Chapter 4: Carbon and its Compounds

Step 1 – Count the valence electrons

• Carbon (atomic number 6) has $$4$$ valence electrons.
• Oxygen (atomic number 8) has $$6$$ valence electrons.
  There are two oxygens: $$2 \times 6 = 12$$ electrons.

Total valence electrons to be shown in the electron-dot (Lewis) structure:

$$4 + 12 = 16$$ electrons

Step 2 – Choose the skeletal arrangement

Because carbon makes more bonds than oxygen, place carbon at the centre:

$$\mathrm{O\;\;C\;\;O}$$ (initial skeleton)

Step 3 – First place single bonds

Each C–O single bond uses one electron pair (two electrons).
Two such bonds use $$2 \times 2 = 4$$ electrons.

Electrons left: $$16 - 4 = 12$$

Step 4 – Complete the octet of the outer atoms (oxygen)

Each oxygen already has $$2$$ electrons in the C–O bond, so each still needs $$6$$ more
($$8 - 2 = 6$$). That is three lone pairs per oxygen.

For two oxygens: $$2 \times 6 = 12$$ electrons — exactly the 12 electrons left.

After this distribution every oxygen has an octet, but the carbon has only $$4$$ shared electrons (from the two single bonds) — an incomplete octet.

Step 5 – Convert lone pairs to multiple bonds to satisfy carbon’s octet

Move one lone pair from each oxygen into the C–O bond region to create a double bond on each side.

Now each bond is a C=O double bond (two shared electron pairs).

• Carbon now has $$4$$ shared pairs $$\times\,2 = 8$$ electrons ⇒ octet satisfied.
• Each oxygen keeps two lone pairs (4 electrons) and shares 4 electrons in the double bond ⇒ octet satisfied.

Step 6 – Draw the final electron-dot structure

Represent each bonding pair by two dots (or a line) and show the lone pairs on oxygen:

• Draw carbon in the middle.
• On both sides of the carbon place two pairs of dots (or two lines) between C and O, indicating the double bond.
• Around each oxygen place its remaining two lone pairs.

Written with dots, one acceptable depiction is:

O : : C : : O
where every “:” stands for a pair of electrons, and the two sets between C and each O form the double bond.

(Any equivalent diagram that shows two bonding pairs between C and each O, plus two lone pairs on every O, earns full marks.)

Goal – Draw the electron (Lewis) dot structure of the molecule $$\mathrm{S_8}$$, which consists of eight sulphur atoms joined in a ring.

Step 1 : Valence electrons on one sulphur atom
Group 16 element ⇒ $$6$$ valence electrons.

Step 2 : Total valence electrons in the molecule
$$8 \text{ atoms}\times 6 = 48 \text{ valence electrons}$$.

Step 3 : Decide the skeleton
The hint says the atoms form a ring, so place the eight S atoms at the corners of a regular octagon and join every neighbouring pair with a single bond.

S – S – S – S – S – S – S – S (the last S closes back to the first).

Eight single bonds use $$8\times 2 = 16$$ electrons.

Step 4 : Distribute remaining electrons as lone pairs
Electrons still to be placed: $$48-16 = 32$$.
Each lone pair = 2 electrons, so $$32/2 = 16$$ lone pairs.
Put two lone pairs on every sulphur atom (a total of $$8\times 2 = 16$$ lone pairs).

Step 5 : Octet check
For each S atom:
• two shared pairs (4 e⁻) + two lone pairs (4 e⁻) = 8 e⁻. ⇒ octet satisfied for every atom.

Final electron–dot structure
Draw an eight-membered ring. Between each adjacent pair of sulphur atoms show one shared pair of dots (or a single line). Around every S write four additional dots (two lone pairs) outside the ring.

(Any neat drawing that meets these conditions is the required Lewis structure of $$\mathrm{S_8}$$.)

Given : The compound is pentane whose molecular formula is $$\mathrm{C_5H_{12}}$$.

Concept : Structural (chain) isomers have the same molecular formula but differ in the actual connectivity of the carbon skeleton. To find every possible isomer we vary the length of the longest (parent) chain and then place the remaining carbon atoms as branches, taking care to avoid repetitions of the same structure drawn in a different way.

Step 1 — Parent chain of $$5$$ carbons

• A continuous chain of all five C-atoms is possible. No carbon atoms are left to branch, so only one structure fits:
    Draw a straight line of five carbons: $$\mathrm{CH_3\!\;–\!CH_2\!–\!CH_2\!–\!CH_2\!–\!CH_3}$$.
    Name: n-pentane (pentane).

Step 2 — Parent chain of $$4$$ carbons

• If the longest chain is $$4$$ carbons, one carbon atom remains to be attached as a side group (a methyl, $$\mathrm{CH_3}$$).
• Placing this branch on C-1 of the chain merely recreates the straight five-carbon chain already counted, so we attach it to the second carbon (C-2). Attaching it to C-3 would give an identical molecule when the chain is re-numbered from the other end. Thus only one new structure arises:
    Draw four carbons in a row and attach a $$\mathrm{CH_3}$$ group to the second carbon: $$\mathrm{(CH_3)_2CH–CH_2–CH_3}$$.
    Name: 2-methylbutane (iso-pentane).

Step 3 — Parent chain of $$3$$ carbons

• If the longest chain contains only $$3$$ carbons, two carbon atoms remain. To keep the longest chain at length $$3$$ the only option is to bond both remaining carbons to the central atom of the chain.
• This yields the compact structure $$\mathrm{(CH_3)_4C}$$ (all four methyl groups attached to one central carbon).
    Name: 2,2-dimethylpropane (neo-pentane).

Step 4 — Check for further possibilities

No other distinct arrangements exist. Any attempt to create a two-carbon parent chain would leave three remaining carbons; at least one would then extend the parent chain beyond two carbons, contradicting the choice. All variations already reduce to one of the three skeletons identified above.

Conclusion

Hence pentane, $$\mathrm{C_5H_{12}}$$, possesses exactly three structural (chain) isomers: n-pentane, 2-methylbutane and 2,2-dimethylpropane.

Step 1 : Electronic configuration of carbon

Atomic number of carbon = 6, therefore its electronic configuration is
$$1s^2\;2s^2\;2p^2$$
The outer (second) shell contains 4 electrons → valency of carbon = 4.

Step 2 : Property I – Tetravalency

• Because carbon has four valence electrons, it can complete its octet by sharing four more electrons with the same or with different atoms.
• Hence a single carbon atom is capable of forming up to four covalent bonds, e.g. $$\mathrm{CH_4}$$ (with four H-atoms), $$\mathrm{CCl_4}$$ (with four Cl-atoms) or $$\mathrm{CH_3CH_3}$$ (sharing with another C as well as H).
• This enormous bonding capacity with many different elements gives rise to a very large variety of compounds.

Step 3 : Property II – Catenation

• Carbon–carbon single bond energy is quite high (~ $$348\;\mathrm{kJ\;mol^{-1}}$$) and the size of the carbon atom is small. These factors make the C–C bond strong and stable.
• Consequently, carbon atoms can repeatedly bond to one another to form chains (straight or branched) and rings containing just a few to many thousands of carbon atoms.
• This self-linking ability, called catenation, vastly multiplies the number of possible skeletons on which different atoms/groups can be attached.

Step 4 : Conclusion

The simultaneous presence of tetravalency (bonding with up to four atoms) and catenation (bonding indefinitely with itself) makes carbon unique, leading to the millions of organic compounds found in nature and synthesized in laboratories.

Step 1 : Identify the type of compound
Cyclopentane is a cyclo-alkane, i.e. a saturated hydrocarbon whose carbon atoms form a closed ring containing five carbon atoms.

Step 2 : Determine the general formula for a cyclo-alkane
For a straight-chain alkane with $$n$$ carbon atoms, the molecular formula is $$\mathrm{C_nH_{2n+2}}$$.
When the chain closes to form a ring, two hydrogen atoms (one from each end) are lost, so the general formula for a cyclo-alkane becomes $$\mathrm{C_nH_{2n}}$$.

Step 3 : Apply n = 5 for cyclopentane
$$n = 5 \;\Rightarrow\; \text{Formula } = \mathrm{C_5H_{2\times5}} = \mathrm{C_5H_{10}}$$

Step 4 : Construct the electron-dot (Lewis) structure

• Arrange the five carbon atoms at the corners of a regular pentagon and join each adjacent pair with a single line (representing a shared pair of electrons, i.e. a C–C single covalent bond).
• Each carbon now has two covalent bonds used (to the neighbouring carbons). A carbon atom needs four covalent bonds in total, so it must form two more single bonds with hydrogen atoms.
• Attach two hydrogen atoms to every carbon, drawing single lines (shared pairs) between C and H.
• Finally, depict the valence electrons as dots:
  • Two dots (one from each atom) make every C–C or C–H single bond.
  • No lone pairs remain on carbon, and hydrogen attains the duplet configuration.

How to draw the diagram (describe for the notebook):
Draw a regular pentagon; place a “C” at each vertex. Draw a single line along every side of the pentagon for the C–C bonds. From every carbon, draw two short lines radiating outward; write “H” at the end of each line. Replace every line with a pair of dots if you wish to show electrons explicitly.

Thus, cyclopentane has the molecular formula $$\mathrm{C_5H_{10}}$$, and its electron-dot structure is a five-membered ring of carbon atoms with two hydrogens bonded to each carbon, all bonds represented by shared pairs of electrons.

Step 1 – Write the chemical change
When ethanol is heated with an oxidising agent such as alkaline $$\mathrm{KMnO_4}$$ or acidified $$\mathrm{K_2Cr_2O_7}$$, it is converted into ethanoic acid.

$$\mathrm{CH_3CH_2OH\;(ethanol)\;\xrightarrow[\;[O]\;]{\mathrm{oxidising\;agent}}\;CH_3COOH\;(ethanoic\;acid)}$$

Step 2 – Recall the meaning of oxidation in organic chemistry

• Addition of oxygen to a molecule, or
• Removal of hydrogen from a molecule

If either of these takes place, the reaction is called an oxidation reaction.

Step 3 – Compare the atomic composition before and after the reaction

The conversion has

• Addition of one extra oxygen atom (from 1 O to 2 O), and
• Removal of two hydrogen atoms (from 6 H to 4 H).

Either change alone satisfies the definition of oxidation; here, both occur simultaneously.

Step 4 – Conclusion
Because ethanol gains oxygen and loses hydrogen during its conversion to ethanoic acid, the process is classified as an oxidation reaction.

Given : A cylinder supplies a mixture of ethyne (acetylene) and pure oxygen for gas welding.

To explain : Why the same ethyne cannot be burnt satisfactorily with ordinary air.

1. Amount of oxidant available

• Ordinary air contains only $$21\,\%$$ oxygen by volume; the rest (mainly $$\mathrm{N_2}$$) does not take part in combustion.
• A separate oxygen cylinder supplies practically $$100\,\%$$ $$\mathrm{O_2}$$.
• Therefore, at the burner tip a much larger volume of air is needed to provide the same moles of $$\mathrm{O_2}$$ as a small jet of pure oxygen.

2. Chemical equations and heat released

(a) Complete combustion with pure oxygen

$$\mathrm{C_2H_2 + 5\,O_2 \;\longrightarrow\; 4\,CO_2 + 2\,H_2O}$$

Heat evolved → very high; the flame temperature reaches about $$3300\;\text{°C}$$, hot enough to melt iron (melting point $$\approx 1535\;\text{°C}$$).

(b) Combustion when only air is available

Because oxygen is limited, the reaction cannot go to completion. The chief products are

$$\mathrm{2\,C_2H_2 + 5\,O_2 \;\longrightarrow\; 4\,CO + 2\,H_2O}$$   (incomplete, sooty flame)

Some unburnt carbon separates as black soot: $$\mathrm{C_2H_2 \;\longrightarrow\; 2\,C\,(s) + H_2}$$.

The flame temperature falls to only $$\approx 1400\;\text{°C}$$— not sufficient for welding steel.

3. Quality of flame

• Luminous, sooty flame with air → deposits carbon on metal; joints become weak.
• Non-luminous, oxy-acetylene flame with oxygen → clean, sharply defined cone; metal edges melt uniformly and fuse.

4. Safety factor

An ethyne–air mixture in the ratio $$\approx 1:13$$ is explosive. Using pure oxygen, the gases mix only at the torch nozzle and burn immediately, reducing the explosion risk inside pipes or hoses.

Hence, ethyne is always burnt with pure oxygen for welding, whereas a mixture of ethyne and ordinary air would give an incomplete, cooler, sooty and potentially explosive flame unsuitable for the process.

Principle 

Carboxylic acids are much stronger acids than alcohols. Hence they react with mild bases such as sodium hydrogen-carbonate (baking soda) or sodium carbonate to give carbon dioxide. Ordinary alcohols are too weakly acidic to do so.

Materials 

• Unknown liquid A
• Unknown liquid B
• Fresh aqueous solution of sodium hydrogen-carbonate (or a pinch of solid $$\mathrm{NaHCO_3}$$)
• Delivery tube fitted to a test-tube (optional, for confirming $$\mathrm{CO_2}$$ with lime water)
• Lime-water (dil. $$\mathrm{Ca(OH)_2}$$)
• Dropper, test tubes, cork

Procedure 

1. Take about 2 mL of liquid A in a clean test-tube.
2. Add 1 mL of freshly prepared $$\mathrm{NaHCO_3}$$ solution (or a pinch of solid) to it.
3. Observe carefully for effervescence. If available, pass the evolved gas through lime-water via a delivery tube.
4. Repeat the same steps with liquid B in another test-tube.

Observations 

Inference 

• Liquid that produces effervescence (and turns lime-water milky) is a carboxylic acid.
• Liquid showing no reaction is an alcohol.

Chemical equations 

For a typical acid, ethanoic acid:

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

The $$\mathrm{CO_2}$$ produced turns lime-water milky:

$$\mathrm{Ca(OH)_2 + CO_2 \rightarrow CaCO_3\downarrow + H_2O}$$

For an alcohol, e.g. ethanol, no reaction occurs:

$$\mathrm{CH_3CH_2OH + NaHCO_3 \nrightarrow}$$ (no observable change)

Conclusion  A simple bicarbonate test, followed if desired by the lime-water confirmation, conclusively distinguishes an alcohol from a carboxylic acid in the school laboratory.

An oxidation–reduction (redox) reaction always involves two components:

• One component gets oxidised.
• The other component brings about that oxidation and is itself reduced.

That second component is called an oxidising agent. A Class 10 pupil can recognise an oxidising agent in any of the following, equivalent, ways:

1. It adds oxygen to another substance.
   Example: In the reaction
   $$\mathrm{2Cu + O_2 \rightarrow 2CuO}$$
   gaseous $$\mathrm{O_2}$$ supplies oxygen to copper, so $$\mathrm{O_2}$$ is the oxidising agent.

2. It removes hydrogen from another substance.
   Example: In the laboratory oxidation of ethanol,
   $$\mathrm{CH_3CH_2OH \xrightarrow[\;heat\;]{acid\,\;K_2Cr_2O_7} CH_3COOH}$$
   acidified $$\mathrm{K_2Cr_2O_7}$$ removes two hydrogens from ethanol; hence $$\mathrm{K_2Cr_2O_7}$$ is the oxidising agent and is itself reduced from Cr(VI) to Cr(III).

3. In electron-bookkeeping terms, it accepts electrons and therefore gets reduced.

Because they bring about oxidation, oxidising agents are also called oxidants. Typical oxidising agents used in organic chemistry at school level include:

• $$\mathrm{K_2Cr_2O_7}$$ (acidified potassium dichromate)
• $$\mathrm{KMnO_4}$$ (acidified or alkaline potassium permanganate)
• $$\mathrm{H_2O_2}$$ (hydrogen peroxide)
• Halogens such as $$\mathrm{Cl_2}$$ and $$\mathrm{Br_2}$$

In every case, the agent itself undergoes reduction while causing some other substance to be oxidised. Hence, the concise definition is:

An oxidising agent is a substance that oxidises another substance by adding oxygen, removing hydrogen or accepting electrons, and in the process the oxidising agent itself gets reduced.

Concept recalled

• Hardness of water is due mainly to $$\mathrm{Ca^{2+}}$$ and $$\mathrm{Mg^{2+}}$$ ions present as hydrogen-carbonates, chlorides or sulphates.
• Soap molecules (sodium or potassium salts of long-chain fatty acids) react with these ions to form insoluble calcium/magnesium salts (“scum”), therefore they do not produce lather in hard water. This lack (or very slow formation) of lather is what we actually observe in the laboratory test for hardness.
• Detergents are sodium salts of long-chain alkyl benzene sulphonates or alkyl sulphates. The calcium or magnesium salts of these sulphonates are themselves soluble in water, so detergents are able to give abundant lather both in soft and in hard water.

Reasoning

Because detergents do not get precipitated by $$\mathrm{Ca^{2+}}$$ or $$\mathrm{Mg^{2+}}$$ ions, the visible behaviour of a detergent solution (ready formation of lather) is the same whether the sample is soft or hard. Hence using a detergent you would observe lather in every case and would not obtain any criterion to distinguish the two types of water.

Conclusion

No. Detergent cannot be used to test the hardness of water. For this purpose one must use an ordinary soap solution, whose lathering behaviour is affected by hardness-causing ions.

Background — what soap does in water

• Soap is usually the sodium or potassium salt of a long-chain fatty acid, e.g. $$\mathrm{C_{17}H_{35}COO^-\,Na^+}$$.
• Each soap ion has
  • a long, non-polar hydrocarbon tail (hydrophobic) and
  • a short, polar carboxylate head (hydrophilic).
• In water the ions arrange themselves into spherical clusters called micelles.
• The tails embed in oily dirt; the heads remain in water, so the dirt particle becomes wrapped by soap and is now dispersed in water.

Why dirt still clings without agitation

• The soap tails can lodge firmly in the grease that is lodged among the fibres of the cloth.
• Although the micelle starts forming, the greasy particle is still held to the cloth by
  • inter-locking of the fibres and
  • weak adhesive forces (van der Waals, hydrogen bonding with dust, etc.).

Role of beating / scrubbing / agitation

1. Mechanical loosening – Striking or rubbing gives kinetic energy that overcomes the weak adhesive forces, so the greased–up micelles detach from the fabric surface.
2. Micelle formation is accelerated – Continuous motion breaks large oil patches into tiny droplets; more surface area lets more soap ions crowd round and stabilise each droplet.
3. Suspension and prevention of redeposition – Stirring keeps the detached, soap-coated dirt dispersed throughout the water until the rinse removes it. If water were still, the particles could settle back on the cloth.

Conclusion

Agitation supplies the mechanical energy needed to pull the soap-encapsulated dirt away from the fabric and keep it dispersed in the wash water; only then can rinsing carry the dirt out, leaving the clothes clean.

Step 1 : Write the structural formula of ethane
Ethane has molecular formula $$\mathrm{C_2H_6}$$. Writing out all valence-shell (single) covalent bonds gives
$$\mathrm{CH_3 \; - \; CH_3}$$
Each dash (–) represents one single covalent bond.

Step 2 : Count the C–H bonds
• In the left $$\mathrm{CH_3}$$ group the carbon is bonded to three hydrogens ⇒ 3 covalent bonds.
• In the right $$\mathrm{CH_3}$$ group the carbon is again bonded to three hydrogens ⇒ 3 more covalent bonds.
Therefore, total C–H bonds = $$3 + 3 = 6$$.

Step 3 : Count the C–C bond
The two carbon atoms are joined to each other by a single covalent bond, so
number of C–C bonds = $$1$$.

Step 4 : Add them up
Total covalent bonds in an ethane molecule = C–H bonds $$+$$ C–C bonds
$$6 + 1 = 7$$ covalent bonds.

Step 5 : Choose the correct option
The correct option is (b) 7 covalent bonds.

Step 1 – Recognise the parent chain
The name "butanone" is built from the root word "but-" and the suffix "-one".

• "but-"  ⇒  the parent hydrocarbon contains $$4$$ carbon atoms (butane).
• "-one"  ⇒  the characteristic suffix used for a ketone.

Step 2 – Locate the functional group indicated by the suffix
The suffix "-one" tells us that the compound possesses a carbonyl group $$\big(\;\mathrm C{=}\mathrm O\;\big)$$ bonded to two carbon atoms; this is exactly the definition of a ketone functional group.

Step 3 – Write (or visualise) the structure
For butanone the condensed structural formula is $$\mathrm{CH_3\! -\! CO\! -\! CH_2\! -\! CH_3}$$.
In this formula the carbonyl carbon ($$\mathrm{CO}$$) lies between two carbon atoms, confirming the presence of a ketone group.

Step 4 – Match with the given options

Therefore, the functional group present in butanone is the ketone group, corresponding to option (c).

Step 1 : Understand what produces a black coating (soot)
Common domestic fuels such as liquefied petroleum gas (LPG), kerosene or bio-gas contain hydrocarbons (e.g. propane, butane, hexane). When any hydrocarbon burns completely, every carbon atom is oxidised to carbon dioxide and every hydrogen atom to water vapour:

$$\mathrm{C_xH_y + O_2 \;\longrightarrow\; CO_2 + H_2O}$$

The flame in this case is blue and no unburnt carbon is left, so the vessel remains clean.

Step 2 : Recall what happens in incomplete combustion
If the supply of oxygen (air) is insufficient, the combustion is incomplete. The reaction pathway then also gives carbon monoxide and fine particles of unburnt carbon (soot):

$$\mathrm{C_xH_y + O_2\;(limited) \;\longrightarrow\; CO + C\,(soot) + H_2O}$$

These tiny solid carbon particles deposit on the cooler surface of the vessel, forming a black layer.

Step 3 : Connect to the given options

• (a) “the food is not cooked completely” – the state of the food does not decide soot formation.
• (b) “the fuel is not burning completely” – matches the explanation above; incomplete combustion produces soot.
• (c) “the fuel is wet” – moisture does not create soot, it only makes ignition harder.
• (d) “the fuel is burning completely” – complete combustion would not blacken the vessel.

Therefore the correct statement is option (b).

Step 1 – Recall the definition of a covalent bond
Covalent bonding is the mutual sharing of one or more pairs of valence electrons between two non-metal atoms so that each attains the stable electronic configuration of the nearest noble gas.

Step 2 – Write the electronic configurations of the atoms taking part

Step 3 – Plan the sharing
The central carbon atom has four valence electrons and therefore needs four more. It satisfies this requirement by forming:

• Three C–H single covalent bonds (one shared pair in each), and
• One C–Cl single covalent bond (one shared pair).

Step 4 – Show the Lewis (electron-dot) picture
Draw the symbols of the atoms; represent valence electrons of carbon as dots (•), of hydrogen as crosses (×), and of chlorine as circles (◦). Arrange them so that the pairs to be shared lie between the relevant atoms:

Describe to the student what to draw:
• Place C in the centre with four single dots around it.
• Surround it by three H symbols, each contributing one cross, placed so that each H shares a pair with one carbon dot.
• Place one Cl symbol; show one circle from Cl sharing with the fourth carbon dot.
• Complete chlorine’s remaining six electrons (three lone pairs) as circles around Cl.

This diagram contains four shared pairs (one in each C–H or C–Cl bond) and the requisite lone pairs on chlorine.

Step 5 – Verify attainment of stable configurations

• After sharing, carbon has $$8$$ electrons in its valence shell: $$\bigl(4\text{ of its own}+4\text{ shared}\bigr)$$ → octet completed.
• Each hydrogen has $$2$$ electrons (its own $$1$$ + the $$1$$ shared with carbon) → duplet completed.
• Chlorine has $$8$$ valence electrons (its own $$7$$ + the $$1$$ shared with carbon) → octet completed.

Step 6 – Conclude on the nature of the bond
In $$\mathrm{CH_3Cl}$$ every C–H and C–Cl linkage is a single covalent bond formed by the sharing of one electron from each bonded atom. Because electrons are shared, not transferred, the molecule is electrically neutral and typically exhibits the general properties of covalent compounds (low melting and boiling points, poor electrical conductivity when not ionised, etc.). Thus the bonding in $$\mathrm{CH_3Cl}$$ clearly illustrates the essential features of a covalent bond.

Step 1 : Understand the meaning of the term

In organic chemistry we come across very large numbers of compounds. To bring some order into their study, chemists arrange them into families called homologous series.

Definition
A homologous series is a family of organic compounds that

• have the same general formula,
• show similar chemical properties because they possess the same functional group,
• differ from the next (or the previous) member by a fixed structural unit, namely one methylene group $$\mathrm{\left(\, -CH_2- \,\right)}$$, and consequently,
• differ in their molecular masses by a constant value of $$14\,\text{u}$$ (the mass of one $$\mathrm{CH_2}$$ group).

Step 2 : Examine a concrete example — the alkane series

Observe the pattern:

• Each successive member increases the carbon number $$n$$ by 1.
• The molecular formula always satisfies $$\mathrm{C_nH_{2n+2}}$$.
• The difference between consecutive molecular masses is
  $$30-16 = 14\,\text{u},\; 44-30 = 14\,\text{u},\; 58-44 = 14\,\text{u}$$ ⇒ the addition of one $$\mathrm{CH_2}$$ unit each time.

Step 3 : Verify similarity in chemical properties

All alkanes undergo complete combustion to give carbon dioxide and water:

For methane    $$\mathrm{CH_4 + 2\,O_2 \;\longrightarrow\; CO_2 + 2\,H_2O}$$

For propane    $$\mathrm{C_3H_8 + 5\,O_2 \;\longrightarrow\; 3\,CO_2 + 4\,H_2O}$$

The type of reaction (combustion) and the nature of products are the same, confirming similar chemical behaviour.

Step 4 : Note the gradation in physical properties

While chemical properties remain alike, physical properties change e.g.

• Boiling points increase smoothly from methane (–161 °C) to ethane (–89 °C) to propane (–42 °C), and so on.
• Density, viscosity, etc., also show regular trends.

Conclusion

A homologous series is, therefore, a systematic grouping of organic compounds fulfilling the above criteria, with the alkane series (methane, ethane, propane, …) serving as a clear illustration.

To distinguish ethanol (ethyl alcohol, $$\mathrm{C_2H_5OH}$$) from ethanoic acid (acetic acid, $$\mathrm{CH_3COOH}$$) we examine both their physical characteristics and simple laboratory reactions.

1. Physical properties

• Odour
  Ethanol has a characteristic pleasant, alcoholic smell, whereas ethanoic acid possesses a sharp vinegar-like smell that stings the nose.
• Boiling point
  The boiling point of ethanol is $$78\,{}^{\circ}\!\mathrm{C}$$; that of ethanoic acid is noticeably higher at $$118\,{}^{\circ}\!\mathrm{C}$$ because its molecules are linked by stronger hydrogen bonds (dimer formation).
• Effect on litmus
  Ethanol is almost neutral: blue or red litmus remains unchanged. Ethanoic acid is weakly acidic and turns blue litmus red.
• Miscibility with water at room temperature
  Both liquids mix in all proportions, but when the mixture is cooled below $$16\,{}^{\circ}\!\mathrm{C}$$ ethanoic acid separates out as ice-like crystals (“glacial” ethanoic acid). Ethanol shows no such freezing separation under these conditions.

2. Chemical tests (reactions)

1. Sodium bicarbonate test
   Add a small amount of solid $$\mathrm{NaHCO_3}$$ to each liquid.
   • With ethanol: no brisk effervescence.
   • With ethanoic acid: immediate fizzing as $$\mathrm{CO_2}$$ gas is liberated:

   $$\mathrm{CH_3COOH + NaHCO_3 \;\longrightarrow\; CH_3COONa + H_2O + CO_2\uparrow}$$

   (Bubble the gas through lime-water; it turns milky, confirming $$\mathrm{CO_2}$$.)

2. Reaction with metallic sodium
   Add a small piece of clean $$\mathrm{Na}$$.
   • Ethanol reacts slowly, giving a gentle effervescence of $$\mathrm{H_2}$$ and forming sodium ethoxide:

     $$\mathrm{2\,C_2H_5OH + 2\,Na \;\longrightarrow\; 2\,C_2H_5ONa + H_2\uparrow}$$

   • Ethanoic acid reacts much faster and more vigorously, because the acidic hydrogen is more readily released:

     $$\mathrm{2\,CH_3COOH + 2\,Na \;\longrightarrow\; 2\,CH_3COONa + H_2\uparrow}$$

3. Neutralisation with a strong base
   Add a few drops of $$\mathrm{NaOH}$$ solution and boil gently.
   • Ethanol shows no visible change.
   • Ethanoic acid is neutralised, giving sodium acetate and water:

   $$\mathrm{CH_3COOH + NaOH \;\longrightarrow\; CH_3COONa + H_2O}$$

   The temperature of the mixture rises (exothermic) and the smell of vinegar disappears.

4. Esterification (smell test)
   Warm each liquid separately with the other (ethanol + ethanoic acid) and a few drops of conc. $$\mathrm{H_2SO_4}$$.
   • If ethanol is absent, no fruity odour appears.
   • If both are present together, a sweet fruity smell of ethyl ethanoate $$\mathrm{CH_3COOC_2H_5}$$ is produced.
     This test therefore probes the presence of each reactant in mixtures; it is not, however, a direct single-substance test.

Summary

Thus, simple laboratory observations—especially the sodium bicarbonate test and litmus test—readily differentiate ethanol from ethanoic acid.

Soap (for example sodium stearate, $$\mathrm{C_{17}H_{35}COO^-Na^+}$$) is an amphiphilic molecule:

• The long hydrocarbon chain $$\mathrm{C_{17}H_{35}}$$ is non-polar (= hydrophobic, water-hating).
• The carboxylate head $$\mathrm{COO^-Na^+}$$ is ionic (= hydrophilic, water-loving).

When a small amount of soap is put into water, each molecule orients itself such that its ionic head is surrounded by water while its hydrocarbon tail stays away from water. If the concentration of soap continues to increase and reaches a certain minimum value called the critical micelle concentration (CMC), the individual molecules can no longer stay isolated. To minimise the contact between their hydrophobic tails and the surrounding polar water, dozens of soap molecules aggregate spontaneously:

• All the hydrocarbon tails crowd together in the interior of the aggregate (away from water).
• All the ionic heads remain on the outside, in contact with water.

This spherical aggregate is called a micelle. Thus micelle formation in water is a direct consequence of the dual nature of soap molecules and the strong polarity of water.

Why not in ethanol?

• Ethanol ($$\mathrm{CH_3CH_2OH}$$) is much less polar than water and can dissolve both the hydrocarbon chain and the ionic head to some extent.
• Because the hydrophobic repulsion is weak in ethanol, the soap molecules remain dispersed individually; the driving force to hide the tails is absent.
• Hence soap does not form micelles in ethanol (or in most other organic solvents).

Step 1  Understand the term “fuel”
A fuel is any substance that, on combustion in air or oxygen, produces a large amount of heat that can be put to practical use.

Step 2  Recall what happens when a carbon compound burns
In the presence of sufficient oxygen, a typical hydrocarbon such as methane burns according to
$$\mathrm{CH_4 + 2\,O_2 \;\longrightarrow\; CO_2 + 2\,H_2O + \text{energy}}$$
The reaction is highly exothermic; that is, the enthalpy of combustion $$\Delta H_c$$ has a large negative value (for methane, $$\Delta H_c \approx -890\;\mathrm{kJ\,mol^{-1}}$$). Similar large magnitudes are obtained for almost all carbon-based fuels (e.g. LPG, petrol, kerosene, diesel, coal).

Step 3  Relate the large heat output to bond energies
• Carbon compounds contain numerous C–H and C–C single bonds.
• During combustion, these comparatively high-energy bonds are replaced by stronger C=O and O–H bonds in $$\mathrm{CO_2}$$ and $$\mathrm{H_2O}$$.
• The formation of stronger bonds releases a lot of energy, hence the high calorific value (heat produced per unit mass or volume).

Step 4  List the practical advantages that follow from this high calorific value

• Small mass, large heat — only a few grams of LPG or petrol are enough to raise the temperature of several litres of water.
• Ease of ignition — carbon fuels have ignition temperatures that can be reached with a simple spark, match-stick or compression (in diesel engines), making them convenient for everyday use.
• Controllability — regulating the air (oxygen) supply can easily adjust the rate of combustion and, therefore, the rate of heat generation.
• Availability and cost — fossil fuels (coal, petroleum, natural gas) are abundant in nature and are relatively inexpensive to extract, refine and transport.
• Storage and transport — solid (coal, coke), liquid (petrol, kerosene) and gaseous (CNG, LPG) forms enable flexible handling for domestic, industrial and vehicular needs.

Step 5  Summarise in one clear sentence
Carbon and its compounds are preferred as fuels because their combustion is highly exothermic, releasing a large amount of heat per unit mass (high calorific value), they ignite easily under controlled conditions, are available in plenty, and can be conveniently stored, transported and handled in different physical forms.

Step 1 : Recall the nature of soap

Common toilet soap is the sodium (or potassium) salt of a long-chain fatty acid, symbolically written as $$\mathrm{RCOO^-\,Na^+}$$ where $$\mathrm{R}$$ represents a long hydrocarbon chain such as $$\mathrm{C_{17}H_{35}}$$.

Step 2 : Recall what makes water ‘hard’

Hard water contains appreciable amounts of dissolved $$\mathrm{Ca^{2+}}$$ and/or $$\mathrm{Mg^{2+}}$$ ions, usually as their bicarbonates, chlorides or sulphates.

Step 3 : Write the reaction between soap and the calcium ion

When soap is added to such water, the divalent metal ion displaces sodium from the soap according to

$$2\,\mathrm{RCOO^-\,Na^+}+\mathrm{Ca^{2+}} \;\longrightarrow\; (\mathrm{RCOO})_2\mathrm{Ca}\downarrow + 2\,\mathrm{Na^+}$$

The product $$(\mathrm{RCOO})_2\mathrm{Ca}$$ is calcium stearate (or in general, calcium fatty acid salt). It is insoluble in water and therefore separates out as a dirty, greyish precipitate.

Step 4 : The same happens with the magnesium ion

$$2\,\mathrm{RCOO^-\,Na^+}+\mathrm{Mg^{2+}} \;\longrightarrow\; (\mathrm{RCOO})_2\mathrm{Mg}\downarrow + 2\,\mathrm{Na^+}$$

Step 5 : How the precipitate appears as scum

• The insoluble calcium/magnesium fatty acid salts float or stick to the container walls.
• This floating, sticky precipitate is called scum.
• Because part of the soap is consumed in forming scum, little or no lather appears until all $$\mathrm{Ca^{2+}}$$/$$\mathrm{Mg^{2+}}$$ ions are removed.

Conclusion

Scum is formed because the $$\mathrm{Ca^{2+}}$$ and $$\mathrm{Mg^{2+}}$$ ions present in hard water react with soap to give insoluble calcium or magnesium salts of the fatty acid, which precipitate out.

Soaps are sodium or potassium salts of long-chain fatty acids, e.g. $$\mathrm{C_{17}H_{35}COONa}$$.

When soap is dissolved in water, the anion $$\mathrm{RCOO^-}$$ undergoes slight hydrolysis:

$$\mathrm{RCOO^- + H_2O \;\longrightarrow\; RCOOH + OH^-}$$

This reaction produces hydroxide ions $$\mathrm{(OH^-)}$$, so the resulting solution is alkaline (basic).

Effect on litmus paper:

• Red litmus: In alkaline medium it turns blue, so red litmus paper becomes blue when dipped in soap solution.
• Blue litmus: Already blue in alkaline medium, so it remains unchanged.

Step 1 – Understand the term “hydrogenation”
Hydrogenation is a chemical reaction in which molecular hydrogen, $$\mathrm{H_2}$$, is added across the double or triple bonds of an unsaturated organic compound (alkene or alkyne) to convert it into a saturated compound (alkane). The reaction is carried out:

• in the presence of a finely divided metal catalyst such as $$\mathrm{Ni}$$, $$\mathrm{Pd}$$ or $$\mathrm{Pt}$$,
• at a moderately high temperature (about $$200 \text{–} 300\,{}^{\circ}\mathrm{C}$$).

For example, the hydrogenation of ethene is written as

$$\mathrm{CH_2=CH_2 + H_2 \xrightarrow[200\,{}^{\circ}C]{Ni} CH_3\!\! -\!\! CH_3}$$

The double bond $$\mathrm{C = C}$$ is broken and each carbon atom gains one hydrogen atom, turning the molecule into the saturated alkane ethane.

Step 2 – Industrial application of hydrogenation

The most important large-scale use of hydrogenation is the conversion of vegetable oils into solid or semi-solid fats such as vanaspati ghee (also called hydrogenated oil).

Vegetable oils are mainly long-chain triglycerides of unsaturated fatty acids that contain one or more $$\mathrm{C = C}$$ bonds. When these oils are reacted with hydrogen in the presence of a nickel catalyst, many of the double bonds are saturated, raising the melting point and giving a product that is solid at room temperature.

Representative equation (simplified):

$$\mathrm{Unsaturated\;oil\; + H_2 \xrightarrow[200\,{}^{\circ}C]{Ni} Saturated\;fat\;(vanaspati)}$$

This hydrogenated fat is widely used in the food industry for making bakery products like pastries, biscuits and for general cooking purposes.

Conclusion
Hydrogenation is therefore both a fundamental organic reaction and a crucial industrial process for turning liquid vegetable oils into usable solid fats.

Key concept → Addition reactions are characteristic of unsaturated hydrocarbons, i.e. those that possess at least one carbon–carbon multiple bond (double or triple). When a reagent such as $$\mathrm{H_2}$$, $$\mathrm{Cl_2}$$, $$\mathrm{Br_2}$$, etc. adds across that multiple bond, the molecule becomes saturated.

Therefore, we must first decide whether each given formula is saturated (alkane) or unsaturated (alkene / alkyne).

Decision → Only unsaturated members $$\mathrm{C_3H_6}$$ (propene) and $$\mathrm{C_2H_2}$$ (ethyne) can undergo addition reactions.

Illustration (optional): For example, in the presence of $$\mathrm{Ni}$$ catalyst, propene adds hydrogen:

$$\mathrm{C_3H_6 + H_2 \xrightarrow[\;\;\;]{Ni,\;473\,K} C_3H_8}$$

The double bond breaks and the product is the corresponding alkane (propane).

Hence, among the given compounds, the ones that undergo addition reactions are:

• $$\mathrm{C_3H_6}$$ (propene)
• $$\mathrm{C_2H_2}$$ (ethyne)

Concept recalled
Saturated hydrocarbons (alkanes) contain only single C–C bonds, while unsaturated hydrocarbons (alkenes and alkynes) possess at least one double or triple bond. The extra π bond(s) in an unsaturated molecule make it capable of undergoing addition reactions that a saturated compound cannot perform under ordinary conditions.

The addition–decolourisation test (Bromine-water test; also called the bromine in $$\mathrm{CCl_4}$$ test)

1. Reagents required: freshly prepared orange bromine solution in $$\mathrm{CCl_4}$$ (or in water), a small quantity of the hydrocarbon sample in an inert solvent if necessary, dry test-tubes with stoppers.
2. Procedure
   1. Pour 1–2 cm³ of the hydrocarbon into a test-tube.
   2. Add 2–3 drops of the bromine solution, stopper and shake gently.
   3. Keep the tube at room temperature, away from direct sunlight (to avoid free-radical substitution in alkanes).
3. Observation
   • If the orange colour of bromine disappears rapidly, the sample is unsaturated.
   • If the colour remains (or fades only extremely slowly on prolonged standing in sunlight), the sample is saturated.
4. Chemical explanation with equations

   Hydrocarbon class	Representative reaction	Result
   Alkene (e.g. ethene)	$$\mathrm{CH_2{=}CH_2 + Br_2 \to CH_2Br{-}CH_2Br}$$ (1,2-dibromoethane)	Bromine adds across the double bond; colour vanishes.
   Alkyne (e.g. ethyne)	$$\mathrm{HC{=}CH + Br_2 \to CHBr{=}CHBr \to CHBr_2{-}CHBr_2}$$ (stepwise)	Two successive additions possible; colour vanishes.
   Alkane (e.g. ethane)	$$\mathrm{CH_3{-}CH_3 + Br_2 \xrightarrow[\text{dark}]{\text{rt}} \text{no reaction}}$$	No addition; colour persists.

5. Alternative reagent mentioned by NCERT: dilute alkaline $$\mathrm{KMnO_4}$$ (Baeyer’s reagent). Its purple colour is discharged by unsaturated hydrocarbons via oxidative addition, while it remains purple with alkanes.

Conclusion
Unsaturated hydrocarbons decolourise bromine water (or alkaline $$\mathrm{KMnO_4}$$) instantly because they undergo rapid addition across their multiple bonds; saturated hydrocarbons do not, hence the reagent’s colour stays.

Structure of a typical soap molecule

• General formula: $$\mathrm{RCOO^-\;Na^+}$$, where $$\mathrm{R}$$ = a long, unbranched hydrocarbon chain (usually $$\mathrm{C_{15} – C_{17}}$$).
• The two structurally different parts behave differently toward water:
  • Tail : the non-polar hydrocarbon chain $$\mathrm{R–}$$ (hydrophobic & lipophilic).
  • Head : the ionic group $$\mathrm{–COO^-\,Na^+}$$ (hydrophilic).

Step-by-step mechanism of cleaning

1. Adsorption on grease
   Oil / grease on fabric is non-polar, so ordinary water cannot wet it. When soap is added, the hydrophobic tails are attracted to (dissolve in) the grease while the ionic heads remain in the surrounding water.
2. Micelle formation
   As more soap molecules arrange themselves, they surround the grease particle. The tails point inward and the heads point outward into the water, giving a spherical aggregate called a micelle (draw a sphere with hydrocarbon tails inside and $$\mathrm{–COO^-}$$ groups on the surface).
3. Emulsification
   Because the micelle’s surface is completely ionic, the whole grease droplet now behaves as if it were soluble in water, i.e. it is emulsified. The repulsion between the similarly charged heads prevents the droplets from coalescing again.
4. Removal by agitation and rinsing
   Rubbing (in hand-washing) or tumbling (in a washing machine) loosens fibres, allowing micelles to lift the dirt from the fabric. When fresh water is poured, the suspended micelles—with the trapped grease inside—are washed away, leaving the surface clean.

Why soap fails in hard water (additional point)

In the presence of $$\mathrm{Ca^{2+}}$$ or $$\mathrm{Mg^{2+}}$$ ions, insoluble calcium / magnesium carboxylates form a curdy precipitate (scum) instead of micelles, so cleaning efficiency drops.