Carbon compounds — covalent bonding, homologous series, functional groups

Question

What is a homologous series? Write the first four members of the alkane series and identify the functional groups in CH $_3$ OH and CH $_3$ COOH.

(CBSE Class 10 — Carbon and its Compounds)

Homologous Series Classification

flowchart TD
    A["Carbon Compound"] --> B{Type of bond?}
    B -->|All single bonds| C["Saturated (Alkane)"]
    B -->|Double bond present| D["Unsaturated (Alkene)"]
    B -->|Triple bond present| E["Unsaturated (Alkyne)"]
    A --> F{Functional group?}
    F -->|"-OH"| G["Alcohol"]
    F -->|"-COOH"| H["Carboxylic Acid"]
    F -->|"-CHO"| I["Aldehyde"]
    F -->|"-CO-"| J["Ketone"]
    F -->|"-Cl, -Br"| K["Haloalkane"]
    C --> L["CH4, C2H6, C3H8, C4H10..."]
    D --> M["C2H4, C3H6, C4H8..."]
    E --> N["C2H2, C3H4, C4H6..."]

Solution — Step by Step

A homologous series is a family of organic compounds that:

Have the same general formula
Differ by CH $_2$ (14 u) from one member to the next
Show a gradual change in physical properties (boiling point, melting point increase)
Have similar chemical properties due to the same functional group

Think of it like a family — all members look similar but each is slightly bigger than the last.

Alkanes have the general formula C $_n$ H $_{2n+2}$ :

Name	Formula	Structure
Methane	CH $_4$	1 carbon, 4 hydrogens
Ethane	C $_2$ H $_6$	2 carbons, 6 hydrogens
Propane	C $_3$ H $_8$	3 carbons, 8 hydrogens
Butane	C $_4$ H $_{10}$	4 carbons, 10 hydrogens

Each successive member adds one CH $_2$ unit: CH $_4$ → C $_2$ H $_6$ (add CH $_2$ ) → C $_3$ H $_8$ (add CH $_2$ ) → C $_4$ H $_{10}$ (add CH $_2$ ).

CH $_3$ OH (Methanol): The functional group is -OH (hydroxyl group). This makes it an alcohol. The IUPAC suffix for alcohols is “-ol.”

CH $_3$ COOH (Acetic acid / Ethanoic acid): The functional group is -COOH (carboxyl group). This makes it a carboxylic acid. The IUPAC suffix is “-oic acid.”

\boxed{\text{CH}_3\text{OH} \rightarrow \text{Alcohol (-OH)}, \quad \text{CH}_3\text{COOH} \rightarrow \text{Carboxylic acid (-COOH)}}

Why This Works

Carbon’s unique ability to form 4 covalent bonds and create long chains (catenation) makes millions of organic compounds possible. A homologous series organises this complexity — all members share the same functional group, so they react similarly. The CH $_2$ difference between members is systematic, making properties predictable.

Covalent bonding (sharing electrons) is why carbon compounds have low melting points and are poor conductors — there are no free ions or electrons.

Alternative Method — Naming from Functional Group

To name any organic compound: (1) count the carbon chain length (meth = 1, eth = 2, prop = 3, but = 4), (2) identify the functional group, (3) add the suffix.

Examples:

2 carbons + -OH = Ethanol
3 carbons + -COOH = Propanoic acid
1 carbon + -CHO = Methanal (formaldehyde)

CBSE asks IUPAC naming almost every year. Memorise the prefixes (meth, eth, prop, but, pent) and the suffixes (-ane for alkane, -ene for alkene, -yne for alkyne, -ol for alcohol, -al for aldehyde, -one for ketone, -oic acid for carboxylic acid). This covers 90% of naming questions.

Common Mistake

Students confuse the general formulas. Alkanes are C $_n$ H $_{2n+2}$ , alkenes are C $_n$ H $_{2n}$ , alkynes are C $_n$ H $_{2n-2}$ . A common error is writing ethene as C $_2$ H $_6$ (that is ethane). Ethene has a double bond, so it is C $_2$ H $_4$ — two fewer hydrogens than ethane. The double bond replaces two hydrogens.