Haloalkanes

Replace one hydrogen of an alkane by a halogen and the molecule stops being inert. The halogen pulls electron density away from the carbon it is attached to, leaving that carbon electron-poor and open to attack by anything carrying a lone pair. That single polarised bond, C–X, is the doorway from hydrocarbons into alcohols, amines, nitriles, ethers and almost everything else in the rest of this book. Haloalkanes are therefore studied less for themselves than as the central junction of organic synthesis.

8.1 Introduction; nomenclature, isomerism and classification of monohaloalkanes

A haloalkane (alkyl halide) is an alkane in which one or more hydrogen atoms have been replaced by halogen atoms. A monohaloalkane contains just one halogen and has the general formula CₙH₂ₙ₊₁X, written R–X.

Nomenclature. In the IUPAC system the halogen is named as a prefix — fluoro-, chloro-, bromo-, iodo- — on the longest chain that contains the halogenated carbon, numbered so that the substituents get the lowest locants. The older common name is simply "alkyl halide".

Structure	IUPAC name	Common name
CH₃Cl	chloromethane	methyl chloride
CH₃CH₂Br	bromoethane	ethyl bromide
CH₃CH₂CH₂I	1-iodopropane	n-propyl iodide
(CH₃)₂CHCl	2-chloropropane	isopropyl chloride
(CH₃)₃CBr	2-bromo-2-methylpropane	tert-butyl bromide
CH₃CH(CH₃)CH₂Cl	1-chloro-2-methylpropane	isobutyl chloride

Classification. Monohaloalkanes are classified by the nature of the carbon carrying the halogen:

Class	Halogen-bearing carbon is joined to	Example
Primary (1°)	one (or no) other carbon	CH₃CH₂CH₂Br
Secondary (2°)	two other carbons	(CH₃)₂CHBr
Tertiary (3°)	three other carbons	(CH₃)₃CBr

This classification is not bookkeeping — it decides the mechanism. Primary halides react by S_N2, tertiary halides by S_N1.

Isomerism. Monohaloalkanes show three kinds of isomerism:

1. Chain isomerism — different carbon skeletons, e.g. 1-bromobutane and 1-bromo-2-methylpropane.
2. Position isomerism — the same skeleton with the halogen on a different carbon, e.g. 1-bromobutane and 2-bromobutane.
3. Optical isomerism — when the halogen-bearing carbon has four different groups, e.g. 2-bromobutane, which exists as a pair of non-superimposable mirror images.

C₄H₉Br has exactly four structural isomers, and they are worth knowing by heart because examiners ask for them year after year.

8.2 Preparation from alkanes, alkenes and alcohols

From alkanes — free radical halogenation

In the presence of ultraviolet light or at 573–773 K, an alkane reacts with chlorine or bromine by a free radical chain mechanism:

CH₄ + Cl₂ --UV light--> CH₃Cl + HCl

Step	Reaction
Initiation	Cl₂ --hv--> 2Cl•
Propagation	CH₄ + Cl• → CH₃• + HCl ; CH₃• + Cl₂ → CH₃Cl + Cl•
Termination	CH₃• + Cl• → CH₃Cl ; 2Cl• → Cl₂ ; 2CH₃• → C₂H₆

The method is cheap but a poor laboratory preparation: the chloromethane formed is itself attacked, giving CH₂Cl₂, CHCl₃ and CCl₄, and a higher alkane gives a mixture of position isomers as well. Reactivity of the halogens falls F₂ > Cl₂ > Br₂ > I₂; fluorination is explosive and iodination is reversible, so only chlorination and bromination are useful. Iodination can be forced by adding an oxidising agent (HIO₃ or conc. HNO₃) to destroy the HI as it forms.

From alkenes

Addition of hydrogen halide — Markovnikov's rule. The hydrogen adds to the doubly bonded carbon that already carries more hydrogens:

CH₃CH=CH₂ + HBr → CH₃CHBrCH₃ (2-bromopropane, major)

The reason is the stability of the intermediate carbocation: the secondary cation CH₃C⁺HCH₃ is more stable than the primary CH₃CH₂C⁺H₂.

Peroxide (Kharasch) effect. In the presence of benzoyl peroxide the addition of HBr only goes the other way, because the mechanism becomes free radical:

CH₃CH=CH₂ + HBr --peroxide--> CH₃CH₂CH₂Br (1-bromopropane)

HCl and HI show no peroxide effect — the H–Cl bond is too strong to break homolytically and the C–I bond is too weak to survive.

Reactivity of the hydrogen halides towards an alkene is HI > HBr > HCl > HF.

From alcohols — the standard laboratory route

This is the method to quote unless a question says otherwise, because it gives one product in good yield.

Reagent	Equation	Note
Conc. HCl + anhyd. ZnCl₂	C₂H₅OH + HCl → C₂H₅Cl + H₂O	ZnCl₂ is a catalyst; the mixture is Lucas reagent
NaBr + conc. H₂SO₄	C₂H₅OH + HBr → C₂H₅Br + H₂O	HBr generated in situ
Red P + I₂	3C₂H₅OH + PI₃ → 3C₂H₅I + H₃PO₃	PI₃ made in the flask
PCl₃	3C₂H₅OH + PCl₃ → 3C₂H₅Cl + H₃PO₃
PCl₅	C₂H₅OH + PCl₅ → C₂H₅Cl + POCl₃ + HCl
SOCl₂	C₂H₅OH + SOCl₂ → C₂H₅Cl + SO₂↑ + HCl↑	Darzen's process — best of all

Thionyl chloride is preferred because both by-products are gases that simply escape, leaving pure haloalkane with nothing to separate.

The ease of replacement of –OH follows the order 3° > 2° > 1°, which is the basis of the Lucas test: a tertiary alcohol gives an oily turbidity at once, a secondary alcohol in about five minutes, and a primary alcohol only on heating.

Halogen exchange. Two named reactions convert one haloalkane into another:

• Finkelstein reaction: R–Cl + NaI --dry acetone--> R–I + NaCl↓ (NaCl is insoluble in acetone, so it precipitates and drives the equilibrium.)
• Swarts reaction: CH₃Br + AgF → CH₃F + AgBr (also with Hg₂F₂ or SbF₃); this is how fluoroalkanes and freons are made.

8.3 Physical properties

Haloalkanes are colourless liquids (the first few members are gases) with a sweetish smell.

Property	Trend	Reason
Boiling point, same alkyl group	RI > RBr > RCl > RF	heavier, more polarisable halogen → stronger van der Waals forces
Boiling point, same halogen	rises with chain length	larger surface area of contact
Effect of branching	branched isomer boils lower	a branched molecule is more spherical, so less surface contact
Density	RI > RBr > RCl; bromides and iodides sink in water	heavy halogen atom
Solubility	insoluble in water, soluble in benzene, ether, alcohol	cannot make hydrogen bonds strong enough to break water's own
Dipole moment	CH₃Cl (1.86 D) > CH₃F (1.85) > CH₃Br (1.79) > CH₃I (1.64)	product of charge separation and bond length

The last row is a favourite short question. Fluorine is the most electronegative atom, yet CH₃F does not have the largest dipole moment, because the C–F bond is so short that the charges are not far enough apart.

8.4 Chemical properties; substitution reactions S_N1 and S_N2

Because the halogen is more electronegative than carbon, the C–X bond is polarised Cδ⁺–Xδ⁻. A nucleophile (an electron-rich species: OH⁻, CN⁻, NH₃, RO⁻, RS⁻, NO₂⁻) attacks the positive carbon and the halide leaves as X⁻ with the bonding pair. The result is nucleophilic substitution, written S_N.

The C–X bond gets weaker and longer down the group, so the leaving group leaves more easily:

Bond	C–F	C–Cl	C–Br	C–I
Bond length (pm)	139	178	193	214
Bond enthalpy (kJ mol⁻¹)	452	351	293	234
Reactivity in substitution	least			greatest

Order of reactivity: R–I > R–Br > R–Cl > R–F. Iodide is the best leaving group because the C–I bond is the weakest, even though iodine is the least electronegative halogen.

The S_N2 mechanism

S_N2 means substitution, nucleophilic, bimolecular. It happens in one step: the nucleophile attacks the carbon from the side exactly opposite to the leaving group (backside attack) while the halide is still leaving. At the half-way point the carbon is joined to five things in a transition state in which the three unchanged groups lie in a plane.

Because the nucleophile comes in from the back, the other three groups are pushed through the plane like an umbrella turning inside out in the wind. The product therefore has the opposite configuration from the reactant. This is Walden inversion, and it is the proof that the mechanism is S_N2.

The S_N1 mechanism

S_N1 means substitution, nucleophilic, unimolecular. It happens in two steps:

• Step 1 (slow, rate-determining). The C–X bond breaks heterolytically on its own, giving a flat, sp²-hybridised carbocation and X⁻.
• Step 2 (fast). The nucleophile attacks the carbocation.

The nucleophile does not appear in the rate law at all, because it joins the reaction only after the slow step is over. Doubling the hydroxide concentration does not change the rate — the single most quotable difference between the two mechanisms.

The carbocation is planar, so the nucleophile can attack either face with equal ease. A single enantiomer of the halide therefore gives a 50 : 50 mixture of both product enantiomers — a racemic mixture. This is racemisation.

Energy profiles and the choice of mechanism

The two mechanisms look completely different on an energy profile. S_N2 has one hump — one transition state, no intermediate. S_N1 has two humps with a well between them: the well is the carbocation, a real (if short-lived) species that can be detected.

	SN1	SN2
Molecularity of slow step	1 (only R–X)	2 (R–X and Nu⁻)
Steps	two	one
Rate law	rate = k[\text{RX}]	rate = k[\text{RX}][\text{Nu}^-]
Intermediate	carbocation	none — only a transition state
Stereochemistry	racemisation	inversion (Walden)
Favoured by	3° > 2° > 1° halide	CH₃X > 1° > 2° > 3°
Nucleophile	weak, low concentration	strong, high concentration
Solvent	polar protic (water, ethanol)	polar aprotic (acetone, DMSO)
Rearrangement	possible (carbocation can shift)	never

8.5 Formation of alcohol, nitrile, amine, ether, thioether, carbylamine, nitrite and nitroalkane

Every reaction below is nucleophilic substitution. Learn the table as one reaction with eight different nucleophiles.

Reagent and conditions	Nucleophile	Product	Class
KOH or NaOH, aqueous, warm	OH⁻	C₂H₅OH	alcohol
moist Ag₂O	OH⁻	C₂H₅OH	alcohol
KCN, alcoholic	⁻C≡N (through carbon)	CH₃CH₂CN, propanenitrile	nitrile
AgCN, alcoholic	⁻N≡C (through nitrogen)	CH₃CH₂NC, ethyl isocyanide	carbylamine (isocyanide)
NH₃, alcoholic, excess, sealed tube	NH₃	C₂H₅NH₂	primary amine
R'ONa, dry (Williamson)	R'O⁻	C₂H₅OC₂H₅	ether
R'SNa or KSH	R'S⁻ / HS⁻	C₂H₅SC₂H₅ / C₂H₅SH	thioether / thiol
KNO₂, aqueous alcohol	O–N=O (through oxygen)	C₂H₅–O–N=O	alkyl nitrite
AgNO₂, alcoholic	⁻NO₂ (through nitrogen)	C₂H₅NO₂	nitroalkane

Written out with equations:

• C₂H₅Br + KOH(aq) → C₂H₅OH + KBr
• C₂H₅Br + KCN → C₂H₅CN + KBr
• C₂H₅Br + AgCN → C₂H₅NC + AgBr
• C₂H₅Br + 2NH₃ → C₂H₅NH₂ + NH₄Br
• C₂H₅Br + C₂H₅ONa → C₂H₅OC₂H₅ + NaBr
• C₂H₅Br + C₂H₅SNa → C₂H₅SC₂H₅ + NaBr
• C₂H₅Br + KNO₂ → C₂H₅ONO + KBr
• C₂H₅Br + AgNO₂ → C₂H₅NO₂ + AgBr

The ammonolysis reaction does not stop at the primary amine: the amine produced is itself a nucleophile, so a mixture of primary, secondary and tertiary amines and finally the quaternary ammonium salt is obtained. A large excess of ammonia favours the primary amine.

8.6 Elimination reaction; reduction; the Wurtz reaction

Dehydrohalogenation and Saytzeff's rule

Heated with alcoholic potassium hydroxide, a haloalkane loses HX and becomes an alkene. The base takes a hydrogen from the carbon next to the one carrying the halogen (the β-carbon), so the reaction is called β-elimination:

CH₃CH₂Br + KOH(alc.) --Δ--> CH₂=CH₂ + KBr + H₂O

When the halide has two different β-carbons, two alkenes are possible. Saytzeff's (Zaitsev's) rule says the major product is the more substituted alkene — the one with more alkyl groups attached to the doubly bonded carbons — because it is the more stable.

The ease of dehydrohalogenation is 3° > 2° > 1°, and for the same alkyl group R–I > R–Br > R–Cl.

Reduction

Reduction replaces the halogen by hydrogen and gives back the parent alkane:

C₂H₅Br + 2[H] --Zn/HCl--> C₂H₆ + HBr

Suitable reducing agents are Zn + HCl, Zn–Cu couple in ethanol, LiAlH₄ in dry ether, and H₂ over a nickel catalyst.

The Wurtz reaction

Two molecules of haloalkane are joined by metallic sodium in dry ether:

2C₂H₅Br + 2Na --dry ether--> C₄H₁₀ + 2NaBr

This is the standard way of climbing from a small alkane to one with twice as many carbon atoms, so it can only make alkanes with an even number of carbons from a single halide. If two different halides are used, three products form (R–R, R–R' and R'–R') and separating them is a nuisance. Tertiary halides give alkenes instead, by elimination.

8.7 Preparation of trichloromethane from ethanol and propanone

Trichloromethane, CHCl₃, is the old anaesthetic chloroform. It is made industrially and in the laboratory by the haloform reaction, in which bleaching powder supplies both the chlorine and the alkali.

Bleaching powder in water provides the two reagents:

CaOCl₂ + H₂O → Ca(OH)₂ + Cl₂

From ethanol — three steps

Step	Reaction	What happens
1. Oxidation	CH₃CH₂OH + Cl₂ → CH₃CHO + 2HCl	ethanol → ethanal
2. Chlorination	CH₃CHO + 3Cl₂ → CCl₃CHO + 3HCl	ethanal → trichloroethanal (chloral)
3. Hydrolysis	2CCl₃CHO + Ca(OH)₂ → 2CHCl₃ + (HCOO)₂Ca	chloral → chloroform + calcium methanoate

From propanone — two steps

Propanone already has the carbonyl group, so no oxidation step is needed and the yield is better. This is the preferred method.

Step	Reaction
1. Chlorination	CH₃COCH₃ + 3Cl₂ → CCl₃COCH₃ + 3HCl
2. Hydrolysis	2CCl₃COCH₃ + Ca(OH)₂ → 2CHCl₃ + (CH₃COO)₂Ca

In both routes the crude chloroform distils over with steam, and is purified by shaking with concentrated sulphuric acid, washing with dilute alkali and water, drying over anhydrous calcium chloride and redistilling at 334 K.

8.8 Chemical properties of trichloromethane

Chloroform is a colourless, sweet-smelling, heavy liquid (density 1.49 g cm⁻³, b.p. 61 °C), almost insoluble in water, non-inflammable, and a good solvent for fats, oils, waxes, resins, rubber and iodine.

Oxidation — the reason chloroform is kept in a dark bottle

In air and sunlight chloroform is slowly oxidised to phosgene (carbonyl chloride), an extremely poisonous gas used as a war gas in 1915:

2CHCl₃ + O₂ --light--> 2COCl₂ + 2HCl

This is why chloroform is stored in dark brown bottles filled to the brim (to keep out light and air) with about 1% ethanol added. The ethanol destroys any phosgene that does form, turning it into harmless diethyl carbonate:

COCl₂ + 2C₂H₅OH → (C₂H₅O)₂CO + 2HCl

Reduction

CHCl₃ + 2[H] --Zn/HCl--> CH₂Cl₂ + HCl

Zinc dust and water give dichloromethane; with excess reducing agent the reduction goes all the way to methane, CHCl₃ + 6[H] → CH₄ + 3HCl.

Action on silver powder

Heated with silver powder, chloroform loses all its chlorine and the two carbon skeletons join, giving ethyne:

2CHCl₃ + 6Ag → C₂H₂↑ + 6AgCl

Action with concentrated nitric acid

CHCl₃ + HNO₃ → CCl₃NO₂ + H₂O

The product, trichloronitromethane, is chloropicrin — a tear gas, and a soil fumigant and insecticide.

Action with propanone

In the presence of dilute potassium hydroxide, chloroform adds across the carbonyl group of propanone:

CHCl₃ + CH₃COCH₃ --dil. KOH--> (CH₃)₂C(OH)CCl₃

The product, 1,1,1-trichloro-2-methylpropan-2-ol, is chloretone, once widely used as a hypnotic and sedative.

Action with aqueous alkali

Boiled with aqueous sodium hydroxide, chloroform is hydrolysed to a salt of methanoic acid:

CHCl₃ + 4NaOH → HCOONa + 3NaCl + 2H₂O

The carbylamine reaction

Warmed with a primary amine and alcoholic KOH, chloroform gives an isocyanide with an appalling smell — the standard test for a primary amine:

C₆H₅NH₂ + CHCl₃ + 3KOH → C₆H₅NC + 3KCl + 3H₂O

Reagent	Product	Name of product
O₂ + light	COCl₂	phosgene
Zn/HCl	CH₂Cl₂	dichloromethane
Ag powder, Δ	C₂H₂	ethyne
conc. HNO₃	CCl₃NO₂	chloropicrin
propanone + dil. KOH	(CH₃)₂C(OH)CCl₃	chloretone
NaOH(aq), boil	HCOONa	sodium methanoate
RNH₂ + alc. KOH	RNC	isocyanide (carbylamine test)

Uses of chloroform. Solvent for fats, oils, rubber, resins and iodine; manufacture of chloretone, chloropicrin and the refrigerant Freon-22 (CHCl₃ + 2HF --SbF₅--> CHClF₂ + 2HCl); a preservative for anatomical specimens; and formerly a general anaesthetic, now abandoned because it damages the liver.

Chapter summary

• A monohaloalkane is CₙH₂ₙ₊₁X. It is 1°, 2° or 3° according to the number of carbons on the halogen-bearing carbon, and shows chain, position and (when that carbon has four different groups) optical isomerism. C₄H₉Br has four structural isomers.
• Preparation: free radical halogenation of alkanes (gives mixtures); Markovnikov addition of HX to alkenes, reversed for HBr by peroxides; and — best — alcohols with conc. HX/ZnCl₂, PCl₃, PCl₅ or SOCl₂ (Darzen's process). Finkelstein (NaI/acetone) and Swarts (AgF) exchange halogens.
• Boiling point rises RF < RCl < RBr < RI and with chain length, and falls with branching. Dipole moments run CH₃Cl > CH₃F > CH₃Br > CH₃I.
• Reactivity in nucleophilic substitution is R–I > R–Br > R–Cl > R–F, because the C–X bond enthalpy falls in that order.
• S_N2: one step, backside attack, rate = k[\text{RX}][\text{Nu}^-], transition state with five groups on carbon, inversion of configuration, favoured by CH₃X > 1° > 2° > 3° and strong nucleophiles.
• S_N1: two steps through a planar carbocation, rate = k[\text{RX}], racemisation, favoured by 3° > 2° > 1°, weak nucleophiles and polar protic solvents. Its energy profile has two humps and an intermediate well; the S_N2 profile has one hump and none.
• The eight nucleophiles: OH⁻ → alcohol; KCN → nitrile but AgCN → isocyanide; NH₃ → amine; R'ONa → ether (Williamson); R'SNa → thioether; KNO₂ → alkyl nitrite but AgNO₂ → nitroalkane. Potassium attacks through C or O, silver through N.
• Aqueous KOH substitutes (alcohol); alcoholic KOH eliminates (alkene), the more substituted alkene being the major product by Saytzeff's rule. Reduction gives the alkane; the Wurtz reaction (2RX + 2Na, dry ether) doubles the chain.
• Chloroform is made from ethanol (3 steps) or propanone (2 steps) with bleaching powder. It is oxidised in air and light to poisonous phosgene, so it is kept in dark bottles with 1% ethanol; with Ag it gives ethyne, with conc. HNO₃ chloropicrin, with propanone chloretone, and with aqueous NaOH sodium methanoate.

Practice questions

Group A — Multiple choice (1 mark each)

1. The number of structural isomers of C₄H₉Br is [1]a2b3c4d5
2. Which haloalkane is most reactive towards nucleophilic substitution? [1]aCH₃CH₂FbCH₃CH₂ClcCH₃CH₂BrdCH₃CH₂I
3. An S_N2 reaction proceeds with [1]aretention of configurationbinversion of configurationcracemisationdno stereochemical change
4. The intermediate in an S_N1 reaction is [1]aa carbanionba free radicalca carbocationda carbene
5. Bromoethane heated with alcoholic KOH gives [1]aethanolbethenecethanedethoxyethane
6. Bromoethane warmed with silver cyanide gives mainly [1]apropanenitrilebethyl isocyanidecethylaminedethanol
7. Chloroform is stored in dark bottles because in light and air it forms [1]achloropicrinbchloretonecphosgenedethyne
8. The reagent that converts ethanol to chloroethane with only gaseous by-products is [1]aPCl₅bPCl₃cSOCl₂dconc. HCl

Answers to Group A

1. (c) — 1-bromobutane, 2-bromobutane, 1-bromo-2-methylpropane and 2-bromo-2-methylpropane.

2. (d) — the C–I bond is the weakest (234 kJ mol⁻¹), so iodide is the best leaving group.

3. (b) — backside attack turns the molecule inside out (Walden inversion).

4. (c) — the slow step is heterolysis of C–X to give a planar carbocation.

5. (b) — alcoholic KOH acts as a base and eliminates HBr.

6. (b) — AgCN is covalent, so attack is through nitrogen and an isocyanide is formed.

7. (c) — 2CHCl₃ + O₂ → 2COCl₂ + 2HCl.

8. (c) — Darzen's process; SO₂ and HCl both escape as gases.

Group B — Short answer (5 marks each)

1. Define a haloalkane. Write the structures, IUPAC names and classification (1°, 2°, 3°) of all the isomers of C₄H₉Br, and state which one is optically active. [5]
2. How is bromoethane prepared from(a) ethene and(b) ethanol? Why is thionyl chloride the best reagent for making chloroethane from ethanol? [5]
3. Explain the S_N2 mechanism with a suitable example. Why does it give inversion of configuration, and why is a tertiary halide unreactive by this route? [5]
4. Distinguish between S_N1 and S_N2 mechanisms under five headings. [5]
5. What happens when bromoethane is treated with(a) KCN,(b) AgCN,(c) AgNO₂,(d) KNO₂,(e) sodium ethoxide? Give equations and name the products. [5]
6. State Saytzeff's rule. Predict and explain the major product when 2-bromobutane is heated with alcoholic KOH. [5]
7. 21.8 g of bromoethane is heated with sodium in dry ether. Name the reaction, write the equation, and calculate the volume of gas produced at STP. [5]
8. How is chloroform prepared from propanone? Why is this route preferred to the one starting from ethanol? [5]

Answers to Group B

2. (a) Ethene + HBr → CH₃CH₂Br; the addition needs no peroxide because ethene is symmetrical, so Markovnikov's rule does not arise. (b) Ethanol is warmed with NaBr and conc. H₂SO₄, which generates HBr in the flask: C₂H₅OH + HBr → C₂H₅Br + H₂O. Thionyl chloride is best for the chloride because C₂H₅OH + SOCl₂ → C₂H₅Cl + SO₂↑ + HCl↑ — both by-products are gases and escape, so the haloalkane is obtained pure with no separation step.

4.

	SN1	SN2
Steps	two	one
Rate law	k[\text{RX}]	k[\text{RX}][\text{Nu}^-]
Intermediate	carbocation	none (transition state only)
Stereochemistry	racemisation	inversion
Favoured by	3° halide, weak nucleophile, protic solvent	1° halide, strong nucleophile, aprotic solvent

6. Saytzeff's rule: in a dehydrohalogenation the alkene with the greater number of alkyl groups on the doubly bonded carbons is the major product, because alkyl groups release electrons into the double bond (hyperconjugation and inductive effect) and stabilise it. 2-Bromobutane can lose a β-hydrogen from C1 or C3. Loss from C3 gives but-2-ene, CH₃CH=CHCH₃, which is disubstituted; loss from C1 gives but-1-ene, CH₂=CHCH₂CH₃, which is monosubstituted. But-2-ene is therefore the major product (about 80%).

7. The Wurtz reaction: 2C₂H₅Br + 2Na → C₄H₁₀ + 2NaBr. n(\text{C}_2\text{H}_5\text{Br}) = 21.8/109 = 0.200\ \text{mol}, so n(\text{C}_4\text{H}_{10}) = 0.100\ \text{mol} and

V = 0.100 \times 22.4 = 2.24\ \text{L at STP}

8. Bleaching powder in water gives Ca(OH)₂ and Cl₂. Then CH₃COCH₃ + 3Cl₂ → CCl₃COCH₃ + 3HCl, and 2CCl₃COCH₃ + Ca(OH)₂ → 2CHCl₃ + (CH₃COO)₂Ca. Propanone is preferred because it already contains the carbonyl group, so the oxidation step needed with ethanol is unnecessary: two steps instead of three, and a higher yield.

Group C — Long answer (8 marks each)

1. (a) Describe the S_N1 and S_N2 mechanisms of hydrolysis of a haloalkane, using suitable examples, and draw the energy profile of each. [5](b) Explain why 2-bromo-2-methylpropane hydrolyses by S_N1 while bromomethane hydrolyses by S_N2. [3]
2. (a) How is trichloromethane prepared from ethanol? Give all three equations. [4](b) What happens when chloroform is treated with (i) air and sunlight, (ii) silver powder, (iii) conc. HNO₃, (iv) aqueous NaOH? Give equations. [4]
3. An organic compound A, C₃H₇Br, reacts with aqueous KOH to give B, and with alcoholic KOH to give C, C₃H₆. A with KCN gives D, and A with sodium in dry ether gives E, C₆H₁₄. Identify A to E, write all the equations, and state the mechanism by which A reacts with aqueous KOH if A is the 2-bromo isomer. [8]

Answers to Group C

1.(b) In 2-bromo-2-methylpropane the halogen-bearing carbon carries three methyl groups. They release electron density by the inductive effect and by hyperconjugation, so the carbocation left behind when Br⁻ departs is strongly stabilised, and the slow heterolysis of step 1 becomes easy — S_N1. The same three methyl groups also crowd the back of the carbon, so backside attack is sterically blocked and S_N2 is impossible. Bromomethane is the opposite case: the CH₃⁺ cation would be very unstable, so S_N1 is ruled out, while the carbon is completely unhindered and the nucleophile can reach it easily — S_N2.

2.(a) CaOCl₂ + H₂O → Ca(OH)₂ + Cl₂; CH₃CH₂OH + Cl₂ → CH₃CHO + 2HCl; CH₃CHO + 3Cl₂ → CCl₃CHO + 3HCl; 2CCl₃CHO + Ca(OH)₂ → 2CHCl₃ + (HCOO)₂Ca.

2.(b) (i) 2CHCl₃ + O₂ → 2COCl₂ + 2HCl (phosgene). (ii) 2CHCl₃ + 6Ag → C₂H₂ + 6AgCl (ethyne). (iii) CHCl₃ + HNO₃ → CCl₃NO₂ + H₂O (chloropicrin). (iv) CHCl₃ + 4NaOH → HCOONa + 3NaCl + 2H₂O (sodium methanoate).

3. A = C₃H₇Br (1-bromopropane or 2-bromopropane). B = propan-1-ol or propan-2-ol, C₃H₇OH: C₃H₇Br + KOH(aq) → C₃H₇OH + KBr. C = propene, CH₃CH=CH₂: C₃H₇Br + KOH(alc.) → CH₃CH=CH₂ + KBr + H₂O. D = butanenitrile (or 2-methylpropanenitrile), C₃H₇CN: C₃H₇Br + KCN → C₃H₇CN + KBr. E = hexane, C₆H₁₄, by the Wurtz reaction: 2C₃H₇Br + 2Na → C₆H₁₄ + 2NaBr.

If A is 2-bromopropane it is a secondary halide. Secondary halides sit on the borderline, but with a strong nucleophile such as concentrated aqueous hydroxide the reaction is mainly S_N2, giving propan-2-ol with inversion at the chiral centre; in a highly ionising solvent with dilute hydroxide it drifts towards S_N1 and partial racemisation.