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AQA A-level Chemistry paper 2 7405/2 Unofficial Markscheme Watch

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    well this paper was not as straightforward as the paper 1 in my opinion.
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    did you get n ethylpropanamide for the name
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    (Original post by AR123456789)
    well this paper was not as straightforward as the paper 1 in my opinion.
    ye idk why people found this one easier
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    I hope the grade boundaries get lowered
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    (Original post by Khanman123)
    Jeeez, I messed up that last question man. 145mg for that other one. I got 15.8 or suttin like that for Ea.
    Same! But I think it's bc they gave us lnA and not A or something
    It's shouldn't be tooo many marks
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    (Original post by eza11)
    i got 0.074 for tangent but its dependent on the line u draw got like 0.38/ 5.0
    Nah that will be outside the range on the mark scheme you should've had somewhere around 0.045 to get the marks
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    (Original post by DanB9)
    Nah that will be outside the range on the mark scheme you should've had somewhere around 0.045 to get the marks
    how tho? what did u get as ur top number?
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    I put LiAlH4 on the last question is that right or was it meant to be HCl?
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    I'm so baffled how are people saying that paper was easy...
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    (Original post by p____)
    I put LiAlH4 on the last question is that right or was it meant to be HCl?
    You're right
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      Would i get marks for the 2 evidences even though i drew my nmr thing wrong 😭
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      For the question about Ea.
      Did you guy use
      lnA as 16.9 or ln16.9.
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      (Original post by Khanman123)
      Jeeez, I messed up that last question man. 145mg for that other one. I got 15.8 or suttin like that for Ea.
      (Original post by brumtown0121)
      U mans know the drill by now, if ur answersmatch, thumbs up the post ygm:

      - 9.07g (first question mass thing)
      - 0.04 (t=0 tangent initial rate)
      - 0.65
      - 0.0281 mol^-1 dm^3 s^-1 (rate constant, k)
      - 50.7 kjmol^-1 (the activation energy)
      - 0.62 (rearrange kc and work out moles of suttin)
      - 1.45x10^-4 (took an L on this question, correct answer is 145mg)
      - 3 peaks in carbon 13 nmr
      - 2 peaks in proton nmr
      - There was a nucleophilic addition-elimination reaction somewhere and the product was N-ethylpropanamide

      - the final 2 questions, basically for the first one, you gotta talk about inductive effect and benzene ring causing lone pair of electron on n of the ammonia, to become partially delocalised onto the ring, making it less available, the one where the ammonia was furthest away from the benzene ring with an alkyl group before it (ch2ch2) was the strognest, due to positive inductive effect. the weakest base was the one where the ch2ch2 was at the end, and ammonia was inbetween the benzene ring and the alkyl group, due to negative inductive effect, and lone pair ultimately fully delocalising onto the ring), the middle one was just the ammonia attached to the ring.
      (I think it was FEG or suttin)

      - for the final one though, for the intermediate you should get a chloro or bromo group attached to the ch2, and it is ultraviolet light + cl2 or br2
      then for the second intermediate, you should have the cl or br replaced by a CN, and reagent = kcn, conditions is aqueous & ethanol as solvent)

      then the final stage (dont know if u had to talk about it), but basically heat the second intermediate under reflux with conc hcl/sn, to form the final product.

      ayt safe
      I got Ea as 15 something as well on the specimen paper i got it wrong because i put it all in the calc at once and for some reason log laws I think it didnt work out. When i did ln k and ln a seperately and then substituted I got it right. I did the same in the exam and got 15 something im pretty I am right on this. Also fam for the mg all you had to do was pv =nrt work out moles times by mr x10? not sure about that one.
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      Did anyone got this for question 7 ?
      so I wrote for the 2 molecules with COOH group which one of them had a OH ( first molecule and 3rd? ) test for carboxylic acid adding an alcohol and con?.H2SO4 both will produce a sweet smell and then test for alcohol using acidified KMnO4(purple-> colourless) or you could use acidified K2Cr2O7 (orange->green) if primary or secondary alcohol present. Tollens(silver mirror)/ Fehlings(red ppt) for the aldehyde( the one at the very right end)
      Second one had a Br group and an OH ? use NaOH and AgNO3 (cream ppt) and test for alcohol
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      (Original post by GangRelated)
      For the question about Ea.
      Did you guy use
      lnA as 16.9 or ln16.9.
      ln16.9
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      For the last question I said react with KCN and then LiAlH4 in ether? Is this wrong?
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      (Original post by p____)
      I put LiAlH4 on the last question is that right or was it meant to be HCl?
      LiALH4 if correct also you could add Pt/Ni catalyst that was probably the condition ?
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      (Original post by Hassankhan2405)
      Did anyone get 870Mg for one of the questions? The Ideal Gas Equation?
      Yeah I got 870mg as well.
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      (Original post by sername)
      Number of moles = 0.789 (maybe for the Kc question?)
      rate constant k = 0.028
      H2/Ni Catalyst or LiAlH4
      The methylbenze should be in excess (in order for 1 substitution only)
      I got EGF (not sure)
      To distinguish between KLMN, add tollen's reagent to get N (only one with silver mirror), NaHCO3 to the remaining KLM to get the one without COOH (L I think, only one WITHOUT effervesence) to KM add K2Cr2O7/c.H2SO4 to identify K (only one with orange to green) and therefore M (by elimination)
      145 mg for P (I think)

      Correct me if I'm wrong
      I was a bit confused how to structure my answer for the distinguish question because most got more than one group each and some were the same as others.
      I just wrote all the tests out tbh. :/
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      (Original post by brumtown0121)
      U mans know the drill by now, if ur answersmatch, thumbs up the post ygm:

      - 9.07g (first question mass thing)
      - 0.04 (t=0 tangent initial rate)
      - 0.65
      - 0.0281 mol^-1 dm^3 s^-1 (rate constant, k)
      - 50.7 kjmol^-1 (the activation energy)
      - 0.62 (rearrange kc and work out moles of suttin)
      - 1.45x10^-4 (took an L on this question, correct answer is 145mg)
      - 3 peaks in carbon 13 nmr
      - 2 peaks in proton nmr
      - There was a nucleophilic addition-elimination reaction somewhere and the product was N-ethylpropanamide

      - the final 2 questions, basically for the first one, you gotta talk about inductive effect and benzene ring causing lone pair of electron on n of the ammonia, to become partially delocalised onto the ring, making it less available, the one where the ammonia was furthest away from the benzene ring with an alkyl group before it (ch2ch2) was the strognest, due to positive inductive effect. the weakest base was the one where the ch2ch2 was at the end, and ammonia was inbetween the benzene ring and the alkyl group, due to negative inductive effect, and lone pair ultimately fully delocalising onto the ring), the middle one was just the ammonia attached to the ring.
      (I think it was FEG or suttin)

      - for the final one though, for the intermediate you should get a chloro or bromo group attached to the ch2, and it is ultraviolet light + cl2 or br2
      then for the second intermediate, you should have the cl or br replaced by a CN, and reagent = kcn, conditions is aqueous & ethanol as solvent)

      then the final stage (dont know if u had to talk about it), but basically heat the second intermediate under reflux with conc hcl/sn, to form the final product.

      ayt safe
      Just realised I flopped some questions I thought I got right...
     
     
     
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