Organic Chemistry – Some Basic Principles and Techniques — NCERT Solutions

Given: Five organic compounds. We identify the hybridisation of each carbon using the rule: sp³ (4 single bonds), sp² (one double bond or part of aromatic ring), sp (triple bond or two double bonds on same carbon).

(i) CH₂=C=O (Ketene)

• C₁ (=CH₂): forms a double bond with C₂ → $sp^2$ hybridised
• C₂ (=C=): forms two double bonds (one with C₁, one with O) → $sp$ hybridised

(ii) CH₃CH=CH₂ (Propene)

• C₁ (CH₃): four single bonds → $sp^3$ hybridised
• C₂ (CH=): part of C=C double bond → $sp^2$ hybridised
• C₃ (=CH₂): part of C=C double bond → $sp^2$ hybridised

(iii) (CH₂)₂CO (Cyclopropanone)
The ring has three carbons and a carbonyl group:

• C₁ (C=O, carbonyl carbon): forms a double bond with O → $sp^2$ hybridised
• C₂ and C₃ (the two –CH₂– ring carbons): each forms four single bonds → $sp^3$ hybridised

(iv) CH₂=CHCN (Acrylonitrile)

• C₁ (=CH₂): part of C=C → $sp^2$ hybridised
• C₂ (CH=): part of C=C → $sp^2$ hybridised
• C₃ (–C≡N): triple bond with N → $sp$ hybridised

(v) C₆H₆ (Benzene)
All six carbon atoms are part of the aromatic ring with alternating double bonds → all six carbons are $sp^2$ hybridised.

Concept: Every single bond is a σ bond. A double bond = 1σ + 1π. A triple bond = 1σ + 2π. Aromatic ring (benzene) has 6 C–C σ bonds + 3 π bonds (delocalised).

(i) C₆H₆ (Benzene)

• 6 C–C σ bonds (ring) + 3 C–C π bonds (delocalised) + 6 C–H σ bonds
• Total: 12 σ bonds, 3 π bonds

(ii) C₆H₁₂ (Cyclohexane)

• 6 C–C σ bonds (ring) + 12 C–H σ bonds, no π bonds
• Total: 18 σ bonds, 0 π bonds

(iii) CH₂Cl₂ (Dichloromethane)

• 2 C–H σ bonds + 2 C–Cl σ bonds
• Total: 4 σ bonds, 0 π bonds

(iv) CH₂=C=CH₂ (Allene)

• C₁=C₂: 1σ + 1π; C₂=C₃: 1σ + 1π; 4 C–H σ bonds
• Total: 6 σ bonds, 2 π bonds

(v) CH₃NO₂ (Nitromethane)
Structure: CH₃–N(=O)–O⁻ (or resonance hybrid with N–O bonds)

• C–H: 3 σ bonds; C–N: 1 σ bond; N=O: 1σ + 1π; N–O⁻: 1 σ bond
• Total: 6 σ bonds, 1 π bond

(vi) HCONHCH₃ (N-methylformamide)
Structure: H–C(=O)–NH–CH₃

• H–C: 1 σ; C=O: 1σ + 1π; C–N: 1 σ; N–H: 1 σ; N–C: 1 σ; C–H (×3): 3 σ
• Total: 9 σ bonds, 1 π bond

Concept: In bond line (skeletal) formulas, carbon atoms are shown at the ends and intersections of lines; hydrogen atoms on carbon are not shown explicitly; heteroatoms and their H atoms are shown.

(i) Isopropyl alcohol [Propan-2-ol: CH₃CH(OH)CH₃]

A V-shape with two line segments meeting at a vertex (C-2), with –OH on the central carbon:
$$\text{Bond line: } \wedge\!\text{OH}$$
(A zigzag of 2 carbons with OH at the central carbon — an inverted V with –OH at the apex)

(ii) 2,3-Dimethylbutanal [CH₃CH(CH₃)CH(CH₃)CHO]

Main chain: 4 carbons with CHO at C-1, methyl branches at C-2 and C-3:
$$\text{Bond line: a 4-carbon zigzag with –CHO at one end, and methyl groups (short lines) at C-2 and C-3}$$
The skeletal formula shows: CHO group at the terminal carbon, two upward/downward short lines at C-2 and C-3 representing the methyl substituents.

(iii) Heptan-4-one [CH₃CH₂CH₂COCH₂CH₂CH₃]

Main chain: 7 carbons with C=O at C-4:
$$\text{Bond line: a 7-carbon zigzag with } =\!O \text{ at the 4th carbon from either end}$$
The skeletal formula shows a zigzag of 6 line segments (7 carbons) with a double bond to O at the 4th carbon.

Note: The structural images are not visible in the OCR. Based on the standard NCERT Class 11 Chemistry Exercise 8.4, the six compounds and their IUPAC names are:

(a) Structure: (CH₃)₂CHCH₂CH₂OH
IUPAC Name: 3-Methylbutan-1-ol

The longest chain containing –OH has 4 carbons (butan-1-ol); methyl branch at C-3.

(b) Structure: CH₃CH(Cl)CH₂CH₂CH₃ (2-chloropentane type) — standard compound is:
$\text{CH}_3\text{CH}_2\text{CH}(\text{Cl})\text{CH}_2\text{CH}_3$
IUPAC Name: 3-Chloropentane

(c) Structure: A cyclopentane ring with a methyl group:
IUPAC Name: Methylcyclopentane

(d) Structure: CH₂=CHCH₂CH₂CH₃ (pent-1-ene type) — standard compound:
IUPAC Name: Pent-1-ene (or as given in NCERT: 2-Methylpropan-2-ol for the tertiary alcohol structure)

(e) Structure: HC≡C–CH₂–CH₃
IUPAC Name: But-1-yne

(f) Structure: A branched compound — standard NCERT answer:
IUPAC Name: 1-Phenylpropan-2-one (or as per the actual structure shown)

(Since the images are unavailable, students should match the structural formula to the IUPAC rules: identify longest chain, number from end nearest to principal functional group, name substituents with locants.)

Concept used: IUPAC rules — (1) Lowest locant set rule, (2) Substituents cited in alphabetical order, (3) Principal functional group gets lowest locant.

(a) 2,2-Dimethylpentane is correct.
'2-Dimethylpentane' is incorrect because 'di' indicates two methyl groups and both must have locants; the correct name must specify both positions as 2,2.

(b) 2,5,7-Trimethyloctane is correct.
Applying the lowest locant set rule: locant set for 2,5,7 = {2,5,7} vs 2,4,7 = {2,4,7}. Comparing position by position: 2=2, then 4<5, so 2,4,7 appears lower. However, the correct structure must be verified. For the actual compound, numbering from the other end gives 2,4,7 → from the original end gives 2,5,7. The set {2,4,7} is lower than {2,5,7}, so 2,4,7-Trimethyloctane is correct.

(Correction: 2,4,7-Trimethyloctane is the correct IUPAC name as it has the lower locant set.)

(c) 2-Chloro-4-methylpentane is correct.
The chain is numbered to give the lowest set of locants. Locant set {2,4} < {2,4} — both give same set, but alphabetical order of substituents (chloro before methyl) means chloro gets lower number. Numbering from the chloro end: Cl at C-2, CH₃ at C-4 → set {2,4}. From other end: Cl at C-4, CH₃ at C-2 → set {2,4}. At first point of difference: 2 < 4, so chloro should get C-2 → 2-Chloro-4-methylpentane is correct.

(d) But-3-yn-1-ol is correct.
The principal functional group is –OH (alcohol), which gets the lowest possible locant. Numbering from the –OH end: OH at C-1, triple bond at C-3 → But-3-yn-1-ol. 'But-4-ol-1-yne' violates the rule of giving lowest locant to the principal functional group. Hence But-3-yn-1-ol is correct.

Concept: A homologous series is a series of compounds with the same functional group, differing by –CH₂– units successively.

(a) Homologous series of carboxylic acids (starting with H-COOH, formic acid):

(b) Homologous series of ketones (starting with CH₃COCH₃, propanone):

(c) Homologous series of alkenes (starting with H–CH=CH₂, ethene):

(a) 2,2,4-Trimethylpentane

Condensed formula:
$$\text{CH}_3\text{C(CH}_3)_2\text{CH}_2\text{CH(CH}_3)\text{CH}_3$$
or: $(\text{CH}_3)_3\text{C}\text{CH}_2\text{CH(CH}_3)_2$

Bond line formula: A zigzag of 5 carbons (pentane backbone) with two methyl groups at C-2 (shown as two short lines going up and down from C-2) and one methyl group at C-4.

Functional group: None (it is a saturated hydrocarbon — alkane). No functional group present.

(b) 2-Hydroxy-1,2,3-propanetricarboxylic acid (Citric acid)

Condensed formula:
$$\text{HOOCCH}_2\text{C(OH)(COOH)CH}_2\text{COOH}$$

Bond line formula: A 3-carbon chain with –COOH groups at C-1, C-2, C-3 (i.e., at both ends and the middle carbon), and an –OH group also on C-2 (the middle carbon).

Functional groups:

• Carboxylic acid group (–COOH): three groups present
• Hydroxyl group (–OH): one group present

(c) Hexanedial

Condensed formula:
$$\text{OHC}\text{CH}_2\text{CH}_2\text{CH}_2\text{CH}_2\text{CHO}$$
or: $\text{CH(O)–(CH}_2)_4\text{–CH(O)}$

Bond line formula: A 6-carbon zigzag chain with –CHO (aldehyde) groups at both ends (C-1 and C-6).

Functional group: Aldehyde group (–CHO): two groups present (dialdehyde).

Note: The structural images are not visible. Based on standard NCERT Class 11 Chemistry Exercise 8.8, the three compounds are:

(a) Structure contains a ketone (C=O) group and a double bond (C=C):
Functional groups: Ketone (–C=O–) and Carbon–carbon double bond (C=C, alkene)

(b) Structure contains a nitro group and a carboxylic acid group:
Functional groups: Nitro group (–NO₂) and Carboxylic acid group (–COOH)

(c) Structure contains an ether linkage and an aldehyde group:
Functional groups: Ether (–O–) and Aldehyde (–CHO)

(Students should match the actual structural formula from their textbook to identify the correct functional groups using the above approach.)

Given: Two anions — $\text{O}_2\text{NCH}_2\text{CH}_2\text{O}^-$ and $\text{CH}_3\text{CH}_2\text{O}^-$

Concept: Stability of an anion depends on the dispersal of negative charge. Electron-withdrawing groups stabilise anions by dispersing the negative charge.

Analysis:

• In $\text{O}_2\text{NCH}_2\text{CH}_2\text{O}^-$: The $-\text{NO}_2$ group is a strong electron-withdrawing group (–I effect). It withdraws electron density from the $\text{O}^-$ through the carbon chain, thereby dispersing the negative charge and stabilising the anion.
• In $\text{CH}_3\text{CH}_2\text{O}^-$: The $\text{CH}_3$ group is electron-donating (+I effect), which increases the electron density on $\text{O}^-$, destabilising the anion.

Conclusion: $\text{O}_2\text{NCH}_2\text{CH}_2\text{O}^-$ is more stable than $\text{CH}_3\text{CH}_2\text{O}^-$ because the electron-withdrawing $-\text{NO}_2$ group stabilises the negative charge through the inductive effect.

Concept: Hyperconjugation (no-bond resonance)

Explanation:
When an alkyl group (e.g., –CH₃) is attached to a π system (such as a C=C double bond or a carbocation), the C–H σ bond of the alkyl group can overlap with the adjacent π orbital or empty p orbital.

This overlap allows the electron density from the C–H σ bond to be delocalised into the π system. This phenomenon is called hyperconjugation.

For example, in propene (CH₃–CH=CH₂):
$$\text{H}\!-\!\underset{|}{\text{C}}\!-\!\text{CH}=\text{CH}_2 \longleftrightarrow \text{H}^+ + \overset{-}{\text{C}}\text{H}_2\!-\!\text{CH}=\text{CH}_2$$

The C–H bonding electrons of the methyl group are donated into the π system, making the alkyl group act as an electron donor (+I and hyperconjugation effects).

Additionally, alkyl groups have a positive inductive effect (+I effect) — they push electrons towards the π system through the σ framework.

Conclusion: Due to hyperconjugation and the +I inductive effect, alkyl groups donate electron density to the attached π system.

Concept: Resonance structures are drawn by shifting electron pairs (lone pairs or π electrons) using curved arrows. The connectivity of atoms remains the same; only electron distribution changes.

(a) C₆H₅OH (Phenol)
The lone pair on oxygen delocalises into the benzene ring:
$$\text{C}_6\text{H}_5\text{OH} \longleftrightarrow \text{ortho and para positions become electron-rich}$$
Resonance contributors:

1. Normal structure with lone pair on O
2. O⁺=C (ring) with negative charge at ortho position
3. O⁺=C (ring) with negative charge at para position
4. Another ortho contributor

Arrow: Curved arrow from lone pair on O → into the ring (C–O bond becomes double bond, ring π bond shifts).

(b) C₆H₅NO₂ (Nitrobenzene)
The π electrons of the ring delocalise into the –NO₂ group:
Resonance contributors show positive charge developing at ortho and para positions of the ring, and negative charge on oxygen of NO₂.

Arrow: Curved arrow from ring π bond → N, then from N=O → O (making O⁻).

(c) CH₃CH=CHCHO (But-2-enal / Crotonaldehyde)
Conjugated system — π electrons delocalise:
$$\text{CH}_3\text{CH}=\text{CH}\!-\!\text{CH}=\!\overset{-}{\text{O}} \longleftrightarrow \text{CH}_3\overset{+}{\text{CH}}\!-\!\text{CH}=\text{CH}\!-\!\text{O}^-$$

Arrow: From C=C π bond → C–C single bond → C=O π bond shifts to O.

Resonance structures:

1. $\text{CH}_3\text{CH}=\text{CH}–\text{CH}=\text{O}$ (main)
2. $\text{CH}_3\overset{+}{\text{CH}}–\text{CH}=\text{CH}–\text{O}^-$

(d) C₆H₅–CHO (Benzaldehyde)
Similar to nitrobenzene — ring π electrons delocalise into the C=O of CHO:
Resonance contributors show positive charge at ortho and para positions of ring, negative charge on O of CHO.

(e) C₆H₅–CH₂· (Benzyl free radical)
The unpaired electron on –CH₂· delocalises into the ring:
$$\text{C}_6\text{H}_5\!-\!\dot{\text{C}}\text{H}_2 \longleftrightarrow \text{ring positions (ortho, para) carry the unpaired electron}$$

Resonance structures show the radical at ortho and para positions of the ring.

(f) CH₃CH=CH·CH₃ (But-2-en-2-yl radical / 1-methylallyl radical)
The radical delocalises through the double bond:
$$\text{CH}_3\text{CH}=\text{CH}\!\cdot\!\text{CH}_3 \longleftrightarrow \text{CH}_3\!\cdot\!\text{CH}–\text{CH}=\text{CHCH}_3$$

Arrow: Curved arrow from C=C π bond → to the carbon bearing the radical.

Resonance structures:

1. $\text{CH}_3\text{CH}=\text{CH}\cdot\text{CH}_3$
2. $\text{CH}_3\cdot\text{CH}–\text{CH}=\text{CHCH}_3$

Electrophiles:

An electrophile (electron-loving species) is a reagent that is electron-deficient and seeks electrons. It accepts an electron pair from a nucleophile to form a new bond.

Characteristics: Electrophiles are either positively charged species or neutral molecules with an electron-deficient atom.

Examples:

• Carbocations: $\text{CH}_3^+$, $(\text{CH}_3)_3\text{C}^+$
• Lewis acids: $\text{BF}_3$, $\text{AlCl}_3$, $\text{FeBr}_3$
• Proton: $\text{H}^+$
• Halogens: $\text{Cl}_2$, $\text{Br}_2$ (act as electrophiles due to polarisation)
• Carbonyl carbon in $\text{C}=\text{O}$ (partial positive charge on C)

Nucleophiles:

A nucleophile (nucleus-loving species) is a reagent that is electron-rich and donates an electron pair to an electrophile to form a new bond.

Characteristics: Nucleophiles are either negatively charged species or neutral molecules with a lone pair of electrons.

Examples:

• Anions: $\text{OH}^-$, $\text{CN}^-$, $\text{Cl}^-$, $\text{Br}^-$, $\text{HS}^-$
• Neutral molecules with lone pairs: $\text{H}_2\text{O}$, $\text{NH}_3$, $\text{ROH}$, $\text{ROR}$
• Carbanions: $\text{CH}_3^-$

Key difference: Electrophiles are electron-pair acceptors; nucleophiles are electron-pair donors.

(a) CH₃COOH + $\mathbf{HO^-}$ → CH₃COO⁻ + H₂O

$\text{HO}^-$ (hydroxide ion) donates an electron pair to the proton of –COOH. It is an electron pair donor.

HO⁻ is a nucleophile.

(b) CH₃COCH₃ + $\mathbf{CN^-}$ → (CH₃)₂C(CN)(OH)

$\text{CN}^-$ (cyanide ion) donates its electron pair to the electrophilic carbonyl carbon of acetone. It is an electron pair donor.

CN⁻ is a nucleophile.

(c) C₆H₆ + $\mathbf{CH_3CO^+}$ → C₆H₅COCH₃

$\text{CH}_3\text{CO}^+$ (acetyl cation / acylium ion) is positively charged and electron-deficient. It accepts the π electron pair from the benzene ring. It is an electron pair acceptor.

CH₃CO⁺ is an electrophile.

(a) CH₃CH₂Br + HS⁻ → CH₃CH₂SH + Br⁻

Here, the nucleophile $\text{HS}^-$ replaces the leaving group $\text{Br}^-$. The –Br is substituted by –SH.

Type: Nucleophilic Substitution Reaction

(b) (CH₃)₂C=CH₂ + HCl → (CH₃)₂ClC–CH₃

HCl adds across the C=C double bond. The double bond is converted to a single bond with addition of H and Cl.

Type: Electrophilic Addition Reaction

(c) CH₃CH₂Br + HO⁻ → CH₂=CH₂ + H₂O + Br⁻

$\text{OH}^-$ removes a β-hydrogen and $\text{Br}^-$ leaves, resulting in formation of a double bond (C=C). A small molecule (HBr equivalent) is eliminated.

Type: Elimination Reaction

(d) (CH₃)₃C–CH₂OH + HBr → (CH₃)₂CBrCH₂CH₂CH₃ + H₂O

The product has a different carbon skeleton compared to the reactant — the methyl group has migrated (1,2-hydride or methyl shift via carbocation rearrangement). This involves rearrangement of the carbon skeleton.

Type: Rearrangement Reaction (involving nucleophilic substitution with carbocation rearrangement)

Note: The structural images are not visible. Based on standard NCERT Class 11 Chemistry Exercise 8.15, the answers are:

(a) The two structures shown are:
$\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{CH}_3$ and $\text{CH}_3\text{CH}(\text{CH}_3)\text{CH}_2\text{CH}_3$

Both have molecular formula $\text{C}_5\text{H}_{12}$ but different connectivity (different carbon skeletons).
Relationship: Structural isomers (chain isomers)

(b) The two structures shown are resonance contributors of the same compound (same connectivity, different electron distribution — e.g., two resonance forms of a carboxylate or similar).
Relationship: Resonance contributors (resonance structures)

(c) The two structures shown are:
Cis and trans forms of but-2-ene (same molecular formula, same connectivity, but different spatial arrangement due to restricted rotation around C=C).
Relationship: Geometrical isomers (cis-trans isomers)

(d) The two structures shown have the same molecular formula but different connectivity.
Relationship: Structural isomers

(Students should verify by examining the actual structures in their textbook.)

Concept:

• Homolysis: Each atom gets one electron from the bond → free radicals formed. Shown by single-headed (fish-hook) curved arrows.
• Heterolysis: One atom gets both electrons → carbocation or carbanion formed. Shown by double-headed curved arrows.

(a) CH₃O–OCH₃ → CH₃O· + ·OCH₃

Each oxygen gets one electron from the O–O bond.

Type: Homolysis

Curved arrow: A single-headed arrow from the O–O bond to each oxygen atom.

Reactive intermediate: Free radicals (CH₃O·, methoxy radicals)

(b) (Image not visible — based on standard NCERT question)

Typically: $\text{CH}_3^+ + \text{Br}^-$ formed from CH₃Br

Both electrons of C–Br bond go to Br.

Type: Heterolysis

Curved arrow: Double-headed arrow from C–Br bond to Br.

Reactive intermediate: Carbocation (CH₃⁺) and bromide ion (Br⁻)

(c) (Image not visible — based on standard NCERT question)

Typically: $\text{CH}_3^- + \text{Br}^+$ or similar

Both electrons of C–Br bond go to C.

Type: Heterolysis

Curved arrow: Double-headed arrow from C–Br bond to C.

Reactive intermediate: Carbanion (CH₃⁻)

(d) (Image not visible)

Based on context, likely homolysis giving free radicals.

Type: Homolysis

Reactive intermediate: Free radicals

Inductive Effect:

The inductive effect is the permanent displacement of electrons along a chain of carbon atoms due to the presence of an electronegative or electropositive atom or group. It operates through σ bonds and decreases with distance.

• Electron-withdrawing groups (–I effect): –F, –Cl, –Br, –NO₂, –CN, –COOH
• Electron-donating groups (+I effect): alkyl groups (–CH₃, –C₂H₅, etc.)

Electromeric Effect:

The electromeric effect is a temporary effect that occurs in the presence of an attacking reagent. It involves the complete transfer of π electrons of a multiple bond to one of the atoms of that bond.

• +E effect: π electrons shift towards the attacking reagent
• –E effect: π electrons shift away from the attacking reagent

This effect operates only in π systems (double or triple bonds) and only when a reagent approaches.

Explanation of acidity orders:

(a) Cl₃CCOOH > Cl₂CHCOOH > ClCH₂COOH

This order is explained by the Inductive Effect (–I effect).

Chlorine is an electronegative atom with a strong –I effect. It withdraws electron density from the –COOH group, making it easier to release H⁺ (increasing acidity). More Cl atoms → stronger –I effect → greater stabilisation of the carboxylate anion → higher acidity.

$$\text{Cl}_3\text{CCOOH (3 Cl atoms)} > \text{Cl}_2\text{CHCOOH (2 Cl atoms)} > \text{ClCH}_2\text{COOH (1 Cl atom)}$$

(b) CH₃CH₂COOH > (CH₃)₂CHCOOH > (CH₃)₃C.COOH

This order is also explained by the Inductive Effect (+I effect).

Alkyl groups have a +I effect — they donate electrons towards the –COOH group, making it harder to release H⁺ (decreasing acidity). More alkyl groups → stronger +I effect → greater destabilisation of the carboxylate anion → lower acidity.

• Propanoic acid: 1 ethyl group
• 2-Methylpropanoic acid: 2 methyl groups (isopropyl)
• 2,2-Dimethylpropanoic acid: 3 methyl groups (tert-butyl) → most electron-donating → least acidic

(a) Crystallisation:

Principle: Crystallisation is based on the difference in the solubilities of the compound and the impurities in a suitable solvent. The impure compound is dissolved in a minimum amount of hot solvent. On cooling, the pure compound crystallises out (as it becomes less soluble at lower temperature) while the impurities remain in solution.

Example: Purification of impure sugar (sucrose) — dissolved in hot water, filtered to remove insoluble impurities, and then cooled to obtain pure sugar crystals.

(b) Distillation:

Principle: Distillation is based on the difference in the boiling points of the components of a liquid mixture. When the mixture is heated, the more volatile component (lower boiling point) vaporises first, and the vapours are condensed and collected separately.

Example: Separation of a mixture of acetone (b.p. 56°C) and water (b.p. 100°C) — acetone distils over first on heating.

(c) Chromatography:

Principle: Chromatography is based on the differential adsorption of the components of a mixture on an adsorbent (stationary phase) as a mobile phase (solvent) moves through it. Components with greater affinity for the stationary phase move more slowly; those with less affinity move faster, leading to separation.

Example: Separation of leaf pigments (chlorophyll, xanthophyll, carotene) by column chromatography using alumina as adsorbent and petroleum ether as mobile phase — different pigments appear as separate coloured bands.

Method: Fractional Crystallisation

Principle: When two compounds have different solubilities in a solvent S, they can be separated by fractional crystallisation. The mixture is dissolved in the minimum amount of hot solvent S. On gradual cooling, the compound with lower solubility crystallises out first, while the more soluble compound remains in solution.

Procedure:

1. Dissolve the mixture of two compounds in the minimum volume of hot solvent S.
2. Allow the solution to cool slowly.
3. The less soluble compound crystallises out first. Filter to collect these crystals.
4. Concentrate the filtrate and cool again to obtain crystals of the more soluble compound.
5. Repeat the process (fractional crystallisation) to obtain pure compounds.

Example: Separation of KNO₃ and NaCl — KNO₃ has much higher solubility at high temperatures but crystallises out readily on cooling, while NaCl solubility changes little with temperature.

Key points:

• Simple distillation: Used when boiling point difference is large (>25°C) and compounds are thermally stable.
• Vacuum distillation: Used for high-boiling or thermally unstable compounds; pressure is reduced so boiling occurs at lower temperature.
• Steam distillation: Used for compounds that are (i) immiscible with water, (ii) volatile in steam, and (iii) thermally unstable at their normal boiling point. The compound distils at a temperature below 100°C.