Homologous Series – Long Answer Questions (CBSE Class 10 Chemistry)

Medium Level (Application & Explanation)

Q1. Define a homologous series. Explain its essential features with two examples from different functional groups.

Answer: A homologous series is a family of organic compounds that share the same functional group, follow a general formula, show similar chemical properties, and differ from each other by a —CH₂— unit (14 u in molecular mass). They display gradual changes in physical properties such as boiling point and solubility. For example:

In the alkane series, methane (CH₄), ethane (C₂H₆), and propane (C₃H₈) differ successively by one —CH₂— group and follow C_nH_2n+2.
In alcohols, methanol (CH₃OH), ethanol (C₂H₅OH), and propanol (C₃H₇OH) share the –OH group and follow C_nH_2n+1OH. These features make learning and predicting properties in organic chemistry simpler, because once the behavior of one member is known, we can anticipate the behavior of others in the same series.

Q2. What role does the functional group play in deciding chemical properties in a homologous series? Illustrate with alcohols, alkenes, and carboxylic acids.

Answer: The functional group is the reactive part of an organic molecule that governs its chemical properties and the types of reactions it undergoes. In a homologous series, all members have the same functional group; hence they show similar chemical behavior.

Alcohols (–OH): All alcohols react with sodium to liberate hydrogen gas, form alkoxides, and undergo esterification with acids. Methanol, ethanol, and propanol behave similarly because of the common –OH group.
Alkenes (C=C): They undergo addition reactions such as hydrogenation and decolorize bromine water due to the presence of the double bond.
Carboxylic acids (–COOH): They turn blue litmus red, react with bases to form salts and water, and form esters with alcohols. Thus, the same functional group is the key reason for similar chemical properties across the series.

Q3. Using general formulas, predict formulas and names of the next members: (i) after propene in alkenes, and (ii) after propanoic acid in carboxylic acids. Show your working.

Answer: General formulas help us extend a series quickly:

For alkenes, the general formula is C_nH_2n. Propene is C₃H₆ (n = 3). The next member will have one extra —CH₂— unit (adding C and 2H), so n = 4. Thus, formula = C₄H₈, named butene (commonly but-1-ene or but-2-ene depending on double bond position).
For carboxylic acids, the general formula is C_nH_2n+1COOH. Propanoic acid is C₂H₅COOH (overall C₃H₆O₂ for the molecule’s main part considering the COOH). The next member is obtained by adding —CH₂— to the alkyl part: C₃H₇COOH, named butanoic acid. In both cases, the molar mass increases by 14 u due to the —CH₂— increment, while the functional group remains unchanged, ensuring similar reactions.

Q4. Explain why boiling point increases down a homologous series. Use the first three alcohols as a case study and relate structure to physical properties.

Answer: In a homologous series, boiling point increases as we move to higher members because molecular size and surface area increase, strengthening intermolecular forces (mainly van der Waals forces). In alcohols, the –OH group enables hydrogen bonding, which is strong; as the carbon chain length grows, dispersion forces add up, requiring more heat to separate molecules.

Methanol (CH₃OH) boils at about 65°C.
Ethanol (C₂H₅OH) boils around 78°C.
Propanol (C₃H₇OH) boils near 97°C. All three share the same functional group and differ by —CH₂—, yet the increasing carbon chain raises the boiling point. Additionally, higher alcohols become less soluble in water and more oily because the non-polar hydrocarbon part dominates over the polar –OH group. Thus, a gradual physical property change is a hallmark of homologous series.

Q5. “Learn one, predict many.” Explain how knowledge of one member of a homologous series helps predict the properties, reactions, and naming of higher members.

Answer: Once you understand a single member of a homologous series, you can predict a lot about its neighbors:

The functional group is the same for all members, so chemical reactions are similar. If you know that ethanol undergoes oxidation to ethanoic acid, you can predict propanol oxidizes to propanoic acid.
The general formula gives a pattern for writing molecular formulas; adding —CH₂— moves you to the next member with a 14 u mass increase.
Physical trends such as boiling point rising or solubility decreasing can be extrapolated for higher members.
IUPAC naming follows a consistent pattern: identify the parent chain (meth-, eth-, prop-, but-, pent-) and attach the suffix that indicates the functional group (–ane, –ene, –yne, –ol, –oic acid). Therefore, knowing one member provides a template for predicting structures, names, and behaviors across the series.

High Complexity (Analytical & Scenario-Based)

Q6. You have two colorless liquids with boiling points 78°C and 97°C. Both react with sodium to release hydrogen but do not decolorize bromine water. Identify the homologous series they likely belong to and justify with tests and reasoning.

Answer: The data suggest both are alcohols:

Boiling points near 78°C and 97°C match ethanol (C₂H₅OH) and propanol (C₃H₇OH), respectively.
Both reacting with sodium to release hydrogen gas indicates the –OH functional group forming alkoxides.
The fact that they do not decolorize bromine water shows they are not unsaturated (so not alkenes).
They differ by —CH₂— and have similar chemical behavior, fitting the homologous series pattern for alcohols (C_nH_2n+1OH). Confirmatory tests:
Perform esterification with acetic acid and conc. H₂SO₄ to form esters with a fruity smell.
Oxidation with acidified KMnO₄ should yield ethanoic acid from ethanol and propanoic acid from propanol. Thus, they are consecutive members of the alcohol homologous series.

Q7. Compare the chemical behavior of alkanes and alkenes across their homologous series. How does the difference in general formula relate to reactivity and typical reactions?

Answer: Alkanes and alkenes belong to distinct homologous series with different general formulas and reactivity:

Alkanes follow C_nH_2n+2 and are saturated (only C–C single bonds). They are relatively less reactive and typically undergo substitution reactions (e.g., chlorination under light) and combustion to CO₂ and H₂O.
Alkenes follow C_nH_2n and are unsaturated due to the C=C bond, making them more reactive. They undergo addition reactions: hydrogenation to alkanes, addition of halogens (decolorizing bromine water), and addition of HCl/HBr.
The missing two hydrogens in alkenes compared to alkanes indicate unsaturation, which provides an electron-rich double bond that reacts readily with electrophiles. Across each series, members differ by —CH₂—, preserving reaction types, but reaction rates and physical properties may vary due to chain length. Thus, functional group presence (none for alkanes, C=C for alkenes) is the decisive factor for reactivity patterns.

Q8. Design an investigation to verify that consecutive members in a homologous series differ by 14 u in molar mass and show a trend in boiling points. Use carboxylic acids as an example.

Answer: Plan:

Select methanoic acid (HCOOH), ethanoic acid (CH₃COOH), and propanoic acid (C₂H₅COOH). Optionally include butanoic acid (C₃H₇COOH).
Calculate molar masses to confirm each higher member increases by 14 u due to the —CH₂— increment in the alkyl part.
Record boiling points from reliable data sources or standard charts. Procedure:
Tabulate compound, formula, calculated molar mass, and boiling point.
Plot two graphs: (i) molar mass vs. member number (should rise linearly by 14 u), and (ii) boiling point vs. member number (should show a gradual increase). Reasoning:
The functional group (–COOH) remains the same, so similar chemistry is expected.
The rise in boiling point is linked to increasing molecular size and stronger intermolecular forces (dispersion forces) alongside hydrogen bonding. Conclusion:
Data should validate both the —CH₂— rule and the gradation in physical properties characteristic of homologous series.

Q9. A mixture contains CH₄, C₂H₂, C₂H₄, and C₂H₅OH. Classify each into its homologous series, justify using general formulas/functional groups, and predict the next member for each series.

Answer: Classification and justification:

CH₄ (methane): an alkane; matches C_nH_2n+2 with n = 1. Next member: C₂H₆ (ethane); differs by —CH₂—.
C₂H₂ (ethyne): an alkyne; matches C_nH_2n−2 with n = 2. Next member: C₃H₄ (propyne).
C₂H₄ (ethene): an alkene; matches C_nH_2n with n = 2. Next member: C₃H₆ (propene).
C₂H₅OH (ethanol): an alcohol; matches C_nH_2n+1OH with n = 2. Next member: C₃H₇OH (propanol). Rationale:
The functional group is the primary criterion: none for alkanes, C=C for alkenes, C≡C for alkynes, and –OH for alcohols.
All series advance by —CH₂—, keeping functional groups constant and establishing predictable chemical behavior and physical trends.