NCERT Exemplar Solutions Class 10 Science Carbon and its Compounds — Multiple Choice Questions Q16–29

NCERT Exemplar Problems, Class 10 Science — Carbon and its Compounds, Multiple Choice Questions 16 to 29. All 14 questions solved.

This is the second half of the unit’s multiple-choice band; Questions 1 to 15 cover bonding, isomerism and the addition and substitution reactions. From here the questions move to naming compounds, recognising homologous series, and the three reactions of ethanol and ethanoic acid that the chapter keeps returning to.

Two tools do nearly all the work.

• The name is the structure. A systematic name has a stem giving the number of carbons (meth- 1, eth- 2, prop- 3, but- 4, pent- 5) and a suffix giving the functional group (-ane, -ene, -yne, -ol, -al, -one, -oic acid). Between them they determine the formula exactly, which settles Questions 25, 26 and 29 without any chemistry.
• A homologous series is a family of compounds with the same functional group whose members differ by a $\mathrm{CH_2}$ unit. Alkanes are $\mathrm{C_nH_{2n+2}}$, alkenes $\mathrm{C_nH_{2n}}$, alkynes $\mathrm{C_nH_{2n-2}}$ — and Question 25 is decided by checking a formula against those.

Key insight. When four structures are offered and only one can be right, count something. Count hydrogens per carbon, count double bonds, count the carbons in the chain. Questions 16, 18, 23 and 24 all look like exercises in recognition and are all in fact exercises in counting.

Choose the correct answer from the given four options in the following questions:

Question 16

Structural formula of benzene is

Solution. Benzene is $\mathrm{C_6H_6}$: a ring of six carbons with one hydrogen on each, and three double bonds alternating with three single bonds around the ring. Two counts settle the question.

Count the hydrogens per carbon. Every carbon in benzene carries exactly one. Structure (b) gives every carbon two, which makes it $\mathrm{C_6H_{12}}$ — cyclohexane, a different compound. Structures (a) and (d) each have one carbon carrying two hydrogens while the others carry one, so their rings are not uniform.

Count the double bonds and check they alternate. Benzene’s three double bonds sit on alternate edges, so no carbon carries two of them and no two adjacent edges are both double. Structure (a) fails this badly: its topmost carbon is joined by a double bond on both sides, which uses all four of its bonds and leaves it with no hydrogen, while a neighbouring carbon has to carry two hydrogens to compensate. Structure (d) has only one double bond in the ring, leaving most of the carbons short.

Only structure (c) has one hydrogen on every carbon and three double bonds in the alternating pattern. Check its tetravalency: each carbon has one bond to a hydrogen, one single bond to a ring neighbour and one double bond to the other — $1 + 1 + 2 = 4$.

Note that (a) also comes to $\mathrm{C_6H_6}$ when you total the hydrogens, so the molecular formula alone does not identify benzene. The arrangement is what the question is about.

Question 17

Ethanol reacts with sodium and forms two products. These are

Solution. Two decisions, and the options cover all four combinations, so both have to be got right.

The gas. Sodium displaces the hydrogen of the $\mathrm{-OH}$ group, exactly as it does from water. The gas is hydrogen, which rules out (b) and (d). Oxygen would have to come from somewhere, and nothing here supplies it.

The salt. Ethanol is $\mathrm{C_2H_5OH}$ and losing that hydrogen leaves the ethoxide ion, $\mathrm{C_2H_5O^-}$:

$$2\mathrm{Na} + 2\mathrm{C_2H_5OH} \rightarrow 2\mathrm{C_2H_5ONa} + \mathrm{H_2}$$

So the product is sodium ethoxide, option (c).

Sodium ethanoate is the trap, and it is a good one. It is $\mathrm{CH_3COONa}$, the salt of ethanoic acid, and it is what you get from sodium and ethanoic acid — a different reactant. Ethanol is an alcohol; its sodium salt is an alkoxide, not a carboxylate.

The parallel with water is worth holding: $2\mathrm{Na} + 2\mathrm{H_2O} \rightarrow 2\mathrm{NaOH} + \mathrm{H_2}$ has exactly the same shape, with the ethyl group in place of one hydrogen.

Question 18

The correct structural formula of butanoic acid is

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

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

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

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

Solution. The name gives three requirements, and each wrong option fails exactly one of them.

But- means four carbon atoms. Structure (b) has five, so it is pentanoic acid.

-an- means the carbon chain is saturated, all single bonds. Structure (a) has a carbon–carbon double bond, so it is an unsaturated acid — but-2-enoic acid, not butanoic acid.

-oic acid means the functional group is carboxyl, $\mathrm{-COOH}$, which is a carbon carrying a doubly bonded oxygen and a hydroxyl group. Structure (c) has the $\mathrm{-OH}$ but no $\mathrm{C{=}O}$ — its fourth carbon carries two hydrogens as well as the $\mathrm{-OH}$ — so it is butanol, an alcohol.

Structure (d) meets all three: four carbons, all single bonds, and a terminal carbon with a double-bonded oxygen and a hydroxyl group. Its condensed formula is $\mathrm{CH_3CH_2CH_2COOH}$, $\mathrm{C_4H_8O_2}$.

The tetravalency check confirms it. In (d) the carboxyl carbon has one bond to the chain, two to the oxygen of the $\mathrm{C{=}O}$ and one to the $\mathrm{-OH}$ — four in all, with no hydrogen of its own, which is why the row of hydrogens below stops at three.

Question 19

Vinegar is a solution of

Solution. Two things have to be right: the solvent and the concentration.

The solvent is water. Vinegar is used in food, and a solution in alcohol would be something quite different — and would not be safe to pour on chips. That removes (a) and (b).

The concentration is 5 to 8 per cent. Between (c) and (d), consider what a 50 per cent solution of acetic acid would be like: it would be strongly corrosive, would burn the mouth, and could not be an ingredient of food. Household vinegar is dilute, around 5 to 8 per cent, which is why it is merely sharp to taste.

It is worth separating two ideas that this question quietly relies on. Acetic acid is a weak acid — it ionises only partly, which is the property of the substance from Question 41 of the previous unit. And vinegar is dilute — a property of the solution. Vinegar is safe to eat for both reasons together.

One more fact worth knowing: pure ethanoic acid is called glacial acetic acid, because it freezes to an ice-like solid just below room temperature, at about $17^\circ\mathrm{C}$.

Question 20

Mineral acids are stronger acids than carboxylic acids because

(i) mineral acids are completely ionised (ii) carboxylic acids are completely ionised (iii) mineral acids are partially ionised (iv) carboxylic acids are partially ionised

Solution. Acid strength is the extent of ionisation in water — how completely the acid gives up its hydrogen ions. That definition answers the question directly.

Mineral acids such as hydrochloric, sulphuric and nitric acid are strong, so they ionise completely:

$$\mathrm{HCl} + \mathrm{H_2O} \rightarrow \mathrm{H_3O^+} + \mathrm{Cl^-}$$

Statement (i) is true and (iii) false.

Carboxylic acids such as ethanoic acid are weak, so they ionise only partially — most of the molecules stay intact, in equilibrium with a small proportion of ions:

$$\mathrm{CH_3COOH} + \mathrm{H_2O} \rightleftharpoons \mathrm{H_3O^+} + \mathrm{CH_3COO^-}$$

Statement (iv) is true and (ii) false.

So (i) and (iv), option (a). Notice that the four statements are two contradictory pairs, so the correct option must take exactly one from each pair — which eliminates (c) and (d) before any chemistry, since each takes both statements about one acid and none about the other.

The visible consequence is that equal concentrations of hydrochloric acid and acetic acid give solutions of different pH, and the acetic acid fizzes more slowly with a carbonate.

Question 21

Carbon forms four covalent bonds by sharing its four valence electrons with four univalent atoms, e.g. hydrogen. After the formation of four bonds, carbon attains the electronic configuration of

Solution. Carbon is $2, 4$ — two electrons in the first shell and four in the second. Forming four covalent bonds gives it a share in four more electrons, so its outer shell now holds $4 + 4 = 8$.

Its configuration becomes $2, 8$, and the noble gas with that configuration is neon.

Helium, option (a), is $2$ — a complete first shell with two electrons. That is the configuration hydrogen attains in $\mathrm{CH_4}$, since hydrogen’s single shell is full at two. So helium is the right answer to a question about the hydrogen in methane, and the wrong answer to this one about the carbon.

Argon is $2, 8, 8$ and krypton $2, 8, 18, 8$; both have far more electrons than carbon could reach by sharing.

The general rule is that an atom completing its octet takes the configuration of the nearest noble gas, and for the second-period elements — carbon, nitrogen, oxygen, fluorine — that is always neon.

Question 22

The correct electron dot structure of a water molecule is

Solution. Work out what the structure must contain before looking at the options.

Oxygen has six valence electrons and each hydrogen has one, so a water molecule has $6 + 1 + 1 = 8$ valence electrons in all. Oxygen needs two more to reach an octet, so it forms two single bonds, one to each hydrogen. That uses four electrons — two bonding pairs — and leaves oxygen with two lone pairs of its own.

So the correct structure shows eight electrons around the oxygen: two pairs between it and the hydrogens, and two pairs standing alone. That is option (c).

Now the others, by counting:

• (a) shows a single electron in each bonding position rather than a pair, giving six electrons in all. A covalent bond is a shared pair, so a single dot between two atoms is not a bond.
• (b) has a pair on one side and a single electron on the other — seven electrons, which is not a possible total, and one bond drawn as a pair and the other not.
• (d) has the two bonding pairs but no lone pairs at all, giving four electrons around the oxygen instead of eight. Oxygen would be four electrons short of its octet.

The electron count is the quickest test here: only (c) totals eight.

Question 23

Which of the following is not a straight chain hydrocarbon?

(a) $\begin{array}{ccccccccc} \mathrm{H_3C} & – & \mathrm{CH_2} & – & \mathrm{CH_2} & – & \mathrm{CH_2} & – & \mathrm{CH_2} \\ & & & & & & & & | \\ & & & & & & & & \mathrm{CH_3} \end{array}$

(b) $\mathrm{H_3C-CH_2-CH_2-CH_2-CH_2-CH_3}$

(c) $\begin{array}{ccccccccc} & & & & & & \mathrm{CH_3} & & \\ & & & & & & | & & \\ \mathrm{H_2C} & – & \mathrm{H_2C} & – & \mathrm{H_2C} & – & \mathrm{CH_2} & & \\ & & & & & & | & & \\ & & & & & & \mathrm{CH_3} & & \end{array}$

(d) $\begin{array}{ccccccc} \mathrm{CH_3} & \diagdown & & & & & \\ & & \mathrm{CH} & – & \mathrm{CH_2} & – & \mathrm{CH_2-CH_3} \\ \mathrm{H_3C} & \diagup & & & & & \end{array}$

Solution. A chain is branched only if some carbon in it is bonded to three or more other carbons. Applying that test is not the same as seeing whether the drawing turns a corner, and this question is built entirely on the difference.

(a) looks branched because the last $\mathrm{CH_3}$ is printed on a second line. Trace the bonds: the first carbon joins one other, then a run of carbons each joined to two, and the final $\mathrm{CH_3}$ joins one. No carbon has three carbon neighbours. It is $\mathrm{CH_3(CH_2)_4CH_3}$ — n-hexane, a straight chain drawn with a bend in it.

(b) is the same molecule written out in a line. Straight chain.

(c) is the same trick as (a), with a $\mathrm{CH_3}$ displaced above the start of the chain and another below its end. Trace it again and it is six carbons in a row — n-hexane once more.

(d) is genuinely different. The carbon written $\mathrm{CH}$ is bonded to $\mathrm{CH_3}$ above, to $\mathrm{H_3C}$ below, and to the $\mathrm{CH_2}$ of the chain — three carbon neighbours, so it is a branch point. The compound is 2-methylpentane, and it is the only branched structure among the four.

The lesson is worth stating: the shape of the drawing carries no information. The same molecule can be drawn straight, bent, or with pieces on separate lines; only the connections count. Trace the bonds rather than reading the layout.

Question 24

Which among the following are unsaturated hydrocarbons?

(i) $\mathrm{H_3C-CH_2-CH_2-CH_3}$

(ii) $\mathrm{H_3C-C \equiv C-CH_3}$

(iii) $\begin{array}{ccccc} \mathrm{H_3C} & – & \mathrm{CH} & – & \mathrm{CH_3} \\ & & | & & \\ & & \mathrm{CH_3} & & \end{array}$

(iv) $\begin{array}{ccccc} \mathrm{H_3C} & – & \mathrm{C} & = & \mathrm{CH_2} \\ & & | & & \\ & & \mathrm{CH_3} & & \end{array}$

Solution. Unsaturated means at least one carbon–carbon double or triple bond. Nothing else matters — not the length of the chain, not whether it is branched.

• (i) is a straight chain with only single bonds — n-butane. Saturated.
• (ii) has a triple bond between the two middle carbons — but-2-yne. Unsaturated.
• (iii) is branched, with a methyl group on the middle carbon, but every bond is single — 2-methylpropane. Saturated.
• (iv) has a double bond and a methyl branch — 2-methylprop-1-ene. Unsaturated.

So (ii) and (iv), option (c).

Structure (iii) is the distractor and is worth dwelling on: branching is not unsaturation. A branched molecule is still saturated if all its bonds are single, and (iii) is in fact the isobutane of Question 5 in the first half of this band, where it was one of the two valid isomers of butane.

Compare (iii) and (iv) directly. They differ by a single bond becoming double and two hydrogens being lost, which changes $\mathrm{C_4H_{10}}$ into $\mathrm{C_4H_8}$ — and that is the whole distinction between the two answers.

Question 25

Which of the following does not belong to the same homologous series?

Solution. Members of a homologous series share a general formula and differ from their neighbours by $\mathrm{CH_2}$. Test each against the alkane formula $\mathrm{C_nH_{2n+2}}$:

Compound	$n$	$2n + 2$	Hydrogens present
$\mathrm{CH_4}$	1	4	4 ✓
$\mathrm{C_2H_6}$	2	6	6 ✓
$\mathrm{C_3H_8}$	3	8	8 ✓
$\mathrm{C_4H_8}$	4	10	8 ✗

The first three are alkanes, and each differs from the one before by exactly $\mathrm{CH_2}$: $\mathrm{CH_4} \to \mathrm{C_2H_6} \to \mathrm{C_3H_8}$. The natural next member would be $\mathrm{C_4H_{10}}$, butane.

$\mathrm{C_4H_8}$ is two hydrogens short of that. It fits $\mathrm{C_nH_{2n}}$ instead, which is the alkene series — it is butene, with a carbon–carbon double bond. Different general formula, different functional group, different series.

There is a quicker route: read the sequence $4, 6, 8, 8$ of hydrogen counts. The first three rise by two each time and the fourth does not rise at all, which breaks the pattern without any formula being needed.

Question 26

The name of the compound $\mathrm{CH_3 – CH_2 – CHO}$ is

Solution. Build the name in two steps: count the carbons for the stem, then identify the functional group for the suffix.

Count the carbons. $\mathrm{CH_3}$, $\mathrm{CH_2}$ and $\mathrm{CHO}$ — the $\mathrm{C}$ of the $\mathrm{CHO}$ counts, so there are three. The stem is prop-, which eliminates (c) and (d) at once, since both are two-carbon names.

Identify the group. $\mathrm{-CHO}$ is the aldehyde group, a carbon with a doubly bonded oxygen and a hydrogen, and the suffix for an aldehyde is -al. So the name is propanal.

Option (b), propanone, has the right stem and the wrong group: -one is a ketone, in which the carbonyl sits in the middle of the chain rather than at the end. Propanone is $\mathrm{CH_3-CO-CH_3}$, acetone.

The distinction is one of position, and it is easy to see once stated: an aldehyde’s carbonyl carbon carries a hydrogen because it is at the end of the chain, and a ketone’s carries two carbons because it is not. Writing the group as $\mathrm{-CHO}$ rather than $\mathrm{-CO-}$ is the signal.

Question 27

The heteroatoms present in $\mathrm{CH_3 – CH_2 – O – CH_2 – CH_2Cl}$ are

(i) oxygen    (ii) carbon    (iii) hydrogen    (iv) chlorine

Solution. A heteroatom is any atom in an organic compound other than carbon and hydrogen. The prefix hetero- means “other”, and the definition is as simple as that.

Carbon and hydrogen make up the hydrocarbon skeleton, so by definition they are never heteroatoms — which removes statements (ii) and (iii), and with them options (a), (b) and (c).

The compound contains four elements: carbon, hydrogen, oxygen and chlorine. The last two are the heteroatoms, so the answer is (i) and (iv), option (d).

The reason the idea matters is that the heteroatom is where the chemistry happens. Here the oxygen sits between two carbon chains as an ether linkage, and the chlorine at the end makes that carbon a haloalkane site. Both are functional groups; the hydrocarbon parts are comparatively inert.

Notice how quickly the question falls once the definition is stated exactly. Two of the four statements name the two elements that are excluded by definition, and only one option avoids both.

Question 28

Which of the following represents saponification reaction?

Solution. Saponification is the reaction that makes soap: an ester heated with an alkali splits into the sodium salt of the acid — which is the soap — and the alcohol. The name comes from sapo, the Latin for soap.

So the identifying marks are an ester on the left and a carboxylate salt plus an alcohol on the right, which is exactly option (d): ethyl ethanoate with sodium hydroxide gives sodium ethanoate and ethanol.

The other three are all real reactions of this chapter, each worth naming so the distinction is clear.

(b) is esterification, and it is the reverse of saponification — an acid and an alcohol combining, with concentrated sulphuric acid as catalyst, to give the ester and water. It is the option most often chosen by mistake, because it also involves an ester; but here the ester is being made, not broken.

(c) is the reaction of a carboxylic acid with sodium, giving the salt and hydrogen. It produces the same salt as (d) but by a different route, and the hydrogen gives it away.

(a) is decarboxylation — sodium ethanoate heated with soda lime, losing carbon dioxide to give methane.

The test that works fastest: look for an ester as a reactant and an alcohol as a product. Only (d) has both.

Question 29

The first member of alkyne homologous series is

Solution. An alkyne contains a carbon–carbon triple bond. A bond between carbons needs two carbon atoms, so no alkyne can have fewer than two — and that single observation answers the question.

The two-carbon alkyne is ethyne, $\mathrm{C_2H_2}$, and it is therefore the first member of the series.

Option (d), methane, has only one carbon, so it could not have a carbon–carbon bond of any kind; it is in fact the first member of the alkane series. Option (b), ethene, has the right number of carbons but a double bond, so it is the first member of the alkene series. Option (c), propyne, is an alkyne but the second member, with three carbons.

Putting the three first members side by side is the cleanest way to hold this:

Series	General formula	First member
Alkane	$\mathrm{C_nH_{2n+2}}$	methane, $\mathrm{CH_4}$
Alkene	$\mathrm{C_nH_{2n}}$	ethene, $\mathrm{C_2H_4}$
Alkyne	$\mathrm{C_nH_{2n-2}}$	ethyne, $\mathrm{C_2H_2}$

Only the alkanes can begin at one carbon, because only they need no carbon–carbon bond at all.

Common mistakes

• Identifying benzene by its molecular formula alone, in Question 16. Option (a) there is also $\mathrm{C_6H_6}$. Benzene is defined by the arrangement: one hydrogen on every carbon and three double bonds on alternate edges.
• Reading the shape of a drawing as branching, in Question 23. Options (a) and (c) are straight-chain hexane printed with a carbon on a second line. Trace the bonds and count how many carbons each carbon touches; the layout carries no information.
• Calling a branched molecule unsaturated, in Question 24. Branching and unsaturation are unrelated. Saturation is about double and triple bonds only.
• Confusing sodium ethoxide with sodium ethanoate, in Questions 17 and 28. The ethoxide comes from the alcohol, the ethanoate from the acid. They look alike and come from different reactants.
• Choosing esterification for saponification, in Question 28. Esterification makes an ester from an acid and an alcohol; saponification breaks an ester with alkali into a soap and an alcohol. They are the same reaction run in opposite directions.
• Giving helium in Question 21. Carbon completing its octet reaches $2, 8$, which is neon. Helium is what the hydrogen atoms in methane attain.
• Drawing a single dot for a bond, in Question 22. A covalent bond is a shared pair, so two dots. Counting the total — eight for water, ten for nitrogen in Question 10 — is the fastest check.
• Forgetting the carbonyl carbon when counting, in Question 26. $\mathrm{CHO}$ contributes a carbon to the chain, so $\mathrm{CH_3CH_2CHO}$ has three carbons and not two.

Practise next

• Exemplar Multiple Choice Questions Q1–15 — bonding, tetravalency and isomerism, which Questions 16, 18 and 24 above all rest on.
• Exemplar Short and Long Answer Questions — where the naming of Questions 25, 26 and 29 has to be done in reverse, from structure to name and back.
• NCERT textbook Chapter 4 — Carbon and its Compounds for the end-of-chapter exercise on homologous series, nomenclature and the reactions of ethanol and ethanoic acid.
• Textbook activities for the experiments behind Questions 19 and 28 — the properties of ethanoic acid and the preparation of soap are done there.