// StartMathJax Script window.MathJax = {loader: {load: [ 'input/asciimath', 'ui/lazy', 'output/chtml', 'ui/menu']} }; (function() { var script = document.createElement('script'); script.src = "https://cdn.jsdelivr.net/npm/mathjax@3/es5/startup.js"; script.async = true; document.head.appendChild(script); })(); ---------- (Different files) ---------- // UpdateTypeset Script config = { attributes: true, childList: true, subtree: true }; // Callback function to execute when mutations are observed callback = (mutationList, observer) => { for (mutation of mutationList) { if (mutation.type === 'childList') { console.log('A child node has been added or removed.'); MathJax.typeset(); } else if (mutation.type === 'attributes') { console.log(`The ${mutation.attributeName} attribute was modified.`); } } }; // Create an observer instance linked to the callback function observer = new MutationObserver(callback); document.onreadystatechange = () => { if (document.readyState === 'complete') { console.log("Loaded fully according to readyState") targetNode = document.getElementById('content-wrapper') console.log(targetNode) // Start observing the target node for configured mutations observer.observe(targetNode, config); } }
top of page

Halogenoalkane Reactions: Past Paper Question Analysis

Updated: Aug 8, 2023

P2 June 2017 Q1

Hi there...

This is the first post in a series 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 www.aqa.org.uk, or click the direct link below, and have a go...

Question 1

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...







I hope that's been helpful. I'll be continuing the series regularly - next up will be Question 2 from the same paper.

Please share your thoughts through the comments box below - I look forward to hearing from you.

Recent Posts

See All


Rated 0 out of 5 stars.
No ratings yet

Add a rating
bottom of page