Carbon and its Compounds — Practice Quiz

C₃H₇COOH

Step 1: CH₃COOH is ethanoic acid — it belongs to the carboxylic acid series (–COOH functional group). Step 2: A homologous series consists of compounds with the same functional group, differing by a –CH₂– unit each time. Step 3: CH₃COOH (2 carbons), C₂H₅COOH (3 carbons), C₃H₇COOH (4 carbons) — each differs by –CH₂– and all have –COOH. Step 4: C₃H₇COOH (butanoic acid) has the –COOH functional group and differs from CH₃COOH by two –CH₂– units, still making it part of the same series. Step 5: C₂H₅OH is an alcohol (–OH group), HCHO is an aldehyde (–CHO), and CH₃CHO is also an aldehyde — none share

Ethene

Step 1: Ethanol (CH₃CH₂OH) undergoes dehydration when heated with excess hot concentrated H₂SO₄ at 443 K. Step 2: Dehydration means removal of water (H₂O) from the molecule. Step 3: When –OH and –H are removed from adjacent carbon atoms in ethanol, water is eliminated and a double bond forms between the two carbon atoms. Step 4: The product is ethene (CH₂=CH₂), an unsaturated hydrocarbon. The reaction is: CH₃CH₂OH → CH₂=CH₂ + H₂O. Step 5: H₂SO₄ acts as a dehydrating agent here, not an oxidising agent. Ethane (option A) would require reduction, not dehydration. Ethanoic acid (option B) forms wh

C₃H₄

Step 1: Alkynes have a triple bond between carbon atoms and follow the general formula CₙH₂ₙ₋₂. Step 2: The first alkyne is ethyne: n=2, formula = C₂H₂₍₂₎₋₂ = C₂H₂. Step 3: The second alkyne: n=3, formula = C₃H₂₍₃₎₋₂ = C₃H₄ (propyne). Step 4: Wait — let us recount: first member is n=2 (ethyne, C₂H₂), second member is n=3 (propyne, C₃H₄), third member is n=4 (butyne, C₄H₆). Step 5: So the third member has n=4: formula = C₄H₂₍₄₎₋₂ = C₄H₆. However, if the series starts counting from the FIRST possible alkyne (n=2, ethyne), the third member is n=4 (C₄H₆). But if we consider propyne (C₃H₄) as the t

Oxidising agent, because it adds oxygen to ethanol

Step 1: Oxidation of an organic compound involves addition of oxygen or removal of hydrogen. Step 2: In this reaction, ethanol (CH₃CH₂OH) is converted to ethanoic acid (CH₃COOH). Step 3: Comparing the two: ethanol has the –OH group, while ethanoic acid has the –COOH group. The carbon has gained an extra oxygen atom — this is oxidation. Step 4: Since alkaline KMnO₄ provides the oxygen for this oxidation, it acts as an oxidising agent. Step 5: KMnO₄ itself is reduced in the process (it loses oxygen/gains electrons), which is why the purple colour of KMnO₄ disappears initially. A reducing agent d