Aqa chem 4/ chem 5 june 2016 thread

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Can someone please explain to me how to work out the rate determining step!! I always get it wrong :frown:

The rate determining step will always have the compounds in the rate equation (or compounds derived from it).
If that doesn't make any sense, could you give an example of a question? I could try to work through it.

Reply 361

bethl303

Original post by okopops

CH3COOH <----> CHCCOO + H+
Think of it as an equilibrium: if a base is added then the OH- react with the H+, which makes eqm to shift to the RHS to replace the H+ lost (so you add on the right )if acid added then eqm shifts to the left to oppose increase of H+ ( so you add on the left)

Thanks so much

Reply 362

emma_1111

Original post by Aethrell

So can anyone confirm whether we need to include δ+/ δ− signs on the carbons/oxygens/halides and whether the M+1 peak is actually on the AQA spec?

Many thanks

What is the M+1 peak?

Reply 363

Aethrell

Original post by 12284

No you don't need to include it.
And yep it is on the spec but I haven't seen a question on it yet.

This came up on the June 2010 paper and the mark scheme said it had 5 peaks.

Wouldn't there be 6 if the M+1 peak was included?

Reply 364

student1856

Can someone please help with 8ci) June 2013. I feel like I know the answer but I don't.

Reply 365

SirRaza97

Original post by student1856

Can someone please help with 8ci) June 2013. I feel like I know the answer but I don't.

Haha i just did the paper! It's the 7 marker one right?

Intermediate - CH3CH2CN
Route A:

Stage 1 - KCN/HCN under ethanolic conditions
Stage 2 - Ni/H+ under ether conditions

Route B - Ammonia in XS

Reply 366

12284

Original post by Aethrell

This came up on the June 2010 paper and the mark scheme said it had 5 peaks.

Wouldn't there be 6 if the M+1 peak was included?

What do you mean?
There is one peak for the molecule with 4 Cl-35, one peak for molecule with 3 Cl-35 and 1 Cl-37, one peak for the molecule with 2 Cl-35 and 2 Cl-37, one peak for the molecule with 1 Cl-35 and 3 Cl-37 and the fifth peak for the molecule with 5 Cl-37.
Have you done the Jan 2012 paper yet?

Reply 367

jammypancake

I have a few questions so would appreciate any answers!
1. How do you work out conjugate base pairs?
2. What is the difference in pOH and pH?
3. What is double equivalence?
4. Difference in racemic mixture and enantiomers?
5. Main chromatography methods we need to learn?
thanks in advance!! much appreciated

Reply 368

jammypancake

Also does anyone else just not understand a single thing about analytical techniques?!!!

Reply 369

Splashoid

Original post by student1856

Can someone please help with 8ci) June 2013. I feel like I know the answer but I don't.

Route A:
- Add KCN. Nucleophilic substituion occurs and the product is Ch3Ch2CN
- Then add H2/Nickel catalyst under high temperature and pressure (catalytic hydrogenation) and i'ts reduced to Ch3Ch2CH2NH2
- equation: CH3Ch2CN + 2H2 = Ch3Ch2CH2NH2

Route B:
- Add excess NH3. Mechanism is nucleophilic substitution. Br is substituted by NH3.

Reply 370

lahigueraxxx

Last year's paper looked like a dream! I'm gutted - no doubt AQA have something nasty in store for us tomorrow :frown:

Reply 371

IvoryToast

Any one else getting the feeling that each year the unit 4 paper gets even harder. Just done last years and only got 70% after I'd been getting 90% in all the others?? Just how bad can they make this years? :/

Reply 372

hopingmedicinae

Can someone aid me in interpreting this- not sure what is means when it says two equivalence points consequently cannot answer the questions

tsr.png13.4KB

Reply 373

Guls

kekule suggested benzene would have alternating c-c and c=c bonds like in the hypothetical cyclohexatriene.
This was disproved as after doing a reduction reaction using Hydrogen and a nickel catalyst. It was suggested that benzene would have an enthalpy change that was thrice that of cyclohexene (3x-120 Kj Mol^-1 so -360). In fact, benzene had an enthalpy change of -208 which suggests that it was 152 Kj Mol^-1more stable.

Spoiler

The old cyclohexatriene idea was scraped and the new model of where the electrons above the carbons formed two rings of delocalised electrons, so no double bonds. Its now known that the lengths of the c-c bonds are in between the average lengths of the c-c and c=c.

Original post by Lilly1234567890

guys, i am slightly confused about benzene. Does its structure have a range of single c-c bonds and c=bond. ? And did Kekulu suggest that the bonds are all of equal length? is that correct.

Reply 374

Aethrell

Original post by 12284

I know that - going by the CGP textbook there would be a 6th peak due to the presence of Carbon 13. Seeing as the answer is 5, I thought it wouldn't be on the syllabus. I was asking for confirmation.

Yes, I've done all of the new spec ones

Reply 375

dnan

Original post by IvoryToast

No, in all honesty I think every paper so far has been easy (jan10 easiest) to moderate ( jan12) with Jun15 being in between, and I hope the paper is like all previous ones, however I can almost guarantee they'll switch it up and make it harder, and unlike previous years.

Reply 376

Guls

Original post by hopingmedicinae

Can someone aid me in interpreting this- not sure what is means when it says two equivalence points consequently cannot answer the questions

Na2CO3 + HCl → NaCl + NaHCO3
NaHCO3 + HCl → NaCl + H2O + CO2
The conjugate acid produced from the first equation is so weak that it can accept another proton to form NaCl creating two concentrations of where the [OH]=[H⁺]

Reply 377