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BedokFunland JC's A Level H2 Chemistry Qns (Part 2)

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    H2 Chemistry @ BedokFunland JC (near VJC & TJC)
    UltimaOnline's Avatar
    13,571 posts since May '05
    • A BedokFunland JC Original H2 Chemistry Challenge

      Considering only the 1st proton dissociation (ie. regardless of proticity), ie. magnitude of Ka1 value, H2SO4 is a stronger acid compared to H2SO3, just as HNO3 is a stronger acid compared to HNO2, which is indeed expected and concordant to the general trend of acidic strength in any given series of oxoacids. However, bucking the trend are the oxoacids of phosphorus, in which H3PO3 is actually a stronger acid compared to H3PO4. There are actually 2 independent reasons for this unusual behavior, the combined effect of both, outweighs the expected general trend for oxoacid families (I assume you know how to explain the much easier within-H2-Chemistry-syllabus question of why H2SO4 is stronger than H2SO3, and why HNO3 is stronger than HNO2... do you?).

      My BedokFunland JC Challenge for you, of course, is : Explain why H3PO3 is a stronger acid compared to H3PO4.

      If you can correctly explain just 1 of the 2 reasons, you're already beyond A grade Distinction level (because even 99% of A graders can't solve this challenge question). If you can correctly explain both of the 2 reasons, you're BedokFunland JC level.

      As with all my BedokFunland JC challenge questions, I won't reveal the correct answers here. My BedokFunland JC students can ask me during tuition, and all other students can go ask your school teacher or private tutor.

  • Moderator
    H2 Chemistry @ BedokFunland JC (near VJC & TJC)
    UltimaOnline's Avatar
    13,571 posts since May '05
    • Originally posted by MapPwner:

      Can I ask a question regarding the 2017 A level H2 Chemistry Specimen Paper 2 Question 6(i),which involves formation of FAME from reaction between a fatty acid,RCO2H,and diazomethane,CH2N2.

      Reaction occurs via two step mechanism:

      -The fatty acid reacts with diazomethane to form a carboxylate ion intermediate in the first step.

      -N2 is formed in the second step

      Suggest the mechanism for this reaction.(4 marks)

      Am i right to say that in the first step of the mechanism,CH2N2 deprotonates RCO2H to form RCOO- and CH3N2 respectively?And then subsequently the C-N bond in CH3N2 breaks and the bond pair is donated to positive charged N atom to form N2,before the RCOO- attacks the CH3 thingy to form RCO2CH3?

       


      Almost correct, but your described mechanism would be 3 steps, when Cambridge already specified 2 steps. The reason for this is the instability of the methyl carbocation, which is why the next step (after the proton transfer) is SN2, not SN1. Ideally, for the student to demonstrate a correct understanding of this reaction, both resonance contributors of diazomethane should be drawn, with the student showing to Cambridge that he/she knows which resonance contributor more appropriately functions as the Bronsted-Lowry base in the 1st step of the reaction mechanism.

  • Moderator
    H2 Chemistry @ BedokFunland JC (near VJC & TJC)
    UltimaOnline's Avatar
    13,571 posts since May '05
  • Moderator
    H2 Chemistry @ BedokFunland JC (near VJC & TJC)
    UltimaOnline's Avatar
    13,571 posts since May '05
    • A BedokFunland JC H2 Chemistry Question

      The amount of CN−(aq) ions in an analyte solution can be determined by titration against Ag+(aq), during which the soluble dicyanoargentate(I) coordination complex is first generated, until sufficient Ag+(aq) is added from the burette to generate the insoluble white precipitate of the silver(I) dicyanoargentate(I) coordination compound. If x mol of Ag+(aq) had been added at first sighting of the white precipitate, what was the amount of CN−(aq) ions present in the analyte solution?

      A) 0.5 x
      B) 1.0 x
      C) 2.0 x
      D) None of the above

  • Moderator
    H2 Chemistry @ BedokFunland JC (near VJC & TJC)
    UltimaOnline's Avatar
    13,571 posts since May '05
    • An Original BedokFunland JC H2 Chemistry Challenge

      image

      https://en.wikipedia.org/wiki/Histidine

      Q1. Explain fully the relative basicities of the 3 N atoms in Histidine (ie. explain which N atom on the R group imidazole ring is more basic, and explain whether the alpha N atom or the R group imidazole ring is more basic).

      Q2. Given pKa1 = 1.60, pKa2 = 5.97 and pKa3 = 9.28, calculate the isoelectric point of Histidine.

      Q3. Calculate the pH at all significant points (ie. initial, 1st maximum buffer capacity, 1st equivalence point, 2nd maximum buffer capacity, 2nd equivalence point, 3rd maximum buffer capacity, 3rd equivalence point) in
      (i) the titration of the fully deprotonated form of Histidine against a strong monoprotic acid added in excess, as well as in
      (ii) the titration of the fully protonated form of Histidine against a strong monoprotic base added in excess.
      Use 0.05M as the molarities for Histidine and both titrants, and 25cm3 as the analyte volume of the Histidine solution for both titrations. Include a sketch of both titration curves, with all significant pH values labeled.

      Q4. By considering the enthalpy and entropy changes during decarboxylation, explain how temperature affects the thermodynamic feasibility and Kc value for the decarboxylation reaction.

      Q5. Suggest and draw out the curved-arrow electron-flow reaction mechanism for the (non-enzyme catalyzed) decarboxylation of Histidine.

      Edited by UltimaOnline 22 Oct `17, 4:20AM
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