Halogenoalkanes (Edexcel A Level Chemistry): Flashcards

In terms of bond enthalpy, the order of C-X bond strength is C-F > C-Cl > C-Br > C-I.

False.

Fluoroalkanes do not react at all because the C-F bond is the strongest C-X bond. Iodoalkanes have the weakest C-I bond and react the fastest.

The halogenoalkane is heated under reflux with an aqueous solution of sodium hydroxide (NaOH) or potassium hydroxide (KOH). The OH^- ion acts as the nucleophile in this hydrolysis reaction.

The halogenoalkane is heated under reflux with an ethanolic solution of potassium cyanide (KCN). It is useful because the CN^- nucleophile extends the carbon chain by one carbon atom.

A primary halogenoalkane is heated under pressure with an ethanolic solution of excess ammonia (NH₃). Excess ammonia is used because the primary amine product is more reactive than ammonia, preventing further substitution reactions.

False.

Ethanolic NaOH causes elimination: an adjacent C−H bond and the C−X bond break, producing an alkene and a hydrogen halide. Aqueous NaOH would cause nucleophilic substitution, producing an alcohol instead.

Silver nitrate in ethanol detects halide ions (X^-) released from a halogenoalkane. A positive result is the formation of a silver halide precipitate: Ag⁺ (aq) + X^- (aq) → AgX (s).

Nucleophilic substitution is a reaction in which a nucleophile attacks an electron-deficient carbon atom, replacing the atom that carries a partial negative charge (the leaving group).

The large difference in electronegativity between carbon and the halogen makes the C-X bond polar. The carbon carries a partial positive charge (C^δ+), making it susceptible to attack by a nucleophile.

In a halogenoalkane, the carbon bonded to the halogen carries a partial positive charge and the halogen carries a partial negative charge due to the polarity of the C-X bond.

The nucleophile is OH^-. It donates a lone pair to the electron-deficient carbon, forming a new C-O bond while the C-X bond breaks, displacing the halide ion and producing an alcohol.

False.

The reaction occurs in two steps. First NH₃ attacks to form [CH₃NH₃]⁺Cl^-. Then a second molecule of NH₃ deprotonates this salt to give methylamine (CH₃NH₂) and NH₄⁺Cl^-.

Step 1: NH₃ acts as a nucleophile, attacking the C^δ+ of CH₃Cl to form [CH₃NH₃]⁺Cl^-.

Step 2: A second NH₃ molecule deprotonates the ammonium salt to give CH₃NH₂ and NH₄⁺Cl^-.

The primary amine product is more reactive than ammonia and can itself undergo further substitution reactions with the halogenoalkane. Using excess NH₃ prevents further substitution and favours formation of the primary amine.

True.

The nucleophile uses its lone pair to form a new bond with the carbon bearing a partial positive charge (C^δ+). The existing C-X bond then breaks heterolytically, with both electrons going to the halogen as X^-.

In the nucleophilic substitution mechanism, the nucleophile replaces the atom carrying the partial negative charge, which departs as a halide ion.

Hydrolysis is the reaction of a halogenoalkane with water (or aqueous alkali) in which the C-X bond breaks and the halogen is replaced by a hydroxyl group.

The silver nitrate reacts with halide ions released during hydrolysis to form a precipitate. The rate at which the precipitate forms indicates the rate of hydrolysis of the halogenoalkane.

In the silver nitrate test, iodoalkanes produce a yellow precipitate, bromoalkanes produce a cream precipitate and chloroalkanes produce a white precipitate.

False.

Iodoalkanes hydrolyse faster because the C-I bond has a lower bond enthalpy than the C-Cl bond, so it is easier to break and releases I^- ions more quickly.

Silver fluoride is soluble in water, so no precipitate forms. This confirms that fluoroalkanes are the least reactive halogenoalkanes.

Tertiary halogenoalkanes hydrolyse the fastest, secondary are in the middle and primary are the slowest. The difference in rate is due to the SN1 and SN2 mechanisms.

The rate of hydrolysis increases down Group 17 because the carbon-halogen bond enthalpy decreases, making the bond easier to break.

True.

The water bath at 50 °C provides a controlled elevated temperature that increases the rate of hydrolysis, allowing precipitates to form in a reasonable time for comparison.

The halogen and the overall molecular formula of the halogenoalkane must remain the same. For example, 1-chlorobutane, 2-chlorobutane and 2-chloro-2-methylpropane are all C₄H₉Cl.

An S_N1 reaction is a two-step nucleophilic substitution in which the rate-determining step depends only on the concentration of the halogenoalkane, forming a carbocation intermediate.

Tertiary halogenoalkanes favour S_N1 because the three alkyl groups stabilise the carbocation intermediate formed when the C-X bond breaks heterolytically.

True.

The S_N2 mechanism is a one-step process: the nucleophile donates a pair of electrons to the δ+ carbon, forming a new bond as the C-X bond breaks simultaneously.

In an S_N1 mechanism, the rate-determining step is the heterolytic breaking of the C-X bond, producing a carbocation intermediate.

In primary halogenoalkanes, the carbon bearing the halogen has only one alkyl group, so a carbocation would be too unstable to form. Instead, the nucleophile attacks directly in a single concerted step.

The C-I bond has the lowest bond enthalpy. It is the easiest to break, so iodoalkanes undergo nucleophilic substitution at the fastest rate.

An S_N2 reaction is a one-step nucleophilic substitution in which the rate-determining step depends on the concentrations of both the halogenoalkane and the nucleophile.

False.

Fluoroalkanes are the least reactive halogenoalkanes in substitution reactions because the C-F bond has the highest bond enthalpy and is the hardest to break, despite being the most polar.

Bond enthalpy decreases from C-F to C-I, so the rate of nucleophilic substitution increases from fluoroalkanes to iodoalkanes.