Write a note on homologous series.

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Q. Write a note on homologous series.

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Concept understanding — Homologous Series and General Formulas

A homologous series is a run of organic compounds that all share one characteristic functional group, and whose successive members differ from each other in molecular formula by exactly one $CH_2$ unit -- for example the alkanes (methane $CH_4$ , ethane $C_2H_6$ , propane $C_3H_8$ ...) or the alcohols (methanol $CH_3OH$ , ethanol $C_2H_5OH$ , propanol $C_3H_7OH$ ...).

Because every member of a series shares the same functional group and the same $CH_2$ -unit spacing, the entire series can be captured by ONE general formula: alkanes are $C_nH_{2n+2}$ , alkenes are $C_nH_{2n}$ , alkynes are $C_nH_{2n-2}$ -- and since an alkadiene (a chain with two C=C double bonds) has the same degree of unsaturation as an alkyne (one C≡C triple bond, or two degrees of unsaturation total), alkadienes share the alkyne formula, $C_nH_{2n-2}$ . This same general-formula logic extends class by class: an aliphatic monohydric alcohol is $C_nH_{2n+1}OH$ (one -OH on a saturated chain); aliphatic ketones are $C_nH_{2n}O$ (one C=O on a saturated chain, needing at least 3 carbons since the carbonyl carbon needs two carbon substituents); aliphatic amines are $C_nH_{2n+3}N$ (one -NH2, or a substituted nitrogen, on a saturated chain); nitroalkanes are $C_nH_{2n+1}NO_2$ for the first six members: $CH_3NO_2$ (nitromethane), $C_2H_5NO_2$ (nitroethane), $C_3H_7NO_2$ (nitropropane), $C_4H_9NO_2$ (nitrobutane), $C_5H_{11}NO_2$ (nitropentane), $C_6H_{13}NO_2$ (nitrohexane).

The carboxylic acid series ( $C_nH_{2n}O_2$ , or $C_nH_{2n+1}COOH$ ) is worth walking through by name and structure for its first four members: formic acid, $HCOOH$ ( $CH_2O_2$ ); acetic acid, $CH_3COOH$ ( $C_2H_4O_2$ ); propanoic acid, $CH_3CH_2COOH$ ( $C_3H_6O_2$ ); and butanoic acid, $CH_3CH_2CH_2COOH$ ( $C_4H_8O_2$ ) -- each member simply one $CH_2$ longer than the last, all sharing the -COOH functional group and the -oic acid suffix.

Because every member of a homologous series shares one functional group, they also share almost identical CHEMICAL behaviour (the same characteristic reactions, happening at the same functional group), even while their PHYSICAL properties (melting point, boiling point, density, solubility) grade smoothly and predictably as the chain lengthens -- and members of a series can typically all be prepared by the same general synthetic methods, which is part of why organising compounds into homologous series is such a powerful simplification.

Step 1. Definition: a homologous series is a series of organic compounds, each containing a characteristic functional group, whose successive members differ from one another in molecular formula by exactly one CH2 group.

Step 2. Examples given in the text: alkanes -- methane (CH4), ethane (C2H6), propane (C3H8); alcohols -- methanol (CH3OH), ethanol (C2H5OH), propanol (C3H7OH).

Step 3. Every member can be captured by one general formula (alkanes CnH2n+2, alkenes CnH2n, alkynes CnH2n-2), and members can typically be prepared by the same general synthetic methods.

Step 4. Members of a series show a regular gradation in PHYSICAL properties (melting point, boiling point, density) as the chain lengthens, while their CHEMICAL properties remain almost identical, since they share the same functional group.

✓Final answer

A run of compounds sharing one functional group, each member one CH2 unit apart, sharing a general formula, similar chemistry, graded physical properties.

Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CBSE 2026Set A1 markMCQ
$Q. The general formula of ether is$
(a) CnH2nO
(b) CnH2n+1O
(c) CnH2n+2O
(d) CnH2nO2
›Reveal solutionSolution

Ethers (R-O-R') are functional isomers of monohydric alcohols, so both share the general formula CnH2n+2O.

An ether has the structure R-O-R' with only single bonds and no rings, i.e. it is fully saturated. Check with the simplest: dimethyl ether is CH3-O-CH3 = C2H6O, which fits CnH2n+2O for n = 2 (2x2+2 = 6). This is the same general formula as the isomeric alcohol (e.g. ethanol C2H6O), confirming ethers and alcohols are functional isomers.

✓Final answer
$(c) CnH2n+2O.$
CBSE 2025Set D1 markMCQ
$Q. The general formula of an amine is$
(a) CnH2n+1N
(b) CnH2n+2N
(c) CnH2n+3N
(d) CnH2nN
›Reveal solutionSolution
A saturated amine has the general formula CnH2n+3N.

A saturated aliphatic hydrocarbon is CnH2n+2. An amine is formed by replacing one hydrogen with an -NH2 (or by attaching nitrogen bearing hydrogens). Adding nitrogen with its extra hydrogens raises the hydrogen count by one relative to the alkane:

Methylamine: CH3NH2 = CH5N -> here n = 1, 2n+3 = 5. Fits CnH2n+3N.

Ethylamine: C2H5NH2 = C2H7N -> n = 2, 2n+3 = 7. Fits.

So the general formula is CnH2n+3N.

✓Final answer
$(c) CnH2n+3N.$
CBSE 2025Set D1 markMCQ
$Q. The general formula of carbonyl compound is$
(a) CnH2nO
(b) CnH2n+2O
(c) CnH2n+1O
(d) none of these
›Reveal solutionSolution

Saturated aldehydes and ketones (carbonyl compounds) share the general formula CnH2nO.

Carbonyl compounds are aldehydes (R-CHO) and ketones (R-CO-R'). For saturated members, the general molecular formula is CnH2nO. For example: HCHO (n=1, CH2O), CH3CHO (n=2, C2H4O), CH3COCH3 (n=3, C3H6O) — each fits CnH2nO. The one degree of unsaturation corresponds to the C=O double bond.

✓Final answer
$(A) CnH2nO is the general formula of carbonyl compounds.$
CBSE 2025Set sz1 markMCQ
$Q. Select the correct one: Which of the following represents the general formula of alkenes?$
(a) CnH2n
(b) CnH2n-2
(c) CnH2n+2
(d) CnH2n+1
›Reveal solutionSolution

The general formula of alkenes (one C=C double bond) is CnH2n.

Saturated open-chain hydrocarbons (alkanes) have the general formula CnH2n+2. Each degree of unsaturation (one double bond or one ring) removes 2 hydrogen atoms from this formula. Alkenes contain exactly one C=C double bond, so their general formula is CnH2n+2 - 2 = CnH2n. (CnH2n-2 is the formula for alkynes/dienes; CnH2n+2 is for alkanes; CnH2n+1 is the general formula of an alkyl group/radical, not a full molecule.)

✓Final answer
$(A) CnH2n is the general formula for alkenes.$
CBSE 2025Set ANNUAL1 markMCQ
$Q. The general formula of alkene is$
(a) CnH2n+1
(b) CnH2n+2
(c) CnH2n-1
(d) CnH2n
›Reveal solutionSolution
Alkenes (one C=C double bond, one degree of unsaturation) follow CnH2n.

The general formulas of the common acyclic hydrocarbon families are:

Alkanes (saturated, no double/triple bonds): $C_nH_{2n+2}$

Alkenes (one C=C double bond): $C_nH_{2n}$

Alkynes (one C≡C triple bond): $C_nH_{2n-2}$

Each degree of unsaturation removes 2 hydrogens relative to the saturated alkane formula, so an alkene (1 degree of unsaturation) has the formula $C_nH_{2n}$ .

✓Final answer
$(d) CnH2n$
CBSE 2024Set D1 markMCQ
$Q. The general molecular formula of saturated monocarboxylic acids is$
(a) CnH2n+2O
(b) CnH2nO
(c) CnH2nO2
(d) CnH2n+1O2
›Reveal solutionSolution
Saturated monocarboxylic acids: CnH2nO2.

A saturated monocarboxylic acid has one -COOH group. Checking with examples:

Formic acid HCOOH → CH2O2 (n=1): C1H2O2 fits CnH2nO2.

Acetic acid CH3COOH → C2H4O2 (n=2): fits CnH2nO2.

Propionic acid C2H5COOH → C3H6O2 (n=3): fits.

So the general molecular formula is CnH2nO2.

✓Final answer
$(c) CnH2nO2 — general formula of saturated monocarboxylic acids.$
CBSE 2022Set ANNUAL1 mark
$Q. The formula of Alkene is ................. (Fill in the blank)$
›Reveal solutionSolution

The general formula of an alkene is CnH2n.

Alkenes are unsaturated hydrocarbons containing one carbon-carbon double bond (C=C). Because of the double bond, an alkene has two fewer hydrogen atoms than the corresponding saturated alkane (CnH2n+2). For example, ethene (n = 2) is C2H4, and propene (n = 3) is C3H6 - both fit CnH2n.

✓Final answer
$CnH2n.$
CBSE 2022Set ANNUAL1 markMCQ
$Q. CnH2n-2 is general formula of :$
(a) Alkanes
(b) Arenes
(c) Alkenes
(d) Alkynes
›Reveal solutionSolution
Each degree of unsaturation removes 2 H atoms from the saturated C $_n$ H $_{2n+2}$ formula; a triple bond (2 degrees) gives C $_n$ H $_{2n-2}$ — the alkyne formula.

Saturated alkanes follow C $_n$ H $_{2n+2}$ (no unsaturation). Each ring or π-bond reduces the H count by 2:

Alkenes (one C=C double bond, one degree of unsaturation): C $_n$ H $_{2n}$ .

Alkynes (one C≡C triple bond, two degrees of unsaturation): C $_n$ H $_{2n-2}$ .

Arenes (aromatic ring) have their own distinct general formulas (e.g. C $_n$ H $_{2n-6}$ for benzenoid arenes), not C $_n$ H $_{2n-2}$ .

✓Final answer
$_n$
CBSE 2021Set A1 markMCQ
$Q. The general formula of monohydric alcohol is$
(a) CnH2n+1OH
(b) CnH2n+2OH
(c) CnH2nOH
(d) None of these
›Reveal solutionSolution

Monohydric alcohol = alkyl-OH = CnH2n+1OH.

A monohydric alcohol contains one hydroxyl (-OH) group attached to a saturated alkyl group. A saturated alkyl group has the formula CnH2n+1 (e.g. CH3-, C2H5-). Attaching -OH gives the general formula CnH2n+1OH, equivalently written CnH2n+2O.

Examples: CH3OH (n=1), C2H5OH (n=2), C3H7OH (n=3).

✓Final answer
$(A) CnH2n+1OH.$
CBSE 2021Set ANNUAL1 markMCQ
$Q. CnH2n is general formula of:$
(a) alkene
(b) alkane
(c) alkyne
(d) arene
›Reveal solutionSolution
Alkenes are unsaturated hydrocarbons with one C=C double bond, giving the general formula $C_nH_{2n}$ — two hydrogens fewer per molecule than the corresponding alkane ( $C_nH_{2n+2}$ ).

Step 1 — Compare the hydrocarbon families:

Alkanes (saturated, all single bonds): $C_nH_{2n+2}$ — e.g. ethane $C_2H_6$ .

Alkenes (one C=C double bond): $C_nH_{2n}$ — e.g. ethene $C_2H_4$ .

Alkynes (one C≡C triple bond): $C_nH_{2n-2}$ — e.g. ethyne $C_2H_2$ .

Step 2 — Why alkenes lose 2 H's relative to alkanes: Forming a double bond between two carbons removes one H from each of those two carbons compared to the fully saturated chain, i.e. 2 fewer H atoms overall, which is exactly the $2n$ vs $2n+2$ pattern.

✓Final answer
$C_nH_{2n}$

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Q.Write a note on homologous series.

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