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PAPER 4

|Year |Paper |Question |Remark |

|2013 (MJ) |MJ (41) |1(b) | |

| | |2(a) iii | |

| | |3 (b) | |

| | |8 (b) (c) | |

| |MJ (42) |1 (a)iii | |

| | |2(b) | |

| | |5(a)iv, b(iv) | |

| | |7 (a) iv | |

| |MJ (43) |2(a) | |

| | |3(b) | |

| | |5(c) | |

| | |8(b) | |

|2013 (ON) |ON (43) |3 b(ii) | |

| | |6 f (ii), (iii) | |

| | |8 (b), (c) | |

| | |9 (b), (c)iii | |

| |ON (41) |4 (c) | |

| | |5 (d)i | |

|2014 (MJ) |MJ(41) |1 b (iv) | |

| | |3 (b)ii, d(iii), iv | |

| | |7 (b)iii,(c) i and ii | |

| | |8 (a) iii | |

| |MJ(42) |7 | |

|2014 (ON) |ON (43) |4(d) ii | |

|2015 (ON) |ON(41) |6 (d)ii | |

| |ON(43) |4(e) ii, (f)ii | |

PAPER 5

|Year |Paper |Question |Remark |

|2010 (MJ) |MJ(51) |2 (d) | |

| | |3 (e), (f) | |

| |MJ(52) |1 a(i), c, d | |

|2010 (ON) |ON(51) |1, 2(F) | |

|2015(ON) |ON(51) |2 (e)i, (f) | |

| |ON(53) |2 | |

Comments:

You don’t ask any questions, so I don’t really know what it is about these things that you have a problem with :s

2013 (MJ)

MJ (41) 1(b)

The Fe2+ and Fe3+ species are aq. Ions so you need an inert conductor to facilitate the transfer of e- from one half cell to the other. Use Pt if in doubt. The HCl solution is 1 mol dm-3 and the iron species are each 1 mol dm-3. The hydrogen half-cell has a platinum electrode in contact with the HCl(aq) and the H2(g) which is being bubbled through. The two half cells are connected by a high resistance voltmeter. State words to the effect of :All solutions at 298K, all gas at 1 atm and all temperatures = 298K.

2(a) iii

This is a commonly asked question by students. It is covered here:

3 (b)That “identify the type of reaction” is also covered at

8 (b) (c) Please see

MJ (42) 1 (a)iii This is equivalent to the MJ P41 Q2. But a different compoiund is used and slightly different concentrations apply. The method of answering the question is the same.

2(b)

There should be no problem with mathematical based questions. Buffer equation can be expressed like this:

[pic]

This tells us that we need 5.66 times more A- than HA.

If total volume is to be 100cm3 we need to add a certain vol of A- with a certain vol of HA. But we don’t know how much.

Use algebra…

Let x = vol of A- to use, and y = vol of HA to use. x+y = 100 and x = 5.66y

Hence 5.66y + y = 100 and so 6.66y = 100 y=15.02cm3 (and x must be 84.98

So the answer is 15cm3 for HA and 85cm3 for salt, A-

If you get stuck, simply apply algebra and work through the problem. Often at the end things will resolve and you can get the answer, and if it doesn’t you should be able to make some progress to pick up some marks.

5(a)iv,

NaOH(aq) is a base and a nucleophile. Nucleophilic substitutions typically take place under conditions of heat. One type of nucleophilic substitution is hydrolysis. We can see from the structure that there is an ester bond present. That will hydrolyze with NaOH under heat (heat under reflux). NaOH(aq) as a base reacts when cold. NaOH(aq) can react with acidid groups. There are two phenol OH’s and one carboxylic acid.

So the answers are 3 and 4 moles respectively (NaOH still acts as a base when hot too!)

2013 MJ P42, 5b(iv)

[pic]

As three OH's groups have alkyl H's beside them and with c.H2SO4plus heat (to 170=C), the necessary requirements for dehydrations have been met. This molecule can undergo three dehydrations. The last remaining OH could be in the 3 or 4 position due to the last dehydration could lose OH from the 4 or the three position.

[pic]

Aq. Br2 is being used so the aliphatic double bond will react as well as the benzene ring. Br2(l) would give dibromo across the double bond, but (aq) Br2 gives the bromohydrin. You could have swapped the Br and OH groups on the (now gone) aliphatic C=C ‘s as you have not learned which positions for OH and Br are most favorable in these cases.

For the OH’s in the ring in G, we have phenol type functionality. Each OH directs into the 2 and 4 positions. With the two OH groups of the phenol in this orientation, ALL positions in the ring are activated towards electrophilic attack.

2013 MJ P42 7 (a) iv

[pic]

In the intro the question, we see the signature of 1 Cl atom, i.e. M and M+2 peaks, with Ratio 3:1 (liwer mass = 3, higher mass = 1)

In part (a) (i), we have worked out the initial No. of C’s is 7.

In part (ii), M+2 indicates there is an isotopic atom with 2 mass units greater. (also the height of M+2 relative to M+, tells us the relative abundance of those isotopes relative to each other) and alsoi the reaction with Na metal mentioned in (ii) tells us an alcohol or a COOH must be present.

(iii) Clearly we have an aromatic ring (peak at >7ppm) This will take up 6 C’s. We also know the ring must be doubly substituted as there are only 4 H’s in the ring.

There’s only one C left. It must be the alcohol or COOH. The Cl must be bonded to the ring.

So the answer for (iii)…. The peaks at 2ppm tell us the C bonded to the ring MUST have two types of H environment on it, neither of which engage in spin-spin coupling (splitting) those groups are not splitting.

(iv) Because of the information above and COOH only has one H so instantly fails to qualify for the above criteria, so we are left with C-OH and the other two positions on the C must be two H atoms, i.e. we have CH2OH So a disubstituted benzene ring, a Cl on the ring and a CH2OH on the ring.

2013 MJ P43 2(a)

This is a similar question to the one in 2013 MJ P41 Q1. The treatment is the same. You have two different concentrations of HCl, each having it’s own graph. You can find the gradient of the tangent at t=o and get an initial rate for the two concentrations. Then use initial rates method to determine the order of HCl. OR you can do gradient at any other time in the graph (but use the same time for each graph) and then work out the order from that.

2013 MJ P43 3(b)This question is covered in wither Desmond’s Q’s of Liyana’s Q’s.

2013 MJ P43 5(c)Na reacts with OH groups. So H2 is given off in ALL cases.

KOH is a base AND nucleophile. It doesn't have the ability to cause any gaseous product for any of the given organic molecules.

Na2CO3 solution reacts with acidic compounds. Alcohols are not acidic and you know Na2CO3 can't react with the water molecule because you are told that you have (AQ) Na2CO3. !!! Phenol isn't acidic enough to react, hence Na2CO3 distinguishes between COOH and phenols.

2013 MJ P43 8(b)

(i) Absorption would increase as COONa can form ION-dipole bonds with water, which are stronger than the hydrogen bond – hydrogen bond between COOH and water.

(ii) Diapers said to be an environmental problem. So they should be biodegradable.

2013 (ON)(43) 3 b(ii)

F is a very small molecule. It has very low shielding for it’s proton number. Hence the bond length is very short. This causes the lone pairs on each F to be close enough to interact. They will repel. Which actually weakens the bond significantly. This lone pair repulsion is not significant in the halides below from F2.

2013 (ON)(43)6 f (ii), (iii)

There is a document on my blog about how to solve this without learning RS absolute configuration methods (which is actually not too hard to learn but is totally not on the syllabus). There is another way to solve it too, but that also isn’t on the syllabus and is a rather long winded and I won’t mention it here. I feel the examiners will NOT put this type of question on again, and I certainly don't think you should spend time on this.



See below for an alternative method.

8 (b), (c)

X-ray spectrscopy is NOT on the syllabus anymore.

2013 (ON)P439 (b), (c)iii

(b) Each C in C60 has 3 sigma bonds (to other C atoms) and an unbonded electron in a p-orbital. It’s a bit like graphite. This unbonded e- can pair up with an e- from a H atom. So C60H60

(c)(iii) Graphene – yes, The electrons can delocalize throughout the whole plane no matter how large it is. Bucky balls no as e- cannot delocalize between different balls.

2013 ON (41) 4 (c)

Not sure if the relevant data is in the 2016 data book. But It will be that the weaker the bond, the more reactive the species is BECAUSE that weak bond can break and react. The more harmful CFC's contain Cl because the C-Cl bond is weaker.

5 (d)i This is what the mark scheme shows:

[pic]

This question has been an 'issue' with students (and teachers) since it came out.

As with ALL diagrams where diagrams of hydrogen bonding involving H2O, you must show:

• assign ( charges on the O and H in water.

• show the hydrogen bond with a dotted line

• lone pair on the O making the hydrogen bond to a H on a different H2O molecule.

(In pure water you need a straight line from the O of a water molecule to the O of a different water molecule, with a H atom bonded to the first O and it lies along the straight line between the two O's. The O-H covalent bond must be shown with a continuous line.)

This answer it's a bit different. You must label the hydrogen bond (actually that's always better than just drawing a single dashed line (the examiner has no idea what you understand by a dashed line). You still need to show the O-H.......O bond but not as a straight line b this makes sense when you recall COOH based molecules dimerise. Even when a H is missing from a COOH, it can still dimerise with water The H of one COOH bonds to an O of the other COOH. Here we are mussing H of COOH is missing here it can still dimerise and this time with water!!!

[pic]

The only thing I dispute is Na+ shown "coordinated" to the O (actually an O- in the diagram). As far as I am aware Na does NOT form dative covalent bonds with any atom *under typical conditions*. I would NOT have written such a thing in an exam. I probably would have shown Na+ beside the O- but I would just have to hope the examiner is learned enough to realize that I was correct and the mark scheme was wrong (P.S. The mark scheme is also wrong in Winter 2015 P41 Qa4, shown an SN1 reaction, as SN2 AND their wrong SN2 is doubly wrong as it shows the nucleophile coming in from the same side as the leaving group whereas in SN2's the incoming nucleophile attacks from directly behind the leaving group!!) whether that would satisfy the e

2014 (MJ) (41) 1 b (iv)

I'll discuss part (iii) here too......

1 b (iii) Whenever you get a question from a past years paper that uses data. Please remember that if you use a different data book, you might get a different answer (as the data may have been changed - for whatever reason!)

The data from the 2016 data book (your data book) is....

Cu2+ + 2e– ⇌ Cu +0.34V

Fe2+ + 2e– ⇌ Fe –0.44V

Youi know that Fe2+/Fe is the most negative reduction potential value therefore will be the one undergoing oxidation. Hence Cu2+/Cu will undergo reduction. We can uee the swap method to get the E(cell .... Swap the most negative reduction potential (i.e. change it's sign) and then, if their e- balance, just add the other equation to it. Simple.

- (-0.44) + ).34 = +0.78 V (don't forget the units)

Apparently, having been taught that method, some students have been told that it's "confusing" (and not that the confusion only lies with it's unfamiliarity to the person who declared it be confusing!!!) hence you may now not like that method. This is fine. Use whatever method YOU find best as you should always do when various methods are available.

1 b (iv) This Q appeared before the Nernst(variant) equation addition to the 2016 syllabus. Actually the 'Nernst' equation makes it bit easier to realize what's happening. (well, if they wrote it properly in the syllabus that is!!)

The convention has it that the most negative cell on the LEFT hand side. (the CIE book does exactly the opposite!!! *sigh*). There is only one exception.... when one of the electrodes is the SHE. Then it is the SHE that is shown on the Left. Solution C in the diagram appears on the Left of the cell diagram, so solution C refers to Fe2+(aq).

Recalling the Fe2+/Fe equation..

Fe2+ + 2e– ⇌ Fe –0.44V

The oxidised form is Fe2+ and the reduced form is Fe, but Fe is a solid so has a concentration of unity i.e. [Fe(s)] = 1 The number of electrons is 2 and the coefficient of Fe2+ in the reduction potential equation is one. REMEMBER THIS VERSION OF THE NERNST EQUATION IS ONLY FOR A HALFCELL.

[pic]

x and y are the coefficient of the respective species in the redox equation

So solution C (Fe2+) concentration increases, The value of the log term will increase but only slightly, so a larger positive value will be added to the E( of the half cell. Adding a positive number to a negative E(, will cause the new E value to be less negative.

2014 (MJ) (41) 3 (b)ii, d(iii), iv

The stereoisomer in Piperic acid in the diagram is shown as being both trans,trans. So simply draw a similar molecule (with the same connectivity!!) with, say, cis,trans.

d(iii)

You need to calculate: Starting pH. (it's actually given to you: "pH of a 0.150 mol dm–3 solution of piperidine is 11.9" Calculate the volume of equivalence, where the moles of HCl added equals the moles of piperidine originally present (check the balanced chemical equation!!!). This is where the vertical region will be. There is only one strong component, the acid. So the vertical region (drawn 4 units long) will be in the acidic region from pH7 to pH3.

You also need to show the final pH. Draw a gently sloping curve from pH3 (where the vertical section ends) to the final pH. To get the final pH, you need to consider how many moles of HCl were added AFTER the equivalence point (this is HCl that has nothing left to react with!) and consider the TOTAL volume of the solution mixture at the end. Students are often bad at doing both of these!

2014 (MJ) (41) 7 (b)iii,(c) i and ii

2014 (MJ) (41) 7 (b)iii

Especially with NMR spectra questions, READ THE WHOLE QUESTION FIRST. Actually, Do this with ALL questions!!!!! ALWAYS READ THE ENTIRE QUESTION FIRST.

Info leading up to this part of the question.... Protons at 1.1 are almost certainly CH3. Peaks aroud 1ppm are almost certainly always CH3 groups. You can see they are split into a triplet. They must be next to 2 H's, probably a CH2 group. CH3 and CH2 groups are incredibly commonly appearing groups at A-level and once again when I say "Groups" here, I do NOT mean functional groups.

So it looks like we CH3-CH2 group.

We have M+ : (M+1)+ peak info. M and (M+1)+ info is usually talking about 12C to 13C ratio.

The number of C's can be deduced.... [pic]

Which gives 3.07, i.e. 3 C's

So there is a "final" C at appears at about 12ppm. This is an "NMR 'magic number'" which corresponds to a carboxylic acid!

CH3-CH2-COOH is therefore the proposed molecule. This fits perfectly with other info. The peak ar 12ppm isn't split. (H's in OH's are never shown as causing splitting at A-level) and the peak gets smaller on addition of D2O. H's in COOH as they are in alcohols and amines are labile.

2014 (MJ) (41) 7 c (i)

For a molecule with 6H and only a very small number of absorptions, two, (i.e. two environments) we are probably going to be looking at molecules with a few of H in the same environment. The quickest way to do this would be to put them in two methyl groups, but have those methyl groups in different environments.

IF you were unable to deduce a molecule here I imagine you made the mistake of only using one oxygen whereas the actually number of oxygen's in G is two.

Molecules with 2 O's in them encounters in 9071 chemistry are esters and diols. Also double bonds to atoms can be replaced by a ring of C'. That should give you a clue as to what possible isomers of G can be offered here. (if you're still stuck...... how about methyl ethanoate)

Q7 part c ii) is actually easy once you've proposed your molecules. Just read off their group type from the data book and offer it as a value in the answer.

2014 (MJ) (41) 8 (a) iii

Questions relating to the actions of drugs are (imho) much less likely to occur in the 2016 syllabus/exams. This Q is basically saying "Which drug will be hydrolyzed in the stomach (where the drug will eventually go if it is taken orally). Do take note, the question is one of these "NEGATIVE" questions. This type of question automatically causes students to answer the question wrongly. Which is why, yet again, it is essentially you read the whole question first - and do so carefully!!! and once you've given an answer read the question again, asking yourself.... DID I ANSWER THE QUESTION ASKED? To answer this question, you are looking for a drug with an ester group or an amide group (or I guess a C(N group??? although C(N's do need a fair bit of heat to hydrolyze them!) B contains an amide so we can discount it. Note the AR-OCH3 group in C is an ether and not an ester! Also not that B contains an ester group too so is "doubly" unsuitable.

2014 MJ(42) 7

electrophoresis = The charge in the amino acid that results from the pH conditions of the buffer solution used. ................
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