Hydrocarbons

Given: Chlorination of methane (free radical reaction).

Concept: Chlorination of methane proceeds via a free radical chain mechanism involving initiation, propagation, and termination steps.

Explanation:

During the chlorination of methane, methyl free radicals ($\cdot CH_3$) are produced in the propagation step:
$$CH_4 + Cl\cdot \rightarrow \cdot CH_3 + HCl$$

In the termination step, two methyl free radicals can combine to form ethane:
$$\cdot CH_3 + \cdot CH_3 \rightarrow CH_3 - CH_3 \text{ (Ethane)}$$

This coupling of two methyl radicals during the termination step accounts for the formation of ethane as a by-product during the chlorination of methane.

Given: Various structural formulas.

(a) $CH_3CH=C(CH_3)_2$

The longest chain containing the double bond has 4 carbons (but-2-ene backbone). The double bond is between C-2 and C-3. There is a methyl group on C-3 (since C-3 already has two methyls — one from the chain and one substituent).

Actually, numbering: $CH_3 - CH = C(CH_3)_2$
- Longest chain: C1–C2=C3–C4 → but-2-ene
- Methyl substituent at C-3

IUPAC Name: 2-Methylbut-2-ene

(b) $CH_2=CH-C\equiv C-CH_3$

Longest chain = 5 carbons with both a double bond (C1=C2) and a triple bond (C3≡C4).
- Name: pent-1-en-3-yne

IUPAC Name: Pent-1-en-3-yne

(c) $CH_3CH=CH-OH$ (as written in OCR — this appears to be an enol, but since the chapter is hydrocarbons, this may be a misprint. Taking the structure as given:)

If the structure is $CH_3-CH=CH-OH$, it is an enol (not a hydrocarbon). However, if interpreted as $CH_3-CH=CH_2$ (propene), the IUPAC name would be Prop-1-ene. Since the OCR shows $CH_3CH=CH-OH$, this is likely Propen-1-ol (prop-1-en-3-ol or prop-2-en-1-ol depending on numbering), but as it falls outside pure hydrocarbons, we note:

IUPAC Name: Prop-1-en-1-ol (if taken literally; likely a misprint in the source for $CH_3CH=CH_2$, i.e., Propene)

(d) $CH_3(CH_2)_4CH(CH_2)_3CH_3$ — with a branch

Expanding: $CH_3-CH_2-CH_2-CH_2-CH_2-CH(-)-CH_2-CH_2-CH_2-CH_3$

The main chain: count the longest continuous chain. The molecule is a decane with a branch. Looking at the structure $CH_3(CH_2)_4CH(CH_2)_3CH_3$:
- One side of the branch: $CH_3(CH_2)_4$ = 5 carbons (C1–C5)
- Branch carbon: C6
- Other side: $(CH_2)_3CH_3$ = 4 carbons (C7–C10)
- Total longest chain = 10 carbons → decane
- The branch at C6 would be... wait, there is no substituent shown explicitly. The structure as written seems to be a straight chain of 10 carbons: $CH_3-(CH_2)_4-CH_2-(CH_2)_3-CH_3$ = decane.

If there is a methyl branch (from the image not visible), assuming the structure is 5-methylnonane or similar. Based on the OCR text $CH_3(CH_2)_4CH(CH_2)_3CH_3$ with an implied substituent from the image:

IUPAC Name: Decane (if no branch) or most likely 5-Methylnonane if a methyl group is at C-5 (from the figure).

(e)/(g) $CH_3CH=CH-CH_2-CH=CH-CH(-C_2H_5)-CH_2-CH=CH_2$

Longest chain containing all three double bonds:
- Numbering the chain: $CH_3-CH=CH-CH_2-CH=CH-CH(-)-CH_2-CH=CH_2$
- Main chain = 10 carbons with double bonds at C2, C5, C9 (numbering from the $CH_3$ end) or C1, C4, C8 from the other end.
- Substituent: $C_2H_5$ (ethyl) at C-7 (if numbered from $CH_3$ end giving lower locants to double bonds).

Double bonds at positions 2, 5, 9 from the $CH_3$ end:
- Locant set: {2, 5, 9}
From the other end: {2, 6, 9} — compare: 2=2, 5<6, so number from $CH_3$ end.

IUPAC Name: 7-Ethyldeca-2,5,9-triene

Given: Molecular formulas with specified degree of unsaturation.

(a) $C_4H_8$ — one double bond (alkenes)

General formula for alkene: $C_nH_{2n}$; for $n=4$: $C_4H_8$ ✓

All possible structural isomers (open chain only, as cyclobutane also fits but is cyclic):

1. But-1-ene:
$$CH_2=CH-CH_2-CH_3$$
IUPAC Name: But-1-ene

2. But-2-ene:
$$CH_3-CH=CH-CH_3$$
IUPAC Name: But-2-ene

3. 2-Methylprop-1-ene (Isobutylene):
$$CH_2=C(CH_3)-CH_3$$
IUPAC Name: 2-Methylprop-1-ene

*(Note: But-2-ene also has cis and trans geometric isomers, but they have the same structural formula.)*

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(b) $C_5H_8$ — one triple bond (alkynes)

General formula for alkyne: $C_nH_{2n-2}$; for $n=5$: $C_5H_8$ ✓

1. Pent-1-yne:
$$CH\equiv C-CH_2-CH_2-CH_3$$
IUPAC Name: Pent-1-yne

2. Pent-2-yne:
$$CH_3-C\equiv C-CH_2-CH_3$$
IUPAC Name: Pent-2-yne

3. 3-Methylbut-1-yne:
$$CH\equiv C-CH(CH_3)-CH_3$$
IUPAC Name: 3-Methylbut-1-yne

Concept: Ozonolysis cleaves the C=C double bond. Each carbon of the double bond becomes a carbonyl carbon. If the carbon has one H attached, it gives an aldehyde; if no H, it gives a ketone.

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(i) Pent-2-ene: $CH_3-CH=CH-CH_2-CH_3$

Cleavage at C2=C3:
- C1–C2 fragment: $CH_3-CHO$ → Ethanal
- C3–C5 fragment: $OHC-CH_2-CH_3$ → Propanal

Products: Ethanal and Propanal

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(ii) 3,4-Dimethylhept-3-ene: $CH_3CH_2-C(CH_3)=C(CH_3)-CH_2CH_2CH_3$

Cleavage at C3=C4:
- C1–C3 fragment: $CH_3CH_2-CO-CH_3$ → Butan-2-one (methyl ethyl ketone)
- C4–C7 fragment: $CH_3-CO-CH_2CH_2CH_3$ → Pentan-2-one

Products: Butan-2-one and Pentan-2-one

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(iii) 2-Ethylbut-1-ene: $CH_2=C(C_2H_5)-CH_2-CH_3$

Cleavage at C1=C2:
- C1 fragment: $CH_2=O$ → Methanal (Formaldehyde)
- C2–C6 fragment: $O=C(C_2H_5)(CH_2CH_3)$ → Both groups on C2 are ethyl, giving Pentan-3-one

Products: Methanal and Pentan-3-one

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(iv) 1-Phenylbut-1-ene: $C_6H_5-CH=CH-CH_2-CH_3$

Cleavage at C1=C2:
- $C_6H_5$ side: $C_6H_5-CHO$ → Benzaldehyde
- Other fragment: $OHC-CH_2-CH_3$ → Propanal

Products: Benzaldehyde and Propanal

Given: Ozonolysis of alkene A gives:
- Ethanal: $CH_3CHO$ (C2 aldehyde — the carbon bearing H came from one side of the double bond)
- Pentan-3-one: $CH_3CH_2-CO-CH_2CH_3$ (C5 ketone — the carbonyl carbon had no H, so it was internal)

Concept: In ozonolysis, the two carbonyl compounds formed come from the two carbons of the C=C bond. Reconnect the carbonyl carbons with a double bond to get the original alkene.

Reconstruction:
- Ethanal: $CH_3-\underline{CH}=O$ → contributes $CH_3-CH=$
- Pentan-3-one: $CH_3CH_2-\underline{C}(=O)-CH_2CH_3$ → contributes $=C(CH_2CH_3)(CH_2CH_3)$...

Wait — pentan-3-one is $CH_3CH_2COCH_2CH_3$, so the carbonyl C is C3, flanked by ethyl groups. This means the double bond carbon from pentan-3-one side had two ethyl groups → it is $=C(C_2H_5)_2$... but that gives 5 carbons on one side.

Actually pentan-3-one: $CH_3-CH_2-CO-CH_2-CH_3$. The C=O carbon (C3) had an ethyl group on each side. So in the alkene, this carbon was $=C(CH_2CH_3)-$ with one ethyl on the chain side.

Reconnecting: $CH_3-CH = C(CH_2CH_3)-CH_2-CH_3$

This is: $CH_3-CH=C(C_2H_5)-CH_2-CH_3$

Longest chain containing the double bond:
- C1: $CH_3$
- C2: $CH=$
- C3: $=C$
- C4: $CH_2$
- C5: $CH_3$ (from the ethyl on C3)
- Branch: $CH_2CH_3$ (ethyl) at C3

Longest chain = hex-2-ene backbone? Let's count: $CH_3-CH=C(-CH_2CH_3)-CH_2-CH_3$
- If we take the chain through C3 and the ethyl: $CH_3-CH=C-CH_2-CH_3$ = 5 carbons (pent-2-ene) with ethyl at C3.
- Or take the longer path: $CH_3-CH=C-CH_2-CH_2-CH_3$... no, the ethyl is $CH_2CH_3$.

Longest chain = 5 carbons: $CH_3-CH=C(C_2H_5)-CH_2-CH_3$ → 3-Ethylpent-2-ene

Structure of A:
$$CH_3-CH=C(C_2H_5)-CH_2-CH_3$$

IUPAC Name: 3-Ethylpent-2-ene

Given:
- Alkene A: 3 C–C σ bonds, 8 C–H σ bonds, 1 C=C π bond
- Ozonolysis gives 2 mol of same aldehyde with molar mass = 44 u

Step 1: Find the aldehyde.

Molar mass of aldehyde = 44 u
If aldehyde is $C_nH_{2n}O$:
$$12n + 2n + 16 = 44 \Rightarrow 14n = 28 \Rightarrow n = 2$$
So aldehyde = $C_2H_4O$ = Ethanal ($CH_3CHO$), molar mass = $12(2)+4+16 = 44$ ✓

Step 2: Reconstruct alkene A.

Since ozonolysis gives 2 moles of ethanal, both carbons of the double bond must give $CH_3CHO$. This means:
$$CH_3-CH=CH-CH_3 \xrightarrow{O_3/Zn,H_2O} 2\,CH_3CHO$$

So A = But-2-ene: $CH_3-CH=CH-CH_3$

Step 3: Verify bond count for but-2-ene.
- C–C σ bonds: C1–C2, C2=C3 (σ part), C3–C4 → 3 C–C σ bonds ✓
- C–H σ bonds: 3(C1) + 1(C2) + 1(C3) + 3(C4) = 8 C–H bonds ✓
- C=C π bond: 1 ✓

IUPAC Name of A: But-2-ene
$$CH_3-CH=CH-CH_3$$

Given: Ozonolysis products are:
- Propanal: $CH_3-CH_2-CHO$ (aldehyde, so the double bond carbon had one H)
- Pentan-3-one: $CH_3CH_2-CO-CH_2CH_3$ (ketone, so the double bond carbon had no H)

Concept: Reconnect the two carbonyl carbons with a C=C double bond.

- From propanal: $CH_3CH_2-\overset{|}{C}H=$ (C3 of propanal becomes one end of double bond)
- From pentan-3-one: $=\overset{|}{C}(CH_2CH_3)_2$...

Pentan-3-one: $CH_3-CH_2-\underset{C3}{C}(=O)-CH_2-CH_3$. The carbonyl carbon (C3) is flanked by two ethyl groups. In the alkene, this becomes $=C(CH_2CH_3)-CH_2CH_3$.

Structural formula of alkene:
$$CH_3CH_2-CH=C(CH_2CH_3)-CH_2CH_3$$

This is: $CH_3CH_2-CH=C(C_2H_5)-CH_2CH_3$

Verification by ozonolysis:
- Left fragment: $CH_3CH_2-CHO$ = Propanal ✓
- Right fragment: $O=C(C_2H_5)(C_2H_5)$ = Pentan-3-one ✓

IUPAC Name: 3-Ethylpent-2-ene

Structural Formula:
$$CH_3-CH_2-CH=C(CH_2CH_3)-CH_2-CH_3$$

Concept: Complete combustion of hydrocarbons in excess oxygen produces $CO_2$ and $H_2O$.

General equation: $C_xH_y + \left(x + \dfrac{y}{4}\right)O_2 \rightarrow x\,CO_2 + \dfrac{y}{2}H_2O$

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(i) Butane ($C_4H_{10}$):
$$2C_4H_{10} + 13O_2 \rightarrow 8CO_2 + 10H_2O$$

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(ii) Pentene ($C_5H_{10}$):
$$2C_5H_{10} + 15O_2 \rightarrow 10CO_2 + 10H_2O$$

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(iii) Hexyne ($C_6H_{10}$):
$$2C_6H_{10} + 17O_2 \rightarrow 12CO_2 + 10H_2O$$

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(iv) Toluene ($C_6H_5CH_3 = C_7H_8$):
$$C_7H_8 + 9O_2 \rightarrow 7CO_2 + 4H_2O$$

Given: Hex-2-ene: $CH_3-CH=CH-CH_2-CH_2-CH_3$

Cis-hex-2-ene (both larger groups on same side):
$$\begin{array}{c} CH_3 \quad CH_2CH_2CH_3 \\ \searrow \swarrow \\ C=C \\ \nearrow \nwarrow \\ H \quad\quad H \end{array}$$

In cis isomer: $CH_3$ and $CH_2CH_2CH_3$ are on the same side of the double bond.

Trans-hex-2-ene (larger groups on opposite sides):
$$\begin{array}{c} CH_3 \quad\quad H \\ \searrow \swarrow \\ C=C \\ \nearrow \nwarrow \\ H \quad CH_2CH_2CH_3 \end{array}$$

In trans isomer: $CH_3$ and $CH_2CH_2CH_3$ are on opposite sides.

Which has higher boiling point?

Cis-hex-2-ene has a higher boiling point.

Reason: The cis isomer is a polar molecule (the bond dipoles do not cancel), resulting in a net dipole moment. This leads to stronger intermolecular dipole–dipole interactions, requiring more energy to overcome. The trans isomer is non-polar (dipoles cancel due to symmetry), so it has weaker intermolecular forces and a lower boiling point.

Answer:

Benzene ($C_6H_6$) is extraordinarily stable due to the phenomenon of resonance (delocalization of π electrons).

Reasons for extra stability:

1. Resonance/Delocalization: Benzene cannot be represented by a single Kekulé structure. The six π electrons are completely delocalized over all six carbon atoms forming a continuous ring of electron cloud above and below the plane of the ring. This delocalization lowers the energy of the molecule significantly.

2. Resonance Energy: The actual benzene is more stable than either of the two Kekulé structures by about 150 kJ/mol. This extra stability is called the resonance energy or delocalization energy.

3. Equal Bond Lengths: All C–C bond lengths in benzene are equal (139 pm), intermediate between a C–C single bond (154 pm) and C=C double bond (134 pm), confirming complete delocalization.

4. Hückel's Rule: Benzene has $(4n+2)\pi$ electrons with $n=1$ (i.e., $6\pi$ electrons), satisfying the aromaticity criterion, which confers special stability.

Due to this aromatic stability, benzene prefers electrophilic substitution over addition reactions, as substitution preserves the aromatic system.

Necessary conditions for aromaticity (Hückel's criteria):

A compound is said to be aromatic if it satisfies all of the following conditions:

1. Planarity: The molecule must be planar (all atoms in the ring lie in the same plane).

2. Complete conjugation: The molecule must have a completely conjugated system of $\pi$ electrons (every atom in the ring must have a $p$-orbital participating in conjugation — i.e., continuous cyclic conjugation).

3. Hückel's Rule — $(4n+2)\pi$ electrons: The cyclic conjugated system must contain $(4n+2)\pi$ electrons, where $n = 0, 1, 2, 3, \ldots$ (a whole number).
- For $n=0$: $2\pi$ electrons (e.g., cyclopropenyl cation)
- For $n=1$: $6\pi$ electrons (e.g., benzene)
- For $n=2$: $10\pi$ electrons (e.g., naphthalene)

Summary: A compound is aromatic if it is cyclic, planar, completely conjugated, and has $(4n+2)\pi$ electrons.

Note: The actual structures in (i), (ii), and (iii) are from figures not fully visible in the OCR. Based on standard NCERT content for this question, the three systems are:
- (i) Cyclopenta-1,3-diene (cyclopentadiene)
- (ii) Cycloocta-1,3,5,7-tetraene (cyclooctatetraene, COT)
- (iii) A cyclopropenyl anion or a non-planar system

Standard NCERT answer:

(i) Cyclopenta-1,3-diene:

This molecule has a $-CH_2-$ group in the ring. The carbon of $CH_2$ is $sp^3$ hybridized and does not have a $p$-orbital available for conjugation. Therefore, the $\pi$ electron system is not continuous/not completely conjugated. Since one of the necessary conditions for aromaticity (complete conjugation) is not met, it is not aromatic.

(ii) Cycloocta-1,3,5,7-tetraene (COT):

COT has $8\pi$ electrons. According to Hückel's rule, an aromatic compound must have $(4n+2)\pi$ electrons. For $8\pi$ electrons: $4n+2=8 \Rightarrow n=1.5$, which is not a whole number. Therefore, COT does not satisfy Hückel's rule. Also, COT is tub-shaped (non-planar), so it is not aromatic (it is anti-aromatic if forced planar, with $4n\pi$ electrons where $n=2$).

(iii) The third system (likely a charged or bicyclic system):

If it is a system where the ring is not planar or the $p$-orbitals cannot overlap continuously, then the condition of planarity and/or complete conjugation is violated. Without continuous overlap of $p$-orbitals, delocalization is not possible, and the compound is not aromatic.

Conclusion: All three systems fail to satisfy one or more of the necessary conditions for aromaticity: planarity, complete conjugation, and $(4n+2)\pi$ electrons.

Concept: Electrophilic aromatic substitution. The order of substitution matters because the first substituent directs the second.

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(i) p-Nitrobromobenzene:

Bromine is an ortho/para director. Nitro group is a meta director.

To get the para product, first introduce $Br$ (o/p director), then nitrate:

Step 1: Bromination of benzene
$$C_6H_6 \xrightarrow{Br_2/FeBr_3} C_6H_5Br \text{ (Bromobenzene)}$$

Step 2: Nitration of bromobenzene (Br directs NO₂ to ortho/para; separate para product)
$$C_6H_5Br \xrightarrow{HNO_3/H_2SO_4} p\text{-}BrC_6H_4NO_2$$

Product: p-Nitrobromobenzene

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(ii) m-Nitrochlorobenzene:

To get meta product, we need a meta director already present. $-NO_2$ is a meta director.

Step 1: Nitration of benzene
$$C_6H_6 \xrightarrow{HNO_3/H_2SO_4} C_6H_5NO_2 \text{ (Nitrobenzene)}$$

Step 2: Chlorination of nitrobenzene ($-NO_2$ directs $Cl$ to meta position)
$$C_6H_5NO_2 \xrightarrow{Cl_2/FeCl_3} m\text{-}ClC_6H_4NO_2$$

Product: m-Nitrochlorobenzene

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(iii) p-Nitrotoluene:

$-CH_3$ is an ortho/para director.

Step 1: Friedel-Crafts alkylation of benzene
$$C_6H_6 \xrightarrow{CH_3Cl/AlCl_3} C_6H_5CH_3 \text{ (Toluene)}$$

Step 2: Nitration of toluene ($-CH_3$ directs $NO_2$ to ortho/para; separate para product)
$$C_6H_5CH_3 \xrightarrow{HNO_3/H_2SO_4} p\text{-}CH_3C_6H_4NO_2$$

Product: p-Nitrotoluene

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(iv) Acetophenone ($C_6H_5COCH_3$):

Friedel-Crafts acylation:
$$C_6H_6 \xrightarrow{CH_3COCl/AlCl_3} C_6H_5-CO-CH_3 \text{ (Acetophenone)}$$

Or using acetic anhydride:
$$C_6H_6 \xrightarrow{(CH_3CO)_2O/AlCl_3} C_6H_5COCH_3$$

Product: Acetophenone

Given: $H_3C-CH_2-C(CH_3)_2-CH_2-CH(CH_3)_2$

Let us label each carbon:

$$\underset{(a)}{CH_3}-\underset{(b)}{CH_2}-\underset{(c)}{C}(\underset{(d)}{CH_3})_2-\underset{(e)}{CH_2}-\underset{(f)}{CH}(\underset{(g)}{CH_3})_2$$

Classification:

| Carbon | Type | Connected to | No. of H atoms |
|--------|------|-------------|----------------|
| (a) $CH_3$ | 1° (Primary) | 1 carbon (b) | 3 H |
| (b) $CH_2$ | 2° (Secondary) | 2 carbons (a, c) | 2 H |
| (c) $C(CH_3)_2$ | 3° (Tertiary) | Wait — C(c) is bonded to (b), (d), (d), and (e) = 4 carbons → Quaternary | 0 H |
| (d) $CH_3$ (×2) | 1° (Primary) | 1 carbon (c) | 3 H each |
| (e) $CH_2$ | 2° (Secondary) | 2 carbons (c, f) | 2 H |
| (f) $CH$ | 3° (Tertiary) | 3 carbons (e, g, g) | 1 H |
| (g) $CH_3$ (×2) | 1° (Primary) | 1 carbon (f) | 3 H each |

Summary:
- Primary (1°) carbons: (a), (d)×2, (g)×2 — each has 3 H atoms
- Secondary (2°) carbons: (b), (e) — each has 2 H atoms
- Tertiary (3°) carbon: (f) — has 1 H atom
- Quaternary carbon: (c) — has 0 H atoms

Answer:

Branching decreases the boiling point of an alkane.

Reason:

The boiling point of alkanes depends on the strength of van der Waals (London dispersion) forces between molecules. These forces depend on the surface area of contact between molecules.

- A straight-chain (unbranched) alkane has a larger surface area, allowing greater contact between molecules, leading to stronger van der Waals forces and a higher boiling point.

- A branched alkane has a more compact, spherical shape with a smaller surface area. This reduces the contact area between molecules, weakening the van der Waals forces, and resulting in a lower boiling point.

Example:
- $n$-Butane (straight chain): b.p. = $-0.5°C$
- Isobutane (2-methylpropane, branched): b.p. = $-11.7°C$

Thus, greater the branching, lower the boiling point.

Given:
- $CH_3-CH=CH_2 + HBr \xrightarrow{\text{no peroxide}} CH_3-CHBr-CH_3$ (2-bromopropane)
- $CH_3-CH=CH_2 + HBr \xrightarrow{\text{benzoyl peroxide}} CH_3-CH_2-CH_2Br$ (1-bromopropane)

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Case 1: Without peroxide — Ionic (Electrophilic Addition) — Markovnikov's Rule

HBr adds according to Markovnikov's rule: $H^+$ adds to the carbon with more H atoms (C1), and $Br^-$ adds to C2 (more substituted carbon, forming more stable secondary carbocation).

Mechanism:

Step 1: Electrophilic attack of $H^+$ on C1:
$$CH_3-CH=CH_2 + H^+ \rightarrow CH_3-\overset{+}{C}H-CH_3 \text{ (2° carbocation, more stable)}$$

Step 2: Nucleophilic attack of $Br^-$:
$$CH_3-\overset{+}{C}H-CH_3 + Br^- \rightarrow CH_3-CHBr-CH_3 \text{ (2-bromopropane)}$$

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Case 2: With benzoyl peroxide — Free Radical Addition — Anti-Markovnikov (Peroxide/Kharasch Effect)

Benzoyl peroxide generates free radicals. $Br\cdot$ (bromine radical) adds first to C1 (less substituted end), forming the more stable secondary carbon radical at C2.

Mechanism:

Initiation:
$$(C_6H_5COO)_2 \xrightarrow{h\nu} 2\,C_6H_5COO\cdot \rightarrow 2\,C_6H_5\cdot + 2CO_2$$
$$C_6H_5\cdot + HBr \rightarrow C_6H_6 + Br\cdot$$

Propagation:

Step 1: $Br\cdot$ adds to C1 (terminal carbon) to give more stable 2° radical:
$$Br\cdot + CH_2=CH-CH_3 \rightarrow Br-CH_2-\overset{\cdot}{C}H-CH_3 \text{ (2° radical, more stable)}$$

Step 2: The radical abstracts H from HBr:
$$Br-CH_2-\overset{\cdot}{C}H-CH_3 + HBr \rightarrow Br-CH_2-CH_2-CH_3 + Br\cdot$$

Termination: Combination of radicals.

Product: $BrCH_2CH_2CH_3$ = 1-bromopropane (Anti-Markovnikov product)

Conclusion: In the presence of peroxides, the reaction follows a free radical mechanism giving the anti-Markovnikov product (1-bromopropane). Without peroxides, ionic mechanism gives the Markovnikov product (2-bromopropane).

Given: 1,2-Dimethylbenzene (o-xylene)

Structure of o-xylene: A benzene ring with $CH_3$ groups at positions 1 and 2.

Ozonolysis of o-xylene:

Kekulé proposed two structures for benzene with alternating single and double bonds. For o-xylene, the two Kekulé structures place the double bond between C1–C2 in one structure and between C2–C3 (i.e., not between C1–C2) in the other.

Kekulé Structure 1 (double bond between C1 and C2, i.e., between the two $CH_3$-bearing carbons):

Ozonolysis of this double bond gives:
$$\text{2 moles of } CH_3-CO-CHO \text{ (methylglyoxal)} + \text{other fragments}$$

Actually, complete ozonolysis of o-xylene (treating the ring as having 3 double bonds per Kekulé structure) gives:

For Kekulé structure with double bond at C1=C2:
- Cleavage at C1=C2 gives a dialdehyde/diketone fragment.

The ozonolysis of o-xylene gives glyoxal ($OHC-CHO$) and methylglyoxal ($CH_3CO-CHO$) as products, consistent with the Kekulé structure having a double bond between C1 and C2.

More precisely, complete ozonolysis of o-xylene gives:
$$\text{o-xylene} \xrightarrow{O_3, \text{then } Zn/H_2O} CH_3CO-CHO + OHC-CHO + CH_3CO-CHO$$

The products obtained are 3-oxobutanal (methylglyoxal, $CH_3COCHO$) and ethanedial (glyoxal, $OHCCHO$), which are consistent with the Kekulé structure.

Support for Kekulé structure:

The ozonolysis products confirm the presence of C=C double bonds at specific positions in the ring as predicted by the Kekulé structure. The fact that two different sets of products can be obtained (corresponding to the two Kekulé structures) supports the concept of resonance between the two Kekulé forms. The actual benzene is a resonance hybrid of both Kekulé structures.

Given: Benzene ($C_6H_6$), $n$-hexane ($C_6H_{14}$), Ethyne ($HC\equiv CH$)

Decreasing order of acidity:
\text{Ethyne} &gt; \text{Benzene} &gt; n\text{-Hexane}

Reason:

Acidic character of a C–H bond depends on the hybridization of the carbon atom. Greater the $s$-character of the hybrid orbital, the closer the electrons are to the nucleus, the more electronegative the carbon, and the more easily it releases $H^+$ (i.e., more acidic).

| Compound | Hybridization of C | $s$-character | Relative Acidity |
|----------|-------------------|---------------|------------------|
| Ethyne ($HC\equiv CH$) | $sp$ | 50% | Highest |
| Benzene ($C_6H_6$) | $sp^2$ | 33.3% | Intermediate |
| $n$-Hexane ($C_6H_{14}$) | $sp^3$ | 25% | Lowest |

- Ethyne ($sp$ carbon, 50% $s$-character): Most electronegative carbon, holds the bonding electrons closest to nucleus → most acidic C–H bond.
- Benzene ($sp^2$ carbon, 33.3% $s$-character): Moderately acidic.
- $n$-Hexane ($sp^3$ carbon, 25% $s$-character): Least electronegative carbon → least acidic.

Decreasing order: Ethyne > Benzene > $n$-Hexane

Answer:

Benzene undergoes electrophilic substitution easily because:

1. Benzene has a high electron density above and below the plane of the ring due to the delocalized $\pi$ electron cloud (6 $\pi$ electrons).
2. This electron-rich $\pi$ system attracts electrophiles ($E^+$) readily.
3. The reaction proceeds via electrophilic substitution (rather than addition) because substitution preserves the aromatic stability (the delocalized $\pi$ system is restored after the reaction), whereas addition would destroy aromaticity.

Benzene undergoes nucleophilic substitution with difficulty because:

1. Nucleophiles are electron-rich species. The electron-rich $\pi$ cloud of benzene repels nucleophiles.
2. There is no low-energy pathway for a nucleophile to attack the electron-dense benzene ring.
3. Nucleophilic substitution would require the ring to become even more electron-rich in the transition state, which is energetically unfavorable.

Conclusion: The high electron density of the aromatic $\pi$ system makes benzene reactive toward electron-seeking electrophiles but unreactive toward electron-donating nucleophiles.

Concept: Benzene ($C_6H_6$) can be obtained by aromatization reactions.

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(i) Ethyne → Benzene:

Ethyne undergoes cyclic trimerization (ethynylation/trimerization) when passed through a red-hot iron tube at 873 K:
$$3\,HC\equiv CH \xrightarrow{\text{Red hot Fe tube, 873 K}} C_6H_6 \text{ (Benzene)}$$

This is the Berthelot reaction (trimerization of acetylene).

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(ii) Ethene → Benzene:

Ethene cannot be directly converted to benzene in one step. The route is:

Step 1: Convert ethene to ethyne by dehydrogenation (or via halogenation and double dehydrohalogenation):
$$CH_2=CH_2 \xrightarrow{\text{Dehydrogenation}} HC\equiv CH$$

Step 2: Trimerization of ethyne:
$$3\,HC\equiv CH \xrightarrow{873\,K, \text{Fe tube}} C_6H_6$$

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(iii) Hexane → Benzene:

$n$-Hexane undergoes dehydrogenation and cyclization (aromatization/reforming) in the presence of a catalyst:
$$CH_3(CH_2)_4CH_3 \xrightarrow{Cr_2O_3/Al_2O_3,\, 773\,K,\, \text{high pressure}} C_6H_6 + 4H_2$$

This process is called catalytic reforming or dehydrocyclization.

Given: Hydrogenation product = 2-methylbutane: $CH_3-CH(CH_3)-CH_2-CH_3$

Structure of 2-methylbutane:
$$CH_3-\underset{C2}{CH}(CH_3)-\underset{C3}{CH_2}-\underset{C4}{CH_3}$$

To find all alkenes that give 2-methylbutane on hydrogenation, we remove $H_2$ from adjacent carbons in all possible ways:

1. 2-Methylbut-1-ene: (remove H₂ from C1 and C2)
$$CH_2=C(CH_3)-CH_2-CH_3$$

2. 2-Methylbut-2-ene: (remove H₂ from C2 and C3)
$$CH_3-C(CH_3)=CH-CH_3$$

3. 3-Methylbut-1-ene: (remove H₂ from C1 and C2 of the main chain, considering the isomer)
$$CH_2=CH-CH(CH_3)-CH_3$$

*(Note: This is a structural isomer — 3-methylbut-1-ene — which on hydrogenation gives 2-methylbutane.)*

Summary of all alkenes:
1. $CH_2=C(CH_3)-CH_2-CH_3$ — 2-Methylbut-1-ene
2. $CH_3-C(CH_3)=CH-CH_3$ — 2-Methylbut-2-ene
3. $CH_2=CH-CH(CH_3)-CH_3$ — 3-Methylbut-1-ene

All three alkenes on hydrogenation ($H_2$/Ni) give 2-methylbutane.

Concept: Electron-donating groups (EDG) activate the benzene ring toward electrophilic substitution (increase reactivity), while electron-withdrawing groups (EWG) deactivate it (decrease reactivity).

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(a) Chlorobenzene, 2,4-Dinitrochlorobenzene, p-Nitrochlorobenzene:

- Chlorobenzene ($C_6H_5Cl$): $-Cl$ is weakly deactivating (EWG by induction, but o/p director by resonance). Moderately reactive.
- p-Nitrochlorobenzene ($p$-$O_2N$-$C_6H_4$-$Cl$): One $-NO_2$ group (strong EWG) deactivates the ring significantly.
- 2,4-Dinitrochlorobenzene ($2,4$-$(NO_2)_2$-$C_6H_3$-$Cl$): Two $-NO_2$ groups (strong EWG) deactivate the ring most strongly.

Decreasing order of reactivity:
\text{Chlorobenzene} &gt; p\text{-Nitrochlorobenzene} &gt; 2,4\text{-Dinitrochlorobenzene}

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(b) Toluene, p-H₃C–C₆H₄–NO₂, p-O₂N–C₆H₄–NO₂:

- Toluene ($C_6H_5CH_3$): $-CH_3$ is an EDG (electron-donating by hyperconjugation and induction) → activates ring → most reactive.
- p-Nitrotoluene ($p$-$CH_3$-$C_6H_4$-$NO_2$): Has both $-CH_3$ (EDG) and $-NO_2$ (EWG). The $-NO_2$ effect dominates → ring is deactivated compared to toluene.
- p-Dinitrobenzene ($p$-$O_2N$-$C_6H_4$-$NO_2$): Two $-NO_2$ groups → strongly deactivated → least reactive.

Decreasing order of reactivity:
\text{Toluene} &gt; p\text{-}CH_3C_6H_4NO_2 &gt; p\text{-}O_2NC_6H_4NO_2

Answer: Toluene will undergo nitration most easily.

Reason:

- Toluene ($C_6H_5CH_3$): The methyl group ($-CH_3$) is an electron-donating group (EDG). It donates electrons to the benzene ring through hyperconjugation and inductive effect, increasing the electron density of the ring. This makes the ring more reactive toward electrophilic attack (by $NO_2^+$, the nitronium ion). → Most reactive.

- Benzene ($C_6H_6$): No substituent; moderate reactivity. → Intermediate reactivity.

- m-Dinitrobenzene ($m$-$(NO_2)_2C_6H_4$): Two nitro groups ($-NO_2$) are strong electron-withdrawing groups (EWG). They withdraw electron density from the ring by both inductive and resonance effects, making the ring electron-poor and strongly deactivated toward electrophilic substitution. → Least reactive.

Order of ease of nitration:
\text{Toluene} &gt; \text{Benzene} &gt; m\text{-Dinitrobenzene}

Given: Ethylation of benzene = Friedel-Crafts alkylation using $C_2H_5Cl$ and a Lewis acid catalyst.

Concept: A Lewis acid accepts an electron pair and helps generate the carbocation (electrophile) from the alkyl halide.

Lewis acids that can be used (other than $AlCl_3$):

- Ferric chloride ($FeCl_3$)
- Boron trifluoride ($BF_3$)
- Zinc chloride ($ZnCl_2$)
- Stannic chloride ($SnCl_4$)

Answer: Any one of the following Lewis acids can be used:
$$FeCl_3,\quad BF_3,\quad ZnCl_2,\quad SnCl_4$$

For example, Ferric chloride ($FeCl_3$) or Boron trifluoride ($BF_3$) can be used as the Lewis acid catalyst for ethylation of benzene.

Given: Wurtz reaction: $2\,R-X + 2Na \xrightarrow{\text{dry ether}} R-R + 2NaX$

Reason:

Wurtz reaction involves coupling of two alkyl halides with sodium metal. If we use two different alkyl halides (say $R-X$ and $R'-X$) to prepare an alkane with an odd number of carbons, the reaction gives a mixture of three products:
1. $R-R$ (from coupling of two $R-X$ molecules)
2. $R'-R'$ (from coupling of two $R'-X$ molecules)
3. $R-R'$ (the desired product, from coupling of $R-X$ and $R'-X$)

Separating the desired product from this mixture is difficult, making the reaction impractical and inefficient for odd-carbon alkanes.

Illustration:

To prepare propane ($C_3H_8$, 3 carbons — odd) using Wurtz reaction, we need to couple $CH_3X$ (methyl halide, 1C) and $C_2H_5X$ (ethyl halide, 2C):

$$CH_3Br + C_2H_5Br + 2Na \xrightarrow{\text{dry ether}} \text{mixture of products}$$

Products formed:
1. $CH_3-CH_3$ (ethane) — from $2\,CH_3Br + 2Na$
2. $C_2H_5-C_2H_5$ (butane) — from $2\,C_2H_5Br + 2Na$
3. $CH_3-C_2H_5$ (propane) — desired product

Since all three products are formed simultaneously, the yield of propane is low and separation is difficult.

Conclusion: Wurtz reaction is not preferred for alkanes with an odd number of carbon atoms because it gives a mixture of products that are difficult to separate.