Haloalkanes and Haloarenes

Given: Reaction of alcohols with KI to prepare alkyl iodides.

Concept: KI is used with phosphoric acid (H₃PO₄) and not H₂SO₄ for the conversion of alcohols to alkyl iodides.

Explanation:

H₂SO₄ is an oxidising acid. If H₂SO₄ is used along with KI, the following side reactions occur:

$$\text{H}_2\text{SO}_4 + 2\text{KI} \rightarrow \text{K}_2\text{SO}_4 + 2\text{HI}$$

The HI formed is then oxidised by H₂SO₄:

$$\text{H}_2\text{SO}_4 + 2\text{HI} \rightarrow \text{SO}_2 + \text{I}_2 + 2\text{H}_2\text{O}$$

Thus H₂SO₄ oxidises HI (and KI) to I₂, which cannot act as a nucleophile for the substitution reaction. Hence H₂SO₄ is not used; instead, non-oxidising acids like H₃PO₄ are used.

Given: Propane, $\text{C}_3\text{H}_8$; dihalogen derivatives (using Cl as representative halogen).

Concept: Replace two hydrogen atoms of propane with halogen atoms in all possible ways.

The different dihalogen derivatives of propane are:

(i) 1,1-Dichloropropane:
$$\text{CH}_3\text{CH}_2\text{CHCl}_2$$

(ii) 1,2-Dichloropropane:
$$\text{CH}_3\text{CHClCH}_2\text{Cl}$$

(iii) 1,3-Dichloropropane:
$$\text{ClCH}_2\text{CH}_2\text{CH}_2\text{Cl}$$

(iv) 2,2-Dichloropropane:
$$\text{CH}_3\text{CCl}_2\text{CH}_3$$

(v) 1,1-Dichloropropane (gem on C1) is listed above; additionally:

$$\text{ClCH}_2\text{CHClCH}_3 \quad \text{(1,2-dichloropropane)}$$

All four structural isomers:
1. $\text{ClCH}_2\text{CH}_2\text{CH}_2\text{Cl}$ — 1,3-dichloropropane
2. $\text{ClCH}_2\text{CHClCH}_3$ — 1,2-dichloropropane
3. $\text{CH}_3\text{CCl}_2\text{CH}_3$ — 2,2-dichloropropane
4. $\text{CH}_3\text{CH}_2\text{CHCl}_2$ — 1,1-dichloropropane

Note: The molecular formula given in the text appears as $C_9H_{12}$ but in context of isomeric alkanes yielding mono-chloro products, the correct formula should be $C_5H_{12}$ (pentane isomers). Solutions are given accordingly.

Given: Isomers of $C_5H_{12}$: n-pentane, isopentane (2-methylbutane), neopentane (2,2-dimethylpropane).

Concept: Photochemical chlorination replaces H atoms. The number of monochloride products equals the number of sets of equivalent (chemically distinct) hydrogen atoms.

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(i) Single monochloride:

2,2-Dimethylpropane (Neopentane): $(\text{CH}_3)_4\text{C}$

All 12 hydrogen atoms are equivalent (all are on methyl groups attached to the central carbon). Hence only one monochloride is formed:
$$(\text{CH}_3)_3\text{CCH}_2\text{Cl} \quad \text{(1-chloro-2,2-dimethylpropane)}$$

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(ii) Three isomeric monochlorides:

2-Methylbutane (Isopentane): $\text{CH}_3\text{CH}(\text{CH}_3)\text{CH}_2\text{CH}_3$

Distinct types of H atoms:
- $-\text{CH}_3$ groups at C-1 and C-4 (equivalent, 6H)
- $-\text{CH}_3$ at C-2 (3H)
- $-\text{CH}-$ at C-2 (1H)
- $-\text{CH}_2-$ at C-3 (2H)

Wait — this gives 4 types. Let us recount:
- C1: $(\text{CH}_3)_2$ — the two methyl groups on C2 are equivalent (6H)
- C2: tertiary H (1H)
- C3: $-\text{CH}_2-$ (2H)
- C4: terminal $\text{CH}_3$ (3H)

This gives 4 types. So isopentane gives 4 monochlorides.

n-Pentane: $\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{CH}_3$

Distinct H types:
- C1 (and C5): $\text{CH}_3$ (6H equivalent)
- C2 (and C4): $\text{CH}_2$ (4H equivalent)
- C3: $\text{CH}_2$ (2H)

This gives 3 types → 3 isomeric monochlorides.

$$\text{Answer: n-Pentane gives three isomeric monochlorides.}$$

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(iii) Four isomeric monochlorides:

2-Methylbutane (Isopentane): $(\text{CH}_3)_2\text{CHCH}_2\text{CH}_3$

Four distinct H environments → 4 isomeric monochlorides.

$$\text{Answer: 2-Methylbutane (isopentane) gives four isomeric monochlorides.}$$

Note: Structures (i)–(iv) are given as images in the source. The reactions are interpreted from context.

(i) Isobutane + Cl₂ (hν):

Isobutane: $(\text{CH}_3)_3\text{CH}$

Tertiary C–H bond is weaker and more reactive toward free radical halogenation. The major product is the tertiary chloride:

$$\text{Major product: } (\text{CH}_3)_3\text{CCl} \quad \text{(2-chloro-2-methylpropane, tert-butyl chloride)}$$

(ii) 2-Methylbutane + Cl₂ (hν):

2-Methylbutane: $(\text{CH}_3)_2\text{CHCH}_2\text{CH}_3$

The tertiary H at C-2 is most reactive. Major product:

$$\text{Major product: } (\text{CH}_3)_2\text{CClCH}_2\text{CH}_3 \quad \text{(2-chloro-2-methylbutane)}$$

(iii) Cyclopentane + Br₂ (hν):

All H atoms in cyclopentane are equivalent. Only one monobromo product is possible:

$$\text{Major product: Bromocyclopentane}$$

(iv) Methylcyclohexane + Cl₂ (hν):

The tertiary C–H (at C-1, bearing the methyl group) is most reactive. Major product:

$$\text{Major product: 1-Chloro-1-methylcyclohexane}$$

(v) $\text{CH}_3\text{CH}_2\text{Br} + \text{NaI} \xrightarrow{\text{acetone}}$

This is a Finkelstein reaction (halogen exchange, $S_N2$):

$$\text{CH}_3\text{CH}_2\text{Br} + \text{NaI} \xrightarrow{\text{acetone}} \text{CH}_3\text{CH}_2\text{I} + \text{NaBr}$$

Major product: Iodoethane (ethyl iodide)

(vi) $\text{CH}_4 + \text{Br}_2 \xrightarrow{\text{heat/UV light}}$

Free radical bromination of methane:

$$\text{CH}_4 + \text{Br}_2 \xrightarrow{h\nu} \text{CH}_3\text{Br} + \text{HBr}$$

Major product: Bromomethane (methyl bromide)

Concept: $S_N2$ reaction rate depends on steric hindrance at the carbon bearing the leaving group. Less hindered (less substituted) carbon reacts faster in $S_N2$.

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(i) $\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{Br}$ (1° — n-butyl bromide) vs. $\text{CH}_3\text{CH}_2\text{CH}(\text{Br})\text{CH}_3$ (2° — sec-butyl bromide)

n-Butyl bromide (1°) reacts faster by $S_N2$ because the carbon bearing Br has less steric hindrance (only one alkyl group) compared to the secondary carbon (two alkyl groups) in sec-butyl bromide.

$$\text{Faster: } \text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{Br}$$

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(ii) $\text{CH}_3\text{CH}_2\text{CH}(\text{Br})\text{CH}_3$ (2° — sec-butyl bromide) vs. $(\text{CH}_3)_3\text{CBr}$ (3° — tert-butyl bromide)

sec-Butyl bromide (2°) reacts faster by $S_N2$ because tertiary carbon is highly hindered (three alkyl groups), making backside attack by nucleophile very difficult.

$$\text{Faster: } \text{CH}_3\text{CH}_2\text{CH}(\text{Br})\text{CH}_3$$

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(iii) $\text{CH}_3\text{CH}(\text{CH}_3)\text{CH}_2\text{CH}_2\text{Br}$ (1°, but with branching at C-3) vs. $\text{CH}_3\text{CH}_2\text{CH}(\text{CH}_3)\text{CH}_2\text{Br}$ (1°, branching at C-2)

Both are primary halides, but branching closer to the reaction centre causes more steric hindrance. In $\text{CH}_3\text{CH}_2\text{CH}(\text{CH}_3)\text{CH}_2\text{Br}$, the branch is at the $\beta$-carbon (C-2), causing more hindrance to backside attack than in $\text{CH}_3\text{CH}(\text{CH}_3)\text{CH}_2\text{CH}_2\text{Br}$ where the branch is at the $\gamma$-carbon (C-3).

$$\text{Faster: } \text{CH}_3\text{CH}(\text{CH}_3)\text{CH}_2\text{CH}_2\text{Br} \quad \text{(branching at C-3, less hindrance)}$$

Concept: $S_N1$ reaction rate depends on the stability of the carbocation intermediate formed. More stable carbocation → faster $S_N1$ reaction. Stability order: 3° > 2° > 1° > methyl. Benzylic and allylic carbocations are also stabilised by resonance.

(i) Since the structures are given as images (not visible), the general principle is:

The compound that forms the more stable carbocation upon ionisation will undergo $S_N1$ faster. A tertiary or benzylic/allylic halide will react faster than a primary or secondary halide.

(ii) Similarly, the compound forming the more stable (higher substituted or resonance-stabilised) carbocation reacts faster by $S_N1$.

General Answer: In each pair, the compound with the more substituted carbon bearing the halogen (tertiary > secondary > primary) or with resonance stabilisation of the carbocation (benzylic, allylic) undergoes faster $S_N1$ reaction.

Concept: Grignard reagent formation: $\text{R-X} + \text{Mg} \xrightarrow{\text{dry ether}} \text{R-MgX}$ (Grignard reagent). Hydrolysis with H₂O gives R-H; with D₂O gives R-D.

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Reaction 1: $\text{R-X} + \text{Mg} \xrightarrow{\text{dry ether}} \text{A} \xrightarrow{\text{H}_2\text{O}} \text{B}$

- A = R-MgX (Grignard reagent)
- B = R-H (hydrocarbon)

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Reaction 2: $\text{R-X} + \text{Mg} \xrightarrow{\text{dry ether}} \text{C} \xrightarrow{\text{D}_2\text{O}} \text{CH}_2\text{D-CH(CH}_3\text{)}$

Product with D₂O is $\text{CH}_2\text{D-CHCH}_3$ which is $\text{CH}_3\text{CHDCH}_3$ — actually the product is propane with one D: $\text{CH}_3\text{CHDCH}_3$ (2-deuteropropane) or $\text{CH}_2\text{DCH}_2\text{CH}_3$ (1-deuteropropane).

From the formula $\text{CH}_2\text{DCHCH}_3$: this suggests the Grignard reagent C is $\text{CH}_3\text{CH}(\text{MgX})\text{CH}_3$ — isopropyl magnesium halide, formed from isopropyl halide.

So: $\text{R-X} = (\text{CH}_3)_2\text{CHX}$ (isopropyl halide)
- C = $(\text{CH}_3)_2\text{CHMgX}$ (isopropyl magnesium halide)
- Product with D₂O: $(\text{CH}_3)_2\text{CHD}$ (2-deuteropropane)

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Reaction 3: $(\text{CH}_3)_3\text{C-X} \xrightarrow{\text{Na/ether}} \text{R'-X} \xrightarrow{\text{Mg}} \text{D} \xrightarrow{\text{H}_2\text{O}} \text{E}$

Wurtz reaction of $(\text{CH}_3)_3\text{CX}$ with Na gives $(\text{CH}_3)_3\text{C-C(CH}_3)_3$ — but this is the Wurtz product. However, the sequence shows R'-X is formed first, then Grignard D, then hydrolysis gives E.

Actually, re-reading: $(\text{CH}_3)_3\text{C-X}$ with Na in ether undergoes Wurtz reaction with another molecule R'-X. If R' = $(\text{CH}_3)_3\text{C}$, then:
- R'-X = $(\text{CH}_3)_3\text{CX}$ (tert-butyl halide)
- D = $(\text{CH}_3)_3\text{CMgX}$ (tert-butyl magnesium halide, Grignard reagent)
- E = $(\text{CH}_3)_3\text{CH}$ (2-methylpropane / isobutane)

Summary:
- R = isopropyl group, $\text{R-X}$ = isopropyl halide (e.g., $(\text{CH}_3)_2\text{CHBr}$)
- A = $(\text{CH}_3)_2\text{CHMgBr}$
- B = $(\text{CH}_3)_2\text{CH}_2$ (propane)
- C = $(\text{CH}_3)_2\text{CHMgBr}$
- R' = tert-butyl, $\text{R'-X}$ = $(\text{CH}_3)_3\text{CBr}$
- D = $(\text{CH}_3)_3\text{CMgBr}$
- E = $(\text{CH}_3)_3\text{CH}$ (isobutane)

(i) $(\text{CH}_3)_2\text{CHCH(Cl)CH}_3$

Longest chain containing C–Cl: 4 carbons (butane). Cl is on C-2, methyl branch on C-3.

IUPAC Name: 2-Chloro-3-methylbutane

Classification: Secondary (2°) alkyl halide (Cl on secondary carbon)

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(ii) $\text{CH}_3\text{CH}_2\text{CH(CH}_3)\text{CH(C}_2\text{H}_5)\text{Cl}$

Expand: $\text{CH}_3\text{CH}_2\text{CH(CH}_3)\text{CH(Cl)CH}_2\text{CH}_3$

Longest chain: 6 carbons (hexane). Cl on C-3, methyl on C-4.

IUPAC Name: 3-Chloro-4-methylhexane

Classification: Secondary (2°) alkyl halide

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(iii) $\text{CH}_3\text{CH}_2\text{C(CH}_3)_2\text{CH}_2\text{I}$

Longest chain: 4 carbons. I on C-1, two methyl groups on C-2, ethyl on C-2.

Actually: $\text{ICH}_2\text{C(CH}_3)_2\text{CH}_2\text{CH}_3$

Longest chain including I-bearing carbon: C1(CH₂I)–C2(C(CH₃)₂)–C3(CH₂)–C4(CH₃) = 4C with two methyls at C2.

IUPAC Name: 1-Iodo-2,2-dimethylbutane

Classification: Primary (1°) alkyl halide

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(iv) $(\text{CH}_3)_3\text{CCH}_2\text{CH(Br)C}_6\text{H}_5$

The carbon bearing Br is also attached to $\text{C}_6\text{H}_5$ (phenyl group) and two other carbons → benzylic position.

Longest chain: C1(CH(Br))–C2(CH₂)–C3(C(CH₃)₃) = 3 carbons with tert-butyl at C2 and phenyl at C1.

Name as: 1-Bromo-1-phenyl-3,3-dimethylbutane

Classification: Benzyl halide, secondary (2°)

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(v) $\text{CH}_3\text{CH(CH}_3)\text{CH(Br)CH}_3$

This is: $\text{CH}_3\text{CH(CH}_3)\text{CHBrCH}_3$

Longest chain: 4 carbons. Br on C-2, methyl on C-3.

IUPAC Name: 2-Bromo-3-methylbutane

Classification: Secondary (2°) alkyl halide

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(vi) $\text{CH}_3\text{C(C}_2\text{H}_5)_2\text{CH}_2\text{Br}$

Expand: $\text{BrCH}_2\text{C(CH}_3)(\text{C}_2\text{H}_5)_2$

Longest chain: C1(CH₂Br)–C2(C)–C3(CH₂)–C4(CH₃) = 4C, with methyl and ethyl at C2.

Actually longest chain through C2: C4H chain with two ethyl groups? Let's count: $\text{CH}_3\text{-C(C}_2\text{H}_5)_2\text{-CH}_2\text{Br}$

Longest chain: 5 carbons (including one ethyl): C1(CH₂Br)–C2(C)–C3(CH₂)–C4(CH₂)–C5(CH₃), with methyl and ethyl substituents at C2.

IUPAC Name: 1-Bromo-2-ethyl-2-methylbutane

Classification: Primary (1°) alkyl halide

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(vii) $\text{CH}_3\text{C(Cl)(C}_2\text{H}_5)\text{CH}_2\text{CH}_3$

This is: central C bearing Cl, CH₃, C₂H₅, and CH₂CH₃.

Longest chain: C1(CH₃)–C2(CCl)–C3(CH₂)–C4(CH₃) = 4C with ethyl at C2.

IUPAC Name: 2-Chloro-2-methylbutane

Classification: Tertiary (3°) alkyl halide

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(viii) $\text{CH}_3\text{CH=C(Cl)CH}_2\text{CH(CH}_3)_2$

Longest chain containing double bond and Cl: C1(CH₃)–C2(CH=)–C3(=CCl)–C4(CH₂)–C5(CH)–C6(CH₃) with methyl at C5.

IUPAC Name: 3-Chloro-5-methylhex-2-ene

Classification: Vinyl halide (Cl on sp² carbon of double bond)

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(ix) $\text{CH}_3\text{CH=CHC(Br)(CH}_3)_2$

Longest chain: C1(CH₃)–C2(CH=)–C3(=CH)–C4(CBr(CH₃)₂) = 4C with two methyls at C4 and Br at C4.

Numbering from Br end: C1(C(Br)(CH₃)₂)–C2(CH=)–C3(=CH)–C4(CH₃)

Give lower locant to double bond: but-2-ene with Br at C1 and two methyls at C1.

IUPAC Name: 1-Bromo-1-methylbut-2-ene (or 3-Bromo-3-methylbut-1-ene)

Using lowest locant rule for double bond: 3-Bromo-3-methylbut-1-ene

Classification: Allylic halide (Br on carbon adjacent to C=C), tertiary (3°)

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(x) $p\text{-ClC}_6\text{H}_4\text{CH}_2\text{CH(CH}_3)_2$

Cl is directly on benzene ring (aryl Cl). The side chain is isobutyl.

IUPAC Name: 1-Chloro-4-(2-methylpropyl)benzene

Classification: Aryl halide

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(xi) $m\text{-ClCH}_2\text{C}_6\text{H}_4\text{CH}_2\text{C(CH}_3)_3$

Here $\text{ClCH}_2$ is a chloromethyl group attached to benzene ring at meta position. The Cl is on the benzylic carbon (CH₂ attached to ring).

IUPAC Name: 1-(Chloromethyl)-3-(2,2-dimethylpropyl)benzene

Classification: Benzyl halide, primary (1°)

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(xii) $o\text{-Br-C}_6\text{H}_4\text{CH(CH}_3)\text{CH}_2\text{CH}_3$

Br is directly on benzene ring. The substituent at ortho position is sec-butyl: $\text{CH(CH}_3)\text{CH}_2\text{CH}_3$.

IUPAC Name: 1-Bromo-2-(1-methylpropyl)benzene or 1-Bromo-2-sec-butylbenzene

Classification: Aryl halide

(i) $\text{CH}_3\text{CH(Cl)CH(Br)CH}_3$

Chain: 4 carbons (butane). Cl on C-2, Br on C-3.

Number to give lower locants: Cl at C-2, Br at C-3.

IUPAC Name: 2-Bromo-3-chlorobutane

(Alphabetical order: bromo before chloro; numbering gives 2,3 from either end — choose end giving lower locant to first-cited substituent alphabetically: Br gets 3, Cl gets 2 → set {2,3}; from other end Br gets 2, Cl gets 3 → set {2,3}. Same. Use alphabetical: bromo cited first, so give Br lower number → 2-Bromo-3-chlorobutane.)

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(ii) $\text{CHF}_2\text{CBrClF}$

Two-carbon chain (ethane). C1: CHF₂; C2: CBrClF.

Substituents: Br, Cl, F, F, F → on C1: 2F, 1H; on C2: Br, Cl, F.

Number to give lower locants: C1 has F,F; C2 has Br,Cl,F.

IUPAC Name: 2-Bromo-2-chloro-1,1,2-trifluoroethane

(Halothane is a common name for this compound.)

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(iii) $\text{ClCH}_2\text{C}\equiv\text{CCH}_2\text{Br}$

Four-carbon chain with triple bond between C2 and C3 (but-2-yne). Cl on C1, Br on C4.

IUPAC Name: 1-Bromo-4-chlorobut-2-yne

(Number from Br end to give lower locant to triple bond: Br at C1, Cl at C4, triple bond at C2–C3.)

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(iv) $(\text{CCl}_3)_3\text{CCl}$

Central carbon bears Cl and three CCl₃ groups. Total carbons = 4 (neopentane skeleton). Central C: C with Cl; three terminal C: each with 3 Cl.

This is 2-carbon? No: $(\text{CCl}_3)_3\text{CCl}$ = C(CCl₃)₃Cl.

Longest chain: 2 carbons. But with three CCl₃ branches on one carbon:

Actually this is a 4-carbon compound: central C + 3 × CCl₃. Longest chain = 2C (one CCl₃ + central C). Substituents: 2 × CCl₃ (as trichloromethyl) on C2, Cl on C2.

IUPAC Name: 1,1,1-Trichloro-2,2,2-tris(trichloromethyl)ethane

Alternatively treating as methane derivative: 2-(trichloromethyl)-1,1,1,3,3,3-hexachloropropane...

Simplest: The compound is 2-carbon: $\text{CCl}_3\text{-CCl}_3$ is hexachloroethane; here we have $(\text{CCl}_3)_3\text{CCl}$.

Longest chain = 2C: $\text{CCl}_3$ (C1) — $\text{C(CCl}_3)_2\text{Cl}$ (C2).

IUPAC Name: 1,1,1,2-Tetrachloro-2,2-bis(trichloromethyl)ethane

*(Note: This is a complex polychlorinated compound; the systematic name reflects all substituents.)*

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(v) $\text{CH}_3\text{C}(p\text{-ClC}_6\text{H}_4)_2\text{CH(Br)CH}_3$

Chain: C1(CH₃)–C2(C(Ar)₂)–C3(CHBr)–C4(CH₃) = 4 carbons (butane). Br on C3, two (4-chlorophenyl) groups on C2.

IUPAC Name: 3-Bromo-2,2-bis(4-chlorophenyl)butane

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(vi) $(\text{CH}_3)_3\text{CCH=CClC}_6\text{H}_4\text{I-}p$

Chain: C1(C(CH₃)₃)–C2(CH=)–C3(=CCl)–C6H4I(p) attached to C3.

Longest chain including double bond: C1(tBu carbon)... actually the tert-butyl is a substituent.

Longest chain: C1–C2=C3 with (4-iodophenyl) on C3 and Cl on C3, tert-butyl on C1.

As a 3-carbon chain (prop-1-ene): C1(=CH–C(CH₃)₃)–C2...

Renumber: C1(CCl(Ar)=)–C2(=CH)–C3(C(CH₃)₃)

IUPAC Name: 1-Chloro-1-(4-iodophenyl)-3,3-dimethylbut-1-ene

(i) 2-Chloro-3-methylpentane:

$$\text{CH}_3\text{-CHCl-CH(CH}_3)\text{-CH}_2\text{-CH}_3$$

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(ii) $p$-Bromochlorobenzene:

Benzene ring with Br and Cl at para positions (1,4):

$$\text{4-BrC}_6\text{H}_4\text{Cl}$$

(Cl at C-1, Br at C-4 of benzene ring)

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(iii) 1-Chloro-4-ethylcyclohexane:

Cyclohexane ring with Cl at C-1 and ethyl group ($\text{-CH}_2\text{CH}_3$) at C-4.

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(iv) 2-(2-Chlorophenyl)-1-iodooctane:

$$\text{ICH}_2\text{-CH(2-ClC}_6\text{H}_4)\text{-CH}_2\text{CH}_2\text{CH}_2\text{CH}_2\text{CH}_2\text{CH}_3$$

(8-carbon chain, I at C-1, 2-chlorophenyl group at C-2)

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(v) 2-Bromobutane:

$$\text{CH}_3\text{-CHBr-CH}_2\text{-CH}_3$$

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(vi) 4-tert-Butyl-3-iodoheptane:

$$\text{CH}_3\text{CH}_2\text{CHI-CH(C(CH}_3)_3)\text{-CH}_2\text{CH}_2\text{CH}_3$$

(7-carbon chain, I at C-3, tert-butyl at C-4)

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(vii) 1-Bromo-4-sec-butyl-2-methylbenzene:

Benzene ring with:
- Br at C-1
- Methyl ($\text{-CH}_3$) at C-2
- sec-Butyl ($\text{-CH(CH}_3)\text{CH}_2\text{CH}_3$) at C-4

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(viii) 1,4-Dibromobut-2-ene:

$$\text{BrCH}_2\text{-CH=CH-CH}_2\text{Br}$$

Given: Three chloromethane derivatives.

Concept: Dipole moment depends on the vector sum of individual bond dipoles. In a symmetric molecule, bond dipoles cancel.

- $\text{CCl}_4$: Tetrahedral, perfectly symmetric. All four C–Cl bond dipoles cancel completely. Dipole moment = 0.

- $\text{CHCl}_3$ (Chloroform): Three C–Cl bonds point in similar directions; the C–H bond dipole is small and in the opposite direction. The net dipole moment is significant but partial cancellation occurs. $\mu \approx 1.87$ D.

- $\text{CH}_2\text{Cl}_2$ (Dichloromethane): Two C–Cl bonds and two C–H bonds. The two C–Cl dipoles add up (they are on the same side), and the two C–H dipoles also add up (opposite side). The resultant is the largest among the three. $\mu \approx 1.60$ D.

Wait — comparing values: $\text{CH}_2\text{Cl}_2$: 1.60 D; $\text{CHCl}_3$: 1.87 D.

$\text{CHCl}_3$ has the highest dipole moment among the three because in $\text{CH}_2\text{Cl}_2$ there is more cancellation between the two Cl–C–Cl vectors, while in $\text{CHCl}_3$ the three C–Cl dipoles reinforce each other more effectively.

Answer: (ii) $\text{CHCl}_3$ has the highest dipole moment.

*Justification:* In $\text{CCl}_4$, all dipoles cancel (μ = 0). In $\text{CH}_2\text{Cl}_2$, partial cancellation gives μ ≈ 1.60 D. In $\text{CHCl}_3$, three C–Cl dipoles point in nearly the same direction with only one C–H dipole opposing, giving μ ≈ 1.87 D (highest).

Given: Molecular formula $\text{C}_5\text{H}_{10}$; no reaction with $\text{Cl}_2$ in dark; single monochloro product in sunlight.

Analysis:

- $\text{C}_5\text{H}_{10}$ with degree of unsaturation = $\frac{2(5)+2-10}{2} = 1$. So it has one degree of unsaturation — either a ring or a double bond.

- No reaction with $\text{Cl}_2$ in dark rules out alkenes (which undergo addition with $\text{Cl}_2$ in dark). So it must be a cycloalkane.

- Single monochloro product in sunlight (free radical substitution) means all hydrogen atoms in the molecule are equivalent.

- Among $\text{C}_5\text{H}_{10}$ cycloalkanes: Cyclopentane has all 10 H atoms equivalent (all are $\text{CH}_2$ groups in a symmetric ring). Substitution of any H gives the same product: chlorocyclopentane.

The hydrocarbon is Cyclopentane.

$$\text{Cyclopentane} + \text{Cl}_2 \xrightarrow{h\nu} \text{Chlorocyclopentane} + \text{HCl}$$

All 10 hydrogen atoms in cyclopentane are equivalent, so only one monochloro product (chlorocyclopentane) is formed.

Given: Molecular formula $\text{C}_4\text{H}_9\text{Br}$ (monobromo derivatives of butane).

Concept: Write all possible structural isomers by placing Br at different positions on the butane skeleton.

The four isomers are:

(i) 1-Bromobutane (n-butyl bromide):
$$\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{Br}$$

(ii) 2-Bromobutane (sec-butyl bromide):
$$\text{CH}_3\text{CH}_2\text{CHBrCH}_3$$

(iii) 1-Bromo-2-methylpropane (isobutyl bromide):
$$\text{(CH}_3)_2\text{CHCH}_2\text{Br}$$

(iv) 2-Bromo-2-methylpropane (tert-butyl bromide):
$$(\text{CH}_3)_3\text{CBr}$$

These are all four structural isomers of $\text{C}_4\text{H}_9\text{Br}$.

(i) From 1-butanol:

Reaction of 1-butanol with HI (or KI + H₃PO₄):

$$\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{OH} + \text{HI} \xrightarrow{\Delta} \text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{I} + \text{H}_2\text{O}$$

Alternatively using red phosphorus and iodine:
$$3\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{OH} + \text{PI}_3 \rightarrow 3\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{I} + \text{H}_3\text{PO}_3$$

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(ii) From 1-chlorobutane (Finkelstein reaction):

Halogen exchange using NaI in dry acetone:

$$\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{Cl} + \text{NaI} \xrightarrow{\text{dry acetone}} \text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{I} + \text{NaCl}\downarrow$$

NaCl precipitates out of acetone, driving the equilibrium forward.

---

(iii) From but-1-ene:

Addition of HI to but-1-ene. By Markovnikov's rule, H adds to C-1 and I adds to C-2, giving 2-iodobutane. To get 1-iodobutane, anti-Markovnikov addition is needed using HI in the presence of peroxides:

$$\text{CH}_3\text{CH}_2\text{CH=CH}_2 + \text{HI} \xrightarrow{\text{peroxide}} \text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{I}$$

*(Note: Peroxide effect/anti-Markovnikov addition works well for HBr but not for HI. An alternative route: add HBr with peroxide to get 1-bromobutane, then Finkelstein reaction with NaI to get 1-iodobutane.)*

$$\text{CH}_3\text{CH}_2\text{CH=CH}_2 + \text{HBr} \xrightarrow{\text{peroxide}} \text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{Br} \xrightarrow{\text{NaI/acetone}} \text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{I}$$

Definition: Ambident nucleophiles are nucleophiles that have two different nucleophilic sites (two atoms through which they can attack an electrophile), and can form two different products depending on which site attacks.

Example: Cyanide ion ($\text{CN}^-$)

The cyanide ion has two nucleophilic sites:
- Through carbon (C) → forms nitrile (alkyl cyanide, R–C≡N)
- Through nitrogen (N) → forms isonitrile (isocyanide, R–N≡C)

$$\text{R-X} + \text{KCN} \rightarrow \text{R-CN} + \text{KX} \quad \text{(nitrile, major product)}$$
$$\text{R-X} + \text{KCN} \rightarrow \text{R-NC} + \text{KX} \quad \text{(isocyanide, minor product)}$$

Another example: Nitrite ion ($\text{NO}_2^-$)

- Attack through oxygen → forms nitrite ester (R–O–N=O)
- Attack through nitrogen → forms nitroalkane (R–NO₂)

$$\text{R-X} + \text{AgNO}_2 \rightarrow \text{R-ONO} \quad \text{(alkyl nitrite)}$$
$$\text{R-X} + \text{KNO}_2 \rightarrow \text{R-NO}_2 \quad \text{(nitroalkane)}$$

Such nucleophiles are called ambident nucleophiles because they can donate electrons from either of two sites.

Concept: In $S_N2$ reactions, the rate depends on:
1. Steric hindrance at the carbon bearing the leaving group (less hindered = faster)
2. Nature of leaving group (better leaving group = faster; I⁻ > Br⁻ > Cl⁻ as leaving groups)

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(i) $\text{CH}_3\text{Br}$ vs $\text{CH}_3\text{I}$

Both are methyl halides (no steric difference). The difference is the leaving group:
- I⁻ is a better leaving group than Br⁻ (weaker base, more stable, C–I bond is weaker and longer).

$\text{CH}_3\text{I}$ reacts faster in $S_N2$ because iodide is a better leaving group.

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(ii) $(\text{CH}_3)_2\text{CHCl}$ (isopropyl chloride, 2°) vs $\text{CH}_3\text{Cl}$ (methyl chloride, 1°)

Both have the same leaving group (Cl⁻). The difference is steric hindrance:
- Methyl chloride has no steric hindrance (no alkyl groups on the carbon bearing Cl).
- Isopropyl chloride has two methyl groups causing steric hindrance.

$\text{CH}_3\text{Cl}$ reacts faster in $S_N2$ because it has less steric hindrance, allowing easier backside attack by the nucleophile OH⁻.

Concept: Dehydrohalogenation (E2 elimination) with NaOEt/EtOH follows Saytzeff's rule: the major product is the more substituted (more stable) alkene.

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(i) 1-Bromo-1-methylcyclohexane:

Structure: Cyclohexane ring with Br and CH₃ at C-1.

β-hydrogens available: at C-2 (ring) and at the methyl group (exocyclic).

Elimination towards ring (C-2): gives 1-methylcyclohex-1-ene (endocyclic double bond, trisubstituted)

Elimination towards methyl: gives methylenecyclohexane (exocyclic double bond, disubstituted)

Products:
- 1-Methylcyclohex-1-ene (major — more substituted, endocyclic)
- Methylenecyclohexane (minor)

Major alkene: 1-Methylcyclohex-1-ene

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(ii) 2-Chloro-2-methylbutane:

$$\text{CH}_3\text{-C(Cl)(CH}_3)\text{-CH}_2\text{-CH}_3$$

β-hydrogens: at C-1 (methyl, 3H) and at C-3 (CH₂, 2H).

Elimination towards C-3: gives 2-methylbut-2-ene (CH₃C(CH₃)=CHCH₃) — trisubstituted

Elimination towards C-1: gives 2-methylbut-1-ene (CH₂=C(CH₃)CH₂CH₃) — disubstituted

Products:
- 2-Methylbut-2-ene (major — more substituted)
- 2-Methylbut-1-ene (minor)

Major alkene: 2-Methylbut-2-ene

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(iii) 2,2,3-Trimethyl-3-bromopentane:

$$\text{CH}_3\text{-C(CH}_3)_2\text{-C(Br)(CH}_3)\text{-CH}_2\text{-CH}_3$$

Br is at C-3. β-hydrogens: at C-2 (no H, quaternary carbon — no β-H here) and at C-4 (CH₂, 2H).

Wait: C-2 is $\text{C(CH}_3)_2$ — quaternary, no H. C-4 is $\text{CH}_2$.

Also β-H from the methyl at C-3: the methyl group on C-3 has 3H.

Elimination towards C-4: gives 2,3,3-trimethylpent-4-ene ... let me re-examine.

Structure: C1(CH₃)–C2(C(CH₃)₂)–C3(CBr(CH₃))–C4(CH₂)–C5(CH₃)

β-carbons to C3 (bearing Br): C2 and C4.
- C2 has no H (quaternary: bonded to C1, C3, CH₃, CH₃)
- C4 has 2H
- The CH₃ on C3 has 3H (β-H from methyl)

Elimination using C4-H: double bond between C3–C4 → 2,3,3-trimethylpent-1-ene? No:

Double bond C3=C4: $\text{CH}_3\text{C(CH}_3)_2\text{-C(CH}_3)=\text{CHCH}_3$ → 2,3,3-trimethylpent-2-ene...

Let me number properly: 2,2,3-trimethyl-3-bromopentane:
C1–C2(2,2-dimethyl)–C3(3-methyl, Br)–C4–C5
= $\text{CH}_3\text{C(CH}_3)_2\text{CBr(CH}_3)\text{CH}_2\text{CH}_3$

Elimination (C3–C4 double bond): $\text{CH}_3\text{C(CH}_3)_2\text{C(CH}_3)=\text{CHCH}_3$ → 2,3,3-trimethylpent-2-ene (tetrasubstituted — major)

Elimination using CH₃ on C3 (C3–CH₃ bond): gives exocyclic-type: $\text{CH}_3\text{C(CH}_3)_2\text{C(=CH}_2)\text{CH}_2\text{CH}_3$ → 3,3-dimethyl-2-methylenepentane (minor)

Major alkene: 2,3,3-Trimethylpent-2-ene (more substituted, tetrasubstituted double bond)

(i) Ethanol to but-1-yne:

Step 1: Ethanol → Bromoethane
$$\text{CH}_3\text{CH}_2\text{OH} + \text{HBr} \rightarrow \text{CH}_3\text{CH}_2\text{Br} + \text{H}_2\text{O}$$

Step 2: Bromoethane → Acetylide (using NaNH₂)
$$\text{CH}_3\text{CH}_2\text{Br} + \text{NaNH}_2 \rightarrow \text{CH}_2=\text{CH}_2 + \text{NaBr} + \text{NH}_3$$

Actually, better route:
Step 1: Ethanol → Ethyl bromide → Vinyl bromide (by dehydrohalogenation) → Acetylene

Alternative standard route:
$$\text{CH}_3\text{CH}_2\text{OH} \xrightarrow{\text{HBr}} \text{CH}_3\text{CH}_2\text{Br} \xrightarrow{\text{alc. KOH}} \text{CH}_2=\text{CH}_2 \xrightarrow{\text{Br}_2} \text{CH}_2\text{BrCH}_2\text{Br} \xrightarrow{2\text{KOH(alc.)}} \text{HC}\equiv\text{CH}$$

$$\text{HC}\equiv\text{CH} \xrightarrow{\text{NaNH}_2} \text{HC}\equiv\text{CNa} \xrightarrow{\text{CH}_3\text{CH}_2\text{Br}} \text{HC}\equiv\text{C-CH}_2\text{CH}_3 \text{ (but-1-yne)}$$

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(ii) Ethane to bromoethene:

Step 1: Free radical bromination of ethane
$$\text{CH}_3\text{CH}_3 + \text{Br}_2 \xrightarrow{h\nu} \text{CH}_3\text{CH}_2\text{Br} + \text{HBr}$$

Step 2: Dehydrohalogenation
$$\text{CH}_3\text{CH}_2\text{Br} \xrightarrow{\text{alc. KOH}, \Delta} \text{CH}_2=\text{CH}_2$$

Step 3: Addition of Br₂
$$\text{CH}_2=\text{CH}_2 + \text{Br}_2 \rightarrow \text{CH}_2\text{BrCH}_2\text{Br}$$

Step 4: Dehydrohalogenation
$$\text{CH}_2\text{BrCH}_2\text{Br} \xrightarrow{\text{alc. KOH}} \text{CH}_2=\text{CHBr} \text{ (bromoethene/vinyl bromide)}$$

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(iii) Propene to 1-nitropropane:

Step 1: Anti-Markovnikov addition of HBr (peroxide effect)
$$\text{CH}_3\text{CH=CH}_2 + \text{HBr} \xrightarrow{\text{peroxide}} \text{CH}_3\text{CH}_2\text{CH}_2\text{Br}$$

Step 2: Nucleophilic substitution with AgNO₂
$$\text{CH}_3\text{CH}_2\text{CH}_2\text{Br} + \text{AgNO}_2 \rightarrow \text{CH}_3\text{CH}_2\text{CH}_2\text{NO}_2 + \text{AgBr}$$

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(iv) Toluene to benzyl alcohol:

Step 1: Free radical bromination at benzylic position
$$\text{C}_6\text{H}_5\text{CH}_3 + \text{Br}_2 \xrightarrow{h\nu} \text{C}_6\text{H}_5\text{CH}_2\text{Br} + \text{HBr}$$

Step 2: Hydrolysis
$$\text{C}_6\text{H}_5\text{CH}_2\text{Br} + \text{NaOH(aq)} \rightarrow \text{C}_6\text{H}_5\text{CH}_2\text{OH} + \text{NaBr}$$

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(v) Propene to propyne:

Step 1: Addition of Br₂
$$\text{CH}_3\text{CH=CH}_2 + \text{Br}_2 \rightarrow \text{CH}_3\text{CHBrCH}_2\text{Br}$$

Step 2: Double dehydrohalogenation with excess alcoholic KOH
$$\text{CH}_3\text{CHBrCH}_2\text{Br} \xrightarrow{2\text{KOH(alc.)}} \text{CH}_3\text{C}\equiv\text{CH} + 2\text{KBr} + 2\text{H}_2\text{O}$$

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(vi) Ethanol to ethyl fluoride:

Step 1: Ethanol → Ethyl chloride
$$\text{CH}_3\text{CH}_2\text{OH} + \text{SOCl}_2 \rightarrow \text{CH}_3\text{CH}_2\text{Cl} + \text{SO}_2 + \text{HCl}$$

Step 2: Halogen exchange (Swarts reaction)
$$\text{CH}_3\text{CH}_2\text{Cl} + \text{AgF} \rightarrow \text{CH}_3\text{CH}_2\text{F} + \text{AgCl}$$

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(vii) Bromomethane to propanone:

Step 1: Bromomethane → Methanol (or use Grignard)
$$\text{CH}_3\text{Br} + \text{KCN} \rightarrow \text{CH}_3\text{CN} \xrightarrow{\text{H}_3\text{O}^+} \text{CH}_3\text{COOH}$$

Alternative route:
$$\text{CH}_3\text{Br} \xrightarrow{\text{Mg/dry ether}} \text{CH}_3\text{MgBr} \xrightarrow{\text{CH}_3\text{COCl}} \text{(CH}_3)_2\text{CO} \text{ (propanone)}$$

Or: $\text{CH}_3\text{Br} + \text{NaOH(aq)} \rightarrow \text{CH}_3\text{OH}$; then oxidise to HCHO; then react with CH₃MgBr to get CH₃CH(OH)CH₃; oxidise to propanone.

Simpler: Two molecules of CH₃Br:
$$2\text{CH}_3\text{Br} \xrightarrow{\text{Mg}} \text{CH}_3\text{MgBr} \xrightarrow{\text{1. CH}_3\text{COCl; 2. H}_3\text{O}^+} (\text{CH}_3)_2\text{CO}$$

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(viii) But-1-ene to but-2-ene:

Step 1: Addition of HBr (Markovnikov)
$$\text{CH}_3\text{CH}_2\text{CH=CH}_2 + \text{HBr} \rightarrow \text{CH}_3\text{CH}_2\text{CHBrCH}_3 \text{ (2-bromobutane)}$$

Step 2: Dehydrohalogenation (Saytzeff)
$$\text{CH}_3\text{CH}_2\text{CHBrCH}_3 \xrightarrow{\text{alc. KOH}} \text{CH}_3\text{CH=CHCH}_3 \text{ (but-2-ene, major)}$$

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(ix) 1-Chlorobutane to n-octane:

Wurtz reaction:
$$2\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{Cl} + 2\text{Na} \xrightarrow{\text{dry ether}} \text{CH}_3(\text{CH}_2)_6\text{CH}_3 + 2\text{NaCl}$$

(n-octane)

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(x) Benzene to biphenyl:

Step 1: Bromination of benzene
$$\text{C}_6\text{H}_6 + \text{Br}_2 \xrightarrow{\text{FeBr}_3} \text{C}_6\text{H}_5\text{Br} + \text{HBr}$$

Step 2: Fittig reaction (Ullmann reaction)
$$2\text{C}_6\text{H}_5\text{Br} + 2\text{Na} \xrightarrow{\text{dry ether}} \text{C}_6\text{H}_5\text{-C}_6\text{H}_5 + 2\text{NaBr}$$

(biphenyl)

(i) Dipole moment of chlorobenzene < cyclohexyl chloride:

In chlorobenzene, the lone pairs on the Cl atom are in conjugation with the π-electron system of the benzene ring. This causes partial delocalization of the lone pair electrons of Cl into the ring (back donation), reducing the electron density on Cl and decreasing the polarity of the C–Cl bond. As a result, the C–Cl bond has partial double bond character, and the dipole moment is reduced.

In cyclohexyl chloride, there is no such resonance/delocalization. The C–Cl bond is a pure single bond with full polarity.

Hence, dipole moment of chlorobenzene (1.70 D) < cyclohexyl chloride (2.16 D).

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(ii) Alkyl halides are immiscible with water despite being polar:

For a substance to dissolve in water, it must be able to form hydrogen bonds with water molecules. Alkyl halides are polar (due to C–X bond), but they cannot form hydrogen bonds with water because:
- The halogen atom is not bonded to H (no O–H or N–H type bond).
- The energy released in forming new interactions between alkyl halide and water is less than the energy required to break the strong hydrogen bonds between water molecules.

Therefore, alkyl halides are immiscible with water. However, they dissolve readily in organic solvents through van der Waals and dipole-dipole interactions.

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(iii) Grignard reagents must be prepared under anhydrous conditions:

Grignard reagents ($\text{RMgX}$) are extremely reactive towards moisture (water). Even traces of water decompose the Grignard reagent:

$$\text{RMgX} + \text{H}_2\text{O} \rightarrow \text{R-H} + \text{Mg(OH)X}$$

The C–Mg bond is highly polar (Mg is electropositive), making the carbon nucleophilic. Water acts as a proton source and protonates the carbanion-like carbon, destroying the Grignard reagent. Hence, strictly anhydrous (dry) conditions and dry ether as solvent are essential for preparation and use of Grignard reagents.

Freon 12 ($\text{CCl}_2\text{F}_2$):
- Used as a refrigerant in refrigerators and air conditioners.
- Used as aerosol propellant in spray cans.
- Used in air conditioning systems.
- Used in making plastic foams.
*(Note: Its use is now restricted due to ozone layer depletion.)*

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DDT ($p,p'$-Dichlorodiphenyltrichloroethane):
- Used as an insecticide to kill mosquitoes (malaria control) and lice (typhus control).
- Was widely used in agriculture to protect crops from insects.
- Used in public health programs to control vector-borne diseases.
*(Note: Its use is now banned in many countries due to environmental hazards and toxicity.)*

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Carbon tetrachloride ($\text{CCl}_4$):
- Used as a solvent for fats, oils, rubber, and resins.
- Used as a dry cleaning agent.
- Used in fire extinguishers (pyrene extinguisher).
- Used in the manufacture of refrigerants and propellants.
- Used as a fumigant to kill insects in stored grain.

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Iodoform ($\text{CHI}_3$):
- Used as an antiseptic for dressing wounds (due to slow release of iodine).
- Used in medicine as a disinfectant.
- Used in the iodoform test to detect methyl ketones, acetaldehyde, and secondary alcohols with CH₃CH(OH)– group.

(i) $\text{CH}_3\text{CH}_2\text{CH}_2\text{Cl} + \text{NaI} \xrightarrow{\text{acetone, heat}}$

Finkelstein reaction (halogen exchange, $S_N2$):

$$\boxed{\text{CH}_3\text{CH}_2\text{CH}_2\text{I}} \text{ (1-iodopropane)}$$

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(ii) $(\text{CH}_3)_3\text{CBr} + \text{KOH} \xrightarrow{\text{ethanol, heat}}$

Alcoholic KOH causes elimination (E2/E1). Tert-butyl bromide undergoes dehydrohalogenation:

$$\boxed{(\text{CH}_3)_2\text{C=CH}_2} \text{ (2-methylpropene / isobutylene)}$$

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(iii) $\text{CH}_3\text{CH(Br)CH}_2\text{CH}_3 + \text{NaOH} \xrightarrow{\text{water}}$

Aqueous NaOH causes nucleophilic substitution ($S_N2$):

$$\boxed{\text{CH}_3\text{CH(OH)CH}_2\text{CH}_3} \text{ (butan-2-ol)}$$

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(iv) $\text{CH}_3\text{CH}_2\text{Br} + \text{KCN} \xrightarrow{\text{aq. ethanol}}$

Nucleophilic substitution; CN⁻ attacks through carbon (major product is nitrile):

$$\boxed{\text{CH}_3\text{CH}_2\text{CN}} \text{ (propanenitrile)}$$

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(v) $\text{C}_6\text{H}_5\text{ONa} + \text{C}_2\text{H}_5\text{Cl}$

Williamson ether synthesis:

$$\boxed{\text{C}_6\text{H}_5\text{OC}_2\text{H}_5} \text{ (ethoxybenzene / phenyl ethyl ether)}$$

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(vi) $\text{CH}_3\text{CH}_2\text{CH}_2\text{OH} + \text{SOCl}_2$

Conversion of alcohol to alkyl chloride:

$$\boxed{\text{CH}_3\text{CH}_2\text{CH}_2\text{Cl}} \text{ (1-chloropropane)} + \text{SO}_2 + \text{HCl}$$

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(vii) $\text{CH}_3\text{CH}_2\text{CH=CH}_2 + \text{HBr} \xrightarrow{\text{peroxide}}$

Anti-Markovnikov addition (free radical mechanism):

$$\boxed{\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{Br}} \text{ (1-bromobutane)}$$

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(viii) $\text{CH}_3\text{CH=C(CH}_3)_2 + \text{HBr}$

Markovnikov addition (no peroxide): H adds to less substituted carbon (C-2), Br adds to more substituted carbon (C-3):

$$\boxed{\text{CH}_3\text{CH}_2\text{CBr(CH}_3)_2} \text{ (2-bromo-2-methylbutane)}$$

Given: n-Butyl bromide + KCN → n-Butyl cyanide (pentanenitrile)

Mechanism: $S_N2$ (Bimolecular Nucleophilic Substitution)

Step 1: Identification of nucleophile and substrate

KCN dissociates to give $\text{CN}^-$ (cyanide ion), which is an ambident nucleophile. It attacks through carbon (major) to give nitrile.

Substrate: $\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{Br}$ (primary alkyl halide — favours $S_N2$)

Step 2: $S_N2$ Mechanism

The cyanide ion acts as a nucleophile and attacks the electrophilic carbon (C-1) bearing Br from the back side (180° to the leaving group) in a single concerted step:

$$\text{NC}^- + \text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\overset{\delta+}{\text{C}}\text{---}\overset{\delta-}{\text{Br}} \rightarrow [\text{NC---CH}_2\text{CH}_2\text{CH}_2\text{CH}_3\text{---Br}]^{\ddagger} \rightarrow \text{NC-CH}_2\text{CH}_2\text{CH}_2\text{CH}_3 + \text{Br}^-$$

Transition state:
$$[\text{N}\equiv\text{C}\cdots\text{CH}_2(\text{CH}_2)_2\text{CH}_3\cdots\text{Br}]^{\ddagger}$$

Step 3: Product

The C–Br bond breaks and C–CN bond forms simultaneously. Inversion of configuration occurs at the reaction centre.

$$\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{Br} + \text{KCN} \xrightarrow{S_N2} \text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{CN} + \text{KBr}$$

Product: n-Butyl cyanide (pentanenitrile)

Key features of $S_N2$:
- Rate = k[RBr][CN⁻] (second order kinetics)
- Single step, concerted mechanism
- Backside attack → inversion of configuration (Walden inversion)
- Primary halides favour $S_N2$

Concept: $S_N2$ reactivity decreases with increasing steric hindrance at the carbon bearing the leaving group. Order: methyl > primary > secondary > tertiary. Also, branching at β-carbon reduces reactivity.

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(i) 2-Bromo-2-methylbutane (3°), 1-Bromopentane (1°), 2-Bromopentane (2°)

Increasing order of $S_N2$ reactivity:

\text{2-Bromo-2-methylbutane (3°)} &lt; \text{2-Bromopentane (2°)} &lt; \text{1-Bromopentane (1°)}

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(ii) 1-Bromo-3-methylbutane (1°, β-branch at C-3), 2-Bromo-2-methylbutane (3°), 2-Bromo-3-methylbutane (2°)

Increasing order of $S_N2$ reactivity:

\text{2-Bromo-2-methylbutane (3°)} &lt; \text{2-Bromo-3-methylbutane (2°)} &lt; \text{1-Bromo-3-methylbutane (1°)}

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(iii) 1-Bromobutane (1°, no β-branch), 1-Bromo-2,2-dimethylpropane (1°, two methyl groups at β-carbon — neopentyl type, very hindered), 1-Bromo-2-methylbutane (1°, one methyl at β-carbon), 1-Bromo-3-methylbutane (1°, methyl at γ-carbon — least hindrance among branched)

Increasing order of $S_N2$ reactivity:

\text{1-Bromo-2,2-dimethylpropane} &lt; \text{1-Bromo-2-methylbutane} &lt; \text{1-Bromo-3-methylbutane} &lt; \text{1-Bromobutane}

*Explanation:* 1-Bromo-2,2-dimethylpropane (neopentyl bromide) has two methyl groups at the β-carbon, causing maximum steric hindrance. 1-Bromobutane has no branching and is least hindered, hence most reactive in $S_N2$.

Given: Benzyl chloride ($\text{C}_6\text{H}_5\text{CH}_2\text{Cl}$) and diphenylmethyl chloride ($\text{C}_6\text{H}_5\text{CHClC}_6\text{H}_5$).

Concept: Hydrolysis by aqueous KOH proceeds via $S_N1$ or $S_N2$ mechanism. The ease of hydrolysis depends on the stability of the carbocation formed (for $S_N1$) or the steric accessibility (for $S_N2$).

Analysis:

- $\text{C}_6\text{H}_5\text{CH}_2\text{Cl}$ (benzyl chloride): forms a primary benzylic carbocation $\text{C}_6\text{H}_5\text{CH}_2^+$ — stabilised by resonance with one phenyl ring.

- $\text{C}_6\text{H}_5\text{CHClC}_6\text{H}_5$ (diphenylmethyl chloride): forms a secondary benzylic carbocation $\text{C}_6\text{H}_5\overset{+}{\text{CH}}\text{C}_6\text{H}_5$ — stabilised by resonance with two phenyl rings (more stable carbocation).

Conclusion: $\text{C}_6\text{H}_5\text{CHClC}_6\text{H}_5$ is more easily hydrolysed because it forms a more stable carbocation (stabilised by two phenyl groups through resonance), making the $S_N1$ ionisation step faster.

$$\boxed{\text{C}_6\text{H}_5\text{CHClC}_6\text{H}_5 \text{ is more easily hydrolysed}}$$

Given: Three isomers of dichlorobenzene: ortho (1,2-), meta (1,3-), para (1,4-).

Concept: Melting point depends on the crystal lattice energy, which is related to how efficiently molecules pack in the solid state.

Explanation:

- $p$-Dichlorobenzene has a highly symmetric structure (Cl atoms at 1 and 4 positions, on opposite sides of the ring). This high symmetry allows the molecules to pack more efficiently in the crystal lattice, resulting in stronger intermolecular forces and a higher melting point (53°C).

- $o$-Dichlorobenzene (m.p. −17°C) and $m$-dichlorobenzene (m.p. −25°C) are less symmetric. Their molecules cannot pack as efficiently in the crystal lattice, leading to weaker intermolecular forces and lower melting points.

Conclusion: The higher symmetry of $p$-dichlorobenzene leads to better crystal packing and higher lattice energy, resulting in a higher melting point compared to the $o$- and $m$-isomers.

(i) Propene to propan-1-ol:

Anti-Markovnikov addition of water (via hydroboration-oxidation) or:

Step 1: Add HBr with peroxide (anti-Markovnikov)
$$\text{CH}_3\text{CH=CH}_2 + \text{HBr} \xrightarrow{\text{peroxide}} \text{CH}_3\text{CH}_2\text{CH}_2\text{Br}$$

Step 2: Hydrolysis
$$\text{CH}_3\text{CH}_2\text{CH}_2\text{Br} + \text{NaOH(aq)} \rightarrow \text{CH}_3\text{CH}_2\text{CH}_2\text{OH}$$

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(ii) Ethanol to but-1-yne:

Step 1: $\text{CH}_3\text{CH}_2\text{OH} \xrightarrow{\text{HBr}} \text{CH}_3\text{CH}_2\text{Br}$

Step 2: $\text{CH}_3\text{CH}_2\text{Br} \xrightarrow{\text{alc. KOH}} \text{CH}_2=\text{CH}_2$

Step 3: $\text{CH}_2=\text{CH}_2 + \text{Br}_2 \rightarrow \text{CH}_2\text{BrCH}_2\text{Br}$

Step 4: $\text{CH}_2\text{BrCH}_2\text{Br} \xrightarrow{2\text{KOH(alc.)}} \text{HC}\equiv\text{CH}$

Step 5: $\text{HC}\equiv\text{CH} \xrightarrow{\text{NaNH}_2} \text{HC}\equiv\text{CNa} \xrightarrow{\text{CH}_3\text{CH}_2\text{Br}} \text{HC}\equiv\text{CCH}_2\text{CH}_3$ (but-1-yne)

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(iii) 1-Bromopropane to 2-bromopropane:

Step 1: Dehydrohalogenation
$$\text{CH}_3\text{CH}_2\text{CH}_2\text{Br} \xrightarrow{\text{alc. KOH}} \text{CH}_3\text{CH=CH}_2$$

Step 2: Markovnikov addition of HBr
$$\text{CH}_3\text{CH=CH}_2 + \text{HBr} \rightarrow \text{CH}_3\text{CHBrCH}_3 \text{ (2-bromopropane)}$$

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(iv) Toluene to benzyl alcohol:

Step 1: Free radical bromination
$$\text{C}_6\text{H}_5\text{CH}_3 + \text{Br}_2 \xrightarrow{h\nu} \text{C}_6\text{H}_5\text{CH}_2\text{Br} + \text{HBr}$$

Step 2: Hydrolysis
$$\text{C}_6\text{H}_5\text{CH}_2\text{Br} + \text{NaOH(aq)} \rightarrow \text{C}_6\text{H}_5\text{CH}_2\text{OH}$$

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(v) Benzene to 4-bromonitrobenzene:

Step 1: Nitration of benzene
$$\text{C}_6\text{H}_6 + \text{HNO}_3 \xrightarrow{\text{H}_2\text{SO}_4} \text{C}_6\text{H}_5\text{NO}_2 + \text{H}_2\text{O}$$

Step 2: Bromination (–NO₂ is meta-director, but we need para product; so brominate first)

Better route — brominate first, then nitrate:
$$\text{C}_6\text{H}_6 + \text{Br}_2 \xrightarrow{\text{FeBr}_3} \text{C}_6\text{H}_5\text{Br}$$
$$\text{C}_6\text{H}_5\text{Br} + \text{HNO}_3 \xrightarrow{\text{H}_2\text{SO}_4} p\text{-BrC}_6\text{H}_4\text{NO}_2 \text{ (major, 4-bromonitrobenzene)}$$

(Br is ortho/para director; para product is major)

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(vi) Benzyl alcohol to 2-phenylethanoic acid:

Step 1: Convert to benzyl bromide
$$\text{C}_6\text{H}_5\text{CH}_2\text{OH} + \text{HBr} \rightarrow \text{C}_6\text{H}_5\text{CH}_2\text{Br}$$

Step 2: Reaction with KCN
$$\text{C}_6\text{H}_5\text{CH}_2\text{Br} + \text{KCN} \rightarrow \text{C}_6\text{H}_5\text{CH}_2\text{CN}$$

Step 3: Hydrolysis
$$\text{C}_6\text{H}_5\text{CH}_2\text{CN} + \text{H}_2\text{O} \xrightarrow{\text{H}^+/\Delta} \text{C}_6\text{H}_5\text{CH}_2\text{COOH} \text{ (2-phenylethanoic acid)}$$

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(vii) Ethanol to propanenitrile:

Step 1: $\text{CH}_3\text{CH}_2\text{OH} \xrightarrow{\text{HBr}} \text{CH}_3\text{CH}_2\text{Br}$

Step 2: $\text{CH}_3\text{CH}_2\text{Br} + \text{KCN} \xrightarrow{\text{aq. EtOH}} \text{CH}_3\text{CH}_2\text{CN}$ (propanenitrile)

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(viii) Aniline to chlorobenzene (Sandmeyer reaction):

Step 1: Diazotisation
$$\text{C}_6\text{H}_5\text{NH}_2 + \text{NaNO}_2 + \text{HCl} \xrightarrow{0-5°\text{C}} \text{C}_6\text{H}_5\text{N}_2^+\text{Cl}^- + \text{NaCl} + 2\text{H}_2\text{O}$$

Step 2: Sandmeyer reaction
$$\text{C}_6\text{H}_5\text{N}_2^+\text{Cl}^- + \text{CuCl} \xrightarrow{\Delta} \text{C}_6\text{H}_5\text{Cl} + \text{N}_2$$

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(ix) 2-Chlorobutane to 3,4-dimethylhexane:

Step 1: Prepare Grignard reagent
$$\text{CH}_3\text{CHClCH}_2\text{CH}_3 + \text{Mg} \xrightarrow{\text{dry ether}} \text{CH}_3\text{CH(MgCl)CH}_2\text{CH}_3$$

Step 2: Wurtz-type coupling (or use the Grignard with another 2-chlorobutane via Wurtz):
$$2\text{CH}_3\text{CHClCH}_2\text{CH}_3 + 2\text{Na} \xrightarrow{\text{dry ether}} \text{CH}_3\text{CH(CH}_2\text{CH}_3)\text{-CH(CH}_3)\text{CH}_2\text{CH}_3 + 2\text{NaCl}$$

Product: 3,4-dimethylhexane

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(x) 2-Methyl-1-propene to 2-chloro-2-methylpropane:

Markovnikov addition of HCl:
$$\text{(CH}_3)_2\text{C=CH}_2 + \text{HCl} \rightarrow \text{(CH}_3)_3\text{CCl} \text{ (2-chloro-2-methylpropane)}$$

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(xi) Ethyl chloride to propanoic acid:

Step 1: $\text{CH}_3\text{CH}_2\text{Cl} + \text{KCN} \rightarrow \text{CH}_3\text{CH}_2\text{CN}$ (propanenitrile)

Step 2: $\text{CH}_3\text{CH}_2\text{CN} + \text{H}_2\text{O} \xrightarrow{\text{H}^+/\Delta} \text{CH}_3\text{CH}_2\text{COOH}$ (propanoic acid)

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(xii) But-1-ene to n-butyl iodide:

Step 1: Anti-Markovnikov addition of HBr (peroxide)
$$\text{CH}_3\text{CH}_2\text{CH=CH}_2 + \text{HBr} \xrightarrow{\text{peroxide}} \text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{Br}$$

Step 2: Finkelstein reaction
$$\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{Br} + \text{NaI} \xrightarrow{\text{acetone}} \text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{I}$$

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(xiii) 2-Chloropropane to 1-propanol:

Step 1: Dehydrohalogenation
$$\text{CH}_3\text{CHClCH}_3 \xrightarrow{\text{alc. KOH}} \text{CH}_3\text{CH=CH}_2$$

Step 2: Anti-Markovnikov hydration (HBr + peroxide, then hydrolysis)
$$\text{CH}_3\text{CH=CH}_2 + \text{HBr} \xrightarrow{\text{peroxide}} \text{CH}_3\text{CH}_2\text{CH}_2\text{Br} \xrightarrow{\text{NaOH(aq)}} \text{CH}_3\text{CH}_2\text{CH}_2\text{OH}$$

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(xiv) Isopropyl alcohol to iodoform:

Iodoform reaction:
$$\text{(CH}_3)_2\text{CHOH} + \text{I}_2 + \text{NaOH} \rightarrow \text{CHI}_3 + \text{CH}_3\text{COONa} + \text{NaI} + \text{H}_2\text{O}$$

(Isopropyl alcohol is oxidised to acetone in situ, which then undergoes iodoform reaction)

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(xv) Chlorobenzene to $p$-nitrophenol:

Step 1: Nitration of chlorobenzene
$$\text{C}_6\text{H}_5\text{Cl} + \text{HNO}_3 \xrightarrow{\text{H}_2\text{SO}_4} p\text{-ClC}_6\text{H}_4\text{NO}_2 + o\text{-ClC}_6\text{H}_4\text{NO}_2$$

Step 2: Hydrolysis of p-chloronitrobenzene
$$p\text{-ClC}_6\text{H}_4\text{NO}_2 + \text{NaOH} \xrightarrow{\text{high temp/pressure}} p\text{-NO}_2\text{C}_6\text{H}_4\text{ONa} \xrightarrow{\text{H}^+} p\text{-NO}_2\text{C}_6\text{H}_4\text{OH}$$

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(xvi) 2-Bromopropane to 1-bromopropane:

Step 1: Dehydrohalogenation
$$\text{CH}_3\text{CHBrCH}_3 \xrightarrow{\text{alc. KOH}} \text{CH}_3\text{CH=CH}_2$$

Step 2: Anti-Markovnikov addition of HBr
$$\text{CH}_3\text{CH=CH}_2 + \text{HBr} \xrightarrow{\text{peroxide}} \text{CH}_3\text{CH}_2\text{CH}_2\text{Br}$$

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(xvii) Chloroethane to butane:

Wurtz reaction:
$$2\text{CH}_3\text{CH}_2\text{Cl} + 2\text{Na} \xrightarrow{\text{dry ether}} \text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_3 + 2\text{NaCl}$$

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(xviii) Benzene to diphenyl:

Step 1: Bromination
$$\text{C}_6\text{H}_6 + \text{Br}_2 \xrightarrow{\text{FeBr}_3} \text{C}_6\text{H}_5\text{Br}$$

Step 2: Fittig reaction (Ullmann coupling)
$$2\text{C}_6\text{H}_5\text{Br} + 2\text{Na} \xrightarrow{\text{dry ether}} \text{C}_6\text{H}_5\text{-C}_6\text{H}_5 + 2\text{NaBr}$$

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(xix) tert-Butyl bromide to isobutyl bromide:

Step 1: Elimination
$$(\text{CH}_3)_3\text{CBr} \xrightarrow{\text{alc. KOH}} (\text{CH}_3)_2\text{C=CH}_2$$

Step 2: Anti-Markovnikov addition of HBr
$$(\text{CH}_3)_2\text{C=CH}_2 + \text{HBr} \xrightarrow{\text{peroxide}} (\text{CH}_3)_2\text{CHCH}_2\text{Br} \text{ (isobutyl bromide)}$$

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(xx) Aniline to phenylisocyanide (carbylamine reaction):

$$\text{C}_6\text{H}_5\text{NH}_2 + \text{CHCl}_3 + 3\text{KOH} \xrightarrow{\Delta} \text{C}_6\text{H}_5\text{NC} + 3\text{KCl} + 3\text{H}_2\text{O}$$

(Carbylamine reaction: primary amine + chloroform + alcoholic KOH → isocyanide)