Halogenoalkanes (Cambridge (CIE) A Level Chemistry): Flashcards
Exam code: 9701
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Define halogenoalkane.
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Define halogenoalkane.
A halogenoalkane is an alkane in which one or more hydrogen atoms have been replaced by a halogen atom. They are classified as primary, secondary or tertiary based on the substitution of the carbon bearing the halogen.
Name three routes by which halogenoalkanes can be produced.
Free-radical substitution of alkanes with Cl2 or Br2 in UV light.
Electrophilic addition of HX or X2 to alkenes.
Substitution of an alcohol using HX, PCl3, PCl5 or SOCl2.
True or False?
A secondary halogenoalkane has the halogen bonded to a carbon that is attached to three other alkyl groups.
False. A secondary halogenoalkane has the halogen on a carbon attached to two alkyl groups. Three alkyl groups defines a tertiary halogenoalkane.
When an alcohol reacts with PCl5 at room temperature, the –OH group is replaced by .........., producing a .......... and HCl.
When an alcohol reacts with PCl5 at room temperature, the –OH group is replaced by –Cl, producing a chloroalkane and HCl.
What is the major product rule when HBr adds to an unsymmetrical alkene to form a halogenoalkane?
The major product has Br bonded to the most substituted carbon (Markovnikov's rule). This is because the more substituted carbon gives the more stable carbocation intermediate.
True or False?
Electrophilic addition of a halogen (X2) to an alkene proceeds with one halogen atom acting as an electrophile and the other as a nucleophile.
True. When X2 approaches the electron-rich C=C bond, one X atom becomes δ+ (the electrophile) and the other becomes X- (the nucleophile) after the X–X bond breaks heterolytically.
A .......... halogenoalkane has the halogen on a carbon bonded to one alkyl group. A .......... halogenoalkane has it bonded to two alkyl groups.
A primary halogenoalkane has the halogen on a carbon bonded to one alkyl group. A secondary halogenoalkane has it bonded to two alkyl groups.
Define tertiary halogenoalkane.
A tertiary halogenoalkane is one in which the halogen is bonded to a carbon atom that is itself attached to three other alkyl groups.
What reagent and conditions convert an alcohol to a chloroalkane using thionyl chloride?
The alcohol is reacted with SOCl2 (thionyl chloride). The reaction replaces the –OH group with –Cl to form the chloroalkane, releasing SO2 and HCl as by-products.
Define nucleophilic substitution.
Nucleophilic substitution is a reaction in which a nucleophile attacks a carbon atom carrying a δ+ charge, replacing the atom or group carrying a δ- charge (the leaving group).
What product is formed when a halogenoalkane reacts with ethanolic KCN under reflux, and why is this reaction useful?
A nitrile is formed (e.g. bromoethane + KCN → propanenitrile). The reaction is useful because it adds one extra carbon atom to the chain, extending the carbon skeleton of the starting material.
True or False?
The hydroxide ion is a better nucleophile than water because it carries a full formal negative charge.
True. OH- has a full negative charge, making it a stronger nucleophile than water, where the oxygen only has a partial negative charge (δ-). Nucleophilic substitution with OH- is therefore much faster than with water.
When bromoethane reacts with excess ethanolic .......... under pressure, the primary product is .......... (an amine).
When bromoethane reacts with excess ethanolic ammonia (NH3) under pressure, the primary product is ethylamine (CH3CH2NH2) (an amine).
Why must ammonia be used in excess in the nucleophilic substitution of a halogenoalkane?
The primary amine product (e.g. ethylamine) can itself act as a nucleophile and attack another halogenoalkane molecule, forming a secondary amine (e.g. diethylamine). Excess NH3 minimises this over-substitution.
True or False?
Iodoalkanes hydrolyse faster than bromoalkanes in aqueous silver nitrate because the C–I bond is shorter and stronger.
False. Iodoalkanes hydrolyse faster because the C–I bond is the weakest and longest of the C–halogen bonds, making it easier to break in nucleophilic substitution.
In the reaction of a halogenoalkane with aqueous NaOH, the nucleophile is .......... and the product is a/an ...........
In the reaction of a halogenoalkane with aqueous NaOH, the nucleophile is OH- and the product is an alcohol.
Define hydrolysis.
Hydrolysis is the breakdown of a molecule by water. In the context of halogenoalkanes, it refers to nucleophilic substitution by water (or OH-) to produce an alcohol and a halide ion.
How can aqueous silver nitrate be used to identify which halogen is present in a halogenoalkane?
Water in the AgNO3 solution hydrolyses the halogenoalkane, releasing halide ions. These react with Ag+ to form coloured precipitates: AgCl (white), AgBr (cream) or AgI (yellow). The rate of precipitate formation also indicates bond strength.
What reagent and conditions are required for the elimination reaction of a halogenoalkane?
The halogenoalkane is heated with ethanolic sodium hydroxide (NaOH dissolved in ethanol). The anhydrous conditions favour elimination to produce an alkene.
True or False?
Heating a halogenoalkane with aqueous NaOH produces an alkene by elimination.
False. Aqueous NaOH favours nucleophilic substitution, producing an alcohol. Ethanolic NaOH (anhydrous) is needed for the elimination reaction that produces an alkene.
Define elimination reaction.
An elimination reaction is one in which a small molecule is lost from a larger organic molecule. In halogenoalkanes, HX is eliminated and a C=C double bond forms, giving an alkene.
In the elimination of bromoethane, a .......... atom and a .......... atom are lost, and a C=C double bond forms to produce ethene.
In the elimination of bromoethane, a hydrogen atom and a bromine atom are lost, and a C=C double bond forms to produce ethene.
What bonds are broken and formed during the elimination of bromoethane with ethanolic NaOH?
The C–Br bond breaks heterolytically, releasing Br-. A C–H bond on the adjacent carbon also breaks. The electrons form a C=C double bond, giving ethene.
True or False?
The small molecule eliminated from a halogenoalkane in an elimination reaction is always water.
False. In elimination reactions of halogenoalkanes, the small molecule lost is a hydrogen halide (e.g. HBr or HCl), not water. Water is lost in the dehydration of alcohols.
The overall equation for elimination of bromoethane is: C2H5Br + NaOH (ethanol) → .......... + NaBr + .......... .
The overall equation for elimination of bromoethane is: C2H5Br + NaOH (ethanol) → C2H4 + NaBr + H2O.
Define ethanolic sodium hydroxide.
Ethanolic sodium hydroxide is NaOH dissolved in ethanol. It provides anhydrous conditions that favour elimination of HX from a halogenoalkane to give an alkene, rather than substitution.
State the key difference in conditions that determines whether elimination or substitution occurs when NaOH reacts with a halogenoalkane.
Elimination uses hot ethanolic NaOH (anhydrous, NaOH in ethanol) and produces an alkene. Substitution uses warm aqueous NaOH and produces an alcohol.
Define SN2 mechanism.
An SN2 mechanism is a one-step nucleophilic substitution in which the nucleophile attacks the δ+ carbon at the same time as the C–X bond breaks. The rate depends on the concentration of both the halogenoalkane and the nucleophile.
Which type of halogenoalkane undergoes SN1 reactions, and why?
Tertiary halogenoalkanes undergo SN1 reactions because the three alkyl groups stabilise the tertiary carbocation intermediate through the inductive effect, making the C–X bond easier to break in the slow rate-determining step.
True or False?
In an SN1 reaction, the rate-determining step is the attack of the nucleophile on the carbocation.
False. The rate-determining step in SN1 is the slow heterolytic fission of the C–X bond to form the carbocation. The nucleophile attack in the second step is fast.
In an SN2 reaction, the nucleophile attacks at the same time as the .......... bond breaks. This gives a ..........-step mechanism.
In an SN2 reaction, the nucleophile attacks at the same time as the C–X bond breaks. This gives a one-step mechanism.
What is the rate-determining step in an SN1 reaction, and what intermediate does it produce?
The rate-determining step is the heterolytic fission of the C–X bond. This is the slow first step and produces a tertiary carbocation. The nucleophile then attacks the carbocation in the fast second step.
True or False?
Secondary halogenoalkanes can undergo both SN1 and SN2 mechanisms.
True. Secondary halogenoalkanes undergo a mixture of SN1 and SN2 reactions, depending on their specific structure and reaction conditions.
In SN1, the '1' means the rate depends on the concentration of the .......... only. In SN2, the '2' means the rate depends on the concentrations of both the halogenoalkane and the ...........
In SN1, the '1' means the rate depends on the concentration of the halogenoalkane only. In SN2, the '2' means the rate depends on the concentrations of both the halogenoalkane and the nucleophile.
Define SN1 mechanism.
An SN1 mechanism is a two-step nucleophilic substitution. In step 1 (slow, rate-determining), the C–X bond breaks to give a carbocation. In step 2 (fast), the nucleophile attacks the carbocation.
Why does a primary halogenoalkane not form a primary carbocation as an intermediate in nucleophilic substitution?
A primary carbocation is much less stable than a tertiary carbocation because it has only one alkyl group donating electron density via the inductive effect. Primary halogenoalkanes therefore follow the SN2 pathway, avoiding the unstable primary carbocation.
Why do iodoalkanes react faster in nucleophilic substitution reactions than fluoroalkanes?
The C–I bond has a lower bond energy (228 kJ mol-1) than the C–F bond (467 kJ mol-1), so it requires less energy to break. The weaker C–I bond is therefore more easily broken during nucleophilic substitution.
True or False?
The C–F bond is the weakest carbon-halogen bond.
False. The C–F bond is the strongest carbon-halogen bond (467 kJ mol-1). The C–I bond is the weakest (228 kJ mol-1), making iodoalkanes the most reactive.
Define bond energy.
Bond energy is the energy required to break one mole of a covalent bond. In halogenoalkanes, the C–X bond energy determines reactivity: a lower bond energy means a weaker bond and faster nucleophilic substitution.
The C–I bond energy is .......... kJ mol-1, and the C–Cl bond energy is .......... kJ mol-1. The weaker bond leads to faster substitution.
The C–I bond energy is 228 kJ mol-1, and the C–Cl bond energy is 346 kJ mol-1. The weaker bond leads to faster substitution.
What precipitates are formed when chloro-, bromo- and iodoalkanes are hydrolysed with aqueous silver nitrate, and what colours are they?
Chloroalkane → AgCl (white precipitate).
Bromoalkane → AgBr (cream precipitate).
Iodoalkane → AgI (yellow precipitate).
True or False?
The order of reactivity of halogenoalkanes in nucleophilic substitution is: fluoroalkanes > chloroalkanes > bromoalkanes > iodoalkanes.
False. The order is iodoalkanes > bromoalkanes > chloroalkanes > fluoroalkanes. Reactivity increases down Group 17 because the C–X bond becomes weaker and easier to break.
In aqueous AgNO3, iodoalkanes form a .......... precipitate (AgI) most .......... because C–I is the weakest C–X bond.
In aqueous AgNO3, iodoalkanes form a yellow precipitate (AgI) most rapidly because C–I is the weakest C–X bond.
Define aqueous silver nitrate test.
An aqueous silver nitrate test is a method that identifies the halogen in a halogenoalkane by hydrolysing it and observing the colour of the silver halide precipitate formed: AgCl (white), AgBr (cream) or AgI (yellow). The rate of formation also indicates reactivity.
Why is bond energy, rather than bond polarity, used to explain the reactivity trend in halogenoalkanes?
Although C–F is the most polar bond, bond energy governs reactivity here. The C–F bond is so strong (467 kJ mol-1) that it resists heterolytic fission, making fluoroalkanes unreactive despite their polarity.
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