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Questions on halogenoalkanes: nucleophilic substitution and rates of hydrolysis
The polar carbon-halogen bond and its bond enthalpy, nucleophiles, nucleophilic substitution of primary halogenoalkanes by hydroxide ions, cyanide ions and ammonia with their conditions and curly-arrow mechanisms, naming nitriles and amines, why excess ammonia is used, and the rate of hydrolysis of chloro-, bromo- and iodoalkanes compared with aqueous silver nitrate in ethanol, making halogenoalkanes from alcohols, hydrolysing nitriles to carboxylic acids, primary against tertiary hydrolysis, SN1 and SN2, and optical activity as evidence for the mechanism.
20 original questions · 66 marks · the halogenoalkanes: nucleophilic substitution and rates of hydrolysis notes · Alkanes and halogenoalkanes · Flashcards for this unit
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Explain the mechanism by which hydroxide ions hydrolyse a primary halogenoalkane, and use the data in the table to explain the order of the rates of hydrolysis of 1-chlorobutane, 1-bromobutane and 1-iodobutane.
Bond Electronegativity of X (C is 2.5) Bond enthalpy / kJ mol⁻¹ C–Cl 3.0 346 C–Br 2.8 290 C–I 2.5 228 Mark scheme
Indicative content
- the C–X bond is polar because the halogen is more electronegative than carbon, so the carbon is δ+ and the halogen δ−
- a lone pair on the oxygen of OH⁻ is donated to the δ+ carbon, shown by a curly arrow from the lone pair to that carbon, forming the C–O bond
- the C–X bond breaks heterolytically, shown by a curly arrow from the bond to X, and X leaves as a halide ion, giving the alcohol
- the rate increases from chloro to bromo to iodo although the bond becomes less polar in that order, so polarity does not decide the rate
- the bond enthalpy falls from 346 to 290 to 228 kJ mol⁻¹, so the C–I bond is the weakest and the easiest to break
- less energy is needed to break the weaker bond, so the activation energy is lower, more collisions succeed and 1-iodobutane is hydrolysed fastest
Levels
- Level 3 (5–6 marks) the polar C–X bond, the attack by the hydroxide lone pair and the heterolytic loss of the halide are all described as the mechanism would be drawn, the rate order is read from the table, bond polarity is rejected as the explanation and bond enthalpy is used to account for the order through activation energy
- Level 2 (3–4 marks) the hydroxide attacks the δ+ carbon and the halide leaves, and the rate order is put down to bond enthalpy; the answer does not dismiss polarity, use the table, or connect the weaker bond to a lower activation energy
- Level 1 (1–2 marks) fragments of the mechanism, with the rate order explained by bond polarity, so that 1-chlorobutane is predicted to react fastest
Outline the mechanism for the reaction of 1-bromopropane with an excess of ammonia.
Mark scheme
- B1 the dipole: δ+ on the carbon bonded to bromine and δ− on the bromine
- B1 a curly arrow from the lone pair on the nitrogen of NH₃ to the δ+ carbon
- B1 a curly arrow from the C–Br bond to the bromine atom
- B1 the intermediate CH₃CH₂CH₂NH₃⁺ drawn with the positive charge on the nitrogen
- B1 a curly arrow from the lone pair of a second NH₃ to an H on the N⁺, and an arrow from that N–H bond to the nitrogen, giving CH₃CH₂CH₂NH₂ and NH₄⁺
Explain why the carbon atom bonded to bromine in 1-bromopropane is attacked by nucleophiles, and identify the nucleophile in its reactions with aqueous sodium hydroxide, with potassium cyanide in ethanol and with ammonia.
Mark scheme
- B1 bromine is more electronegative than carbon, so the C–Br bond is polar, with the carbon δ+ and the bromine δ−
- B1 a nucleophile is an electron pair donor, and its lone pair is attracted to the δ+ carbon and forms a new bond to it
- B1 with aqueous sodium hydroxide the nucleophile is the hydroxide ion, OH⁻
- B1 with potassium cyanide it is the cyanide ion, CN⁻, through the lone pair on carbon, and with ammonia it is NH₃, through the lone pair on nitrogen
Give the reagent and conditions for converting bromoethane into propanenitrile and into ethylamine, and explain how the name propanenitrile follows from the reaction.
Mark scheme
- B1 potassium cyanide dissolved in ethanol and water, heated under reflux; reject aqueous KCN alone
- B1 the cyanide carbon joins the chain, so CH₃CH₂CN has three carbons counting the C of the C≡N: propanenitrile
- B1 an excess of ammonia, dissolved in ethanol
- B1 heated in a sealed tube, under pressure, so the ammonia cannot escape
Describe how the rates of hydrolysis of 1-chlorobutane, 1-bromobutane and 1-iodobutane can be compared using silver nitrate solution.
Mark scheme
- B1 equal volumes of each halogenoalkane are mixed with ethanol in separate tubes and warmed to the same temperature in a water bath
- B1 equal volumes of aqueous silver nitrate, at the same temperature, are added and a clock is started
- B1 the time for a precipitate to appear is recorded; a shorter time means a faster hydrolysis
- B1 the iodo compound gives a yellow precipitate first, the bromo a cream one later and the chloro a white one last
Calculate the rate of hydrolysis of 1-iodopropane and of 1-chloropropane relative to 1-bromopropane, taking the rate as proportional to 1/time. The table gives the times for a precipitate to appear when each was warmed to 50 °C with ethanol and aqueous silver nitrate.
Halogenoalkane Time for precipitate to appear / s 1-chloropropane 1900 1-bromopropane 160 1-iodopropane 25 Mark scheme
- M1 rate is proportional to 1/t, so the relative rate of each compound is 160 ÷ its own time
- M1 1-iodopropane: 160 ÷ 25; 1-chloropropane: 160 ÷ 1900
- A1 1-iodopropane reacts 6.4 times as fast as 1-bromopropane, a ratio with no unit, to two significant figures
- A1 1-chloropropane reacts 0.084 times as fast, to two significant figures; allow 0.08 or 1/12
Draw the mechanism for the reaction of bromoethane with hydroxide ions.
Mark scheme
- B1 the dipole: δ+ on the carbon bonded to bromine and δ− on the bromine
- B1 a curly arrow from a lone pair on the oxygen of OH⁻ to the δ+ carbon; allow the arrow from the minus sign, reject an arrow from the H
- B1 a curly arrow from the C–Br bond to the bromine atom, giving CH₃CH₂OH and Br⁻
Outline the mechanism for the reaction of 1-chlorobutane with cyanide ions, and name the organic product.
Mark scheme
- B1 a curly arrow from the lone pair on the carbon of the cyanide ion to the δ+ carbon bonded to chlorine
- B1 the dipole on the C–Cl bond, and a curly arrow from the C–Cl bond to the chlorine atom
- B1 the product is CH₃CH₂CH₂CH₂CN, pentanenitrile; reject butanenitrile, which omits the carbon of the C≡N
Explain why 1-iodobutane is hydrolysed faster than 1-chlorobutane, even though the C–Cl bond is more polar.
Mark scheme
- B1 the rate is determined by the C–X bond enthalpy, because the bond must break for substitution to occur
- B1 the C–I bond (228 kJ mol⁻¹) is weaker than the C–Cl bond (346 kJ mol⁻¹)
- B1 less energy is needed to break C–I, so the activation energy is lower and the reaction is faster
Explain why ethanol is added to the mixture in the silver nitrate hydrolysis test, and why water rather than hydroxide ions is used as the nucleophile.
Mark scheme
- B1 halogenoalkanes do not mix with the aqueous silver nitrate solution
- B1 ethanol is a solvent for both, so they form one solution and the water can react with the halogenoalkane
- B1 hydroxide ions would react with silver ions to precipitate silver oxide, hiding the halide precipitate
Calculate the percentage yield of propan-1-ol when 1-bromopropane is heated under reflux with excess aqueous sodium hydroxide, using the masses in the table.
Substance Mr Mass / g 1-bromopropane used 122.9 6.15 propan-1-ol obtained 60.0 2.10 Mark scheme
- M1 amount of 1-bromopropane = 6.15 ÷ 122.9 = 0.05004 mol
- M1 the reaction is 1 : 1, so the theoretical mass of propan-1-ol is 0.05004 × 60.0 = 3.002 g
- A1 percentage yield = 2.10 ÷ 3.002 × 100 = 69.9%, to three significant figures; error carried forward from M1
Suggest the structure and name of the product formed when 2-bromopropane is heated with potassium cyanide in ethanol and water, and say how many carbon atoms it contains.
Mark scheme
- B1 CH₃CH(CN)CH₃, the cyanide group replacing bromine on the middle carbon
- B1 2-methylpropanenitrile, since the longest chain including the nitrile carbon has three carbons with a methyl branch
- B1 four carbon atoms in all, one more than 2-bromopropane, because the new bond is C–C
Suggest why an excess of ammonia is used to make ethylamine from bromoethane, and what forms if it is not.
Mark scheme
- B1 ethylamine has a lone pair on nitrogen, so it is also a nucleophile
- B1 it can react with more bromoethane to give diethylamine, and further substitution can follow
- B1 with ammonia in excess, bromoethane is more likely to meet ammonia than ethylamine, so the primary amine is the main product
Compare the reaction of bromoethane with hydroxide ions with its reaction with ammonia.
Mark scheme
- B1 in both, a lone pair on the nucleophile attacks the δ+ carbon and the C–Br bond breaks, releasing Br⁻
- B1 hydroxide is negatively charged and gives the alcohol in one step; ammonia is neutral and first gives a positively charged ammonium intermediate that must lose H⁺
- B1 hydroxide is used in warm aqueous solution; ammonia in excess, in ethanol, in a sealed tube
Explain why hydrolysis of a single enantiomer of a chiral tertiary halogenoalkane gives a product that does not rotate plane-polarised light.
Mark scheme
- B1 a tertiary halogenoalkane reacts by SN1: the C–X bond breaks first to give a carbocation
- B1 the carbocation is planar, so the nucleophile attacks either face with equal probability
- B1 equal amounts of the two enantiomers form, a racemic mixture, whose rotations cancel
Suggest why 2-bromo-2-methylpropane gives a precipitate with aqueous silver nitrate and ethanol much sooner than 1-bromobutane, and why these two compounds make a fair comparison.
Mark scheme
- B1 both are C₄H₉Br and break a C–Br bond, so the halogen and the bond enthalpy are the same and only the type of carbon differs
- B1 the tertiary compound reacts by a different route: the C–Br bond breaks first to give a tertiary carbocation
- B1 the carbocation is stabilised by its three alkyl groups pushing electron density towards it, so it forms readily and bromide ions are released faster
Suggest a two-step route from bromoethane to propanoic acid, giving the reagent and conditions for each step and the intermediate.
Mark scheme
- B1 step 1: potassium cyanide in ethanol and water, heated under reflux
- B1 the intermediate is propanenitrile, CH₃CH₂CN
- B1 step 2: dilute hydrochloric acid, heated under reflux, hydrolyses the nitrile to CH₃CH₂COOH
Explain why 1-fluorobutane gives no precipitate in the silver nitrate hydrolysis test.
Mark scheme
- B1 the C–F bond is very strong, 467 kJ mol⁻¹, the strongest carbon-halogen bond
- B1 it is not broken under these conditions, so no fluoride ions are released to react with silver ions
Suggest why the reaction of a halogenoalkane with cyanide ions is useful when making a compound with a longer carbon chain.
Mark scheme
- B1 a new C–C bond forms, so the carbon chain grows by one atom
- B1 the nitrile group can then be converted into other groups, such as a carboxylic acid or an amine
Give an equation for the conversion of ethanol into chloroethane with thionyl chloride, and give one advantage of this reagent.
Mark scheme
- B1 CH₃CH₂OH + SOCl₂ → CH₃CH₂Cl + SO₂ + HCl
- B1 both by-products are gases that leave the flask, so the product is easily separated
Practise halogenoalkanes: nucleophilic substitution and rates of hydrolysis one question at a time