Carbon and its Compounds — NCERT Solutions

Concept: In electron dot (Lewis) structures, valence electrons of each atom are shown as dots. Shared pairs form bonds; unshared pairs are lone pairs.

(a) Ethanoic acid ($\mathrm{CH_3COOH}$):

Structure: $\mathrm{CH_3 - C(=O) - O - H}$

$$\begin{array}{ccccc} H & & \ddot{O}: & & \\ | & & \| & & \\ H-\overset{\displaystyle|}{C}-C & & & & \\ | & & \dot{O}-H & & \\ H & & & & \end{array}$$

Written out clearly:

• Carbon 1 (methyl): bonded to 3 H atoms and 1 C atom (4 single bonds, octet complete).
• Carbon 2 (carbonyl): bonded to C1 by single bond, to one O by double bond ($\mathrm{C=O}$), and to –OH by single bond (octet complete).
• The –OH oxygen has 2 lone pairs.
• The $\mathrm{C=O}$ oxygen has 2 lone pairs.

$$\mathrm{H}\!:\!\overset{\displaystyle H}{\underset{\displaystyle H}{C}}\!:\!\overset{\displaystyle :O:}{\underset{}{C}}\!:\!\overset{\displaystyle}{\underset{}{O}}\!:\!H$$

Full electron dot representation:
$$\left[H-\overset{H\atop|}{\underset{|}{C}}-H\right] - \overset{\ddot{O}}{\|} C - \overset{..}{O} - H$$

(b) $\mathrm{H_2S}$:

• Sulphur has 6 valence electrons; each H has 1.
• S shares one electron with each H, forming 2 S–H bonds. S has 2 lone pairs remaining.

$$H : \overset{..}{\underset{..}{S}} : H$$

This is similar to water. S has 2 bonding pairs and 2 lone pairs; each H has a duplet.

(c) Propanone ($\mathrm{CH_3COCH_3}$):

• Structure: $\mathrm{CH_3 - C(=O) - CH_3}$
• Central carbon is double-bonded to oxygen and single-bonded to two methyl carbons.
• Each methyl carbon is bonded to 3 H atoms.

$$H\!-\!\overset{H\atop|}{\underset{|}{C}}\!-\!H \;-\; \overset{\ddot{O}}{\|} C \;-\; H\!-\!\overset{H\atop|}{\underset{|}{C}}\!-\!H$$

Oxygen has 2 lone pairs; all carbons have complete octets; all hydrogens have duplets.

(d) $\mathrm{F_2}$:

• Each fluorine has 7 valence electrons and needs 1 more.
• The two F atoms share one pair of electrons, forming a single F–F covalent bond. Each F has 3 lone pairs.

$$:\!\overset{..}{\underset{..}{F}}\!:\!:\!\overset{..}{\underset{..}{F}}\!:$$

or

$$:F - F:$$

where each F has 3 lone pairs (6 non-bonding electrons) and shares 1 bonding pair.

Definition: A homologous series is a group (series) of organic compounds that:

1. Have the same general formula.
2. Have the same functional group.
3. Show similar chemical properties.
4. Show a gradual change in physical properties (such as melting point, boiling point, density) with increasing molecular mass.
5. Each successive member differs from the previous one by a $-\mathrm{CH_2}-$ unit (methylene group), i.e., by a molecular mass of 14 u.

Example — Homologous series of alkanes:

• General formula: $\mathrm{C_nH_{2n+2}}$
• Each member differs from the next by $\mathrm{CH_2}$ (14 u).
• All members undergo similar reactions (e.g., combustion, substitution).
• Boiling points increase gradually from methane to butane.

Conclusion: The homologous series concept helps us predict the properties of unknown members from the known properties of other members of the series.

Given: Two compounds — Ethanol ($\mathrm{C_2H_5OH}$) and Ethanoic acid ($\mathrm{CH_3COOH}$).

Differences based on Physical Properties:

Differences based on Chemical Properties:

1. Litmus test:

• Ethanol: Does not change the colour of either red or blue litmus (neutral).
• Ethanoic acid: Turns blue litmus red (acidic in nature).

2. Reaction with sodium carbonate/sodium bicarbonate ($\mathrm{Na_2CO_3}$ / $\mathrm{NaHCO_3}$):

• Ethanol: Does not react with $\mathrm{Na_2CO_3}$ or $\mathrm{NaHCO_3}$; no effervescence.
• Ethanoic acid: Reacts vigorously with $\mathrm{Na_2CO_3}$ and $\mathrm{NaHCO_3}$ to produce brisk effervescence of $\mathrm{CO_2}$ gas:

$$\mathrm{2CH_3COOH + Na_2CO_3 \rightarrow 2CH_3COONa + H_2O + CO_2\uparrow}$$
$$\mathrm{CH_3COOH + NaHCO_3 \rightarrow CH_3COONa + H_2O + CO_2\uparrow}$$

3. Reaction with sodium metal:

• Both react with sodium to release hydrogen gas, but ethanoic acid reacts more vigorously due to its acidic nature.

4. Esterification:

• Ethanoic acid reacts with ethanol in the presence of a few drops of concentrated $\mathrm{H_2SO_4}$ (catalyst) to form a sweet-smelling ester (ethyl ethanoate):

$$\mathrm{CH_3COOH + C_2H_5OH \underset{\Delta}{\overset{H^+}{\rightleftharpoons}} CH_3COOC_2H_5 + H_2O}$$

• Ethanol alone does not undergo this reaction without the acid.

Conclusion: The simplest test to differentiate them is the litmus test and the sodium bicarbonate test — ethanoic acid turns blue litmus red and produces $\mathrm{CO_2}$ with $\mathrm{NaHCO_3}$, while ethanol does neither.

Why micelle formation takes place in water:

A soap molecule (e.g., sodium stearate, $\mathrm{C_{17}H_{35}COONa}$) has two distinct parts:

1. A long hydrophobic (water-repelling) hydrocarbon tail.
2. A hydrophilic (water-attracting) ionic/polar head ($-\mathrm{COO^-Na^+}$).

When soap is dissolved in water:

• The hydrophilic heads are attracted to water molecules and tend to remain in the aqueous phase.
• The hydrophobic tails try to avoid water and tend to cluster together away from water.
• To minimise the contact of hydrophobic tails with water, many soap molecules arrange themselves into a spherical cluster called a micelle, where:
• The hydrophobic tails point inward (towards the centre, away from water).
• The hydrophilic heads point outward (towards the water).

This arrangement is thermodynamically stable in water, which is why micelles form spontaneously.

Will micelles form in ethanol?

No, micelles will NOT be formed in ethanol (or other organic solvents).

Reason: Ethanol is an organic solvent that can dissolve both the hydrophobic tail and the hydrophilic head of the soap molecule. Since both parts of the soap molecule are soluble in ethanol, there is no tendency for the hydrophobic tails to cluster together to avoid the solvent. Hence, the driving force for micelle formation is absent, and micelles do not form in ethanol.

Carbon and its compounds are widely used as fuels because of the following reasons:

1. High calorific value: Carbon and its compounds (such as coal, petroleum, LPG, CNG, wood) release a large amount of heat energy on combustion, making them excellent fuels.

2. Complete combustion: Carbon compounds burn in the presence of oxygen to produce carbon dioxide and water, releasing a large amount of energy:

$$\mathrm{C + O_2 \rightarrow CO_2 + \text{Heat and Light}}$$
$$\mathrm{CH_4 + 2O_2 \rightarrow CO_2 + 2H_2O + \text{Heat and Light}}$$

3. Easy availability: Carbon-based fuels (coal, petroleum, natural gas) are abundantly available in nature.

4. Easy storage and transport: Carbon compounds are easy to store and transport in solid, liquid, or gaseous form.

5. Ease of ignition: Most carbon compounds ignite easily and burn smoothly.

6. Moderate ignition temperature: They have a moderate ignition temperature — not too low (safe to handle) and not too high (easy to ignite).

Conclusion: The combination of high energy output, easy availability, ease of storage and transport, and moderate ignition temperature makes carbon and its compounds ideal fuels for most applications.

Hard water: Water that contains dissolved salts of calcium ($\mathrm{Ca^{2+}}$) and magnesium ($\mathrm{Mg^{2+}}$) — such as their chlorides, sulphates, and hydrogencarbonates — is called hard water.

Soap: Soap is a sodium or potassium salt of a long-chain fatty acid, e.g., sodium stearate ($\mathrm{C_{17}H_{35}COONa}$).

Formation of scum:

When soap is added to hard water, the sodium ions of soap are replaced by calcium or magnesium ions. The calcium and magnesium salts of fatty acids are insoluble in water and precipitate out as a white, curdy solid called scum.

The chemical reactions are:
$$\mathrm{2C_{17}H_{35}COONa + CaCl_2 \rightarrow (C_{17}H_{35}COO)_2Ca\downarrow + 2NaCl}$$
$$\mathrm{2C_{17}H_{35}COONa + MgCl_2 \rightarrow (C_{17}H_{35}COO)_2Mg\downarrow + 2NaCl}$$

The insoluble calcium stearate or magnesium stearate formed is the scum.

Effect: The formation of scum wastes a large amount of soap (as it is converted to insoluble scum instead of forming lather) and leaves a grey, sticky deposit on clothes, making washing difficult. A large amount of soap must be used before lather is formed and effective cleaning begins.

Given: Soap is a sodium or potassium salt of a long-chain fatty acid. When dissolved in water, it undergoes hydrolysis to give a weakly alkaline solution.

Observation:

• Red litmus paper: When red litmus paper is dipped in soap solution, it turns blue. This indicates that the soap solution is alkaline (basic) in nature.
• Blue litmus paper: When blue litmus paper is dipped in soap solution, it remains blue (no change). This further confirms that the solution is alkaline.

Reason: Soap is a salt of a strong base (NaOH or KOH) and a weak acid (fatty acid). When dissolved in water, it undergoes hydrolysis:
$$\mathrm{C_{17}H_{35}COONa + H_2O \rightleftharpoons C_{17}H_{35}COOH + NaOH}$$
The NaOH produced makes the solution alkaline (basic), which turns red litmus blue.

Conclusion: Soap solution is alkaline — it turns red litmus blue and has no effect on blue litmus.

Hydrogenation:

Hydrogenation is the process of addition of hydrogen to an unsaturated organic compound (containing C=C double bonds or C≡C triple bonds) in the presence of a catalyst (usually finely divided nickel, palladium, or platinum) to convert it into a saturated compound.

General reaction:
$$\mathrm{R-CH=CH-R' + H_2 \xrightarrow{Ni/\Delta} R-CH_2-CH_2-R'}$$

Example:
$$\mathrm{CH_2=CH_2 + H_2 \xrightarrow{Ni,\,\Delta} CH_3-CH_3}$$
$$\text{(Ethene)} \quad\quad\quad\quad\quad\quad\quad\quad \text{(Ethane)}$$

Industrial Application:

The most important industrial application of hydrogenation is the conversion of vegetable oils (unsaturated) into vegetable ghee (vanaspati ghee) (saturated fats).

• Vegetable oils contain long-chain unsaturated fatty acids (with C=C double bonds) and are liquid at room temperature.
• When hydrogen gas is passed through vegetable oil at about $200^\circ\mathrm{C}$ in the presence of finely divided nickel catalyst, the double bonds are broken and hydrogen is added across them.
• The resulting product is a solid or semi-solid fat (vanaspati ghee) which is saturated.

$$\mathrm{Vegetable\;oil\;(unsaturated) + H_2 \xrightarrow{Ni,\,200^\circ C} Vegetable\;ghee\;(saturated)}$$

This process is used extensively in the food industry to manufacture vanaspati ghee from cheaper vegetable oils.

Concept: Addition reactions are characteristic of unsaturated hydrocarbons — those containing carbon-carbon double bonds (alkenes) or triple bonds (alkynes). Saturated hydrocarbons (alkanes) do not undergo addition reactions; they undergo substitution reactions instead.

Analysis of each compound:

Answer: Only $\mathrm{C_3H_6}$ (propene) and $\mathrm{C_2H_2}$ (ethyne) undergo addition reactions, as they are unsaturated hydrocarbons.

Example reactions:
$$\mathrm{C_3H_6 + H_2 \xrightarrow{Ni} C_3H_8}$$
$$\mathrm{C_2H_2 + 2H_2 \xrightarrow{Ni} C_2H_6}$$

Note: The question lists $\mathrm{C_2H_8}$ which is not a valid hydrocarbon formula; it is likely a misprint for $\mathrm{C_2H_2}$ (ethyne), which has been solved above.

Test: Bromine water test (or Bromine in carbon tetrachloride test)

Procedure:

1. Take two test tubes. Add a few drops of bromine water (or bromine dissolved in $\mathrm{CCl_4}$, which is orange-brown in colour) to each.
2. Pass the saturated hydrocarbon (e.g., ethane) through one test tube and the unsaturated hydrocarbon (e.g., ethene) through the other.

Observation:

• Saturated hydrocarbon (e.g., $\mathrm{C_2H_6}$): The orange-brown colour of bromine water is not decolourised (no change). Saturated hydrocarbons do not react with bromine water under normal conditions.
• Unsaturated hydrocarbon (e.g., $\mathrm{C_2H_4}$ or $\mathrm{C_2H_2}$): The orange-brown colour of bromine water is immediately decolourised (turns colourless). This is because bromine adds across the double or triple bond:

$$\mathrm{CH_2=CH_2 + Br_2 \rightarrow CH_2Br-CH_2Br}$$
$$\text{(Ethene, colourless)} \quad \text{(Bromine, orange-brown)} \quad \text{(1,2-dibromoethane, colourless)}$$

Conclusion:

• Decolourisation of bromine water → Unsaturated hydrocarbon.
• No decolourisation of bromine water → Saturated hydrocarbon.

Alternative test: Alkaline potassium permanganate ($\mathrm{KMnO_4}$) solution (purple) is also decolourised by unsaturated hydrocarbons but not by saturated ones.

Structure of soap molecule:
A soap molecule (e.g., sodium stearate, $\mathrm{C_{17}H_{35}COONa}$) has two parts:

1. Hydrophobic tail: A long non-polar hydrocarbon chain that is repelled by water but attracted to oil/grease.
2. Hydrophilic head: A polar ionic group ($-\mathrm{COO^-Na^+}$) that is attracted to water.

Mechanism of cleaning action (step-by-step):

Step 1 — Adsorption on dirt:
When a soap solution is applied to a dirty (oily/greasy) surface (cloth or skin), the hydrophobic tails of soap molecules dissolve into the oil/grease droplets (like dissolves like), while the hydrophilic heads remain in the water phase, pointing outward.

Step 2 — Micelle formation:
Many soap molecules arrange themselves around each oil/grease droplet in a spherical cluster called a micelle:

• The hydrophobic tails point inward, surrounding the oil/grease droplet at the centre.
• The hydrophilic heads point outward into the water.

The oil/grease is thus trapped inside the micelle.

Step 3 — Emulsification:
The micelles, being negatively charged on the outside (due to $-\mathrm{COO^-}$ groups), repel each other and remain dispersed in water without coagulating. This forms a stable emulsion of oil in water.

Step 4 — Rinsing:
When the cloth is rinsed with water (and agitated), the micelles containing the trapped dirt are washed away with the water, leaving the cloth clean.

Diagram (description):

• Centre of micelle: Oil/grease droplet (non-polar dirt)
• Middle layer: Hydrophobic tails of soap molecules pointing inward
• Outer layer: Hydrophilic heads of soap molecules pointing outward into water

Conclusion: The dual nature of the soap molecule (hydrophobic tail + hydrophilic head) enables it to surround oily dirt in micelles and carry it away in water, thus cleaning the surface.