CBSE Class 12 Chemistry Haloalkane Haloarenes Exam Notes

CBSE Class 12 Chemistry Haloalkane Haloarenes Exam Notes. Please refer to the examination notes which you can use for preparing and revising for exams. These notes will help you to revise the concepts quickly and get good marks.

Ans. (a) & (c) With four different alkyl group.

These Class 12 Chemistry exam notes on Chapter 4, Haloalkanes and Haloarenes, cover important reasoning questions, SN1 and SN2 mechanisms with a comparison table, the name reactions (Sandmeyer, Finkelstein, Swarts, Wurtz, Friedel-Crafts, Wurtz-Fittig and Fittig) and solved conversions with reagents and products.

Topics Covered

1. Reasonings
2. IUPAC naming
3. Mechanisms
4. Name reactions
5. Conversions

REASONINGS

Q1. In the following pairs of halogen compounds, which undergoes the SN2 reaction faster?

(a) Cyclohexylmethyl chloride (C₆H₁₁CH₂Cl) and chlorocyclohexane
(b) The iodo compound and the chloro compound of the same alkyl chain
(c) C₄H₉Br and C₂H₅CH(Br)CH₃
(d) (CH₃)₃CCl and CH₃Cl

[Figure: Structures for Q1(a) and (b) are skeletal drawings. In (a) a cyclohexane ring carries CH₂Cl, and a second ring carries Cl directly. In (b) a zigzag carbon chain carries I, and a second zigzag chain carries Cl.]

Ans. (a) Cyclohexylmethyl chloride (1° alkyl halide) > chlorocyclohexane (2° alkyl halide)
(b) The iodo compound (bigger size, better leaving group) > the chloro compound
(c) C₄H₉Br (1°) > C₂H₅CH(Br)CH₃
(d) (CH₃)₃CCl < CH₃Cl (1° alkyl halide)

Q2. Predict the order of reactivity of the following compounds in SN1 and SN2 reactions.

(a) The four isomeric bromobutanes
(b) C₆H₅CH₂Br, C₆H₅CH(C₆H₅)Br, C₆H₅CH(CH₃)Br, C₆H₅C(CH₃)(C₆H₅)Br

Ans. (a) For the SN1 reaction (reactivity depends on stability of the carbocation: 3° > 2° > 1°):
CH₃CH₂CH₂CH₂Br < (CH₃)₂CHCH₂Br < CH₃CH₂CH(Br)CH₃ < (CH₃)₃C-Br
For the SN2 reaction (less steric hindrance, more reactive):
CH₃CH₂CH₂CH₂Br > (CH₃)₂CHCH₂Br > CH₃CH₂CH(Br)CH₃ > (CH₃)₃C-Br
(b) For the SN1 reaction: C₆H₅C(CH₃)(C₆H₅)Br > C₆H₅CH(C₆H₅)Br > C₆H₅CH(CH₃)Br > C₆H₅CH₂Br
For the SN2 reaction: C₆H₅CH₂Br > C₆H₅CH(CH₃)Br > C₆H₅CH(C₆H₅)Br > C₆H₅C(CH₃)(C₆H₅)Br

Q3. Identify the chiral and achiral molecules in each of the following pairs.

(a) CH₃-CH(OH)Br with H, CH₃, OH and Br on the central carbon
(b) CH₃-CBr₂H with H, CH₃, Br and Br on the central carbon
(c) CH₃-CH(OH)-CH₂-CH₂-CH₃ (OH on the second carbon)
(d) CH₃CH₂-CH(OH)-CH₂CH₃ (OH on the middle carbon)

Ans. (a) and (c) are chiral, because the carbon carries four different groups.

Q4. Although chlorine is an electron withdrawing group, it is ortho-para directing in electrophilic aromatic substitution reactions. Why?

Ans. Chlorine shows the -I effect, which destabilises the intermediate carbocation, but the +R effect of Cl stabilises the intermediate carbocation. The reactivity is controlled by the stronger -I effect, while the orientation is controlled by the weaker resonance effect.

Q5. Haloalkanes react with KCN to form RCN but with AgCN form RNC. Why?

Ans. CN is an ambident group (it can be attached from both ends). KCN is ionic, so the reaction occurs through carbon, since C-C bonds are stronger. AgCN is covalent, so only the nitrogen electron pair is available for bond formation.

Q6. RCl is hydrolysed to ROH slowly, but the reaction is rapid if a catalytic amount of KI is added to the reaction mixture. Why?

Ans. I^- is a stronger nucleophile, so it replaces Cl to form RI. RI is then hydrolysed more readily to give ROH, because I^- is a better leaving group than Cl^-.

Q7. Haloalkanes give nucleophilic substitution while haloarenes give electrophilic substitution reactions.

Ans. Haloalkanes are more polar than haloarenes, so their carbon has more positive character and they give nucleophilic substitution. Haloarenes have a benzene ring which shows resonance, so they give electrophilic substitution due to the -I effect and +R effect.

Q8. Why is sulphuric acid not used during the reaction of alcohols with KI?

Ans. Sulphuric acid oxidises KI to I₂.

Q9. Arrange on the basis of increasing boiling point.

(a) Bromomethane, bromoform, chloromethane, dibromomethane
(b) 1-Chloropropane, isopropyl chloride, 1-chlorobutane

Ans. (a) Chloromethane < bromomethane < dibromomethane < bromoform (boiling point increases with molecular weight)
(b) Isopropyl chloride (branched) < 1-chloropropane < 1-chlorobutane (more branching means less surface area and a lower boiling point)

Q10. Which of the following has the highest dipole moment: CH₃Cl, CH₂Cl₂, CCl₄?

[Figure: Three tetrahedral structures with bond dipole arrows pointing towards Cl: CH₂Cl₂ (two Cl, two H), CHCl₃ (three Cl, one H) and CCl₄ (four Cl), joined by greater than signs. The printed order is CH₂Cl₂ > CHCl₃ > CCl₄.]

Ans. The dipole moment decreases as the number of chlorine atoms increases, because the bond dipoles partly cancel: CH₃Cl (1.87 D) > CH₂Cl₂ (1.60 D) > CHCl₃ (1.04 D) > CCl₄ (zero, symmetrical). Of the three compounds asked, CH₃Cl has the highest dipole moment. The printed order CH₂Cl₂ > CHCl₃ > CCl₄ is correct for those three compounds.

Q11. The dipole moment of chlorobenzene is lower than that of cyclohexyl chloride.

Ans. The cyclohexyl group is an electron donating group, so it increases the electron density at the Cl atom and makes the bond more polar. In chlorobenzene the benzene ring acts as an electron withdrawing, resonating ring, so its dipole moment is lower.

Q12. Alkyl halides, though polar, are not soluble in water.

Ans. Alkyl halides cannot form hydrogen bonds with water molecules.

Q13. RMgX should be prepared under anhydrous conditions.

Ans. RMgX is very reactive and reacts with moisture present in the apparatus or in the starting material.

Q14. Out of C₆H₅CH₂Cl and C₆H₅CHClC₆H₅, which is easily hydrolysed by aqueous KOH?

Ans. C₆H₅CHClC₆H₅ is a 2° alkyl halide and is more reactive under the SN1 reaction. Under the SN2 reaction the 1° alkyl halide (C₆H₅CH₂Cl) is hydrolysed more easily.

Q15. p-Dichlorobenzene has a higher melting point and lower solubility than o-dichlorobenzene.

Ans. p-Dichlorobenzene has a symmetrical structure, so its melting point is high and its solubility is low.

Q16. Why is SOCl₂ used for the preparation of alkyl halides from ROH?

Ans. On reaction between SOCl₂ and ROH, we get RCl along with HCl(g) and SO₂(g), which escape as gases, so the product is easily obtained pure.

Q17. Arrange the following in order of increasing ease of nucleophilic substitution reaction: 4-nitrochlorobenzene, chlorobenzene, 2,4,6-trinitrochlorobenzene, 2,4-dinitrochlorobenzene.

Ans. Chlorobenzene < 4-nitrochlorobenzene < 2,4-dinitrochlorobenzene < 2,4,6-trinitrochlorobenzene. Equivalently, 2,4,6-trinitrochlorobenzene > 2,4-dinitrochlorobenzene > 4-nitrochlorobenzene > chlorobenzene in ease of reaction.

Further questions listed in the handout:
- Among the aromatic compounds having molecular formula C₇H₇Cl, how many isomers are possible? Write structures and names.
- Identify all the possible monochloro structural isomers that would be expected to form on free radical chlorination of (CH₃)₂CHCH₂CH₃.
- Differentiate between (a) chiral and achiral, (b) retention and inversion.

Q18. What are enantiomers and a racemic mixture?

Q19. Allyl chloride is hydrolysed more readily than n-propyl chloride. Vinyl chloride is hydrolysed more slowly than ethyl chloride.

Q20. Explain: the reaction between an alkyl halide and aqueous KOH yields an alcohol, but with alcoholic KOH it yields alkenes.

Q21. Which one is the better nucleophile, Br^- or I^-?

Q22. Among the isomeric alkanes of molecular formula C₅H₁₂, identify the one that on photochemical chlorination yields (i) a single monochloride, (ii) three isomeric monochlorides, (iii) four isomeric monochlorides. Uses of freon 12, DDT, carbon tetrachloride and iodoform.

MECHANISMS

SN2 Mechanism:
^-OH + CH₃-Cl → [HO---CH₃---Cl]^- (transition state, the HO and Cl placed on opposite sides of carbon, with the three H atoms in one plane) → HO-CH₃ + Cl^-

[Figure: Hydroxide attacks the carbon of CH₃Cl from the side opposite to Cl. The central transition state is shown in square brackets with dotted bonds from carbon to OH and to Cl, and the product CH₃OH shows the three H atoms flipped over (inversion).]

SN1 Mechanism:
Step 1 (slow): (CH₃)₃CBr ⇌ (CH₃)₃C⁺ + Br^-
Step 2 (fast): (CH₃)₃C⁺ + ^-OH → (CH₃)₃COH

COMPARISON OF SN2 AND SN1 REACTIONS

MECHANISM OF THE REACTION nBuBr + KCN (EtOH-H₂O) → nBuCN

K⁺CN^- + CH₃-CH₂-CH₂-CH₂-Br → CH₃-CH₂-CH₂-CH₂-CN + KBr
(n-Butyl bromide gives n-Butyl cyanide)

[Figure: Curved arrow from the CN^- ion to the carbon bearing Br (carrying δ+), and a second curved arrow from the C-Br bond to the Br atom (carrying δ-).]

NAME REACTIONS

1. Sandmeyer's Reaction:
Benzenediazonium halide (C₆H₅N₂⁺X^-) with Cu₂X₂ → halobenzene (C₆H₅X) + N₂

2. Finkelstein Reaction:
RX + NaI → RI + NaX (solvent - acetone)

3. Swarts Reaction:
CH₃Br + AgF → CH₃F + AgBr

4. Wurtz Reaction:
RX + 2Na + RX → R-R + 2NaX

5. Friedel-Crafts Reaction:
Alkylation: chlorobenzene + CH₃Cl with anhydrous AlCl₃ → 1-chloro-2-methylbenzene (minor) + 1-chloro-4-methylbenzene (major)
Acylation: chlorobenzene + CH₃COCl with anhydrous AlCl₃ → 2-chloroacetophenone (minor) + 4-chloroacetophenone (major)

6. Wurtz-Fittig Reaction:
Aryl halide (C₆H₅X) + Na + RX, in ether → C₆H₅R + NaX

7. Fittig Reaction:
Aryl halide (C₆H₅X) + Na + RX (aryl halide), in ether → biphenyl + 2NaX

CONVERSIONS

Q1. Identify A, B, C, D, E, R and R¹ in the following:

(i) Bromocyclohexane + Mg, dry ether → A; A with H₂O → B
(ii) R-Br + Mg, dry ether → C; C with D₂O → CH₃CHDCH₃
(iii) (CH₃)₃C-C(CH₃)₃ (2,2,3,3-tetramethylbutane) is obtained from R¹-X with Na/ether; R¹-X + Mg → D; D with H₂O → E

Ans. (i) Bromocyclohexane + Mg (dry ether) → cyclohexylmagnesium bromide (A); on reaction with H₂O this gives cyclohexane (B) + Mg(OH)Br.
(ii) Since D of D₂O gets attached to the carbon atom to which MgBr is attached, C is CH₃CH(MgBr)CH₃ (isopropylmagnesium bromide). Therefore the compound R-Br is CH₃CHBrCH₃ (2-bromopropane).
CH₃CHBrCH₃ + Mg (dry ether) → CH₃CH(MgBr)CH₃ (C); with D₂O → CH₃CHDCH₃
(iii) The halide R¹-X is (CH₃)₃C-X (tert-butyl halide). Therefore compound D is (CH₃)₃C-MgBr (tert-butylmagnesium bromide), and E is (CH₃)₃CH (isobutane).

Q2. How will you bring about the following conversions?

(i) Ethanol to but-1-yne
(ii) Ethane to bromoethene
(iii) Propene to 1-nitropropane
(iv) Toluene to benzyl alcohol
(v) Propene to propyne
(vi) Ethanol to ethyl fluoride
(vii) Bromomethane to propanone
(viii) But-1-ene to but-2-ene
(ix) 1-Chlorobutane to n-octane
(x) Benzene to biphenyl

Ans. (i) CH₃CH₂OH (ethanol) → CH₃CH₂Cl (chloroethane) + SO₂ + HCl, using SOCl₂, pyridine
HC≡CH + NaNH₂ → HC≡C^-Na⁺ (liq. NH₃)
CH₃CH₂-Cl + HC≡C^-Na⁺ → CH₃CH₂C≡CH (but-1-yne) + NaCl

(ii) CH₃-CH₃ (ethane) → CH₃-CH₂Br (bromoethane) + HBr, using Br₂/UV light or heat
CH₃CH₂Br → CH₂=CH₂ + HBr, using KOH (alc.), heat
CH₂=CH₂ → BrCH₂-CH₂Br (vic-dibromide), using Br₂/CCl₄
BrCH₂-CH₂Br → CH₂=CHBr (bromoethene), using KOH (alc.), heat

(iii) CH₃-CH=CH₂ (propene) + HBr, peroxide → CH₃-CH₂-CH₂Br (1-bromopropane)
CH₃CH₂CH₂Br + AgNO₂ → AgBr + CH₃-CH₂-CH₂NO₂ (1-nitropropane)

(iv) Do yourself.

(v) CH₃-CH=CH₂ (propene) → CH₃-CHBr-CH₂Br (1,2-dibromopropane), using Br₂/CCl₄
CH₃-CHBr-CH₂Br → CH₃-C≡CH (propyne), using NaNH₂, liq. NH₃

(vi) CH₃-CH₂-OH (ethanol) → CH₃-CH₂-Cl (chloroethane), using PCl₅ (by-products POCl₃ and HCl)
CH₃-CH₂-Cl + AgF → CH₃-CH₂-F (ethyl fluoride)

(vii) CH₃-Br (bromomethane) + KCN (alc.) → CH₃-CN (acetonitrile)
CH₃-CN + CH₃MgBr → (CH₃)₂C=NMgBr
(CH₃)₂C=NMgBr, on hydrolysis with H₃O⁺ → CH₃-CO-CH₃ (propanone)

(viii) CH₃CH₂CH=CH₂ (but-1-ene) + HBr (Markovnikov addition) → CH₃CH₂CHBrCH₃ (2-bromobutane)
2-Bromobutane, with KOH (alc.), heat (loss of HBr) → CH₃CH=CHCH₃ (but-2-ene)

(ix) 2CH₃CH₂CH₂CH₂-Cl (1-chlorobutane) + 2Na, dry ether → CH₃CH₂CH₂CH₂CH₂CH₂CH₂CH₃ (n-octane) + 2NaCl

(x) Benzene → bromobenzene, using Br₂/Fe, dark
Bromobenzene + 2Na, dry ether → diphenyl (biphenyl) + 2NaBr

Important reactions

1. CH₃CH₂CH₂Cl (1-chloropropane) + NaI, acetone, heat (Finkelstein reaction) → CH₃CH₂CH₂I (1-iodopropane) + NaCl
2. (CH₃)₃CBr (2-bromo-2-methylpropane) + KOH, ethanol, heat (dehydrohalogenation) → CH₃-C(CH₃)=CH₂ (2-methylpropene) + KBr + H₂O
3. CH₃CH(Br)CH₂CH₃ (2-bromobutane) + NaOH, water → CH₃CH(OH)CH₂CH₃ (butan-2-ol) + NaBr
4. CH₃CH₂Br (bromoethane) + KCN, aq. ethanol (nucleophilic substitution) → CH₃CH₂CN (propanenitrile) + KBr
5. C₆H₅ONa (sodium phenoxide) + C₂H₅Cl (chloroethane), Williamson synthesis → C₆H₅-O-C₂H₅ (phenetole) + NaCl
6. CH₃CH₂CH₂OH (1-propanol) + SOCl₂ → CH₃CH₂CH₂Cl (1-chloropropane) + SO₂ + HCl
7. CH₃CH₂CH=CH₂ (but-1-ene) + HBr, peroxide (anti-Markovnikov addition) → CH₃CH₂CH₂CH₂-Br (1-bromobutane)
8. CH₃CH=C(CH₃)₂ (2-methylbut-2-ene) + HBr (Markovnikov addition) → CH₃-CH₂-C(Br)(CH₃)-CH₃ (2-bromo-2-methylbutane)

Q3. How can the following conversions be carried out?

(i) Propene to propan-1-ol
(ii) Ethanol to but-1-yne
(iii) 1-Bromopropane to 2-bromopropane
(iv) Toluene to benzyl alcohol
(v) Benzene to 4-bromonitrobenzene
(vi) Benzyl alcohol to 2-phenylethanoic acid
(vii) Ethanol to propanenitrile
(viii) Aniline to chlorobenzene
(ix) 2-Chlorobutane to 3,4-dimethylhexane
(x) 2-Methyl-1-propene to 2-chloro-2-methylpropane
(xi) Ethyl chloride to propanoic acid
(xii) But-1-ene to n-butyl iodide
(xiii) 2-Chloropropane to 1-propanol
(xiv) Isopropyl alcohol to iodoform
(xv) Chlorobenzene to p-nitrophenol
(xvi) 2-Bromopropane to 1-bromopropane
(xvii) Chloroethane to butane
(xviii) Benzene to diphenyl
(xix) tert-Butyl bromide to isobutyl bromide
(xx) Aniline to phenyl isocyanide

Ans. (i) CH₃-CH=CH₂ (propene) + HBr/peroxide (anti-Markovnikov addition) → CH₃-CH₂-CH₂-Br (1-bromopropane); then aq. KOH/heat (nucleophilic substitution) → CH₃-CH₂-CH₂-OH (propan-1-ol)

(ii) CH₃CH₂OH (ethanol) → CH₃CH₂Br (ethyl bromide), using red P/Br₂
CH₃CH₂Br + Mg/ether → CH₃CH₂MgBr (ethylmagnesium bromide)
CH₃CH₂MgBr + ethylene oxide (H₂C-CH₂ with O bridge), ether (Grignard reaction) → [CH₃CH₂CH₂CH₂OMgBr] (addition product); with H⁺/H₂O → CH₃CH₂CH₂CH₂OH (1-butanol)
1-Butanol + red P/I₂ → CH₃CH₂CH₂CH₂-I (iodobutane)
Iodobutane + KOH (alc.)/heat (dehydrohalogenation) → CH₃CH₂CH=CH₂ (but-1-ene)
But-1-ene + Br₂/CCl₄ (electrophilic addition) → CH₃CH₂CHBr-CH₂Br (1,2-dibromobutane)
1,2-Dibromobutane + NaNH₂, liq. NH₃, 196 K (dehydrohalogenation) → CH₃CH₂C≡CH (but-1-yne)

(iii) CH₃-CH₂-CH₂-Br (1-bromopropane) + KOH (alc.)/heat (dehydrohalogenation) → CH₃-CH=CH₂ (propene); then HBr (Markovnikov addition) → CH₃-CHBr-CH₃ (2-bromopropane)

(iv) Toluene + Cl₂/UV light or heat → C₆H₅CH₂Cl (benzyl chloride); then aq. KOH/heat (nucleophilic substitution) → C₆H₅CH₂OH (benzyl alcohol)

(v) Benzene + Br₂/FeBr₃, dark → bromobenzene; then HNO₃/H₂SO₄ (nitration) → 4-bromonitrobenzene

(vi) Benzyl alcohol + PCl₅ (-POCl₃, -HCl) → C₆H₅CH₂Cl; then KCN, aq. ethanol (-KCl) → C₆H₅CH₂CN (benzyl cyanide); then H⁺/H₂O (hydrolysis) → C₆H₅CH₂COOH (2-phenylethanoic acid)

(vii) CH₃-CH₂-OH (ethanol) + red P/Br₂ → CH₃-CH₂-Br (bromoethane); then KCN, aq. ethanol → CH₃-CH₂-CN (propanenitrile)

(viii) Aniline + NaNO₂ + 2HCl, 273-278 K (-NaCl, -2H₂O) → benzenediazonium chloride (C₆H₅N₂⁺Cl^-); then Cu₂Cl₂ → chlorobenzene + N₂

(ix) 2CH₃-CHCl-CH₂-CH₃ (2-chlorobutane) + 2Na/dry ether (Wurtz reaction) → CH₃-CH₂-CH(CH₃)-CH(CH₃)-CH₂-CH₃ (3,4-dimethylhexane) + 2NaCl

(x) CH₃-C(CH₃)=CH₂ (2-methylpropene) + HCl (Markovnikov addition) → (CH₃)₃C-Cl (2-chloro-2-methylpropane)

(xi) CH₃-CH₂-Cl (ethyl chloride) + KCN, aq. ethanol (nucleophilic substitution) → CH₃-CH₂-CN (propanenitrile) + KCl; then H⁺/H₂O (hydrolysis) → CH₃-CH₂-COOH (propanoic acid)

(xii) CH₃-CH₂-CH=CH₂ (but-1-ene) + HBr/peroxide (anti-Markovnikov addition) → CH₃-CH₂-CH₂-CH₂-Br (1-bromobutane); then NaI, dry acetone (Finkelstein reaction) → CH₃-CH₂-CH₂-CH₂-I (n-butyl iodide)

(xiii) CH₃-CHCl-CH₃ (2-chloropropane) + KOH (alc.)/heat (dehydrohalogenation) → CH₃-CH=CH₂ (propene) + HCl; then HBr/peroxide (anti-Markovnikov addition) → CH₃-CH₂-CH₂-Br (1-bromopropane); then aq. KOH/heat (nucleophilic substitution) → CH₃-CH₂-CH₂-OH (1-propanol)

(xiv) CH₃-CH(OH)-CH₃ (isopropyl alcohol) + CrO₃ (oxidation) → CH₃-CO-CH₃ (propanone); then NaOI (iodoform reaction) → CH₃COONa + CHI₃ (iodoform)

(xv) Chlorobenzene + HNO₃/H₂SO₄ (nitration) → p-chloronitrobenzene (major product) + m-chloronitrobenzene + o-chloronitrobenzene. Note: p-nitrophenol is obtained from p-chloronitrobenzene by heating with aqueous NaOH at 443 K followed by acidification.

(xvi) CH₃-CHBr-CH₃ (2-bromopropane) + KOH (alc.)/heat (dehydrohalogenation) → CH₃-CH=CH₂ (propene) + HBr; then HBr/peroxides (anti-Markovnikov addition) → CH₃-CH₂-CH₂-Br (1-bromopropane)

(xvii) 2CH₃-CH₂-Cl (chloroethane) + 2Na/dry ether (Wurtz reaction) → CH₃-CH₂-CH₂-CH₃ (butane) + 2NaCl

(xviii) Benzene + Br₂/FeBr₃ → bromobenzene; then 2Na/dry ether (Fittig reaction) → biphenyl + 2NaBr

(xix) (CH₃)₃C-Br (tert-butyl bromide) + KOH (alc.)/heat (dehydrohalogenation) → CH₃-C(CH₃)=CH₂ (2-methylpropene); then HBr/peroxide (anti-Markovnikov addition) → (CH₃)₂CH-CH₂-Br (isobutyl bromide)

(xx) Aniline + CHCl₃ + 3KOH, heat (carbylamine reaction) → C₆H₅NC (phenyl isocyanide) + 3KCl + 3H₂O

Q4. Primary alkyl halide C₄H₉Br (a) reacted with alcoholic KOH to give compound (b). Compound (b) is reacted with HBr to give (c) which is an isomer of (a). When (a) is reacted with sodium metal it gives compound (d), C₈H₁₈, which is different from the compound formed when n-butyl bromide is reacted with sodium. Give the structural formula of (a) and write the equations for all the reactions.

Ans. 2CH₃CH₂CH₂CH₂Br (n-butyl bromide) + 2Na/dry ether (Wurtz reaction) → CH₃CH₂CH₂CH₂CH₂CH₂CH₂CH₃ (n-octane) + 2NaBr
2(CH₃)₂CHCH₂Br (isobutyl bromide, a) + 2Na/dry ether (Wurtz reaction) → (CH₃)₂CHCH₂CH₂CH(CH₃)₂ (2,5-dimethylhexane, d) + 2NaBr
(CH₃)₂CHCH₂Br (a) + alcoholic KOH → CH₃-C(CH₃)=CH₂ (2-methylpropene, b)
CH₃-C(CH₃)=CH₂ (b) + HBr (Markovnikov addition) → (CH₃)₃C-Br (2-bromo-2-methylpropane, c, an isomer of a)