01 Sep, 10:06PM in sunny Singapore!

Recent Posts by UltimaOnline

Subscribe to Recent Posts by UltimaOnline

  • Moderator
    A Level Chem, O Level Chem + Phys / Bio @ BedokFunland JC
    UltimaOnline's Avatar
    5,641 posts since May '05
  • Moderator
    A Level Chem, O Level Chem + Phys / Bio @ BedokFunland JC
    UltimaOnline's Avatar
    5,641 posts since May '05
  • Moderator
    A Level Chem, O Level Chem + Phys / Bio @ BedokFunland JC
    UltimaOnline's Avatar
    5,641 posts since May '05
  • Moderator
    A Level Chem, O Level Chem + Phys / Bio @ BedokFunland JC
    UltimaOnline's Avatar
    5,641 posts since May '05
    • Azly J Nor (Teh Tarik Party) shares about his role as a pioneer member of SMRT Ltd (Feedback)

      https://www.facebook.com/notes/718297924981280/

      Excerpt :
      I developed the concept of Teh Tarik Economics to represent the philosophical idea of determinism where every human decisions and actions are based from preceding circumstances, both cognitive and metaphysical, to cause a future chain of events shaped out of our individual motives and desires. Such events are then influenced through Psychological Totalism - the idealogical concept of total control over human behaviour and thought.

  • Moderator
    A Level Chem, O Level Chem + Phys / Bio @ BedokFunland JC
    UltimaOnline's Avatar
    5,641 posts since May '05
  • Moderator
    A Level Chem, O Level Chem + Phys / Bio @ BedokFunland JC
    UltimaOnline's Avatar
    5,641 posts since May '05
  • Moderator
    A Level Chem, O Level Chem + Phys / Bio @ BedokFunland JC
    UltimaOnline's Avatar
    5,641 posts since May '05
  • Moderator
    A Level Chem, O Level Chem + Phys / Bio @ BedokFunland JC
    UltimaOnline's Avatar
    5,641 posts since May '05
  • Moderator
    A Level Chem, O Level Chem + Phys / Bio @ BedokFunland JC
    UltimaOnline's Avatar
    5,641 posts since May '05
    • Originally posted by Mrworry:

      Xiexie~

      Btw I am not sure why but the answer given for first part is CH3COCOH and second part is CH3OCH2CHO. @.@ 

      Are my answers accepted? 


      Yes, your answers are fine. The given answer is fine for the 2nd qn, but wrong for the 1st qn (count the no. of H atoms).

  • Moderator
    A Level Chem, O Level Chem + Phys / Bio @ BedokFunland JC
    UltimaOnline's Avatar
    5,641 posts since May '05
    • Originally posted by Mrworry:

      c3h6o2 which does not produce triodomethane but reduce fehling soln

      CH2(OH)CH2OCH 

      Does this fulfill the req, if not why? Thanks 


      Your structure is in error. Ensure that you satisfy the valencies of O and C such that all atoms have a stable octet but no formal charge.

      Update : MrWorry, you probably got the correct structure, but you expressed the condensed structural formula wrongly. You should write it as follows (ie. the way I wrote it below).

      The correct answer should be : CH2(OH)CH2CHO

      Edited by UltimaOnline 30 Aug `15, 5:03PM
  • Moderator
    A Level Chem, O Level Chem + Phys / Bio @ BedokFunland JC
    UltimaOnline's Avatar
    5,641 posts since May '05
    • Originally posted by Mrworry :

      A compound C3H6O2 which produces CHI3 and reduce Fehling solution. Does CH2(OH)COCH3 fulfill this requirement?

      UltimaOnline aka BedokFunland JC replied :

      Yes it does. This qn would kill A level H2 Chem students though ;Þ

      Originally posted by Mrworry :

      c3h6o2 which does not produce triodomethane but reduce fehling soln

      CH2(OH)CH2OCH 

      Does this fulfill the req, if not why?

      UltimaOnline aka BedokFunland JC replied :

      Your structure is in error. Ensure that you satisfy the valencies of O and C such that all atoms have a stable octet but no formal charge.

      Update : MrWorry, you probably got the correct structure, but you expressed the condensed structural formula wrongly. You should write it as follows (ie. the way I wrote it below).

      The correct answer should be : CH2(OH)CH2CHO

      Originally posted by Mrworry:

      Xiexie~

      Btw I am not sure why but the answer given for first part is CH3COCOH and second part is CH3OCH2CHO. @.@ 

      Are my answers accepted? 


      UltimaOnline aka BedokFunland JC replied :

      Yes, your answers are fine. The given answer is fine for the 2nd qn, but wrong for the 1st qn (count the no. of H atoms).

      Edited by UltimaOnline 30 Aug `15, 5:08PM
  • Moderator
    A Level Chem, O Level Chem + Phys / Bio @ BedokFunland JC
    UltimaOnline's Avatar
    5,641 posts since May '05
    • Originally posted by Mrworry:

      Hi

      A compound C3H6O2 which produces CHI3 and reduce Fehling solution. Does CH2(OH)COCH3 fulfill this requirement? Thanks 


      Yes it does. This qn would kill A level H2 Chem students though ;Þ

  • Moderator
    A Level Chem, O Level Chem + Phys / Bio @ BedokFunland JC
    UltimaOnline's Avatar
    5,641 posts since May '05
    • Q1. Writing the mechanism is the minimum required to get the marks. Better to draw out the curved arrows. For FRS, Cambridge doesn't require it and Singapore JCs don't teach the full curved arrow mechanisms, simply because they don't think H2 Chem students are smart enough to draw the full mechanisms, hence it's reserved for H3 Chem, Olympiad Chem, and *ahem* BedokFunland JC students to draw out the full curved arrow mechanisms for FRS and other reactions (eg. nucleophilic acyl substitutions) not required by the basic H2 Chem syllabus.

      Q2. There will indeed be 12 chiral C atoms. You missed out the new sp3 C atoms that used to be the sp2 C atoms of the alkene.

      Q4. By looking at the rate equation (which will usually be given for the H2 syllabus). If the kinetic orders for each reactant aren't given or can't be deduced (eg. table of kinetic data), then the question must be such that you can make reasonable assumptions and deductions (comparing the stabilities of the reactants vs products, and concordantly the Ea required), but these are beyond the expectations of the H2 syllabus, and particularly for this reaction, it's gonna be a pretty close toss up between the 3 elementary steps, so if Cambridge asks this, they'll definitely give additional clues (most probably kinetics order data), and/or make the different Ea required for the 3 elementary steps more obviously varying and distinctive.

      Q5. Option D (and for that matter, C as well) is wrong because it focuses on the C-C bond instead of the C-H bond.



      Edited by UltimaOnline 30 Aug `15, 1:36AM
  • Moderator
    A Level Chem, O Level Chem + Phys / Bio @ BedokFunland JC
    UltimaOnline's Avatar
    5,641 posts since May '05
  • Moderator
    A Level Chem, O Level Chem + Phys / Bio @ BedokFunland JC
    UltimaOnline's Avatar
    5,641 posts since May '05
  • Moderator
    A Level Chem, O Level Chem + Phys / Bio @ BedokFunland JC
    UltimaOnline's Avatar
    5,641 posts since May '05
    • You're right that the position of equilibrium has shifted to the RHS, generating a *slightly* greater number of moles of RHS species at equilibrium.

      However, because the volume has doubled (ie. *greatly* increased), hence even with the *slight* increase in moles of RHS species, the new equilibrium molarities of the RHS species still decrease (though the decrease is not as much as the LHS species, whose moles and molarities *both* decrease).

      Originally posted by Light5:

      Q.) http://gceguide.com/papers/A%20Levels/Chemistry%20(9701)/9701_s15_qp_11.pdf

      In q 6 of the above paper, if volume is increasing from 1 to 2 , then pressure should decrease therefore equilibrium must shift to the RHS to increase No of Moles of PCl3 and Cl2 and decrease No of Moles of PCl5. This is reflected in both answer choices A and B yet the examiner has given B as the answer in the marking scheme? Can you please clarify this?

       

       


  • Moderator
    A Level Chem, O Level Chem + Phys / Bio @ BedokFunland JC
    UltimaOnline's Avatar
    5,641 posts since May '05
    • You're right that the position of equilibrium has shifted to the RHS, generating a *slightly* greater number of moles of RHS species at equilibrium.

      However, because the volume has doubled (ie. *greatly* increased), hence even with the *slight* increase in moles of RHS species, the new equilibrium molarities of the RHS species still decrease (though the decrease is not as much as the LHS species, whose moles and molarities *both* decrease).

      Originally posted by Light5:

      Q.) http://gceguide.com/papers/A%20Levels/Chemistry%20(9701)/9701_s15_qp_11.pdf

      In q 6 of the above paper, if volume is increasing from 1 to 2 , then pressure should decrease therefore equilibrium must shift to the RHS to increase No of Moles of PCl3 and Cl2 and decrease No of Moles of PCl5. This is reflected in both answer choices A and B yet the examiner has given B as the answer in the marking scheme? Can you please clarify this?

       

       


  • Moderator
    A Level Chem, O Level Chem + Phys / Bio @ BedokFunland JC
    UltimaOnline's Avatar
    5,641 posts since May '05
    • Hi Gohby,

      Q1. You proceed by drawing (or if you're good enough, visualizing) the mechanism, ie. with curved arrows, and not just writing the mechanism, as poorly taught in Singapore JCs to H2 students (the majority of whom just blindly memorize the written mechanisms without deeper understanding). If you're good enough, always draw out the full mechanism in the A level exams. Then the answer to this MCQ will become understandable and self-evident.

      Q2. There will indeed be 12 chiral C atoms. You missed out the new sp3 C atoms that used to be the sp2 C atoms of the alkene.

      Q3. Thiols nucleophilically substitutes away the halide leaving group more readily (ie. lower Ea and greater yield at equilibrium) than alcohols do, as shown in reaction 1. Thiols are able to reduce Br2 to Br-, a reaction not seen with alcohols, as shown in reaction 2.

      Q4. Step 3 may not actually be the true rate determining step. Don't trust JC prelim papers' qns and ans. Regardless, in step 3 the reactant behaves as a Bronsted-Lowry base rather than a nucleophile, and hence steric hindrance (ie. being bulky) is irrelevant. Also, for the reactant in step 2, the -ve charge isn't only on the C atom, but is delocalized by resonance over to the carbonyl O atom (that's why it was possible to deprotonate the aldehyde in the first place).

      Edited by UltimaOnline 30 Aug `15, 1:36AM
  • Moderator
    A Level Chem, O Level Chem + Phys / Bio @ BedokFunland JC
    UltimaOnline's Avatar
    5,641 posts since May '05
  • Moderator
    A Level Chem, O Level Chem + Phys / Bio @ BedokFunland JC
    UltimaOnline's Avatar
    5,641 posts since May '05
    • Originally posted by Mrworry:

      Hi again :D

      For Organic chem made easy page 258,

      Example 8.4, part b)

      Why cannot it be:

      1. Add conc H2SO4(aq) 170C

      2.Cl2 (aq), heat

      3. Add Ethanolic KCN, heat

      4. HCl (aq), reflux

      xie2=3



      Your proposed synthesis pathway will score some marks, but not full marks.

      You'll be penalized *mostly* because your proposed synthesis pathway isn't the shortest possible (ie. George Chong's answer). Always give the shortest possible pathway, because using a longer-than-necessary synthesis pathway will result in a lower atom economy and/or equilibrium yield, and hence will always be penalized by Cambridge. (But hey, if you can't think of any shorter pathway, getting some marks is certainly preferable to getting no marks, so go ahead and just write the shortest pathway you can think of, even if it's 1 or 2 steps longer than specified by the qn).

      In addition (this is a minor point), although your step 2 will indeed obtain the required intermediate (used in your step 3) as the major product, but nonetheless there will still be a mixture of products, which should still be avoided in synthesis qns whenever possible (it's perfectly fine to use it when there's no other choice, but in this qn, there *is* another, even shorter pathway, ie. George Chong's answer).

      But I emphasize that the main problem with your answer is that it isn't the shortest possible synthesis pathway. If it had the same no. of steps as the shortest possible pathway, then Cambridge (in most cases) wouldn't penalize you for using electrophilic addition of aqueous halogen. No worries.

      On this note, using halogenation via free radical substitution will always get you penalized for synthesis qns (as it generates a variety of products, hence low overall atom economy which means wasteful on resources), unless there is absolutely no other choice for that qn.

      Edited by UltimaOnline 27 Aug `15, 3:00PM
  • Moderator
    A Level Chem, O Level Chem + Phys / Bio @ BedokFunland JC
    UltimaOnline's Avatar
    5,641 posts since May '05
    • You really love George Chong's book, doncha? He's busy climbing Mont Blanc right now even as we type, but up there in the lonely peaks of Mont Blanc, he'll surely appreciate the support of a loyal fan! ;Þ

      Your proposed synthesis pathway will score some marks, but not full marks.

      You'll be penalized *mostly* because your proposed synthesis pathway isn't the shortest possible (ie. George Chong's answer). Always give the shortest possible pathway, because using a longer-than-necessary synthesis pathway will result in a lower atom economy and/or equilibrium yield, and hence will always be penalized by Cambridge. (But hey, if you can't think of any shorter pathway, getting some marks is certainly preferable to getting no marks, so go ahead and just write the shortest pathway you can think of, even if it's 1 or 2 steps longer than specified by the qn).

      In addition (this is a minor point), although your step 2 will indeed obtain the required intermediate (used in your step 3) as the major product, but nonetheless there will still be a mixture of products, which should still be avoided in synthesis qns whenever possible (it's perfectly fine to use it when there's no other choice, but in this qn, there *is* another, even shorter pathway, ie. George Chong's answer).

      But I emphasize that the main problem with your answer is that it isn't the shortest possible synthesis pathway. If it had the same no. of steps as the shortest possible pathway, then Cambridge (in most cases) wouldn't penalize you for using electrophilic addition of aqueous halogen. No worries.

      On this note, using halogenation via free radical substitution will always get you penalized for synthesis qns (as it generates a variety of products, hence low overall atom economy which means wasteful on resources), unless there is absolutely no other choice for that qn.

      Originally posted by Mrworry:

      Hi again :D

      For Organic chem made easy page 258,

      Example 8.4, part b)

      Why cannot it be:

      1. Add conc H2SO4(aq) 170C

      2.Cl2 (aq), heat

      3. Add Ethanolic KCN, heat

      4. HCl (aq), reflux

      xie2=3


      Edited by UltimaOnline 27 Aug `15, 2:59PM
  • Moderator
    A Level Chem, O Level Chem + Phys / Bio @ BedokFunland JC
    UltimaOnline's Avatar
    5,641 posts since May '05
    • Originally posted by Light5:

      Sorry if the interhalogen question was vague....i actually constructed the question by myself because cambridge has a knack of asking weird questions from time to time. They have already tested displacement reactions of interhalogen compounds so i wondered if they could test a question like this as well.

      In q1(c) , we would obviously have a colourless test tube as all nitrates are soluble and AgCl is also soluble but what would be the products of the nucleophilic attack of NH3 and H2O on BrNO3?

      In q13(from the CIE paper), how can i predict the temperature required for decomposition of the 3 fire retardants since there are 2 Group 2 elements in one fire retardant.

      Thanks Again.


      No prob.

      Q1c) NH2Br + HNO3, and BrOH + HNO3, all of which are colourless.

      Q13. Each metal carbonate will decompose according to their own temperature, and you can treat each mineral as a mixture of two separate metal carbonates. Eg. For Mg3Ca(CO3)4, magnesium carbonate decomposes first, followed by calcium carbonate. Hence BaCa(CO3)2 is the worst fire retardant because Ba2+ has the lowest cationic charge density and thus requires the highest temperature to decompose, so by the time sufficient CO2(g) is released, your house would have burnt down completely.

  • Moderator
    A Level Chem, O Level Chem + Phys / Bio @ BedokFunland JC
    UltimaOnline's Avatar
    5,641 posts since May '05
    • Originally posted by Light5:

      Sorry if the interhalogen question was vague....i actually constructed the question by myself because cambridge has a knack of asking weird questions from time to time. They have already tested displacement reactions of interhalogen compounds so i wondered if they could test a question like this as well.

      In q1(c) , we would obviously have a colourless test tube as all nitrates are soluble and AgCl is also soluble but what would be the products of the nucleophilic attack of NH3 and H2O on BrNO3?

      In q13(from the CIE paper), how can i predict the temperature required for decomposition of the 3 fire retardants since there are 2 Group 2 elements in one fire retardant.

      Thanks Again.


      No prob.

      Q1c) NH2Br + HNO3, and BrOH + HNO3, all of which are colourless.

      Q13. Each metal carbonate will decompose according to their own temperature, and you can treat each mineral as a mixture of two separate metal carbonates. Eg. For Mg3Ca(CO3)4, magnesium carbonate decomposes first, followed by calcium carbonate. Hence BaCa(CO3)2 is the worst fire retardant because Ba2+ has the lowest cationic charge density and thus requires the highest temperature to decompose, so by the time sufficient CO2(g) is released, your house would have burnt down completely.

  • Moderator
    A Level Chem, O Level Chem + Phys / Bio @ BedokFunland JC
    UltimaOnline's Avatar
    5,641 posts since May '05
    • Originally posted by Light5:

      Q1.)

      (a) State the colour you would expect for the interhalogen compound BrCl

      (b.) Predict the equation of the reaction between BrCl (interhalogen) and AgNO3.

      (c.) Describe what you would observe when the products of (a) are added to Dilute NH3 in a test tube.

      Q2.) http://gceguide.com/papers/A%20Levels/Chemistry%20(9701)/9701_s15_qp_11.pdf

         Please explain how to solve Q13 and Q24 from the above paper. 

      Mark Scheme :         http://gceguide.com/papers/A%20Levels/Chemistry%20(9701)/9701_s15_ms_11.pdf

      Thank You !


      Where did you get the interhalogen qn from? Is it a CIE 'A' level Chem qn, or a school's exam qn (if so, which country is the school based in)?

      Q1a) Somewhere between the colours of Cl2 (yellow-green) and Br2 (reddish-brown).

      Q1b) Although AgBr is less soluble than AgCl, but due to the greater electronegativity of Cl over Br, we have Br+ (not Br-), hence the products would be AgCl(s) and (covalent) BrNO3. BrCl is a strong oxidizing agent, but neither Ag+ nor NO3- (with N already in it's most positive OS) can be readily oxidized (it is difficult to oxidize Ag+ and O atoms having OS of -2).

      Q1c) The AgCl ppt would dissolve in NH3(aq), as the soluble ionic diamminesilver(I) coordination complex is generated. The Br atom of BrNO3 would also be nucleophilically attacked by NH3 and H2O, but the reaction products are not readily observable (hence irrelevant to the question).

      Q13. This qn tests you on the thermal stabilities of Group 2 carbonates (determined by cationic charge densities). As a fire retardant, you would want two things : greater moles of CO2(g) generated per mole of retardant, and lower thermal decomposition temperature to release the CO2(g) more readily.

      Q24. The mechanism involves 2 x nucleophilic substitution (first SN2 on the less sterically hindered terminal C atom, followed by deprotonation, then intramolecular SN2 or SN1 attack on the 5th C atom, followed by deprotonation) of the Br atoms on the 1st and 5th C atoms, by the NH3 nucleophile, to generate the required product Coniine.

      Edited by UltimaOnline 27 Aug `15, 2:13AM
  • Moderator
    A Level Chem, O Level Chem + Phys / Bio @ BedokFunland JC
    UltimaOnline's Avatar
    5,641 posts since May '05
    • Originally posted by Light5:

      Q1.)

      (a) State the colour you would expect for the interhalogen compound BrCl

      (b.) Predict the equation of the reaction between BrCl (interhalogen) and AgNO3.

      (c.) Describe what you would observe when the products of (a) are added to Dilute NH3 in a test tube.

      Q2.) http://gceguide.com/papers/A%20Levels/Chemistry%20(9701)/9701_s15_qp_11.pdf

         Please explain how to solve Q13 and Q24 from the above paper. 

      Mark Scheme :         http://gceguide.com/papers/A%20Levels/Chemistry%20(9701)/9701_s15_ms_11.pdf

      Thank You !


      Where did you get the interhalogen qn from? Is it a CIE 'A' level Chem qn, or a school's exam qn (if so, which country is the school based in)?

      Q1a) Somewhere between the colours of Cl2 (yellow-green) and Br2 (reddish-brown).

      Q1b) Although AgBr is less soluble than AgCl, but due to the greater electronegativity of Cl over Br, we have Br+ (not Br-), hence the products would be AgCl(s) and (covalent) BrNO3. BrCl is a strong oxidizing agent, but neither Ag+ nor NO3- (with N already in it's most positive OS) can be readily oxidized (it is difficult to oxidize Ag+ and O atoms having OS of -2).

      Q1c) The AgCl ppt would dissolve in NH3(aq), as the soluble ionic diamminesilver(I) coordination complex is generated. The Br atom of BrNO3 would also be nucleophilically attacked by NH3 and H2O, but the reaction products are not readily observable (hence irrelevant to the question).

      Q13. This qn tests you on the thermal stabilities of Group 2 carbonates (determined by cationic charge densities). As a fire retardant, you would want two things : greater moles of CO2(g) generated per mole of retardant, and lower thermal decomposition temperature to release the CO2(g) more readily.

      Q24. The mechanism involves 2 x nucleophilic substitution (first SN2 on the less sterically hindered terminal C atom, followed by deprotonation, then intramolecular SN2 or SN1 attack on the 5th C atom, followed by deprotonation) of the Br atoms on the 1st and 5th C atoms, by the NH3 nucleophile, to generate the required product Coniine.

      Edited by UltimaOnline 27 Aug `15, 2:13AM