Carbon and its Compounds — Important Questions

Answer: (c) 7. Ethane has six C-H bonds and one C-C bond, giving a total of 7 covalent bonds. Each carbon completes its octet by sharing electrons, consistent with the tetravalency of carbon.

Answer: (b) $-CH_2$ and 14 u. Consecutive members of a homologous series differ by one $-CH_2-$ group, whose mass is $12 + (2\times 1) = 14$ u. This is why the series shows a regular gradation in physical properties.

Answer: (c) Addition of hydrogen in the presence of a nickel catalyst. Vegetable oils are unsaturated; adding $H_2$ across their C=C double bonds (hydrogenation, an addition reaction) saturates them into solid fats. This is why replacing saturated ghee with unsaturated oils in the diet is recommended for health.

Answer: (b) A colourless, odourless gas that turns lime water milky. Ethanoic acid reacts with sodium hydrogencarbonate to release carbon dioxide: $CH_3COOH + NaHCO_3 \rightarrow CH_3COONa + H_2O + CO_2$. The $CO_2$ turns lime water milky, confirming a carboxylic acid.

Answer: (a) Both A and R are true and R is the correct explanation of A. Carbon's tetravalency lets each atom bond with four other atoms, and catenation lets carbon atoms link into long chains, branches and rings. Together these directly account for the vast number of carbon compounds.

Carbon has 4 valence electrons. To form ions it would have to either lose 4 electrons (forming $C^{4+}$) or gain 4 electrons (forming $C^{4-}$).

(i) Removing four electrons requires a very large amount of energy, and the resulting $C^{4+}$ with only two electrons holding six protons would be unstable.

(ii) Adding four extra electrons is difficult because it is hard for the small nucleus (6 protons) to hold 10 electrons.

Therefore carbon attains a stable octet by sharing its four electrons, i.e. by covalent bonding.

Chemical test (bromine water test): Add bromine water (orange/brown) to each compound. An unsaturated hydrocarbon decolourises bromine water because it undergoes an addition reaction across its C=C/C≡C bond; a saturated hydrocarbon does not decolourise it under these conditions.

Flame: Saturated hydrocarbons burn with a clean blue (non-sooty) flame, whereas unsaturated hydrocarbons burn with a yellow, sooty (smoky) flame because of their higher percentage of carbon.

(a) A homologous series is a series of carbon compounds having the same functional group in which any two adjacent members differ by a $-CH_2-$ group (14 u) and which can be represented by a single general formula.

(b) Two characteristics:
(i) All members show similar chemical properties because they have the same functional group.
(ii) Physical properties (e.g. melting point, boiling point, solubility) show a regular gradation with increasing molecular mass.

(c) General formula of alkenes: $C_nH_{2n}$ (e.g. ethene $C_2H_4$).

(a) Identifying test — reaction with sodium hydrogencarbonate: Add a small amount of $NaHCO_3$ to each liquid. Ethanoic acid produces brisk effervescence of $CO_2$ (which turns lime water milky); ethanol shows no reaction. (Alternatively, litmus: ethanoic acid turns blue litmus red; ethanol has no effect.)

(b) On warming ethanoic acid with ethanol in the presence of a few drops of concentrated $H_2SO_4$:

$CH_3COOH + C_2H_5OH \xrightarrow{conc.\ H_2SO_4} CH_3COOC_2H_5 + H_2O$

The product is ethyl ethanoate (an ester), which has a sweet fruity smell, and the reaction is called esterification.

(a) Hard water is water that contains dissolved salts of calcium and magnesium ($Ca^{2+}$, $Mg^{2+}$).

(b) Soaps are sodium/potassium salts of long-chain carboxylic acids. In hard water they react with $Ca^{2+}$/$Mg^{2+}$ ions to form an insoluble precipitate (scum) instead of lather, wasting soap. Detergents are ammonium/sulphonate salts whose calcium and magnesium salts remain soluble, so they lather and clean effectively even in hard water.

(c) Advantage: detergents clean well in hard water. Disadvantage: many detergents are non-biodegradable and cause water pollution.

(a) Cleansing action of soap: Most dirt is oily/greasy and does not dissolve in water. A soap molecule has a hydrophobic (water-repelling) hydrocarbon tail and a hydrophilic (water-attracting) ionic head. When soap is added to water, the tails attach to the grease while the ionic heads face the surrounding water, forming spherical clusters called micelles with the oily dirt trapped inside. Agitation lifts the grease off the fabric and holds it suspended in water as an emulsion, so it can be rinsed away.

(b) Labelled diagram of a soap micelle:

(c) Dual nature: One end of the molecule (the ionic head) is hydrophilic and dissolves in water, while the other end (the long hydrocarbon tail) is hydrophobic and dissolves in oil/grease. Because the two ends behave oppositely towards water, the molecule is said to have a dual nature, and this is exactly what allows it to remove oily dirt.

(a)(i) Reaction with sodium — ethanol reacts with sodium to release hydrogen gas:

$2C_2H_5OH + 2Na \rightarrow 2C_2H_5ONa + H_2\uparrow$

(product: sodium ethoxide; a brisk evolution of $H_2$ confirms the -OH group.)

(ii) Reaction with hot conc. $H_2SO_4$ — heating ethanol at about 443 K with excess concentrated sulphuric acid causes dehydration to ethene:

$C_2H_5OH \xrightarrow[443\ K]{conc.\ H_2SO_4} C_2H_4 + H_2O$

Here concentrated $H_2SO_4$ acts as a dehydrating agent.

(b) Denatured alcohol: Ethanol meant for industrial use is made unfit for drinking by adding small amounts of poisonous substances such as methanol (and often dyes). This is done to prevent its misuse as a beverage and to avoid tax on drinking alcohol.

(c) Weak acid: Ethanoic acid ionises only partially in water, producing a low concentration of $H^+$ ions, whereas a mineral acid like HCl ionises almost completely (strong acid). Consequently, for the same concentration, ethanoic acid reacts more slowly, and a universal indicator/pH shows it is less acidic (higher pH) than HCl. It still turns blue litmus red and liberates $CO_2$ from carbonates, confirming it is an acid — just a weak one.

(a) Compound (IV) $C_2H_4$ (ethene) decolourises bromine water because it is unsaturated (contains a C=C double bond) and undergoes an addition reaction with bromine, removing the colour.

(b) Compound (III) $CH_3COOH$ (ethanoic acid) gives brisk effervescence with $NaHCO_3$; the gas evolved is carbon dioxide ($CO_2$) — this is a confirmatory test for the carboxylic acid group.

(c) Compound (II) $C_2H_5OH$ contains the hydroxyl (alcohol, -OH) group; compound (III) $CH_3COOH$ contains the carboxylic acid (-COOH) group.

(d) Add solid sodium hydrogencarbonate to each: ethanoic acid (III) gives effervescence of $CO_2$, whereas ethanol (II) shows no reaction. (Alternatively, ethanoic acid turns blue litmus red; ethanol does not.)