ScienceCBSE Class 10 · Carbon and its Compounds

Homologous Series and Nomenclature of Carbon Compounds

Q: What is a homologous series in Class 10?

A series of carbon compounds with the same functional group, in which successive members differ by a CH_2 unit. Members have similar chemical properties and gradually changing physical properties.

Q: How do you name carbon compounds?

Count the carbons to get the root (meth, eth, prop, but...), then add a prefix (chloro, bromo) or suffix (-ol, -al, -one, -oic acid) for the functional group, or -ene/-yne for double or triple bonds.

Q: Why do members of a homologous series have similar chemical properties?

Chemical properties depend on the functional group, and all members share the same functional group.

Q: Which functional groups are in the syllabus?

Halogens (chloro, bromo), alcohol, aldehyde, ketone, and the carboxylic acid group of ethanoic acid, plus double and triple bonds in alkenes and alkynes.

Q: What is the smallest ketone?

Propanone, CH_3COCH_3, because the C=O group must sit between two other carbons.

Updated 14 October 2026 · Fact-checked

A homologous series is a family of carbon compounds with the same functional group, where each member differs from the next by a CH2 unit (14 u). Members share chemical properties and show a gradual change in physical properties. Names are built from the carbon-chain root plus a prefix or suffix for the functional group.

Each alkane differs from the next by one CH2 unit (14 u), and all fit CnH2n+2, while boiling point climbs steadily along the series.

Illustration drawn for QuestionBuddy.

Understand Homologous Series and Nomenclature

Hydrocarbons and their families. Compounds of only carbon and hydrogen are hydrocarbons. Saturated ones (only single C–C bonds) are alkanes; those with a double bond are alkenes; those with a triple bond are alkynes. NCERT's table of alkanes runs methane $C H_{4}$ , ethane $C_{2} H_{6}$ , propane $C_{3} H_{8}$ , butane $C_{4} H_{10}$ , pentane $C_{5} H_{12}$ , hexane $C_{6} H_{14}$ .

Homologous series. Look at successive alkanes: $C H_{4}$ and $C_{2} H_{6}$ differ by $C H_{2}$ ; so do $C_{2} H_{6}$ and $C_{3} H_{8}$ . A difference of one $C H_{2}$ unit means a difference of $12 + 2 = 14$ u in molecular mass. The same is true of alkenes ( $C_{2} H_{4}$ , $C_{3} H_{6}$ , $C_{4} H_{8}$ ...) and of alcohols ( $C H_{3} O H$ , $C_{2} H_{5} O H$ , $C_{3} H_{7} O H$ , $C_{4} H_{9} O H$ ). A series of compounds in which the same functional group replaces hydrogen in a carbon chain, with successive members differing by $C H_{2}$ , is a homologous series. General formulae:

alkanes $C_{n} H_{2 n + 2}$
alkenes $C_{n} H_{2 n}$ ( $n = 2, 3, 4...$ )
alkynes $C_{n} H_{2 n - 2}$

In a series, chemical properties stay similar because they are decided by the functional group. Physical properties change gradually with molecular mass: melting and boiling points rise, and solubility in a given solvent shows a similar gradation.

Functional groups. An atom other than C or H in the chain is a heteroatom. A heteroatom or group of atoms that gives a compound its characteristic properties, whatever the chain length, is a functional group:

halo: $- C l$ , $- B r$ (haloalkane)
alcohol: $- O H$
aldehyde: $- C H O$
ketone: $> C = O$ (carbonyl inside the chain)
carboxylic acid: $- C O O H$

Isomers. Butane can have its four carbons in a straight chain or a branched chain; both are $C_{4} H_{10}$ . Compounds with the same molecular formula but different structures are structural isomers. Pentane has three.

Naming (nomenclature).
1. Count the carbons and pick the root: 1 meth, 2 eth, 3 prop, 4 but, 5 pent, 6 hex (as in methane, ethane, propane...).
2. Show the functional group as a prefix or suffix: chloro-/bromo- (prefix); -ol alcohol; -al aldehyde; -one ketone; -oic acid carboxylic acid.
3. If the suffix starts with a vowel, drop the final 'e' of the alkane name: propane − e + one = propanone; propane − e + ol = propanol.
4. For unsaturated chains replace 'ane' by 'ene' (double bond) or 'yne' (triple bond): propene, propyne.
The carbon of the $- C H O$ or $- C O O H$ group is counted in the chain: $C H_{3} C H O$ has two carbons, so it is ethanal; $C H_{3} C O O H$ is ethanoic acid. A ketone needs at least three carbons, since its $C = O$ must sit inside the chain; the smallest is propanone.

Homologous Series and Nomenclature: formulas and key facts

Alkanes: $C_{n} H_{2 n + 2}$; Methane, ethane, propane, butane...
Alkenes: $C_{n} H_{2 n}$; Start at n = 2: ethene $C_{2} H_{4}$ .
Alkynes: $C_{n} H_{2 n - 2}$; Start at n = 2: ethyne $C_{2} H_{2}$ .
Difference between successive members: $- C H_{2} - unit = 12 + 2 = 14 u$; Atomic masses: C 12 u, H 1 u.
Name pattern: $prefix + root + suffix$; Bromo + propane; propan + one; propan + oic acid.
Suffixes: $- o l (O H), - a l (C H O), - o n e (C = O), - o i c a c i d (C O O H), - e n e, - y n e$; Halogens are prefixes: chloro-, bromo-.

How to solve Homologous Series and Nomenclature questions

For naming and 'draw the structure' questions, use this routine.

1Count all carbon atoms in the longest chain, including the carbon of $- C H O$ or $- C O O H$ . Choose the root (meth, eth, prop, but, pent, hex).
2Identify any functional group or multiple bond.
3Halogen: add chloro- or bromo- before the alkane name (bromoethane).
4Suffix group: drop the final 'e' if the suffix starts with a vowel and add -ol, -al, -one or -oic acid.
5Double or triple bond: change -ane to -ene or -yne.
6To draw a structure from a name, reverse the steps: draw the chain, place the group, then fill each carbon's remaining valencies with hydrogen so every C has four bonds.
7For series questions, use the general formula and 14 u per $C H_{2}$ step.

Homologous Series and Nomenclature in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Homologous series and gradation of properties

Suppose your family uses LPG cylinders, and you read that the gas inside is mainly butane and propane, while the fuel in a camping lighter is also butane. You wonder why these gases can be stored as liquids under pressure but methane cannot be so easily.

Show that methane, propane and butane belong to the same homologous series, find the difference in molecular mass between propane and butane, and explain the trend in boiling points along the series.

Try it first, then show the solutionSolution

Formulae: $C H_{4}$ , $C_{3} H_{8}$ , $C_{4} H_{10}$ . All fit $C_{n} H_{2 n + 2}$ (n = 1, 3, 4), so they are alkanes in one homologous series.
Molecular masses: $C_{3} H_{8} = 3 \times 12 + 8 = 44$ u; $C_{4} H_{10} = 4 \times 12 + 10 = 58$ u; difference 14 u, one $C H_{2}$ unit.
In a homologous series, boiling points rise as molecular mass increases.
So butane and propane have higher boiling points than methane (111 K) and turn liquid under pressure more readily, while their chemical property of burning as fuels stays similar.

Answer: All three are alkanes ( $C_{n} H_{2 n + 2}$ ); propane and butane differ by 14 u ( $C H_{2}$ ). Boiling points rise with molecular mass, so the heavier butane and propane liquefy more easily than methane.

Memory hook: Add a CH2, add 14 u, raise the boiling point.

Tricks to remember Homologous Series and Nomenclature

Short memory aids. Say each one aloud once, then use it on a question the same day.

Roots: 'Monkeys Eat Peeled Bananas'
Meth 1, Eth 2, Prop 3, But 4; then Pent 5, Hex 6 like pentagon and hexagon.
Use it when: Choosing the root name.
Suffix vowels
ol, al, one, oic acid all start with a vowel, so drop the 'e': propan-ol, ethan-al.
Use it when: Getting spellings right.
2n+2, 2n, 2n−2
Each extra bond between carbons removes two hydrogens: alkane, alkene, alkyne.
Use it when: Writing and checking general formulae.

Homologous Series and Nomenclature: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Name these compounds · 2 marks · Very short answerBoard level

Name: (a) $C H_{3} C H_{2} B r$ (b) $H C H O$ (c) $C H_{3} C O C H_{3}$ (d) $C H_{3} C H_{2} C O O H$

Try it first, then show the solutionSolution

(a) Two carbons, bromo group: bromoethane (½ mark).
(b) One carbon, aldehyde: methanal (½ mark).
(c) Three carbons, ketone: propanone (½ mark).
(d) Three carbons including the COOH carbon, carboxylic acid: propanoic acid (½ mark).

Answer: (a) Bromoethane (b) Methanal (c) Propanone (d) Propanoic acid

Example 2: Draw structures from names · 3 marks · Short answerHarder board level

Draw the structures of (a) ethanoic acid, (b) butanone, (c) hexanal, and state the functional group in each.

Try it first, then show the solutionSolution

(a) Two carbons, one of them the COOH carbon: $C H_{3} - C O O H$ ; carboxylic acid group (1 mark).
(b) Four carbons with C=O inside the chain on carbon 2: $C H_{3} - C O - C H_{2} - C H_{3}$ ; ketone group (1 mark).
(c) Six carbons with CHO at the end: $C H_{3} - C H_{2} - C H_{2} - C H_{2} - C H_{2} - C H O$ ; aldehyde group (1 mark).

Answer: (a) $C H_{3} C O O H$ , –COOH. (b) $C H_{3} C O C H_{2} C H_{3}$ , ketone. (c) $C H_{3} (C H_{2})_{4} C H O$ , –CHO.

Example 3: Molecular mass and series · 3 marks · Short answerHarder board level

(a) An alkane has molecular mass 72 u. Find its formula and name. (b) An alkene has molecular mass 56 u. Find its formula and name. (c) Write the formula of the next member of each series and its molecular mass. (C = 12 u, H = 1 u)

Try it first, then show the solutionSolution

(a) $C_{n} H_{2 n + 2}$ : $12 n + 2 n + 2 = 72$ , so $14 n = 70$ , $n = 5$ : $C_{5} H_{12}$ , pentane. Check: $60 + 12 = 72$ (1 mark).
(b) $C_{n} H_{2 n}$ : $14 n = 56$ , $n = 4$ : $C_{4} H_{8}$ , butene. Check: $48 + 8 = 56$ (1 mark).
(c) Add $C H_{2}$ (14 u): $C_{6} H_{14}$ , 86 u (hexane); $C_{5} H_{10}$ , 70 u (pentene) (1 mark).

Answer: (a) $C_{5} H_{12}$ , pentane. (b) $C_{4} H_{8}$ , butene. (c) $C_{6} H_{14}$ (86 u) and $C_{5} H_{10}$ (70 u).

Example 4: Isomers of pentane · 5 marks · Long answerHarder board level

(a) What are structural isomers? (b) Draw the carbon skeletons of all structural isomers of pentane, $C_{5} H_{12}$ . (c) Can methane, ethane or propane have structural isomers? Explain.

Try it first, then show the solutionSolution

(a) Compounds with the same molecular formula but different structures (1 mark).
(b) Straight chain: C–C–C–C–C (1 mark).
(b) One branch: a four-carbon chain with a $C H_{3}$ on the second carbon (1 mark).
(b) Two branches: a three-carbon chain with two $C H_{3}$ groups on the middle carbon. So pentane has three isomers (1 mark).
(c) No. With one, two or three carbons there is only one way to join the carbon atoms, so only one structure is possible; isomers begin with butane (1 mark).

Answer: (a) Same formula, different structures. (b) Three skeletons: straight, one branch, two branches on the middle carbon. (c) No; with three or fewer carbons only one skeleton is possible.

Example 5: Spot the wrong name · 3 marks · Short answerHOTS / competency

A student names three compounds: (i) $C H_{3} C H O$ as 'methanal', (ii) $C H_{3} C H = C H_{2}$ as 'propane', (iii) $C H_{3} C H_{2} O H$ as 'ethaneol'. Correct each name and give the rule broken.

Try it first, then show the solutionSolution

(i) The CHO carbon counts in the chain, giving two carbons: ethanal (1 mark).
(ii) A double bond means -ene, not -ane: propene (1 mark).
(iii) Suffix '-ol' starts with a vowel, so drop the final 'e': ethanol (1 mark).

Answer: (i) Ethanal (ii) Propene (iii) Ethanol.

Where students lose marks in Homologous Series and Nomenclature

The slips examiners see most often, and what to write instead.

Mistake: Not counting the CHO or COOH carbon
The functional group looks separate from the chain.
Wrong working
$C H_{3} C O O H$ = methanoic acid.
Right working
$C H_{3} C O O H$ = ethanoic acid (two carbons).
Do this: Count every carbon, including the one in the functional group.
Usually costs: ½ to 1 mark per name.
Mistake: Keeping the 'e' before a vowel suffix
Students add the suffix to the full alkane name.
Wrong working
Propaneone, ethaneol.
Right working
Propanone, ethanol.
Do this: Drop the final 'e' before -ol, -al, -one, -oic acid.
Usually costs: ½ mark.
Mistake: Wrong general formula
Alkene and alkyne formulae are swapped.
Wrong working
Alkenes: $C_{n} H_{2 n - 2}$ .
Right working
Alkenes: $C_{n} H_{2 n}$ ; alkynes: $C_{n} H_{2 n - 2}$ .
Do this: Remember: each extra C–C bond removes two hydrogens.
Usually costs: 1 mark in series questions.
Mistake: Saying physical properties stay the same in a series
Students over-apply 'members are similar'.
Wrong working
All alcohols have the same boiling point.
Right working
Chemical properties are similar; physical properties show a gradation with molecular mass.
Do this: Separate chemical (similar) from physical (gradual change).
Usually costs: 1 mark.

Exam tips

Write the general formula and show the calculation when asked to identify a member from its mass.
In structure drawings, give every carbon exactly four bonds; examiners check this.
State both features of a homologous series: same functional group and $C H_{2}$ (14 u) difference.
Nomenclature in your syllabus covers alkanes, alkenes, alkynes and compounds with halogen, alcohol, aldehyde, ketone (and carboxylic acid); simple unbranched names are enough.
For isomer questions, draw only carbon skeletons unless full structures are asked.

Practice questions on Homologous Series and Nomenclature

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
Which of the following is the next member of the homologous series after $C_{3} H_{4}$ ?
(a) $C_{4} H_{6}$ (b) $C_{4} H_{8}$ (c) $C_{4} H_{10}$ (d) $C_{3} H_{6}$
Hint
$C_{3} H_{4}$ fits $C_{n} H_{2 n - 2}$ . Add $C H_{2}$ .
Answer
(a) $C_{4} H_{6}$ (butyne).
Q2 · 1 mark · Assertion-reasonHarder board level
Assertion (A): $C H_{3} O H$ and $C_{2} H_{5} O H$ have similar chemical properties.
Reason (R): Members of a homologous series have the same functional group.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
What decides chemical properties in a series?
Answer
(a). Both have the –OH group, which decides their chemical properties.
Q3 · 3 marks · Short answerBoard level
Write the names and formulae of the first three members of the homologous series of (a) alcohols and (b) carboxylic acids. Calculate the difference in molecular mass between successive members.
Hint
Count the functional-group carbon in acids.
Answer
(a) Methanol $C H_{3} O H$ , ethanol $C_{2} H_{5} O H$ , propanol $C_{3} H_{7} O H$ . (b) Methanoic acid $H C O O H$ , ethanoic acid $C H_{3} C O O H$ , propanoic acid $C_{2} H_{5} C O O H$ . Successive members differ by $C H_{2}$ , i.e. 14 u.
Q4 · 4 marks · Case-basedHOTS / competency
Read the passage and answer.
Suppose a student lists these compounds: P $C_{2} H_{6}$ , Q $C_{3} H_{6}$ , R $C_{4} H_{10}$ , S $C_{3} H_{7} O H$ , T $C_{2} H_{5} O H$ , U $C_{4} H_{8}$ .
(a) Which two compounds belong to the alkene series? (1 mark)
(b) Name S and T and state why they are in the same series. (1 mark)
(c) Arrange P and R in order of boiling point and explain. Would P and R react differently with oxygen? Justify. (2 marks)
Hint
Use general formulae and the gradation of properties.
Answer
(a) Q ( $C_{3} H_{6}$ , propene) and U ( $C_{4} H_{8}$ , butene) fit $C_{n} H_{2 n}$ . (b) S propanol, T ethanol; both have the –OH group and differ by $C H_{2}$ . (c) P (ethane) < R (butane); boiling point rises with molecular mass in a series. They burn alike (both give $C O_{2}$ and water), because chemical properties are similar within a series.
Q5 · 3 marks · Short answerHOTS / competency
Bromopentane has the formula $C_{5} H_{11} B r$ . Using a straight five-carbon chain only, how many different structures are possible depending on where the bromine is attached? Explain, and state whether these are structural isomers.
Hint
Positions 1 and 5 are the same by symmetry; so are 2 and 4.
Answer
Three: Br on an end carbon (1, same as 5), on carbon 2 (same as 4), or on the middle carbon 3. They have the same formula but different structures, so they are structural isomers. (Branched chains give more.)
Q6 · 3 marks · Short answerHarder board level
Write the general formula of alkynes and use it to find (a) the formula of the alkyne with 5 carbons, (b) its molecular mass, (c) its name. Then explain why the first member of the alkene and alkyne series has n = 2, not n = 1.
Hint
A double or triple bond needs two carbons.
Answer
$C_{n} H_{2 n - 2}$ . (a) $C_{5} H_{8}$ . (b) $60 + 8 = 68$ u. (c) Pentyne. A double or triple bond is between two carbon atoms, so the smallest alkene or alkyne must have two carbons.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.General formula of alkanes?Show answer
$C_{n} H_{2 n + 2}$ .
2.Difference in mass between successive members of a series?Show answer
14 u (one $C H_{2}$ unit).
3.Suffix for a ketone?Show answer
-one, as in propanone.
4.What is a heteroatom?Show answer
An atom other than carbon or hydrogen (such as O, N, S, Cl) that replaces hydrogen in a chain.
5.How many structural isomers does butane have?Show answer
Two.
6.Name of $C H_{3} C H_{2} C H O$ ?Show answer
Propanal.

Homologous Series and Nomenclature: frequently asked questions

What is a homologous series in Class 10?

A series of carbon compounds with the same functional group, in which successive members differ by a CH_2 unit. Members have similar chemical properties and gradually changing physical properties.

How do you name carbon compounds?

Count the carbons to get the root (meth, eth, prop, but...), then add a prefix (chloro, bromo) or suffix (-ol, -al, -one, -oic acid) for the functional group, or -ene/-yne for double or triple bonds.

Why do members of a homologous series have similar chemical properties?

Chemical properties depend on the functional group, and all members share the same functional group.

Which functional groups are in the syllabus?

Halogens (chloro, bromo), alcohol, aldehyde, ketone, and the carboxylic acid group of ethanoic acid, plus double and triple bonds in alkenes and alkynes.

What is the smallest ketone?

Propanone, CH_3COCH_3, because the C=O group must sit between two other carbons.