top of page
new sciteach chem small.png

Halogenoalkane Reactions

2017 Paper Two

Question One
Looking at past AQA A-level Chemistry questions, I wanted to go beyond the answers and focus on thought processes, ideas and tips that will help in examinations.

The series will help you to spot weaknesses and help with revision, or it can serve as addenda to your notes.If you haven't gone through the paper, please look up the questions at, or click the direct link below, and have a go...

AQA A-level CHEMISTRY Paper 2 Organic and Physical ChemistryMonday 19 June 2017

This question is in 6 parts with 13 marks available

Initially, it’s a good idea to take a good look at the diagram to settle yourself and anticipate what you will be asked.

The starting point is a primary halogenoalkane (1-bromopropane) that is polar (has areas of 𝛿+ and 𝛿-).
NaOH gives OH– ions in solution. OH– and NH3 are both nucleophilic (have a lone pair of electrons).
In reaction 2, Br is substituted – no other products? This will be the same for the formation of J.
Reaction 3 gives an alkene (propene) and is an elimination (loss of HBr).
So, here come the questions...

OH is a nucleophile so will take the place of Br. The question asks for the displayed formula:

If you produce a skeletal or symbol formula you lose the mark. You need to include the O – H and not show it as – OH

Hopefully, you’ve recognised this as a nucleophilic substitution
For the second mark, the essential condition is excess ammonia.

Notes: A propylammonium ion is formed initially and, in a second step, free ammonia will remove H+ from this, leaving the primary amine. Ammonia in excess ensures that propylamine is the major product – production of a secondary or tertiary amine through multiple substitutions is restricted.

Make sure you can recall the conditions needed when answering questions about mechanisms - this comes up a lot and examiners have noted that many students don't answer this well.
Try to keep full values throughout your calculations and only round at the end (the initial values indicate that your answer should be to 3 significant figures).

1 mole of the halogenoalkane gives a mole of the amine…

moles of halogenoalkane = mass/Mr = 25.2/122.9 = 0.205…
moles for 75% of the amine = 0.205… x 0.75 = 0.153…
mass of amine = Mr of amine x moles of amine = 59 x 0.153...
= 9.07g (to 3 sig.fig.)

This is with the calculations carried through on a calculator. If you rounded to 3 significant figures throughout, your answer would be 9.09g.

Double-check your calculated values for the Mr.

Examiners noted that 60% of students failed to gain both marks here by drawing a primary amine or a non-skeletal structure.

A skeletal formula is required. The main thing to latch onto is the ‘different conditions’ bit of the question. We already know that excess ammonia will lead to a primary amine as a major product, and ‘different conditions’ lead us to conclude that multiple substitutions are more likely. The question wants the ‘classification’ which gives this idea credence.

Interestingly, there’s an odd number of carbons and 3x the number of carbons and hydrogens than in the original halogenoalkane – looks like we have a tertiary amine…

To get to an alkene, an elimination is required – in effect, the removal of hydrogen bromide. The reagent will be concentrated sodium or potassium hydroxide - don't just write 'hydroxide'.

Water can’t be involved since that would favour the formation of an alcohol through substitution.

So, for the conditions, we want ethanolic, not aqueous, sodium or potassium hydroxide. (The mark scheme gives NaOH/ ethanol or KOH / ethanol)

From the product, we can see this is a straightforward elimination with OH behaving as a Bronsted-Lowry base (proton acceptor) - this gets you one mark. The rest of the marks are all about how you draw curly arrows.

Good practice when you draw a nucleophile is to make sure lone pairs are included along with any charge it may have. Make sure the lone pair is on the correct element – in this case, the oxygen of the OH. Double-check that you have Br as the halogen.

So, to get the rest of the marks, draw a curly arrow from the lone pair on the oxygen of the hydroxide. Your arrow’s path will be directed towards a hydrogen – but which one?

It will be the one attached to the carbon that is the neighbour of the carbon bonded to the halogen... phew! (see diagram).

The next mark is for a curly arrow from the C–H bond to the C–­C bond with the final mark for a curly arrow from the C–Br bond onto the Br. Remember, follow the movement of electrons – always draw curly arrows starting from lone pairs or bonds.
I hope that's been helpful.
bottom of page