Halogenoalkanes, Alcohols — Diagnostic Tests

Unit Tests

UT-1: SN1 vs SN2 Mechanism Comparison

Question:

Consider the reaction of 1-bromobutane and 2-bromo-2-methylpropane with sodium hydroxide.

(a) For each halogenoalkane, state whether the reaction proceeds by SN1 or SN2 and explain why.

(b) Write the rate equation for each reaction and explain how it relates to the mechanism.

(c) Explain the role of the solvent in each mechanism. Why is aqueous ethanol commonly used for both types of reaction?

Solution:

(a) 1-bromobutane (primary halogenoalkane): Reacts via SN2. Primary carbocations are too unstable to form, so the reaction proceeds via a concerted mechanism where the nucleophile attacks as the leaving group departs. The transition state involves a pentacoordinate carbon.

2-bromo-2-methylpropane (tertiary halogenoalkane): Reacts via SN1. The tertiary carbocation formed after loss of Br$^-$ is stabilised by the electron-donating effect of three methyl groups. The carbocation is a viable intermediate.

(b)

1-bromobutane (SN2): $\text{rate} = k[\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{Br}][\text{OH}^-]$

Both the halogenoalkane and the nucleophile are involved in the rate-determining (and only) step. This is second order overall.

2-bromo-2-methylpropane (SN1): $\text{rate} = k[(\text{CH}_3)_3\text{CBr}]$

Only the halogenoalkane is involved in the slow step (loss of Br$^-$ to form the carbocation). The nucleophile ($\text{OH}^-$) reacts in a fast subsequent step. This is first order overall.

(c) Aqueous ethanol is a polar protic solvent, which can dissolve both the ionic nucleophile (NaOH, $\text{OH}^-$) and the organic halogenoalkane. The polarity stabilises the transition state in both mechanisms:

• SN2: The polar solvent stabilises the negatively charged nucleophile and the partial charges in the transition state.
• SN1: The polar solvent stabilises the charged carbocation intermediate and the leaving group ($\text{Br}^-$).

The protic nature (ability to donate hydrogen bonds) is particularly important for SN1, where hydrogen bonding helps stabilise the developing ions.

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UT-2: Elimination Reactions and Competition with Substitution

Question:

2-bromobutane can undergo both substitution and elimination with sodium hydroxide.

(a) Write equations for both the substitution and elimination reactions, identifying the products.

(b) State the conditions that favour elimination over substitution and explain why.

(c) Explain why a tertiary halogenoalkane undergoes elimination more readily than a primary halogenoalkane.

Solution:

(a) Substitution (SN1 or SN2 depending on conditions):

$\text{CH}_3\text{CHBrCH}_2\text{CH}_3 + \text{OH}^- \to \text{CH}_3\text{CH}(\text{OH})\text{CH}_2\text{CH}_3 + \text{Br}^-$

Product: butan-2-ol

Elimination (E1 or E2):

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

Product: but-2-ene (major) and possibly but-1-ene (minor, via removal of a proton from the terminal carbon)

(b) Conditions favouring elimination:

• Concentrated NaOH (or $\text{KOH}$ in ethanol): A high concentration of base favours elimination because the base abstracts a proton (a bimolecular process that benefits from high [base])
• High temperature: Elimination has a higher activation energy than substitution (the transition state for elimination is less stable), so higher temperatures favour elimination (the proportion of product from elimination increases)
• Use of ethanol as solvent rather than water: Ethanol favours elimination because it is less polar than water and stabilises the ionic intermediates of substitution less effectively

Conditions favouring substitution:

• Dilute aqueous NaOH
• Lower temperature
• Water as the main solvent (more polar, stabilises ionic intermediates)

(c) Tertiary halogenoalkanes undergo elimination more readily because:

1. Carbocation stability (E1): Tertiary carbocations are more stable, so the E1 pathway (loss of leaving group first) is accessible. The carbocation can then lose a proton to give the alkene.

2. Steric hindrance (E2): In tertiary halogenoalkanes, the carbon bearing the halogen is surrounded by three alkyl groups, making it sterically hindered. The nucleophile finds it difficult to approach the carbon (disfavouring SN2), but it can readily abstract a proton from an adjacent carbon (favoured in E2).

3. Number of $\beta$-hydrogens: Tertiary halogenoalkanes have more $\beta$-hydrogens available for elimination, increasing the statistical probability of elimination occurring.

4. Saytzeff's rule: Elimination from tertiary halogenoalkanes preferentially gives the more substituted (more stable) alkene, which provides a thermodynamic driving force.

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UT-3: Oxidation of Alcohols and Distinguishing Isomers

Question:

Three alcohols with the molecular formula $\text{C}_4\text{H}_{10}\text{O}$ are: butan-1-ol (primary), butan-2-ol (secondary), and 2-methylpropan-2-ol (tertiary).

(a) Describe the products formed when each alcohol is oxidised with acidified potassium dichromate(VI) under (i) gentle conditions (distillation) and (ii) vigorous conditions (reflux).

(b) A student has an unknown alcohol that is one of these three. Describe a simple chemical test to distinguish between them.

(c) 2-methylpropan-2-ol does not undergo oxidation with acidified dichromate. Explain why.

Solution:

(a)

Butan-1-ol (primary):

• (i) Gentle oxidation (distillation): butanal ($\text{CH}_3\text{CH}_2\text{CH}_2\text{CHO}$)
• (ii) Vigorous oxidation (reflux): butanoic acid ($\text{CH}_3\text{CH}_2\text{CH}_2\text{COOH}$)

Primary alcohols can be oxidised to aldehydes (which can be further oxidised to carboxylic acids). Distillation removes the volatile aldehyde before further oxidation.

Butan-2-ol (secondary):

• (i) and (ii): butanone ($\text{CH}_3\text{COCH}_2\text{CH}_3$)

Secondary alcohols are oxidised to ketones only. Ketones are resistant to further oxidation because there is no C--H bond adjacent to the carbonyl for the oxidising agent to attack. Both gentle and vigorous conditions give the same product.

2-methylpropan-2-ol (tertiary):

• (i) and (ii): No reaction -- tertiary alcohols cannot be oxidised by acidified dichromate.

(b)

Step 1: Add acidified potassium dichromate(VI) (orange solution) and warm.

• Butan-1-ol: The solution turns green (Cr$^{3+}$ formed), and the smell of butanal (or butanoic acid under reflux) is detected.
• Butan-2-ol: The solution turns green (Cr$^{3+}$ formed), but the product (butanone) has a different smell from the aldehyde/acid.
• 2-methylpropan-2-ol: The solution remains orange (no oxidation occurs).

Step 2: To distinguish butan-1-ol from butan-2-ol, test the oxidation product with Tollens' reagent:

• Aldehyde (from butan-1-ol): forms a silver mirror
• Ketone (from butan-2-ol): no reaction

Alternatively, test with Fehling's solution: aldehyde gives a red precipitate; ketone gives no change.

(c) Tertiary alcohols cannot be oxidised because oxidation requires the removal of a hydrogen atom from the carbon bearing the $-$OH group. In a tertiary alcohol, this carbon is bonded to three other carbons and has no C--H bond to remove. The oxidation mechanism involves formation of a C$=$O double bond, which requires losing a hydrogen from the alcohol carbon alongside the $-$OH group. Since tertiary alcohols have no such hydrogen, oxidation is not possible under normal conditions. The only way to break down a tertiary alcohol is via complete combustion or harsh conditions that break C--C bonds.

Integration Tests

IT-1: Reaction Pathway with Mechanism Analysis (with Kinetics)

Question:

1-chlorobutane is converted to butanoic acid via a three-step synthesis.

(a) Propose the three steps, giving reagents, conditions, and equations.

(b) For the step involving the formation of the alcohol, explain whether the mechanism is SN1 or SN2 and describe the evidence that supports this.

(c) The first step has a rate constant $k = 4.0 \times 10^{-5}\,\text{mol}^{-1}\text{ dm}^3\text{ s}^{-1}$ at $25\,^\circ\text{C}$ and $1.6 \times 10^{-4}\,\text{mol}^{-1}\text{ dm}^3\text{ s}^{-1}$ at $35\,^\circ\text{C}$. Calculate the activation energy for this step.

Solution:

(a)

Step 1: 1-chlorobutane to butan-1-ol (nucleophilic substitution)

$\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{Cl} + \text{NaOH}(aq) \to \text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{OH} + \text{NaCl}$

Conditions: Reflux with dilute aqueous NaOH.

Step 2: Butan-1-ol to butanal (oxidation)

$\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{OH} \xrightarrow{[\text{O}], \text{distillation}} \text{CH}_3\text{CH}_2\text{CH}_2\text{CHO} + \text{H}_2\text{O}$

Conditions: Acidified potassium dichromate(VI), distillation.

Step 3: Butanal to butanoic acid (further oxidation)

$\text{CH}_3\text{CH}_2\text{CH}_2\text{CHO} \xrightarrow{[\text{O}], \text{reflux}} \text{CH}_3\text{CH}_2\text{CH}_2\text{COOH}$

Conditions: Acidified potassium dichromate(VI), reflux.

(b) The substitution of 1-chlorobutane with NaOH proceeds via SN2. Evidence:

1. The rate equation would be second order: $\text{rate} = k[\text{1-chlorobutane}][\text{OH}^-]$ (both species involved in rate-determining step).
2. Primary halogenoalkanes cannot form stable carbocations, so SN1 is not feasible.
3. The product has inverted stereochemistry at the chiral centre (if one were present) -- Walden inversion.
4. The rate constant has units of $\text{mol}^{-1}\text{ dm}^3\text{ s}^{-1}$, confirming second-order kinetics.

(c) Using the Arrhenius equation:

$\ln\frac{k_2}{k_1} = \frac{E_a}{R}\left(\frac{1}{T_1} - \frac{1}{T_2}\right)$

$\ln\frac◆LB◆1.6 \times 10^{-4}◆RB◆◆LB◆4.0 \times 10^{-5}◆RB◆ = \frac{E_a}{8.31}\left(\frac{1}{298} - \frac{1}{308}\right)$

$\ln(4.0) = \frac{E_a}{8.31} \times 1.096 \times 10^{-4}$

$1.386 = \frac◆LB◆E_a \times 1.096 \times 10^{-4}◆RB◆◆LB◆8.31◆RB◆$

$E_a = \frac◆LB◆1.386 \times 8.31◆RB◆◆LB◆1.096 \times 10^{-4}◆RB◆ = \frac◆LB◆11.52◆RB◆◆LB◆1.096 \times 10^{-4}◆RB◆ = 105100\,\text{J mol}^{-1} = 105\,\text{kJ mol}^{-1}$

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IT-2: Halogenoalkane Reactivity and Leaving Group Ability (with Electrochemistry)

Question:

The reactivity of halogenoalkanes in nucleophilic substitution follows the order: $\text{RI} \gt \text{RBr} \gt \text{RCl} \gt \text{RF}$.

(a) Explain this trend in terms of bond enthalpy and polarisability.

(b) The hydrolysis of 1-bromopropane with NaOH is a second-order reaction. An experiment shows that doubling the concentration of 1-bromopropane doubles the rate. Explain what would happen to the rate if the temperature is increased from $25\,^\circ\text{C}$ to $50\,^\circ\text{C}$, assuming the activation energy is $85\,\text{kJ mol}^{-1}$.

(c) Explain why fluoroalkanes are much less reactive than iodoalkanes, despite the C--F bond being the most polar.

Solution:

(a) The trend in reactivity is explained by two factors:

Bond enthalpy: C--I ($234\,\text{kJ mol}^{-1}$) $\lt$ C--Br ($285\,\text{kJ mol}^{-1}$) $\lt$ C--Cl ($339\,\text{kJ mol}^{-1}$) $\lt$ C--F ($485\,\text{kJ mol}^{-1}$). The weaker the bond, the easier it is to break in the rate-determining step. C--I bonds are weakest and break most readily.

Polarisability: The iodide ion is large and has a diffuse electron cloud that is easily polarised. In the transition state, the C--I bond is stretched and the developing negative charge on iodine is better stabilised by its polarisability. Fluorine is small and has a concentrated charge that is not easily polarised.

The combined effect of weaker bonds and greater polarisability makes iodoalkanes the most reactive.

(b) Using the Arrhenius equation:

$\frac{k_{50}}{k_{25}} = e^{\frac{E_a}{R}\left(\frac{1}{T_1} - \frac{1}{T_2}\right)} = e^{\frac{85000}{8.31}\left(\frac{1}{298} - \frac{1}{323}\right)} = e^{10229 \times 2.583 \times 10^{-4}} = e^{2.642} = 14.0$

The rate increases by a factor of approximately 14. This dramatic increase shows the exponential dependence of rate on temperature.

(c) Despite the C--F bond being the most polar (largest electronegativity difference), fluoroalkanes are the least reactive because:

1. Bond strength: The C--F bond is very strong ($485\,\text{kJ mol}^{-1}$), making it very difficult to break. Bond enthalpy dominates over bond polarity in determining reactivity.

2. Polarisability: Fluoride is very small with a tightly held electron cloud. It is the least polarisable halide ion, meaning it cannot stabilise the partial negative charge that develops on the leaving group in the transition state.

3. Orbital overlap: The small size of fluorine allows very effective overlap with carbon orbitals, creating a strong, short bond that resists heterolytic fission.

The C--F bond polarity is misleading: a more polar bond does not mean a better leaving group. Leaving group ability depends on how well the leaving group can stabilise the negative charge after departure, which requires polarisability (diffuse charge), not electronegativity (concentrated charge).

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IT-3: Multi-Step Synthesis Involving Alcohol and Halogenoalkane Interconversions (with Organic Chemistry)

Question:

Propose a synthesis of but-2-ene starting from ethanol. Your synthesis should involve at least three steps.

(a) Write equations for each step, including reagents and conditions.

(b) Explain the mechanism of the key elimination step in your synthesis.

(c) A student suggests converting ethanol directly to ethene then dimerising to but-2-ene. Evaluate the feasibility of this approach.

Solution:

(a)

Step 1: Ethanol to bromoethane (substitution)

$\text{CH}_3\text{CH}_2\text{OH} \xrightarrow{\text{PBr}_3\text{ or conc. H}_2\text{SO}_4\text{/NaBr}} \text{CH}_3\text{CH}_2\text{Br}$

Conditions: $\text{PBr}_3$ or concentrated $\text{H}_2\text{SO}_4$ with NaBr.

Step 2: Bromoethane to ethyne (double elimination -- not suitable for A-Level). Alternative route:

Step 1 (revised): Ethanol to ethene (elimination)

$\text{CH}_3\text{CH}_2\text{OH} \xrightarrow{\text{conc. H}_2\text{SO}_4, 170\,^\circ\text{C}} \text{CH}_2=\text{CH}_2 + \text{H}_2\text{O}$

Step 2: Ethene to bromoethane (addition of HBr)

$\text{CH}_2=\text{CH}_2 + \text{HBr} \to \text{CH}_3\text{CH}_2\text{Br}$

Step 3: Bromoethane to ethanol (substitution)

$\text{CH}_3\text{CH}_2\text{Br} + \text{NaOH}(aq) \xrightarrow{\text{reflux}} \text{CH}_3\text{CH}_2\text{OH} + \text{NaBr}$

This doesn't get us to but-2-ene. Let me revise:

Better route using Wurtz-type or Grignard approach (not A-Level):

A-Level feasible route:

Step 1: Ethanol to ethene (elimination)

$\text{CH}_3\text{CH}_2\text{OH} \xrightarrow{\text{conc. H}_2\text{SO}_4, 170\,^\circ\text{C}} \text{CH}_2=\text{CH}_2 + \text{H}_2\text{O}$

Step 2: Ethene to 1,2-dibromoethane (addition)

$\text{CH}_2=\text{CH}_2 + \text{Br}_2 \to \text{CH}_2\text{BrCH}_2\text{Br}$

Step 3: 1,2-dibromoethane to but-2-ene (double elimination)

$\text{CH}_2\text{BrCH}_2\text{Br} + 2\text{KOH}(\text{ethanolic}) \xrightarrow{\text{heat}} \text{CH}_3\text{CH}=\text{CHCH}_3 + 2\text{KBr} + 2\text{H}_2\text{O}$

Conditions: Hot, concentrated, ethanolic KOH (elimination conditions). Two successive eliminations occur.

(b) The key step is the elimination of 1,2-dibromoethane. With hot ethanolic KOH:

First elimination: KOH abstracts a proton from one carbon while Br$^-$ leaves from the adjacent carbon (E2 mechanism), forming bromoethene ($\text{CH}_2=\text{CHBr}$).

Second elimination: A second equivalent of KOH abstracts the proton from the remaining $-$CHBr group while Br$^-$ leaves, forming buta-1,3-diene or rearranging.

Actually, for 1,2-dibromoethane, double elimination gives ethyne (acetylene), not but-2-ene. This approach does not work.

Corrected A-Level feasible synthesis:

The most straightforward A-Level route to but-2-ene from ethanol is not possible in exactly 3 clean steps. The student's suggestion of dimerising ethene is evaluated below.

(c) The student's approach of converting ethanol to ethene then dimerising to but-2-ene is not feasible at A-Level:

• Ethene can be produced from ethanol (dehydration with conc. $\text{H}_2\text{SO}_4$)
• However, dimerisation of ethene to but-2-ene is not a standard A-Level reaction. Industrial dimerisation requires specific catalysts (e.g., nickel complexes) and gives predominantly but-1-ene, not but-2-ene. The reaction conditions are beyond A-Level chemistry.
• Additionally, ethene dimerisation produces but-1-ene, which would then need isomerisation to but-2-ene.

A more practical A-Level approach: start from butan-2-ol (dehydration with $\text{H}_2\text{SO}_4$) to give but-2-ene directly. But if starting from ethanol, the synthesis requires building a 4-carbon chain, which at A-Level level is best done through a Grignard reaction (not on all syllabi) or by using ethanoic acid and reducing.

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Additional Practice Problems

UT-4: Distinguishing Alcohols

Question: Describe a chemical test to distinguish between propan-1-ol, propan-2-ol, and 2-methylpropan-2-ol. State the observations for each.

Solution:

Use acidified potassium dichromate(VI) ($\mathrm{K}_2\mathrm{Cr}_2\mathrm{O}_7/\mathrm{H}^+$) as the oxidising agent (1 mark):

• Propan-1-ol (primary alcohol): Oxidised to propanal (orange to green, can be distilled off before further oxidation). If heated under reflux, oxidised to propanoic acid (orange to green) (1 mark).

• Propan-2-ol (secondary alcohol): Oxidised to propanone (acetone). Orange to green colour change. The ketone product cannot be further oxidised (1 mark).

• 2-Methylpropan-2-ol (tertiary alcohol): No reaction. The solution remains orange. Tertiary alcohols are resistant to oxidation because there is no hydrogen atom on the carbon bearing the OH group (1 mark).

Alternatively, the Lucas test (conc. $\mathrm{HCl}$ with $\mathrm{ZnCl}_2$) distinguishes by reaction rate:

• Tertiary: immediate cloudiness (forms tertiary chloride, insoluble)
• Secondary: cloudiness within 5 minutes
• Primary: no cloudiness at room temperature (1 mark).

UT-5: Elimination Conditions

Question: Explain why using a concentrated solution of $\mathrm{NaOH}$ in ethanol at high temperature favours elimination over substitution for 2-bromobutane.

Solution:

Elimination requires $\mathrm{OH}^-$ to act as a base (abstracting a $\beta$-hydrogen) rather than as a nucleophile (attacking the carbon) (1 mark).

A concentrated solution of $\mathrm{NaOH}$ provides a high concentration of $\mathrm{OH}^-$, increasing the rate of both substitution and elimination. However, the ethanol solvent is a polar protic solvent that stabilises the charged transition state of elimination and does not solvate the nucleophile as effectively as water would (1 mark).

High temperature favours elimination because elimination has a higher activation energy than substitution (it involves breaking a C--H bond in addition to the C--Br bond). According to the Arrhenius equation, reactions with higher $E_a$ are more sensitive to temperature increases, so the rate of elimination increases more than the rate of substitution as temperature rises (1 mark).