Halogenoalkanes

Halogenoalkanes (also called alkyl halides) are compounds in which one or more hydrogen atoms of an alkane have been replaced by a halogen atom. The general formula is $\mathrm{R-X}$, where $\mathrm{X}$ is F, Cl, Br, or I. The C--X bond is polar (halogens are more electronegative than carbon), making the carbon electrophilic and susceptible to attack by nucleophiles.

Classification

Halogenoalkanes are classified by the carbon bearing the halogen:

• Primary ($1^\circ$): The halogen-bearing carbon is attached to one other carbon (e.g. $\mathrm{CH}_3\mathrm{CH}_2\mathrm{Br}$).
• Secondary ($2^\circ$): Attached to two other carbons (e.g. $(\mathrm{CH}_3)_2\mathrm{CHBr}$).
• Tertiary ($3^\circ$): Attached to three other carbons (e.g. $(\mathrm{CH}_3)_3\mathrm{CBr}$).

Nucleophilic Substitution

Nucleophilic substitution is the replacement of the halogen (the leaving group) by a nucleophile. Two distinct mechanisms operate depending on the substrate structure, nucleophile strength, and solvent.

The SN2 Mechanism

The subscript N2 stands for "substitution, nucleophilic, bimolecular." It is a one-step, concerted mechanism in which bond-making and bond-breaking occur simultaneously.

Rate equation: Rate $= k[\mathrm{R-X}][\mathrm{Nu}^-]$

The rate depends on the concentration of both the halogenoalkane and the nucleophile. This is confirmed experimentally by observing that doubling either concentration doubles the rate.

Mechanism for the reaction of bromoethane with hydroxide:

$$\mathrm{OH}^- + \mathrm{CH}_3\mathrm{CH}_2\mathrm{Br} \to \mathrm{CH}_3\mathrm{CH}_2\mathrm{OH} + \mathrm{Br}^-$$

1. The nucleophile ($\mathrm{OH}^-$) approaches the carbon bearing the leaving group ($\mathrm{Br}$) from the side opposite to the C--Br bond (backside attack).
2. A pentacoordinate transition state forms in which the carbon is simultaneously partially bonded to both the incoming nucleophile and the outgoing leaving group.
3. The C--Br bond breaks fully as the C--OH bond forms fully.

Stereochemistry: Inversion of configuration at the carbon centre (Walden inversion). If the carbon is chiral, the product has the opposite absolute configuration to the reactant. The transition state has trigonal bipyramidal geometry.

Favouring factors:

• Primary halogenoalkanes (minimal steric hindrance to backside attack).
• Strong nucleophiles (e.g. $\mathrm{OH}^-$, $\mathrm{CN}^-$, $\mathrm{NH}_3$).
• Polar aprotic solvents (e.g. acetone, DMSO, DMF), which solvate cations but not anions, keeping the nucleophile "naked" and reactive.

The SN1 Mechanism

The subscript N1 stands for "substitution, nucleophilic, unimolecular." It is a two-step mechanism involving a carbocation intermediate.

Rate equation: Rate $= k[\mathrm{R-X}]$

The rate depends only on the concentration of the halogenoalkane. The nucleophile does not appear in the rate equation because it participates only in the fast second step.

Mechanism for the reaction of 2-bromo-2-methylpropane with hydroxide:

Step 1 (slow, rate-determining): Heterolytic fission of the C--Br bond to form a carbocation:

$$(\mathrm{CH}_3)_3\mathrm{CBr} \to (\mathrm{CH}_3)_3\mathrm{C}^+ + \mathrm{Br}^-$$

Step 2 (fast): Nucleophilic attack on the carbocation:

$$(\mathrm{CH}_3)_3\mathrm{C}^+ + \mathrm{OH}^- \to (\mathrm{CH}_3)_3\mathrm{COH}$$

Stereochemistry: Racemisation at the chiral centre. The planar carbocation intermediate can be attacked from either face with equal probability, producing a 50:50 mixture of enantiomers (racemate). In practice, some retention of configuration is often observed because the leaving group partially shields one face (ion pair effect).

Favouring factors:

• Tertiary halogenoalkanes (stable carbocation intermediate).
• Weak nucleophiles (e.g. $\mathrm{H}_2\mathrm{O}$, $\mathrm{ROH}$).
• Polar protic solvents (e.g. water, ethanol), which stabilise the carbocation and the leaving group through solvation.

SN2 vs SN1 Summary

Feature	SN2	SN1
Steps	One (concerted)	Two (stepwise)
Rate law	$k[\mathrm{RX}][\mathrm{Nu}]$	$k[\mathrm{RX}]$
Stereochemistry	Inversion	Racemisation
Substrate preference	Primary $>$ secondary	Tertiary $>$ secondary
Nucleophile	Strong, concentrated	Weak, dilute
Solvent	Polar aprotic	Polar protic
Intermediate	None (transition state)	Carbocation

Secondary halogenoalkanes can react by either mechanism, depending on conditions.

Reactivity Trends

Effect of the Halogen

Bond dissociation enthalpy: C--F ($485\,\mathrm{kJ/mol}$) $>$ C--Cl ($339\,\mathrm{kJ/mol}$) $>$ C--Br ($276\,\mathrm{kJ/mol}$) $>$ C--I ($238\,\mathrm{kJ/mol}$).

Reactivity in nucleophilic substitution: $\mathrm{RI} \gt \mathrm{RBr} \gt \mathrm{RCl} \gg \mathrm{RF}$.

The C--I bond is weakest and iodine is the best leaving group (least basic, largest, most polarisable). Fluorine forms an extremely strong bond and is a very poor leaving group ($\mathrm{F}^-$ is a strong base).

Effect of Substrate Structure (SN2)

Primary $>$ secondary $>$ tertiary. Steric hindrance impedes the backside attack required for SN2. Tertiary halogenoalkanes essentially do not undergo SN2.

Effect of Substrate Structure (SN1)

Tertiary $>$ secondary $>$ primary. The rate-determining step forms a carbocation, whose stability increases with substitution (hyperconjugation and inductive effects). Primary carbocations are too unstable to form under normal conditions, so primary halogenoalkanes do not undergo SN1.

Elimination Reactions

When a strong, bulky base (e.g. $\mathrm{OH}^-$ in ethanol, $\mathrm{KOH}$ in ethanol) reacts with a halogenoalkane at elevated temperature, elimination competes with substitution:

$$\mathrm{CH}_3\mathrm{CHBrCH}_3 + \mathrm{OH}^- \to \mathrm{CH}_3\mathrm{CH}=\mathrm{CH}_2 + \mathrm{H}_2\mathrm{O} + \mathrm{Br}^-$$

Factors Favouring Elimination Over Substitution

• High temperature: Elimination has a higher activation energy than substitution (breaking two $\sigma$ bonds vs one), so increasing temperature favours elimination (Arrhenius equation: $E_a$ effects dominate at higher $T$).
• Strong, bulky base: Bulky bases (e.g. $\mathrm{KO}^\mathrm{t}\mathrm{Bu}$, potassium tert-butoxide) are hindered from the backside attack required for SN2 and instead abstract a proton.
• Tertiary substrate: Tertiary halogenoalkanes are heavily sterically hindered for SN2 and form stable carbocations for E1/E2.

Zaitsev's Rule

When multiple elimination products are possible, the major product is the more substituted alkene (the alkene with the greater number of alkyl groups on the C=C double bond). This is because more substituted alkenes are thermodynamically more stable (hyperconjugation).

For 2-bromobutane: elimination gives a mixture of but-1-ene (minor) and but-2-ene (major).

Environmental Impact of CFCs and Ozone Depletion

Chlorofluorocarbons (CFCs)

CFCs (e.g. $\mathrm{CCl}_3\mathrm{F}$, CFC-11; $\mathrm{CCl}_2\mathrm{F}_2$, CFC-12) were widely used as refrigerants, propellants, and blowing agents. Their advantages (non-toxic, non-flammable, chemically inert in the troposphere) led to massive global production.

The chemical inertness that makes CFCs useful also makes them environmentally persistent. They survive in the troposphere for decades and eventually diffuse to the stratosphere, where they are photolysed by UV radiation of wavelength $\lt 315\,\mathrm{nm}$.

The Ozone Depletion Mechanism

Initiation: UV photolysis of a CFC releases a chlorine radical:

$$\mathrm{CCl}_3\mathrm{F} \xrightarrow{\mathrm{UV}} \mathrm{CCl}_2\mathrm{F} + \mathrm{Cl}^\bullet$$

Propagation -- catalytic destruction of ozone:

$$\mathrm{Cl}^\bullet + \mathrm{O}_3 \to \mathrm{ClO}^\bullet + \mathrm{O}_2$$ $$\mathrm{ClO}^\bullet + \mathrm{O} \to \mathrm{Cl}^\bullet + \mathrm{O}_2$$

Net reaction: $\mathrm{O}_3 + \mathrm{O} \to 2\mathrm{O}_2$

The chlorine radical is regenerated in the second step, acting as a catalyst. A single chlorine atom can destroy up to $100,000$ ozone molecules before being removed from the catalytic cycle.

Consequences

Stratospheric ozone absorbs harmful UV-B radiation ($280$--$315\,\mathrm{nm}$). Ozone depletion increases UV-B flux at the Earth's surface, causing increased rates of skin cancer, cataracts, and damage to marine phytoplankton (the base of the oceanic food chain).

The Antarctic ozone hole (first detected in 1985) forms during the Antarctic spring due to heterogeneous reactions on polar stratospheric cloud (PSC) surfaces that convert inactive chlorine reservoirs ($\mathrm{HCl}$, $\mathrm{ClONO}_2$) into active $\mathrm{Cl}_2$ and $\mathrm{HOCl}$.

Montreal Protocol

The Montreal Protocol (1987, amended multiple times) phased out the production of CFCs. Substitute compounds include HCFCs (hydrochlorofluorocarbons, which have some chlorine but also hydrogen, making them more reactive in the troposphere and less likely to reach the stratosphere) and HFCs (hydrofluorocarbons, which contain no chlorine and do not deplete ozone but are potent greenhouse gases).

Green Chemistry Alternatives

The Montreal Protocol is widely regarded as the most successful international environmental agreement. Key milestones:

• 1987: Original protocol -- 50% reduction in CFC production by 2000.
• 1990 (London Amendment): Complete phase-out of CFCs by 2000.
• 1992 (Copenhagen Amendment): Accelerated phase-out; HCFCs to be phased out by 2030.
• 2007: Accelerated HCFC phase-out for developing countries.

Current replacements include:

Compound class	Example	Ozone impact	Global warming impact
HFCs	$\mathrm{CF}_3\mathrm{CH}_2\mathrm{F}$ (HFC-134a)	Zero (no chlorine)	High (GWP $\approx 1430$)
HFOs (hydrofluoroolefins)	$\mathrm{CF}_3\mathrm{CF}=\mathrm{CH}_2$ (HFO-1234yf)	Zero	Low (GWP $\approx 4$)
Natural refrigerants	$\mathrm{NH}_3$, $\mathrm{CO}_2$, hydrocarbons	Zero	Zero or low

HFOs are the latest generation of refrigerants, designed to have both zero ozone depletion potential and very low global warming potential.

Hydrolysis of Halogenoalkanes in Detail

Experimental Determination of Rate

The rate of hydrolysis of halogenoalkanes can be followed by measuring the rate of appearance of halide ions (e.g. using $\mathrm{AgNO}_3$ to precipitate the halide):

$$\mathrm{R-X} + \mathrm{H}_2\mathrm{O} \to \mathrm{R-OH} + \mathrm{X}^-$$

Adding $\mathrm{AgNO}_3(aq)$ produces a precipitate ($\mathrm{AgCl}$, $\mathrm{AgBr}$, or $\mathrm{AgI}$) whose appearance can be timed. The time for the precipitate to appear is inversely proportional to the rate of hydrolysis.

Halogenoalkane	Time for $\mathrm{AgNO}_3$ precipitate	Relative rate
1-chlorobutane	Slowest	1
1-bromobutane	Intermediate	50
1-iodobutane	Fastest	10,000

The rate differences correlate with C--X bond strength: $\mathrm{C}-\mathrm{I} \lt \mathrm{C}-\mathrm{Br} \lt \mathrm{C}-\mathrm{Cl}$.

Effect of Solvent on SN1 vs SN2

The choice of solvent can determine which mechanism operates:

• Polar protic solvents (water, ethanol, carboxylic acids) stabilise ions through hydrogen bonding. They stabilise the carbocation intermediate and the leaving group, favouring SN1.
• Polar aprotic solvents (acetone, DMSO, DMF, acetonitrile) solvate cations strongly but do not solvate anions well. The nucleophile remains relatively unsolvated and highly reactive, favouring SN2.

Ozone Depletion and the Montreal Protocol

The Catalytic Cycle

In the stratosphere, UV radiation breaks CFCs into chlorine radicals:

$$\mathrm{CFCl}_3 \xrightarrow{\mathrm{UV}} \mathrm{CFCl}_2\cdot + \mathrm{Cl}\cdot$$

The chlorine radical catalytically destroys ozone:

$$\mathrm{Cl}\cdot + \mathrm{O}_3 \to \mathrm{ClO}\cdot + \mathrm{O}_2$$ $$\mathrm{ClO}\cdot + \mathrm{O}_3 \to \mathrm{Cl}\cdot + 2\mathrm{O}_2$$

Net reaction: $2\mathrm{O}_3 \to 3\mathrm{O}_2$

One chlorine radical can destroy approximately $100\,000$ ozone molecules before being removed from the cycle (by forming $\mathrm{HCl}$, which diffuses back to the troposphere).

Why CFCs Are Particularly Damaging

CFCs are inert in the troposphere (no reactions with $\mathrm{OH}$ radicals or water). They have atmospheric lifetimes of 50--100 years, allowing them to reach the stratosphere intact. Once there, UV photolysis releases the chlorine radical. The combination of stability in the lower atmosphere and reactivity in the upper atmosphere makes CFCs uniquely harmful.

Montreal Protocol Milestones

Year	Milestone
1985	Vienna Convention for the Protection of the Ozone Layer
1987	Montreal Protocol signed -- 50% cut in CFC production by 2000
1990	London Amendment -- complete phase-out of CFCs by 2000
1992	Copenhagen Amendment -- accelerated phase-out to 1996
2007	Accelerator Amendment -- HCFC phase-out by 2030 for developed nations
2016	Kigali Amendment -- phase-down of HFCs (potent greenhouse gases)

Green Chemistry Alternatives

Compound	ODP	GWP	Notes
$\mathrm{CFCl}_3$ (CFC-11)	1.0	4,660	Banned
$\mathrm{CF}_2\mathrm{Cl}_2$ (CFC-12)	1.0	10,900	Banned
$\mathrm{CHF}_2\mathrm{Cl}$ (HCFC-22)	0.05	1,810	Being phased out
$\mathrm{CH}_2\mathrm{FCF}_3$ (HFC-134a)	0	1,430	Replacement; high GWP
$\mathrm{CF}_3\mathrm{CH}=\mathrm{CH}_2$ (HFO-1234yf)	0	4	Next-generation; low GWP

ODP = Ozone Depletion Potential (relative to CFC-11). GWP = Global Warming Potential (100-year, relative to $\mathrm{CO}_2$).

Hydrolysis Rate Data

Halogenoalkane	Relative rate of hydrolysis with $\mathrm{NaOH}(aq)$
$\mathrm{CH}_3\mathrm{CH}_2\mathrm{CH}_2\mathrm{I}$	Fastest
$\mathrm{CH}_3\mathrm{CH}_2\mathrm{CH}_2\mathrm{Br}$	Fast
$\mathrm{CH}_3\mathrm{CH}_2\mathrm{CH}_2\mathrm{Cl}$	Slow
$\mathrm{CH}_3\mathrm{CH}_2\mathrm{CH}_2\mathrm{F}$	Very slow (practically inert)

Solvent Effects on SN1 vs SN2

Solvent type	Effect on SN1	Effect on SN2
Polar protic (water, ethanol)	Favours -- stabilises carbocation and leaving group	Disfavoured -- solvates nucleophile, reducing its reactivity
Polar aprotic (acetone, DMSO, DMF)	Disfavoured -- carbocation poorly stabilised	Favoured -- nucleophile not solvated, remains highly reactive

This explains why $\mathrm{NaOH}$ in water favours substitution, while $\mathrm{NaOH}$ in ethanol favours elimination.

Common Pitfalls

1. Confusing SN1 and SN2 conditions. Memorise the table: primary/SN2, tertiary/SN1, strong nucleophile/SN2, weak nucleophile/SN1, polar aprotic/SN2, polar protic/SN1.

2. Writing the wrong stereochemical outcome. SN2 gives inversion; SN1 gives racemisation. These are distinct predictions that can be tested experimentally.

3. Assuming CFCs react in the troposphere. CFCs are inert in the troposphere. Their environmental damage occurs only in the stratosphere, where UV radiation has sufficient energy to break the C--Cl bond.

4. Forgetting Zaitsev's rule in elimination. When asked to predict the major elimination product, identify the most substituted alkene.

5. Writing $\mathrm{OH}^-$ in ethanol for SN2. $\mathrm{OH}^-$ in aqueous solution favours substitution (SN2 for primary substrates). $\mathrm{OH}^-$ in ethanol at elevated temperature favours elimination. The solvent matters.

Practice Problems

Problem 1

For each of the following reactions, identify the mechanism (SN1, SN2, E1, or E2) and predict the major organic product:

(a) $\mathrm{CH}_3\mathrm{CH}_2\mathrm{CH}_2\mathrm{Br}$ with $\mathrm{NaOH}$ in aqueous solution at $25^\circ\mathrm{C}$.

(b) $(\mathrm{CH}_3)_3\mathrm{CBr}$ with $\mathrm{NaOH}$ in ethanol at $80^\circ\mathrm{C}$.

(c) $\mathrm{CH}_3\mathrm{CHBrCH}_3$ with $\mathrm{KOH}$ in ethanol at $60^\circ\mathrm{C}$.

Solution:

(a) Primary halogenoalkane, strong nucleophile, aqueous solution (polar protic solvent), low temperature. SN2. Product: $\mathrm{CH}_3\mathrm{CH}_2\mathrm{CH}_2\mathrm{OH}$ (propan-1-ol).

(b) Tertiary halogenoalkane, strong base, ethanol solvent, high temperature. E2 (elimination strongly favoured over SN1 because of the high temperature and strong base). Product: $\mathrm{CH}_3\mathrm{C}(\mathrm{CH}_3)=\mathrm{CH}_2$ (2-methylpropene).

(c) Secondary halogenoalkane, strong base in ethanol, elevated temperature. Both SN2 and E2 compete. At elevated temperature, E2 predominates. Product (by Zaitsev's rule): $\mathrm{CH}_3\mathrm{CH}=\mathrm{CHCH}_3$ (but-2-ene, the more substituted alkene) as major product, with some $\mathrm{CH}_3\mathrm{CH}_2\mathrm{CH}=\mathrm{CH}_2$ (but-1-ene) as minor product.

Problem 2

Explain why the rate of hydrolysis of 1-chlorobutane is much slower than that of 1-iodobutane under the same conditions.

Solution:

The mechanism is SN2 (primary substrate). The rate-determining step involves breaking the C--X bond. The C--Cl bond dissociation enthalpy ($339\,\mathrm{kJ/mol}$) is significantly greater than the C--I bond dissociation enthalpy ($238\,\mathrm{kJ/mol}$). The stronger C--Cl bond has a higher activation energy for cleavage, giving a slower reaction rate.

Additionally, $\mathrm{Cl}^-$ is a poorer leaving group than $\mathrm{I}^-$ because $\mathrm{Cl}^-$ is a stronger base (more willing to accept a proton, less willing to depart with its lone pair). The larger, more polarisable $\mathrm{I}^-$ stabilises the departing negative charge more effectively.

Problem 3

Propose a synthesis of butanenitrile ($\mathrm{CH}_3\mathrm{CH}_2\mathrm{CH}_2\mathrm{CN}$) starting from butan-1-ol. State all reagents and conditions.

Solution:

Step 1: Convert butan-1-ol to 1-bromobutane using $\mathrm{PBr}_3$ (or concentrated $\mathrm{HBr}$):

$$\mathrm{CH}_3\mathrm{CH}_2\mathrm{CH}_2\mathrm{CH}_2\mathrm{OH} \xrightarrow{\mathrm{PBr}_3} \mathrm{CH}_3\mathrm{CH}_2\mathrm{CH}_2\mathrm{CH}_2\mathrm{Br}$$

Step 2: Nucleophilic substitution with cyanide ion:

$$\mathrm{CH}_3\mathrm{CH}_2\mathrm{CH}_2\mathrm{CH}_2\mathrm{Br} \xrightarrow{\mathrm{KCN},\,\text{ethanol, reflux}} \mathrm{CH}_3\mathrm{CH}_2\mathrm{CH}_2\mathrm{CH}_2\mathrm{CN}$$

The mechanism is SN2 (primary halogenoalkane). The cyanide ion attacks from the opposite side of the C--Br bond, giving the nitrile with an extended carbon chain (one extra carbon). $\mathrm{KCN}$ is used instead of $\mathrm{NaCN}$ because $\mathrm{KCN}$ is more soluble in ethanol. The reaction is heated under reflux to increase the rate.

Problem 4

Explain why tertiary halogenoalkanes undergo elimination rather than substitution with a strong base in ethanol, whereas primary halogenoalkanes undergo substitution. Discuss the role of the solvent.

Solution:

Tertiary halogenoalkanes form stable carbocation intermediates (SN1/E1). With a strong base ($\mathrm{OH}^-$) at elevated temperature, the elimination pathway (E1 or E2) is favoured because:

1. The carbocation intermediate is readily formed (stable tertiary carbocation).
2. The base is present in high concentration and can abstract a $\beta$-hydrogen competitively.
3. At elevated temperature, elimination is entropically favoured (produces more particles: alkene + water vs alcohol).

Primary halogenoalkanes cannot form stable carbocations (the primary carbocation is too high in energy). They react via a concerted SN2 mechanism in which the nucleophile attacks as the leaving group departs. In a polar protic solvent (water), the nucleophile ($\mathrm{OH}^-$) is the strongest available nucleophile and substitution predominates.

Role of solvent:

• Aqueous solution (polar protic): Favours substitution. The solvent stabilises the transition state and the ionic products. The nucleophile ($\mathrm{OH}^-$) is well-solvated but still the strongest nucleophile present.
• Ethanol (polar protic but less polar than water): Favours elimination. The lower polarity reduces the nucleophilicity of $\mathrm{OH}^-$ and the solvation of the carbocation, while the higher temperature (required for ethanol as solvent) shifts the competition towards elimination.
• Polar aprotic (e.g. DMSO, acetone): Favours SN2. The nucleophile is not solvated and retains its full reactivity.

Worked Examples: SN2 Mechanism in Detail

Example 1: Reaction of 1-bromobutane with NaOH

Reaction: $\mathrm{CH}_3\mathrm{CH}_2\mathrm{CH}_2\mathrm{CH}_2\mathrm{Br} + \mathrm{NaOH}(aq) \to \mathrm{CH}_3\mathrm{CH}_2\mathrm{CH}_2\mathrm{CH}_2\mathrm{OH} + \mathrm{NaBr}$

Mechanism (SN2):

Step 1: The $\mathrm{OH}^-$ ion approaches the carbon bearing the bromine from the side opposite to the C--Br bond (backside attack at $180^\circ$ to the leaving group).

Step 2: As the C--O bond begins to form, the C--Br bond weakens and elongates. A transition state is formed in which the central carbon is partially bonded to both $\mathrm{OH}$ and $\mathrm{Br}$, with the three remaining groups arranged in a trigonal planar geometry.

Transition state: $[\mathrm{HO}\cdots\mathrm{CH}_2\mathrm{CH}_2\mathrm{CH}_2\mathrm{CH}_3\cdots\mathrm{Br}]^{\ddagger}$

Step 3: The C--Br bond breaks completely as the C--O bond forms completely. The product butan-1-ol is formed with inversion of configuration at the carbon centre (Walden inversion).

Rate equation: Rate $= k[\mathrm{CH}_3\mathrm{CH}_2\mathrm{CH}_2\mathrm{CH}_2\mathrm{Br}][\mathrm{OH}^-]$

The reaction is second-order overall (first-order with respect to each reactant).

Example 2: Multi-Step Synthesis via Nucleophilic Substitution

Target: Synthesise 2-aminobutane from butan-2-ol.

Step 1: Convert butan-2-ol to 2-bromobutane using concentrated HBr:

$\mathrm{CH}_3\mathrm{CH}(\mathrm{OH})\mathrm{CH}_2\mathrm{CH}_3 + \mathrm{HBr} \to \mathrm{CH}_3\mathrm{CHBrCH}_2\mathrm{CH}_3 + \mathrm{H}_2\mathrm{O}$

Step 2: Nucleophilic substitution with excess ammonia:

$\mathrm{CH}_3\mathrm{CHBrCH}_2\mathrm{CH}_3 + \mathrm{NH}_3 \to \mathrm{CH}_3\mathrm{CH}(\mathrm{NH}_2)\mathrm{CH}_2\mathrm{CH}_3 + \mathrm{HBr}$

The excess ammonia ensures that the primary amine is the major product, minimising further alkylation to secondary and tertiary amines.

Example 3: Demonstrating the Rate Law Experimentally

Aim: To determine the rate equation for the hydrolysis of 1-bromobutane.

Method:

1. Prepare solutions of 1-bromobutane in ethanol at known concentrations.
2. Prepare solutions of $\mathrm{NaOH}$ at known concentrations.
3. Mix the solutions and start a timer.
4. At regular intervals, quench aliquots in ice-cold water and titrate the remaining $\mathrm{OH}^-$ with standardised $\mathrm{HCl}$ using phenolphthalein indicator.
5. Plot $[\mathrm{OH}^-]$ against time for various initial concentrations.

Expected results:

• When $[\mathrm{1\text{-}bromobutane}]$ is doubled (keeping $[\mathrm{OH}^-]$ constant), the initial rate doubles.
• When $[\mathrm{OH}^-]$ is doubled (keeping $[\mathrm{1\text{-}bromobutane}]$ constant), the initial rate doubles.
• This confirms: Rate $= k[\mathrm{1\text{-}bromobutane}][\mathrm{OH}^-]$ (second-order overall).

Example 4: Competition Between SN1 and E1

Reaction: 2-bromo-2-methylpropane with ethanol at $50^\circ\mathrm{C}$.

$\mathrm{(CH}_3)_3\mathrm{CBr} \xrightarrow{\mathrm{EtOH},\,50^\circ\mathrm{C}} \mathrm{(CH}_3)_2\mathrm{C}=\mathrm{CH}_2 + \mathrm{(CH}_3)_3\mathrm{COCH}_2\mathrm{CH}_3$

Both products form via the same carbocation intermediate:

$(\mathrm{CH}_3)_3\mathrm{CBr} \xrightarrow{\text{slow}} (\mathrm{CH}_3)_3\mathrm{C}^+ + \mathrm{Br}^-$

E1 pathway (major): The carbocation loses a proton to the solvent:

$(\mathrm{CH}_3)_3\mathrm{C}^+ \xrightarrow{-\mathrm{H}^+} (\mathrm{CH}_3)_2\mathrm{C}=\mathrm{CH}_2 \text{ (2-methylpropene)}$

SN1 pathway (minor): The carbocation is attacked by ethanol:

$(\mathrm{CH}_3)_3\mathrm{C}^+ + \mathrm{CH}_3\mathrm{CH}_2\mathrm{OH} \to (\mathrm{CH}_3)_3\mathrm{COCH}_2\mathrm{CH}_3 \text{ (2-ethoxy-2-methylpropane)}$

At elevated temperature, elimination predominates (higher activation energy pathway favoured by Arrhenius equation).

Example 5: Hydrolysis Rate Comparison Using AgNO3

Aim: Compare the rates of hydrolysis of 1-chlorobutane, 1-bromobutane, and 1-iodobutane.

Method:

1. Prepare three test tubes each containing $5\,\mathrm{cm}^3$ of ethanol and $1\,\mathrm{cm}^3$ of the halogenoalkane.
2. Place all three in a water bath at $60^\circ\mathrm{C}$.
3. Add $2\,\mathrm{cm}^3$ of $\mathrm{AgNO}_3(aq)$ to each test tube simultaneously.
4. Record the time for the first appearance of a precipitate.

Results:

Halogenoalkane	Time for precipitate	Identity of precipitate
1-chlorobutane	Very slow (minutes)	$\mathrm{AgCl}$ (white)
1-bromobutane	Moderate (seconds)	$\mathrm{AgBr}$ (cream)
1-iodobutane	Fast (immediate)	$\mathrm{AgI}$ (yellow)

Conclusion: The rate of hydrolysis increases in the order $\mathrm{C}\text{-}\mathrm{Cl} \lt \mathrm{C}\text{-}\mathrm{Br} \lt \mathrm{C}\text{-}\mathrm{I}$, consistent with the decreasing bond enthalpy: $\mathrm{C}\text{-}\mathrm{Cl} \,(339\,\mathrm{kJ/mol}) \gt \mathrm{C}\text{-}\mathrm{Br} \,(276\,\mathrm{kJ/mol}) \gt \mathrm{C}\text{-}\mathrm{I} \,(238\,\mathrm{kJ/mol})$.

Example 6: Using KCN for Chain Extension

Target: Convert bromoethane to butanoic acid (extending the carbon chain by two carbons).

Step 1: Nucleophilic substitution with $\mathrm{KCN}$:

$\mathrm{CH}_3\mathrm{CH}_2\mathrm{Br} + \mathrm{KCN} \to \mathrm{CH}_3\mathrm{CH}_2\mathrm{CN} + \mathrm{KBr}$

Step 2: Hydrolysis of the nitrile to carboxylic acid (reflux with dilute $\mathrm{HCl}$):

$\mathrm{CH}_3\mathrm{CH}_2\mathrm{CN} + 2\mathrm{H}_2\mathrm{O} + \mathrm{HCl} \xrightarrow{\text{reflux}} \mathrm{CH}_3\mathrm{CH}_2\mathrm{COOH} + \mathrm{NH}_4\mathrm{Cl}$

Mechanism of Step 1 (SN2): The $\mathrm{CN}^-$ ion attacks the electrophilic carbon of bromoethane from the opposite side of the C--Br bond. The C--Br bond breaks as the C--CN bond forms, giving propanenitrile. The chain has been extended by one carbon.

Note: Using $\mathrm{KCN}$ rather than $\mathrm{NaCN}$ is preferred because $\mathrm{KCN}$ has greater solubility in ethanol, the typical reaction solvent. The use of $\mathrm{KCN}$ (or $\mathrm{NaCN}$) in ethanol ensures the cyanide ion is available as a free nucleophile.

Example 7: SN1 Stereochemistry

Reaction: Hydrolysis of (R)-2-bromobutane in aqueous ethanol.

$\mathrm{CH}_3\mathrm{CHBrCH}_2\mathrm{CH}_3 + \mathrm{H}_2\mathrm{O} \to \mathrm{CH}_3\mathrm{CH}(\mathrm{OH})\mathrm{CH}_2\mathrm{CH}_3 + \mathrm{HBr}$

Mechanism (SN1):

Step 1 (slow): Heterolytic cleavage of the C--Br bond forms a planar secondary carbocation:

$\mathrm{CH}_3\mathrm{CHBrCH}_2\mathrm{CH}_3 \to \mathrm{CH}_3\overset◆LB◆+◆RB◆◆LB◆\mathrm{C}◆RB◆\mathrm{HCH}_2\mathrm{CH}_3 + \mathrm{Br}^-$

Step 2 (fast): Water attacks the planar carbocation from either face with equal probability:

$\mathrm{CH}_3\overset◆LB◆+◆RB◆◆LB◆\mathrm{C}◆RB◆\mathrm{HCH}_2\mathrm{CH}_3 + \mathrm{H}_2\mathrm{O} \to \mathrm{CH}_3\mathrm{CH}(\mathrm{OH}_2^+)\mathrm{CH}_2\mathrm{CH}_3$

Step 3: Deprotonation gives the alcohol:

$\mathrm{CH}_3\mathrm{CH}(\mathrm{OH}_2^+)\mathrm{CH}_2\mathrm{CH}_3 \to \mathrm{CH}_3\mathrm{CH}(\mathrm{OH})\mathrm{CH}_2\mathrm{CH}_3 + \mathrm{H}^+$

Stereochemistry: The product is a racemic mixture of (R)-butan-2-ol and (S)-butan-2-ol. The planar carbocation intermediate can be attacked with equal probability from either face, resulting in racemisation.

Practical Techniques for Halogenoalkane Reactions

Required Practical: Investigating the Hydrolysis of Halogenoalkanes (AQA RP 8)

Objective: To compare the rates of hydrolysis of primary halogenoalkanes with different halogen atoms.

Apparatus: Test tubes, water bath, thermometer, stopwatch, pipettes, 1-chlorobutane, 1-bromobutane, 1-iodobutane, ethanol, silver nitrate solution ($0.1\,\mathrm{mol/dm}^3$).

Safety: Halogenoalkanes are irritants. Wear eye protection and gloves. Ethanol is flammable. Silver nitrate stains skin.

Procedure:

1. Add $2\,\mathrm{cm}^3$ of ethanol to each of three test tubes labelled Cl, Br, I.
2. Add 5 drops of the appropriate halogenoalkane to each tube.
3. Place all three tubes in a water bath maintained at $60^\circ\mathrm{C}$.
4. Add $1\,\mathrm{cm}^3$ of $\mathrm{AgNO}_3(aq)$ to each tube and start the stopwatch.
5. Record the time for the first permanent appearance of a precipitate in each tube.

Key control variables: Temperature (use the same water bath), volume of ethanol (same solvent environment), concentration of $\mathrm{AgNO}_3$ (same in each tube), volume of halogenoalkane.

Sources of error:

• Subjectivity in judging the first appearance of precipitate. Use a white tile behind the test tubes and define the endpoint clearly.
• Temperature fluctuations. Use a thermostatically controlled water bath.
• Evaporation of volatile halogenoalkanes. Keep test tubes stoppered between readings.

Purification of Organic Products After Nucleophilic Substitution

After a nucleophilic substitution reaction, the crude product typically contains unreacted starting materials, solvent, inorganic salts, and the desired product. A standard purification procedure:

1. Separation: Transfer the reaction mixture to a separating funnel. If the product is organic, add water and a suitable organic solvent (e.g. diethyl ether). Shake and allow the layers to separate. Collect the organic layer.
2. Washing: Wash the organic layer with water to remove water-soluble impurities, then with sodium hydrogencarbonate solution to remove any acidic impurities (caution: $\mathrm{CO}_2$ evolution), then with brine (saturated $\mathrm{NaCl}$) to remove residual water.
3. Drying: Add a drying agent (anhydrous $\mathrm{MgSO}_4$ or $\mathrm{Na}_2\mathrm{SO}_4$) to the organic layer. Swirl and allow to stand for 10 minutes. Filter off the drying agent.
4. Distillation: Purify the product by simple distillation, collecting the fraction that boils at the expected boiling point of the product.

Exam-Style Questions with Full Mark Schemes

Q1 (6 marks)

Bromoethane reacts with aqueous sodium hydroxide to form ethanol.

(a) Write the mechanism for this reaction. (4 marks)

(b) Explain how the rate of this reaction would change if the concentration of both reactants were doubled. (2 marks)

Mark Scheme:

(a) 4 marks:

• Curly arrow from lone pair/negative charge on $\mathrm{OH}^-$ to the C atom bonded to Br (1 mark).
• Curly arrow from the middle of the C--Br bond to the Br atom (1 mark).
• Correct transition state or intermediate showing partial bonds (1 mark).
• Correct product $\mathrm{CH}_3\mathrm{CH}_2\mathrm{OH}$ and $\mathrm{Br}^-$ (1 mark).

(b) 2 marks:

• The reaction is second-order (first-order with respect to each reactant) (1 mark).
• Doubling both concentrations increases the rate by a factor of $2 \times 2 = 4$ (1 mark).

Q2 (5 marks)

Explain why the hydrolysis of 2-chloro-2-methylpropane is faster than the hydrolysis of 1-chlorobutane, even though the C--Cl bond in 1-chlorobutane is weaker.

Mark Scheme:

5 marks:

• 2-chloro-2-methylpropane reacts via SN1 (1 mark).
• 1-chlorobutane reacts via SN2 (1 mark).
• The SN1 rate-determining step involves formation of a tertiary carbocation, which is stabilised by hyperconjugation and inductive effects from three methyl groups (1 mark).
• The primary carbocation required for 1-chlorobutane to react via SN1 is too unstable to form (1 mark).
• Therefore, 1-chlorobutane must react via SN2, which has a higher activation energy than the SN1 pathway available to the tertiary halogenoalkane (1 mark).

Q3 (7 marks)

A student carried out an experiment to determine the order of reaction with respect to hydroxide ions for the hydrolysis of 1-bromobutane. The following data were obtained at constant temperature, with the concentration of 1-bromobutane kept constant:

$[\mathrm{OH}^-]$ ($\mathrm{mol\,dm^{-3}}$)	Initial rate ($\mathrm{mol\,dm^{-3}\,s^{-1}}$)
0.020	$4.0 \times 10^{-5}$
0.040	$8.0 \times 10^{-5}$
0.060	$1.2 \times 10^{-4}$
0.080	$1.6 \times 10^{-4}$

(a) Deduce the order of reaction with respect to hydroxide ions. (2 marks)

(b) Calculate the value of the rate constant, including units. (3 marks)

(c) State the overall order of the reaction. (1 mark)

(d) State and explain the effect on the rate of doubling the concentration of 1-bromobutane only. (1 mark)

Mark Scheme:

(a) 2 marks: When $[\mathrm{OH}^-]$ doubles, the rate doubles (e.g. $0.020 \to 0.040$: rate $4.0 \to 8.0$). This first-order relationship holds for all data pairs (1 mark). Therefore the reaction is first-order with respect to $\mathrm{OH}^-$ (1 mark).

(b) 3 marks: From rate $= k[\mathrm{1\text{-}bromobutane}]^1[\mathrm{OH}^-]^1$:

$k = \frac◆LB◆\text{rate}◆RB◆◆LB◆[\mathrm{1\text{-}bromobutane}][\mathrm{OH}^-]◆RB◆$

Using the first data point and assuming $[\mathrm{1\text{-}bromobutane}] = c$:

$k = \frac◆LB◆4.0 \times 10^{-5}◆RB◆◆LB◆c \times 0.020◆RB◆ = \frac◆LB◆2.0 \times 10^{-3}◆RB◆◆LB◆c◆RB◆$

Units: $\frac◆LB◆\mathrm{mol\,dm^{-3}\,s^{-1}}◆RB◆◆LB◆\mathrm{mol\,dm^{-3}} \times \mathrm{mol\,dm^{-3}}◆RB◆ = \mathrm{mol^{-1}\,dm^3\,s^{-1}}$ (1 mark for correct calculation, 1 mark for correct units, 1 mark for method).

(c) 1 mark: Second-order overall (first-order with respect to each reactant).

(d) 1 mark: The rate would double (1 mark) because the rate equation is first-order with respect to 1-bromobutane (the rate is directly proportional to its concentration).

Q4 (6 marks)

Describe an experiment to compare the rates of hydrolysis of 1-chlorobutane, 1-bromobutane, and 1-iodobutane. Explain why the results would differ.

Mark Scheme:

6 marks:

Method (3 marks):

• Equal volumes of ethanol and equal volumes of each halogenoalkane in separate test tubes (1 mark).
• Heat in the same water bath at the same temperature (1 mark).
• Add equal volumes of $\mathrm{AgNO}_3(aq)$ simultaneously and time the appearance of precipitate (1 mark).

Explanation (3 marks):

• C--I bond is weakest ($238\,\mathrm{kJ/mol}$) so hydrolysis is fastest (1 mark).
• C--Cl bond is strongest ($339\,\mathrm{kJ/mol}$) so hydrolysis is slowest (1 mark).
• The rate-determining step involves breaking the C--X bond, so bond enthalpy determines the rate (1 mark).

Q5 (5 marks)

Propose a mechanism for the reaction of 2-bromo-2-methylpropane with warm aqueous ethanol, explaining why a mixture of products is formed.

Mark Scheme:

5 marks:

• Rate-determining step: heterolytic fission of C--Br bond to form tertiary carbocation and $\mathrm{Br}^-$ (1 mark for equation, 1 mark for identification as slow step).
• Substitution product: nucleophilic attack by ethanol on the carbocation, followed by proton loss to give the ether (1 mark).
• Elimination product: loss of a proton from the carbocation to give 2-methylpropene (1 mark).
• The carbocation intermediate is attacked by either ethanol (substitution) or loses a proton (elimination), giving a mixture (1 mark).

Retrosynthesis with Halogenoalkanes

Retrosynthetic analysis involves working backwards from the target molecule to identify suitable starting materials and reagents. Halogenoalkanes are versatile intermediates in retrosynthesis because they can be prepared from alcohols and can be converted to many other functional groups.

General Retrosynthetic Disconnections for Halogenoalkanes

Target	Disconnection	Equivalent synthons	Forward reagents
Alcohol $\mathrm{R-OH}$	$\mathrm{C}\text{-}\mathrm{OH} \to \mathrm{C}\text{-}\mathrm{X}$	$\mathrm{R-X}$ + $\mathrm{OH}^-$	$\mathrm{NaOH}(aq)$, SN2
Nitrile $\mathrm{R-CN}$	$\mathrm{C}\text{-}\mathrm{CN} \to \mathrm{C}\text{-}\mathrm{X}$	$\mathrm{R-X}$ + $\mathrm{CN}^-$	$\mathrm{KCN}$, ethanol
Amine $\mathrm{R-NH_2}$	$\mathrm{C}\text{-}\mathrm{NH_2} \to \mathrm{C}\text{-}\mathrm{X}$	$\mathrm{R-X}$ + $\mathrm{NH_3}$	Excess $\mathrm{NH_3}$
Ether $\mathrm{R-O-R'}$	$\mathrm{C}\text{-}\mathrm{O} \to \mathrm{C}\text{-}\mathrm{X}$	$\mathrm{R-X}$ + $\mathrm{R'O}^-$	Williamson synthesis
Alkene $\mathrm{C=C}$	$\mathrm{C=C} \to \mathrm{C-C-X}$	Elimination of HX	$\mathrm{KOH}$ in ethanol, $\Delta$

Worked Retrosynthesis Example

Target: 3-phenylpropan-1-amine ($\mathrm{C}_6\mathrm{H}_5\mathrm{CH}_2\mathrm{CH}_2\mathrm{CH}_2\mathrm{NH}_2$)

Retrosynthetic analysis:

1. Disconnect the $\mathrm{C}\text{-}\mathrm{NH}_2$ bond: the amine comes from the nitrile $\mathrm{C}_6\mathrm{H}_5\mathrm{CH}_2\mathrm{CH}_2\mathrm{CN}$ via reduction with $\mathrm{LiAlH}_4$.

2. Disconnect the $\mathrm{C}\text{-}\mathrm{CN}$ bond: the nitrile comes from 3-phenyl-1-bromopropane ($\mathrm{C}_6\mathrm{H}_5\mathrm{CH}_2\mathrm{CH}_2\mathrm{CH}_2\mathrm{Br}$) via SN2 with $\mathrm{KCN}$.

3. Disconnect the $\mathrm{C}\text{-}\mathrm{Br}$ bond: the bromoalkane comes from 3-phenylpropan-1-ol via $\mathrm{PBr}_3$.

4. The alcohol comes from the reduction of 3-phenylpropanal ($\mathrm{C}_6\mathrm{H}_5\mathrm{CH}_2\mathrm{CH}_2\mathrm{CHO}$) with $\mathrm{NaBH}_4$.

5. The aldehyde comes from Friedel-Crafts acylation of benzene with propanoyl chloride ($\mathrm{CH}_3\mathrm{CH}_2\mathrm{COCl}$), followed by Clemmensen reduction.

Forward synthesis:

$\mathrm{C}_6\mathrm{H}_6 \xrightarrow{\mathrm{CH}_3\mathrm{CH}_2\mathrm{COCl},\,\mathrm{AlCl}_3} \mathrm{C}_6\mathrm{H}_5\mathrm{COCH}_2\mathrm{CH}_3 \xrightarrow[\mathrm{Zn(Hg)}]{\mathrm{HCl}} \mathrm{C}_6\mathrm{H}_5\mathrm{CH}_2\mathrm{CH}_2\mathrm{CH}_3$

$\mathrm{C}_6\mathrm{H}_5\mathrm{CH}_2\mathrm{CH}_2\mathrm{CH}_3 \xrightarrow{\mathrm{Br}_2,\,h\nu} \mathrm{C}_6\mathrm{H}_5\mathrm{CH}_2\mathrm{CHBrCH}_3 + \mathrm{C}_6\mathrm{H}_5\mathrm{CHBrCH}_2\mathrm{CH}_3$

Note: Radical bromination gives a mixture. A more controlled route would use the following:

$\mathrm{C}_6\mathrm{H}_5\mathrm{CH}_2\mathrm{CH}_2\mathrm{COOH} \xrightarrow{\mathrm{LiAlH}_4} \mathrm{C}_6\mathrm{H}_5\mathrm{CH}_2\mathrm{CH}_2\mathrm{CH}_2\mathrm{OH} \xrightarrow{\mathrm{PBr}_3} \mathrm{C}_6\mathrm{H}_5\mathrm{CH}_2\mathrm{CH}_2\mathrm{CH}_2\mathrm{Br} \xrightarrow{\mathrm{KCN}} \mathrm{C}_6\mathrm{H}_5\mathrm{CH}_2\mathrm{CH}_2\mathrm{CN} \xrightarrow{\mathrm{LiAlH}_4} \mathrm{C}_6\mathrm{H}_5\mathrm{CH}_2\mathrm{CH}_2\mathrm{CH}_2\mathrm{NH}_2$

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tip

Diagnostic Test Ready to test your understanding of Halogenoalkanes? The diagnostic test contains the hardest questions within the A-Level specification for this topic, each with a full worked solution.

Unit tests probe edge cases and common misconceptions. Integration tests combine Halogenoalkanes with other chemistry topics to test synthesis under exam conditions.

See Diagnostic Guide for instructions on self-marking and building a personal test matrix.