aqa a level chemistry paper 2 help questions

Hi, is a diester just a molecule with two ester bonds? For 7.1, the wording of the question completely confuses me. Most of the tests given in the ms surely wouldn't distinguish between all four. For example, adding sodium hydrogencarbonate and seeing effervescence occur would tell us that either k or m was present, but we wouldn't know which? Also, it says add AgNO3 and observe a cream precipitate - would you surely not have to add NaOH first and then add the AgNO3? How do you do 9.1? For 10.4, how does R have two triplets - how do you know which adjacent carbon to look at? Many thanks https://filestore.aqa.org.uk/sample-papers-and-mark-schemes/2017/june/AQA-74052-QP-JUN17.PDF

Let's take one at a time:

Yes, a diester is a molecule containing two ester groups. For example, animal fat is a triester.

If you add silver nitrate solution, you are actually adding Ag⁺(aq) ions. These form a cream ppt with any bromide ions present. Tertiary bromo compounds actually exist in equilibrium with the corresponding carbocation and bromide ions, so you could assume that there are some present.

You are correct that both K and M give positive tests with carbonate ions. However, the question says minimum number of tests. That implies that you may need more than one.

For the DNA you have to learn the fundamental structure:
DNA

10.4. There are three unique environments for the carbon atoms and therefore three signals. It does not say triplet, it says three signals.

thanks. for 10.5, this is something that has confused me for a long time now. If there are more than one adjacent carbons, how do you know which to look at and add one to to find the splitting pattern?

Also in regards to the diester, does this mean than a diamine is something that contains two amine groups in (2 nh2s)? How about a diamide?

Also, in regards to tertiary alcohols, if the oh is bonded to a c that is bonded to two other carbons and one halogen, is this tertiary?

Splitting patterns are only relevant in ¹H NMR (at this level)

If there are hydrogen atoms on adjacent (non-equivalent) carbon atoms then both split the central hydrogens, but with different coupling constants.



"di" means two of whatever immediately follows the multiplier.



No, tertiary refers to the carbon skeleton.

I'm still confused on q10.5 and finding the splitting pattern. Like for the h in ch2, it's adjacent to c=o and ch2. So which do we consider an how do you know? If we consider c=o, then it's a singlet, but if we consider ch2, it's a triplet?

Hydrogen atoms are only split by other non-equivalent hydrogen atoms.

The CH2 hydrogen atoms cannot be split by a carbonyl group, only by the adjacent CH2 hydrogens.

what do you mean by "only split by other non equivalent hydrogen atoms"? what does this actually mean?

CH3CH3

There is no splitting because the hydrogen atoms are in the same environment

CH3CH2-

The CH3 hydrogens are split into a triplet by the CH2 hydrogens AND
The CH2 hydrogens are split into a quartet by the CH3 protons.

The splitting occurs because the protons are in a different environment at each carbon.

CH3CH2CH3

The CH3 protons are in the same environment.

The six CH3 protons split the CH2 into a septet. The CH2 splits the six CH3 hydrogens into a triplet.

CH3COCH3

The six (equivalent) protons form a singlet as they have no adjacent protons

I still dont get how this relates to the question and how you know which carbon to use