2014 H2 Chemistry Paper 2, Paper 3 and Paper 1 discussions
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Hi,
For qn33. Since SO4 is much bigger than M, shouldnt Hsoln LE be counted as insignificant compated to hydration energy. Since the increasing size of M will only affect LE minimally as SO4 is alr so big. My answer was D...
Qn37. My answer was C because the conditions given didny specify excess conc NH3. So polyalkylation likely to take place and primary amines formed will undergo further nucleo sub
Thanks!
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Hi
For question 15, doesn't black phosphorus conduct electricity in solid state ?
Thanks!
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For the past years tys I'd done, I realise cambridge dont really care about ambiguity. For example, H2 and nickel can happen at rtp without heating to 200degrees, phenol can form ester without forming phenoxide first etc, unlike what we are told in sch. So I dont think the conditions of 'excess' NH3 'heat in a sealed tube' is 'required' for it to be the answer
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Hi may i know why Q21 is A and not C :) and is 33/40 (mcq) and 63/80 (p3) and 60/72 (p2) plus practicals that went fine enuf for an A judging frm ur experience :) and thx for the effort in replying us :)
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Originally posted by Darkcygnet:
Hi
For question 15, doesn't black phosphorus conduct electricity in solid state ?
Thanks!
That's only 1 out of many different allotropes of P. -
Originally posted by Richard22:
For the past years tys I'd done, I realise cambridge dont really care about ambiguity. For example, H2 and nickel can happen at rtp without heating to 200degrees, phenol can form ester without forming phenoxide first etc, unlike what we are told in sch. So I dont think the conditions of 'excess' NH3 'heat in a sealed tube' is 'required' for it to be the answer
Cambridge isn't always 100% consistent (afterall, it's not always the same exact Cambridge dude writing all the mark schemes for all the questions for all the papers for all the decades... if it was, he might have already gone on strike for being severely overworked and underpaid). Some years Cambridge may not require "excess", some years they may require it.. So when calculating your own marks, it's safer to err on the side of caution. -
Originally posted by enorh12:
Hi,
For qn33. Since SO4 is much bigger than M, shouldnt Hsoln LE be counted as insignificant compated to hydration energy. Since the increasing size of M will only affect LE minimally as SO4 is alr so big. My answer was D...
Qn37. My answer was C because the conditions given didny specify excess conc NH3. So polyalkylation likely to take place and primary amines formed will undergo further nucleo sub
Thanks!
Q33. You're already implicitly considering both lattice dissociation enthalpy as well as hydration enthalpy, in determining solution enthalpy.
Q34. Even if polyalkylation takes place, this is still one of the possible (and immediate) products. -
Originally posted by rynal:
Hi may i know why Q21 is A and not C :) and is 33/40 (mcq) and 63/80 (p3) and 60/72 (p2) plus practicals that went fine enuf for an A judging frm ur experience :) and thx for the effort in replying us :)
Q21. C is the transition state for SN2, but the diagram indicates SN1.
Based on the overall weightage (see pg 4 below), if your overall score is 77% or greater, you'll most likely score A grade. But take your school teacher's or private tutor's answers with a pinch of salt, and self-mark on the side of caution (so if you think you scored > 77% but turns out you didn't get an A grade, your actual score is most likely < 77%).
https://www.seab.gov.sg/content/syllabus/alevel/2015Syllabus/9647_2015.pdf -
Originally Posted by Runepout :
Yup... I feel that this year's paper is on the whole much easier than past few years paper because there are less out of syllabus questions/challenging questions???
P2 & P3 are of similar difficulty as previous years (most students don't realize the real correct answers for a couple of distinction level P2 & P3 sub-questions, require going beyond the H2 syllabus, as it did in recent years 2010 to 2014; whether Cambridge accepts the simplified H2 version that JCs teach, is a separate point); it's P1 that is significantly easier.
Hence my upward revision for the A grade cut-off overall (ie. P1 + P2 + P3 + Spa) % from (my previous estimate of) 75% to (my final estimate of) 77%. When self-marking or self-estimating your score, err on the side of caution, both because you never know when you might have missed some careless mistake you didn't even know you made, as well as because I personally feel the standard H2 answers taught by Singapore JCs and other tuition teachers to be lacking (when it comes to the couple of distinction level P2 & P3 sub-questions I mentioned previously). -
Runepout posted :
Ahhhh ok... fingers crossed I guess >.
In life, always hope for the best, but be prepared for the worst.
JC grads should now be perusing over their options for the next phase of their lives, and begin seriously planning for their choice of University and degree course.
Ok fine, take a well-deserved break and enjoy yourselves for now. But do give your future some serious and careful deliberation. Don't choose the wrong Uni course for the wrong reasons, or you may regret it for the rest of your life.
That said, Happy Holidays!
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Hi, with regards to Q29, may I ask why is the answer B?
My own interpretation of the question is that aqueous bromine
- Adds 1 bromine atom across the alkene
- The carboxylic group is unaffected
- The -OH group is 2,4 directing while the electron-rich (alkene/carboxylic acid) side chain either reinforce the 2,4 position(with respect to the -OH group) or enables the other 2 position to be attacked by Bromine.
Thus, I arrive at the answer of 3 or 5 brome atoms being incorporated, with C being my answer as 3 is not part of the choices.Thank you for the help in advance. :)
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Originally posted by oayihzeet:
Hi, with regards to Q29, may I ask why is the answer B?
My own interpretation of the question is that aqueous bromine
- Adds 1 bromine atom across the alkene
- The carboxylic group is unaffected
- The -OH group is 2,4 directing while the electron-rich (alkene/carboxylic acid) side chain either reinforce the 2,4 position(with respect to the -OH group) or enables the other 2 position to be attacked by Bromine.
Thus, I arrive at the answer of 3 or 5 brome atoms being incorporated, with C being my answer as 3 is not part of the choices.Thank you for the help in advance. :)
There are many possible products, but Cambridge specified maximum possible number of Br atoms.
Although Br and OH is expected to be added across the alkene double bond when aqueous bromine is used, but as a minor product, it's also possible for 2 Br atoms to be added across instead. So that's 2.
The remaining 2 Br atoms come from the electrophilic aromatic substitution of 2 H+ by 2 Br+ electrophile cations on both ortho positions with respect to the strongest activator group, the phenolic OH group which donates electrons by resonance.
In contrast, the typical alkyl group only donates electrons by induction, which is a much weaker effect and insufficient for halogenation via electrophilic aromatic substitution.
Interestingly for this particular question, the alkyl group actually withdraws electrons both by induction (due to higher % s character of the sp2 C atom of the alkene group), as well as by resonance delocalization of electrons over to the electron-withdrawing COOH group as there is conjugation throughout the entire molecule from one end to the other.
So that's a total of 4 Br atoms, as the maximum possible number. (Cambridge knew that most Singapore students would automatically assume Br and OH added across the double bond, so Cambridge very kindly eliminated 3 Br atoms as a possible option). -
Hi. why is the answer for Qn 34 B? shouldn't the answer be D instead? SHE consists of H2 gas at 1 atm bubbling over platinum electrode coated with finely divided Pt which is dipped into 1 mol per dm^3 H+ (aq) solution at 298K. In this case, only 1 is wrong as the pressure and temperature is incorrect. Options 2 and 3 are correct. So the answer should be D! please correct me if I'm wrong.
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H2so4 dibasic give 2 moldm3 h+
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For h2so4 supposed to use 0.5moldm^-3
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Guys I know this is a chem discussion page but can someone post their answers for Bio MCQ today? :S Think I'm gonna get 33/34 crying