Organic Chemistry – Some Basic Principles and Techniques

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):

| Member | Formula | Name |
|--------|---------|------|
| 1 | HCOOH | Methanoic acid (Formic acid) |
| 2 | CH₃COOH | Ethanoic acid (Acetic acid) |
| 3 | CH₃CH₂COOH | Propanoic acid |
| 4 | CH₃CH₂CH₂COOH | Butanoic acid |
| 5 | CH₃CH₂CH₂CH₂COOH | Pentanoic acid |

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

| Member | Formula | Name |
|--------|---------|------|
| 1 | CH₃COCH₃ | Propan-2-one (Acetone) |
| 2 | CH₃COCH₂CH₃ | Butan-2-one |
| 3 | CH₃COCH₂CH₂CH₃ | Pentan-2-one |
| 4 | CH₃COCH₂CH₂CH₂CH₃ | Hexan-2-one |
| 5 | CH₃COCH₂CH₂CH₂CH₂CH₃ | Heptan-2-one |

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

| Member | Formula | Name |
|--------|---------|------|
| 1 | CH₂=CH₂ | Ethene |
| 2 | CH₃CH=CH₂ | Propene |
| 3 | CH₃CH₂CH=CH₂ | But-1-ene |
| 4 | CH₃CH₂CH₂CH=CH₂ | Pent-1-ene |
| 5 | CH₃CH₂CH₂CH₂CH=CH₂ | Hex-1-ene |

(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.

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(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

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(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)

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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.

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(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.

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(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

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(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

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(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

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(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)

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(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)

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(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)

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(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)

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(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⁻)

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(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⁻)

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(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.

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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)} &gt; \text{Cl}_2\text{CHCOOH (2 Cl atoms)} &gt; \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.

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(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.

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(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.

| Feature | Simple Distillation | Distillation under Reduced Pressure (Vacuum Distillation) | Steam Distillation |
|---------|--------------------|---------------------------------------------------------|-------------------|
| Principle | Difference in boiling points of miscible liquids | Boiling point decreases at reduced pressure | Immiscible liquid distils below its normal boiling point in presence of steam |
| Conditions | Normal atmospheric pressure | Reduced pressure (vacuum) | Steam is passed through the compound |
| Used for | Separating miscible liquids with sufficiently different boiling points | Purifying liquids that decompose at their normal boiling point | Purifying organic compounds that are immiscible with water and have high boiling points |
| Example | Separation of acetone and water | Purification of glycerol (b.p. 290°C, decomposes before boiling at normal pressure) | Purification of aniline (b.p. 184°C) — distils at ~98°C with steam |

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.

Lassaigne's Test (Sodium Fusion Test):

The organic compound is fused with metallic sodium. The covalently bonded elements (N, S, halogens) are converted into their ionic forms (water-soluble sodium salts) in the sodium fusion extract (Lassaigne's extract).

Chemistry:

(1) Detection of Nitrogen:
$$\text{Na} + \text{C} + \text{N} \xrightarrow{\Delta} \text{NaCN}$$
The extract is boiled with $\text{FeSO}_4$ and then acidified with $\text{H}_2\text{SO}_4$:
$$\text{FeSO}_4 + 2\text{NaCN} \rightarrow \text{Fe(CN)}_2 + \text{Na}_2\text{SO}_4$$
$$\text{Fe(CN)}_2 + 4\text{NaCN} \rightarrow \text{Na}_4[\text{Fe(CN)}_6] \text{ (sodium hexacyanoferrate(II))}$$
$$3\text{Na}_4[\text{Fe(CN)}_6] + 4\text{FeCl}_3 \rightarrow \text{Fe}_4[\text{Fe(CN)}_6]_3 + 12\text{NaCl}$$
Prussian blue colour confirms nitrogen.

(2) Detection of Sulphur:
$$2\text{Na} + \text{S} \rightarrow \text{Na}_2\text{S}$$
- Lead acetate test: $\text{Na}_2\text{S} + (\text{CH}_3\text{COO})_2\text{Pb} \rightarrow \text{PbS}\downarrow + 2\text{CH}_3\text{COONa}$
Black precipitate of PbS confirms sulphur.
- Sodium nitroprusside test: $\text{Na}_2\text{S} + \text{Na}_2[\text{Fe(CN)}_5\text{NO}] \rightarrow \text{Na}_4[\text{Fe(CN)}_5\text{NOS}]$
Violet/purple colour confirms sulphur.

(3) Detection of Halogens:
$$\text{Na} + \text{X} \rightarrow \text{NaX} \quad (\text{X} = \text{Cl, Br, I})$$
The extract is acidified with dilute $\text{HNO}_3$ and treated with $\text{AgNO}_3$:
- White precipitate of AgCl (soluble in NH₃) → Chlorine
- Pale yellow precipitate of AgBr (sparingly soluble in NH₃) → Bromine
- Yellow precipitate of AgI (insoluble in NH₃) → Iodine

Note: If N and S are both present, $\text{NaCNS}$ (sodium thiocyanate) is formed. The extract must be boiled with dilute $\text{H}_2\text{SO}_4$ before testing for halogens to decompose $\text{NaCNS}$ (which would otherwise give a false positive with $\text{FeCl}_3$).

| Feature | Dumas Method | Kjeldahl's Method |
|---------|-------------|------------------|
| Principle | The organic compound is heated with excess CuO in a stream of CO₂. Nitrogen in the compound is converted to free N₂ gas. The volume of N₂ is measured over KOH solution (which absorbs CO₂). | The organic compound is digested with concentrated H₂SO₄ (with K₂SO₄ and CuSO₄ as catalysts). Nitrogen is converted to (NH₄)₂SO₄. The digest is made alkaline with NaOH and the NH₃ liberated is distilled into a known volume of standard acid. The excess acid is back-titrated. |
| Reaction | $\text{C, H, N compound} + \text{CuO} \xrightarrow{\Delta} \text{CO}_2 + \text{H}_2\text{O} + \text{N}_2$ | $\text{N (organic)} \xrightarrow{\text{conc. H}_2\text{SO}_4} (\text{NH}_4)_2\text{SO}_4 \xrightarrow{\text{NaOH}} \text{NH}_3 \xrightarrow{\text{H}_2\text{SO}_4} (\text{NH}_4)_2\text{SO}_4$ |
| Measurement | Volume of N₂ gas collected over KOH | Volume of standard acid neutralised by NH₃ |
| Applicability | Applicable to all nitrogen-containing compounds | Not applicable to compounds containing N in nitro (–NO₂), azo (–N=N–) groups or ring nitrogen (pyridine) |
| Formula | $\%\text{N} = \dfrac{28 \times V \times P}{22400 \times W} \times 100$ (where V = vol. of N₂, P = pressure, W = mass of compound) | $\%\text{N} = \dfrac{1.4 \times M \times V}{W}$ (where M = molarity of acid, V = vol. of acid used by NH₃, W = mass of compound) |

Estimation of Halogens — Carius Method:

Principle: The organic compound is heated with fuming nitric acid in a sealed tube (Carius tube) in the presence of silver nitrate. The halogen is oxidised and converted to silver halide (AgX), which is filtered, dried and weighed.

Reaction:
$$\text{Organic halide} + \text{AgNO}_3 \xrightarrow{\text{fuming HNO}_3, \Delta} \text{AgX} \downarrow$$

Calculation:
$$\%\text{X} = \frac{\text{Atomic mass of X}}{\text{Molar mass of AgX}} \times \frac{\text{Mass of AgX}}{\text{Mass of compound}} \times 100$$

---

Estimation of Sulphur — Carius Method:

Principle: The organic compound is heated with fuming nitric acid in a Carius tube. Sulphur is oxidised to sulphuric acid, which is then precipitated as barium sulphate by adding excess barium chloride solution. The precipitate is filtered, dried and weighed.

Reaction:
$$\text{S (organic)} \xrightarrow{\text{fuming HNO}_3} \text{H}_2\text{SO}_4 \xrightarrow{\text{BaCl}_2} \text{BaSO}_4 \downarrow$$

Calculation:
$$\%\text{S} = \frac{32}{233} \times \frac{\text{Mass of BaSO}_4}{\text{Mass of compound}} \times 100$$

---

Estimation of Phosphorus:

Principle: The organic compound is heated with fuming nitric acid. Phosphorus is oxidised to phosphoric acid ($\text{H}_3\text{PO}_4$), which is precipitated as ammonium phosphomolybdate by adding ammonium molybdate. Alternatively, it is precipitated as magnesium ammonium phosphate ($\text{MgNH}_4\text{PO}_4$), which on ignition gives magnesium pyrophosphate ($\text{Mg}_2\text{P}_2\text{O}_7$).

Reaction:
$$\text{P (organic)} \xrightarrow{\text{fuming HNO}_3} \text{H}_3\text{PO}_4 \xrightarrow{\text{Mg}^{2+}/\text{NH}_3} \text{MgNH}_4\text{PO}_4 \xrightarrow{\text{ignition}} \text{Mg}_2\text{P}_2\text{O}_7$$

Calculation:
$$\%\text{P} = \frac{62}{222} \times \frac{\text{Mass of Mg}_2\text{P}_2\text{O}_7}{\text{Mass of compound}} \times 100$$

Paper Chromatography — Principle:

Paper chromatography is a type of partition chromatography in which the stationary phase is water (or another liquid) held in the fibres of the paper, and the mobile phase is a suitable organic solvent or solvent mixture.

Principle: The separation is based on the differential partitioning of the components of a mixture between the stationary phase (water adsorbed on paper) and the mobile phase (solvent). Components that are more soluble in the mobile phase travel faster (higher $R_f$ value), while those more attracted to the stationary phase travel slower (lower $R_f$ value).

Procedure:
1. A small spot of the mixture is applied near the bottom of the chromatography paper.
2. The paper is suspended in a closed chamber with the solvent (mobile phase) at the bottom.
3. The solvent rises up the paper by capillary action, carrying the components with it.
4. Different components travel different distances depending on their partition coefficients.
5. The paper is removed, dried, and the spots are visualised (by UV light or spraying with a reagent).

$R_f$ value (Retardation factor):
$$R_f = \frac{\text{Distance travelled by the component}}{\text{Distance travelled by the solvent front}}$$

Each compound has a characteristic $R_f$ value under given conditions, which helps in identification.

Example: Separation of amino acids or leaf pigments using paper chromatography.

Reason:

The sodium fusion extract (Lassaigne's extract) may contain sodium cyanide (NaCN) and sodium sulphide (Na₂S) if the organic compound also contains nitrogen and sulphur.

If silver nitrate is added directly:
- NaCN would give a white precipitate of AgCN
- Na₂S would give a black precipitate of Ag₂S

These precipitates would interfere with and mask the precipitate of silver halide (AgX), giving a false positive result for halogens.

Therefore, dilute nitric acid is added before adding AgNO₃:
$$\text{NaCN} + \text{HNO}_3 \rightarrow \text{HCN} \uparrow + \text{NaNO}_3$$
$$\text{Na}_2\text{S} + 2\text{HNO}_3 \rightarrow \text{H}_2\text{S} \uparrow + 2\text{NaNO}_3$$

This decomposes NaCN and Na₂S, removing the interfering ions. Now, when AgNO₃ is added, only the silver halide precipitate is formed, giving an unambiguous test for halogens.

Note: HNO₃ used must be dilute; concentrated HNO₃ would oxidise the halide ions and give a false negative.

Reason:

In organic compounds, nitrogen, sulphur and halogens are present in covalent form (bonded to carbon). They cannot be directly detected by ionic reactions in aqueous solution.

For example:
- Nitrogen is present as C–N or C≡N (covalent bond)
- Sulphur is present as C–S (covalent bond)
- Halogens are present as C–X (covalent bond)

These covalently bonded elements do not give ionic reactions directly.

Fusion with metallic sodium converts these covalently bonded elements into their water-soluble ionic forms (sodium salts):
$$\text{C} + \text{N} + \text{Na} \rightarrow \text{NaCN (ionic)}$$
$$\text{S} + 2\text{Na} \rightarrow \text{Na}_2\text{S (ionic)}$$
$$\text{X} + \text{Na} \rightarrow \text{NaX (ionic)}$$

These ionic compounds dissolve in water to give the sodium fusion extract (Lassaigne's extract), which can then be tested by standard ionic reactions for the detection of N, S and halogens.

Conclusion: Sodium fusion is necessary to convert covalently bonded elements into ionic (water-soluble) forms so that they can be detected by ionic reactions.

Suitable Technique: Sublimation

Reason: Camphor is a substance that sublimes readily (converts directly from solid to vapour on heating without passing through the liquid state), whereas calcium sulphate (CaSO₄) does not sublime.

Procedure: When the mixture is heated gently, camphor sublimes and its vapours are collected on a cold surface (inverted funnel with cotton plug), where they solidify. Calcium sulphate remains behind as a residue.

Result: Pure camphor is obtained as the sublimate, and calcium sulphate remains in the dish.

Conclusion: Sublimation is the suitable technique for separating camphor from calcium sulphate.

Explanation:

In steam distillation, the organic liquid (which is immiscible with water) is distilled in the presence of steam (water vapour).

According to Dalton's Law of Partial Pressures, when two immiscible liquids are present together, the total vapour pressure of the system is the sum of the individual vapour pressures of the two components:
$$P_{\text{total}} = P_{\text{water}} + P_{\text{organic liquid}}$$

The mixture boils when the total vapour pressure equals the atmospheric pressure ($P_{\text{atm}}$):
$$P_{\text{water}} + P_{\text{organic liquid}} = P_{\text{atm}}$$

Since P_{\text{total}} = P_{\text{water}} + P_{\text{organic liquid}} &gt; P_{\text{organic liquid}} alone, the total pressure reaches atmospheric pressure at a lower temperature than the boiling point of the pure organic liquid.

Therefore, the organic liquid vaporises (distils) at a temperature below its normal boiling point and also below 100°C (the boiling point of water).

Example: Aniline (b.p. 184°C) steam distils at about 98°C.

Answer: No, CCl₄ will not give a white precipitate of AgCl on heating with silver nitrate.

Reason:

In CCl₄ (carbon tetrachloride), the C–Cl bonds are covalent in nature. The chlorine atoms are not present as free chloride ions (Cl⁻).

For AgCl to precipitate, free Cl⁻ ions must be present in solution:
$$\text{Ag}^+ + \text{Cl}^- \rightarrow \text{AgCl} \downarrow$$

Since CCl₄ is a non-polar covalent compound, it does not ionise in solution and does not release Cl⁻ ions. Moreover, CCl₄ is chemically very stable and does not react with AgNO₃ under normal conditions to release chloride ions.

Conclusion: CCl₄ does not give a white precipitate of AgCl with AgNO₃ because it does not ionise to give free Cl⁻ ions. The C–Cl bond in CCl₄ is covalent and non-ionic.

Reason:

In the estimation of carbon and hydrogen in an organic compound, the compound is burnt in excess oxygen. The products are:
- $\text{CO}_2$ (from carbon)
- $\text{H}_2\text{O}$ (from hydrogen)

The water is first absorbed by anhydrous calcium chloride (or anhydrous $\text{P}_2\text{O}_5$), and then the $\text{CO}_2$ is absorbed by potassium hydroxide (KOH) solution.

KOH is used because:
$$\text{CO}_2 + 2\text{KOH} \rightarrow \text{K}_2\text{CO}_3 + \text{H}_2\text{O}$$

KOH is a strong alkali that reacts completely and quantitatively with $\text{CO}_2$. The increase in mass of the KOH tube gives the mass of $\text{CO}_2$ absorbed, from which the percentage of carbon is calculated:
$$\%\text{C} = \frac{12}{44} \times \frac{\text{Mass of CO}_2}{\text{Mass of compound}} \times 100$$

KOH is preferred over other bases because it is a strong, non-volatile base that absorbs $\text{CO}_2$ completely and does not absorb water (water is absorbed separately by CaCl₂ before the KOH tube).

Reason:

In the lead acetate test for sulphur, the sodium fusion extract (containing Na₂S) is acidified and then treated with lead acetate solution:
$$\text{Na}_2\text{S} + (\text{CH}_3\text{COO})_2\text{Pb} \rightarrow \text{PbS} \downarrow (\text{black}) + 2\text{CH}_3\text{COONa}$$

If sulphuric acid (H₂SO₄) were used:
$$\text{Na}_2\text{S} + \text{H}_2\text{SO}_4 \rightarrow \text{Na}_2\text{SO}_4 + \text{H}_2\text{S} \uparrow$$

The H₂S gas would escape, and no sulphide ions would remain in solution to react with lead acetate. Also, the sulphate ions from H₂SO₄ would react with lead acetate to form a white precipitate of lead sulphate (PbSO₄), which would interfere with the test:
$$\text{Pb}^{2+} + \text{SO}_4^{2-} \rightarrow \text{PbSO}_4 \downarrow (\text{white, interfering precipitate})$$

Acetic acid (CH₃COOH) is used because:
1. It is a weak acid and does not decompose Na₂S completely to release H₂S.
2. It does not introduce sulphate ions, so there is no interference.
3. It provides mildly acidic conditions sufficient for the test.

Conclusion: Acetic acid is used to avoid loss of H₂S and to prevent interference from sulphate ions that would be introduced by H₂SO₄.

Given:
- % C = 69%
- % H = 4.8%
- % O = 100 – 69 – 4.8 = 26.2%
- Mass of compound = 0.20 g

Step 1: Find mass of C and H in 0.20 g of compound.

$$\text{Mass of C} = \frac{69}{100} \times 0.20 = 0.138 \text{ g}$$

$$\text{Mass of H} = \frac{4.8}{100} \times 0.20 = 0.0096 \text{ g}$$

Step 2: Calculate mass of CO₂ produced.

All carbon is converted to CO₂:
$$\text{C} + \text{O}_2 \rightarrow \text{CO}_2$$
$$12 \text{ g C} \rightarrow 44 \text{ g CO}_2$$
$$0.138 \text{ g C} \rightarrow \frac{44}{12} \times 0.138 = \frac{44 \times 0.138}{12}$$
$$= \frac{6.072}{12} = 0.506 \text{ g CO}_2$$

Step 3: Calculate mass of H₂O produced.

All hydrogen is converted to H₂O:
$$2\text{H}_2 + \text{O}_2 \rightarrow 2\text{H}_2\text{O}$$
$$2 \text{ g H} \rightarrow 18 \text{ g H}_2\text{O}$$
$$0.0096 \text{ g H} \rightarrow \frac{18}{2} \times 0.0096 = 9 \times 0.0096 = 0.0864 \text{ g H}_2\text{O}$$

Results:
$$\boxed{\text{Mass of CO}_2 = 0.506 \text{ g}}$$
$$\boxed{\text{Mass of H}_2\text{O} = 0.0864 \text{ g}}$$

Given:
- Mass of organic compound = 0.50 g
- Volume of H₂SO₄ = 50 mL, Molarity = 0.5 M
- Volume of NaOH for back titration = 60 mL, Molarity = 0.5 M

Step 1: Find moles of H₂SO₄ taken.
$$n(\text{H}_2\text{SO}_4) = 0.5 \times \frac{50}{1000} = 0.025 \text{ mol}$$

Step 2: Find moles of NaOH used for residual acid.
$$n(\text{NaOH}) = 0.5 \times \frac{60}{1000} = 0.030 \text{ mol}$$

Step 3: Find moles of H₂SO₄ neutralised by NaOH (residual acid).
$$\text{H}_2\text{SO}_4 + 2\text{NaOH} \rightarrow \text{Na}_2\text{SO}_4 + 2\text{H}_2\text{O}$$
$$n(\text{H}_2\text{SO}_4)_{\text{residual}} = \frac{0.030}{2} = 0.015 \text{ mol}$$

Step 4: Find moles of H₂SO₄ that reacted with NH₃.
$$n(\text{H}_2\text{SO}_4)_{\text{reacted with NH}_3} = 0.025 - 0.015 = 0.010 \text{ mol}$$

Step 5: Find moles of NH₃ (and hence N).
$$\text{H}_2\text{SO}_4 + 2\text{NH}_3 \rightarrow (\text{NH}_4)_2\text{SO}_4$$
$$n(\text{NH}_3) = 2 \times 0.010 = 0.020 \text{ mol}$$

Since each mole of NH₃ contains 1 mole of N:
$$n(\text{N}) = 0.020 \text{ mol}$$

Step 6: Calculate mass of N.
$$\text{Mass of N} = 0.020 \times 14 = 0.28 \text{ g}$$

Step 7: Calculate percentage of N.
$$\%\text{N} = \frac{0.28}{0.50} \times 100 = \boxed{56\%}$$

Given:
- Mass of organic compound = 0.3780 g
- Mass of AgCl obtained = 0.5740 g
- Molar mass of AgCl = 108 + 35.5 = 143.5 g/mol
- Atomic mass of Cl = 35.5 g/mol

Formula:
$$\%\text{Cl} = \frac{\text{Atomic mass of Cl}}{\text{Molar mass of AgCl}} \times \frac{\text{Mass of AgCl}}{\text{Mass of compound}} \times 100$$

Calculation:
$$\%\text{Cl} = \frac{35.5}{143.5} \times \frac{0.5740}{0.3780} \times 100$$

$$= \frac{35.5}{143.5} \times 1.5185 \times 100$$

$$= 0.2474 \times 1.5185 \times 100$$

$$= 0.2474 \times 151.85$$

$$= 37.57\%$$

$$\boxed{\%\text{Cl} \approx 37.57\%}$$

Given:
- Mass of organic compound = 0.468 g
- Mass of BaSO₄ obtained = 0.668 g
- Molar mass of BaSO₄ = 137 + 32 + 64 = 233 g/mol
- Atomic mass of S = 32 g/mol

Formula:
$$\%\text{S} = \frac{\text{Atomic mass of S}}{\text{Molar mass of BaSO}_4} \times \frac{\text{Mass of BaSO}_4}{\text{Mass of compound}} \times 100$$

Calculation:
$$\%\text{S} = \frac{32}{233} \times \frac{0.668}{0.468} \times 100$$

$$= \frac{32}{233} \times 1.4274 \times 100$$

$$= 0.13734 \times 1.4274 \times 100$$

$$= 0.13734 \times 142.74$$

$$= 19.60\%$$

$$\boxed{\%\text{S} \approx 19.6\%}$$

Correct Answer: (c) sp²–sp³

Justification:

The compound is: $\text{CH}_2=\text{CH}–\text{CH}_2–\text{CH}_2–\text{C}\equiv\text{CH}$

Numbering the carbons:
- C₁ = CH₂= (part of C=C) → $sp^2$
- C₂ = =CH– (part of C=C) → $sp^2$
- C₃ = –CH₂– (single bonds only) → $sp^3$
- C₄ = –CH₂– (single bonds only) → $sp^3$
- C₅ = –C≡ (part of C≡C) → $sp$
- C₆ = ≡CH (part of C≡C) → $sp$

The C₂–C₃ bond is between:
- C₂: $sp^2$ hybridised (part of the double bond)
- C₃: $sp^3$ hybridised (saturated carbon)

Therefore, the pair of hybridised orbitals is $sp^2–sp^3$.

Answer: (c)

Correct Answer: (b) Fe₄[Fe(CN)₆]₃

Justification:

In Lassaigne's test for nitrogen:
1. Sodium fusion converts N to NaCN.
2. NaCN reacts with FeSO₄ to form sodium hexacyanoferrate(II): $\text{Na}_4[\text{Fe(CN)}_6]$
3. On acidification with H₂SO₄, Fe²⁺ is partially oxidised to Fe³⁺.
4. Fe³⁺ reacts with $[\text{Fe(CN)}_6]^{4-}$ to form ferric hexacyanoferrate(II) — Prussian blue:
$$4\text{FeCl}_3 + 3\text{Na}_4[\text{Fe(CN)}_6] \rightarrow \text{Fe}_4[\text{Fe(CN)}_6]_3 \downarrow + 12\text{NaCl}$$

$\text{Fe}_4[\text{Fe(CN)}_6]_3$ is Prussian blue (ferric ferrocyanide).

Answer: (b)

Correct Answer: (b) (CH₃)₃C⁺

Justification:

Stability of carbocations increases with increasing number of alkyl groups attached to the positively charged carbon (due to +I inductive effect and hyperconjugation of alkyl groups):

\text{Tertiary} &gt; \text{Secondary} &gt; \text{Primary} &gt; \text{Methyl}

Analysis of options:
- (a) (CH₃)₃C–CH₂⁺: Primary carbocation (the positive charge is on –CH₂⁺, a primary carbon)
- (b) (CH₃)₃C⁺: Tertiary carbocation (positive charge on carbon attached to three methyl groups) — most stable
- (c) CH₃CH₂CH₂⁺: Primary carbocation
- (d) CH₃CH⁺CH₂CH₃: Secondary carbocation

Answer: (b) (CH₃)₃C⁺ is the most stable carbocation.

Correct Answer: (d) Chromatography

Justification:

Chromatography is the most versatile, sensitive and modern technique for the separation, purification and identification of organic compounds. It can separate even very small quantities of closely related compounds (including those with similar physical properties) that cannot be separated by crystallisation, distillation or sublimation. It is widely used in research, pharmaceuticals, forensics and industry.

Answer: (d)

Correct Answer: (b) Nucleophilic substitution

Justification:

In this reaction:
- The substrate is CH₃CH₂I (ethyl iodide)
- KOH(aq) provides OH⁻ (hydroxide ion), which is a nucleophile (electron pair donor)
- OH⁻ attacks the electrophilic carbon bearing the iodine
- The iodide ion (I⁻) is displaced as the leaving group
- The product is CH₃CH₂OH (ethanol)

This is a classic nucleophilic substitution (SN) reaction where the nucleophile (OH⁻) replaces the leaving group (I⁻).

Answer: (b)