? (d) How can a solution of lime water be prepared in the laboratory?

Solution.

(a) The calcium hydroxide solution turns milky. The milkiness is a fine suspension of insoluble calcium carbonate, and it is the characteristic test for carbon dioxide.

(b) In tube A, the acid attacks the carbonate and carbon dioxide is released:

$$2\mathrm{CH_3COOH} + \mathrm{Na_2CO_3} \rightarrow 2\mathrm{CH_3COONa} + \mathrm{H_2O} + \mathrm{CO_2}$$

Note the coefficient of 2 on the acid — sodium carbonate has two sodiums to replace.

In tube B, the gas reacts with the lime water:

$$\mathrm{Ca(OH)_2} + \mathrm{CO_2} \rightarrow \mathrm{CaCO_3}\downarrow + \mathrm{H_2O}$$

If gas continues to be passed, the milkiness disappears as the carbonate dissolves to the soluble hydrogencarbonate:

$$\mathrm{CaCO_3} + \mathrm{H_2O} + \mathrm{CO_2} \rightarrow \mathrm{Ca(HCO_3)_2}$$

(c) No. Ethanol is an alcohol, not an acid — it is neutral to litmus and does not release carbon dioxide from a carbonate:

$$\mathrm{C_2H_5OH} + \mathrm{Na_2CO_3} \rightarrow \text{no reaction}$$

With no gas evolved, the lime water in tube B stays clear. This is in fact the standard laboratory test for telling an alcohol from a carboxylic acid, and it is a good deal more reliable than smell.

(d) Lime water is a solution of calcium hydroxide. Dissolve calcium oxide (quicklime) in water, stir, and allow the mixture to stand; then decant the clear supernatant liquid, which is the lime water. The undissolved solid is left behind, because calcium hydroxide is only slightly soluble.

$$\mathrm{CaO} + \mathrm{H_2O} \rightarrow \mathrm{Ca(OH)_2}$$

Question 54

How would you bring about the following conversions? Name the process and write the reaction involved.

(a) ethanol to ethene. (b) propanol to propanoic acid.

Solution. Each conversion is one of the named reactions from the table at the top of this page, and naming it correctly is most of the answer.

(a) Ethanol to ethene is a dehydration. Heat the ethanol with excess concentrated sulphuric acid at 443 K:

$$\mathrm{CH_3CH_2OH} \xrightarrow[\text{443 K}]{\text{conc. } \mathrm{H_2SO_4}} \mathrm{CH_2{=}CH_2} + \mathrm{H_2O}$$

A molecule of water is removed and the two carbons left short of a bond form a double bond. A saturated alcohol becomes an unsaturated alkene. This is Questions 36 and 39 again.

(b) Propanol to propanoic acid is an oxidation. Heat the propanol with an oxidising agent — alkaline potassium permanganate, or acidified potassium dichromate:

$$\mathrm{CH_3CH_2CH_2OH} \xrightarrow[\text{Heat}]{\text{Alkaline } \mathrm{KMnO_4}} \mathrm{CH_3CH_2COOH}$$

The terminal $\mathrm{-CH_2OH}$ gains an oxygen and loses two hydrogens to become $\mathrm{-COOH}$.

The contrast between the two is worth drawing out, because both start from an alcohol. Dehydration removes water and shortens nothing — the carbon count is unchanged and a double bond appears. Oxidation adds oxygen — the carbon count is again unchanged, but the functional group becomes a carboxyl. Which one you get depends entirely on the reagent, not on the alcohol.

Question 55

Draw the possible isomers of the compound with molecular formula $\mathrm{C_3H_6O}$ and also give their electron dot structures.

Solution. Three carbons, six hydrogens and one oxygen. Saturated three-carbon compounds with one oxygen would be $\mathrm{C_3H_8O}$, so $\mathrm{C_3H_6O}$ is two hydrogens short — there must be one double bond, and with an oxygen present the natural place for it is a carbonyl, $\mathrm{C{=}O}$.

A carbonyl can sit in two places on a three-carbon chain, and that gives the two isomers.

At the end of the chain — an aldehyde, propanal, $\mathrm{CH_3{-}CH_2{-}CHO}$:

$\begin{array}{ccccccc} & \mathrm{H} & & \mathrm{H} & & \mathrm{O} & \\ & | & & | & & \| & \\ \mathrm{H}- & \mathrm{C} & – & \mathrm{C} & – & \mathrm{C} & -\mathrm{H} \\ & | & & | & & & \\ & \mathrm{H} & & \mathrm{H} & & & \end{array}$

In the middle of the chain — a ketone, propanone (acetone), $\mathrm{CH_3{-}CO{-}CH_3}$:

$\begin{array}{ccccccc} & \mathrm{H} & & \mathrm{O} & & \mathrm{H} & \\ & | & & \| & & | & \\ \mathrm{H}- & \mathrm{C} & – & \mathrm{C} & – & \mathrm{C} & -\mathrm{H} \\ & | & & & & | & \\ & \mathrm{H} & & & & \mathrm{H} & \end{array}$

Electron dot structures. Both have $3 \times 4 + 6 \times 1 + 6 = 24$ valence electrons. In each, every $\mathrm{C-H}$ and $\mathrm{C-C}$ bond is one shared pair, the $\mathrm{C{=}O}$ is two shared pairs, and the oxygen carries two lone pairs of its own. Writing them out:

Propanal — $\mathrm{H\!:\!\overset{H}{\underset{H}{C}}\!:\!\overset{H}{\underset{H}{C}}\!:\!\overset{H}{C}\!::\!\ddot{O}}$

Propanone — $\mathrm{H\!:\!\overset{H}{\underset{H}{C}}\!:\!C(::\!\ddot{O})\!:\!\overset{H}{\underset{H}{C}}\!:\!H}$

Count the oxygen in either: two shared pairs in the double bond plus two lone pairs is eight electrons — a complete octet. Each carbon likewise has four shared pairs and no lone pairs.

Note that the carbonyl carbon in propanal carries a hydrogen while the one in propanone carries two carbons. That single difference is the whole distinction between an aldehyde and a ketone, and it is what Questions 32 and 35 test.

Question 56

Explain the given reactions with the examples

(a) Hydrogenation reaction   (b) Oxidation reaction   (c) Substitution reaction   (d) Saponification reaction   (e) Combustion reaction

Solution.

(a) Hydrogenation. An unsaturated hydrocarbon adds hydrogen across its double or triple bond in the presence of a nickel catalyst, becoming saturated. It is a particular kind of addition reaction, and it is how vegetable oil is turned into vanaspati.

$$\mathrm{R_2C{=}CR_2} + \mathrm{H_2} \xrightarrow{\text{Ni}} \mathrm{R_2CH{-}CHR_2}$$

(b) Oxidation. An alcohol gains oxygen and loses hydrogen to become a carboxylic acid, using an oxidising agent such as alkaline potassium permanganate.

$$\mathrm{CH_3CH_2OH} \xrightarrow[\text{Heat}]{\text{Alkaline } \mathrm{KMnO_4}} \mathrm{CH_3COOH}$$

(c) Substitution. An atom of a saturated hydrocarbon is replaced by another atom, with the displaced one leaving as part of a second product. Chlorine substitutes for hydrogen in methane in the presence of sunlight.

$$\mathrm{CH_4} + \mathrm{Cl_2} \xrightarrow{\text{Sunlight}} \mathrm{CH_3Cl} + \mathrm{HCl}$$

(d) Saponification. An ester is hydrolysed by an alkali to the sodium salt of the acid — the soap — and the alcohol.

$$\mathrm{CH_3COOC_2H_5} + \mathrm{NaOH} \rightarrow \mathrm{CH_3COONa} + \mathrm{C_2H_5OH}$$

(e) Combustion. A carbon compound burns in oxygen, releasing a large amount of heat and light and giving carbon dioxide and water.

$$\mathrm{CH_4} + 2\mathrm{O_2} \rightarrow \mathrm{CO_2} + 2\mathrm{H_2O} + \text{Heat and light}$$

Two of these five are worth distinguishing carefully, because they are the pair most often confused. Addition — including hydrogenation — gives one product and needs an unsaturated starting material; substitution gives two products and works on a saturated one. Counting the products tells you which you are looking at.

Question 57

An organic compound A on heating with concentrated $\mathrm{H_2SO_4}$ forms a compound B which on addition of one mole of hydrogen in presence of Ni forms a compound C. One mole of compound C on combustion forms two moles of $\mathrm{CO_2}$ and 3 moles of $\mathrm{H_2O}$. Identify the compounds A, B and C and write the chemical equations of the reactions involved.

Solution. The only quantitative clue is at the end of the chain of reactions, so work backwards from C.

Finding C. One mole of C gives 2 moles of $\mathrm{CO_2}$ and 3 moles of $\mathrm{H_2O}$ on combustion. Every carbon in C ends up in one $\mathrm{CO_2}$, so C has 2 carbons; every two hydrogens end up in one $\mathrm{H_2O}$, so C has $3 \times 2 = $ 6 hydrogens. That gives $\mathrm{C_2H_6}$ — ethane.

Finding B. C was formed by adding one mole of hydrogen to B, so B has two hydrogens fewer: $\mathrm{C_2H_4}$ — ethene. That is consistent with the reagent, since hydrogen adds over a nickel catalyst only across a double bond.

Finding A. B was formed by heating A with concentrated sulphuric acid, which is dehydration — so A is an alcohol with one water more than ethene: $\mathrm{C_2H_4} + \mathrm{H_2O} = \mathrm{C_2H_6O}$ — ethanol, $\mathrm{C_2H_5OH}$.

The three equations:

$$\mathrm{C_2H_5OH} \xrightarrow[\text{443 K}]{\text{conc. } \mathrm{H_2SO_4}} \mathrm{C_2H_4} + \mathrm{H_2O}$$

$$\mathrm{C_2H_4} + \mathrm{H_2} \xrightarrow{\text{Ni}} \mathrm{C_2H_6}$$

$$2\mathrm{C_2H_6} + 7\mathrm{O_2} \rightarrow 4\mathrm{CO_2} + 6\mathrm{H_2O} + \text{Heat and light}$$

Check the last one against the question: dividing through by 2 gives one mole of ethane producing two moles of $\mathrm{CO_2}$ and three of $\mathrm{H_2O}$, exactly as stated. ✓

The method is the one to carry away: combustion products count the atoms. One $\mathrm{CO_2}$ per carbon and one $\mathrm{H_2O}$ per two hydrogens turns a combustion result straight into a molecular formula.

Common mistakes

• Forgetting the coefficient 2 in Question 38. Hydrogen gas is diatomic, so two molecules of ethanol are needed: $2\mathrm{Na} + 2\mathrm{C_2H_5OH} \rightarrow 2\mathrm{C_2H_5ONa} + \mathrm{H_2}$.
• Missing a carbon when naming, in Questions 31 and 55. The carbon of a $\mathrm{-COOH}$ or $\mathrm{-CHO}$ group is part of the chain. Question 31(c) has seven carbons, not six.
• Confusing $\mathrm{-CHO}$ with $\mathrm{CO}$, in Question 35. $\mathrm{CHO}$ carries a hydrogen and must be at the end of a chain (aldehyde); $\mathrm{CO}$ carries none and sits inside it (ketone). The same $\mathrm{C{=}O}$ gives two different classes depending on position.
• Assuming saponification always releases ethanol, as the printed key’s Question 50(b) does. It releases whichever alcohol the ester was made from — the key’s own Question 52(d) produces methanol.
• Calling alkaline $\mathrm{KMnO_4}$ a catalyst, in Question 46(c). It is consumed; the purple colour fades as it is reduced. Only the Ni and the $\mathrm{H_2SO_4}$ in that question are catalysts.
• Missing isomers in Question 45. Work down from the longest chain — six in a row, then five with one branch, then four with two. Hexane has exactly five isomers; stopping at three or four is the usual result of guessing rather than enumerating.
• Heating the ester mixture on a direct flame, in Question 51. Ethanol vapour catches fire. A water bath is not a detail of the method; it is the safety requirement the question expects you to state.
• Writing one coefficient too few in Question 53(b). Sodium carbonate has two sodium atoms, so it needs two molecules of ethanoic acid — unlike sodium hydrogencarbonate in Question 47, which needs one.

Practise next

• Exemplar Multiple Choice Questions Q1–15 — bonding, tetravalency and isomerism, which Questions 45, 50 and 55 above rest on.
• Exemplar Multiple Choice Questions Q16–29 — nomenclature and homologous series in recognition form, the fastest drill for Questions 31, 32 and 35.
• NCERT textbook Chapter 4 — Carbon and its Compounds for the end-of-chapter exercise on ethanol, ethanoic acid and soaps.
• Textbook activities for the experiments behind Questions 47, 51 and 53 — the carbonate test, the ester preparation and the soap experiments are all performed there.