2016 H2 Chemistry JC1 & 2 students post your questions here
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[Olympics] - Rio official who said, "oops, the pool turned green coz we added too much chloride, it's not our fault, it's chemistry's fault" obviously failed his A level Chemistry. Can you figure out what's chemically wrong with his explanation?
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probably due to wrong/imbalanced composition/make-up/balance of chemicals
one of the reasons of the green (dis)coloration could be due to green or yellow coloured ions(yellow colored ions because yellow ions + blue backdrop = greenish water)
few examples of green and yellow ions may be found here
https://en.m.wikipedia.org/wiki/Color_of_chemicals
how the pollution of chemicals got there, not so sure. one of the reasons, poor pool management, sanitation and washings
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might be due to the following reasons too :
due to bacteria growth
water treatment process is negatively affected hence the pool water quality
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Originally posted by UltimaOnline:
Halogenation via electrophilic aromatic substitution.
As long as you correctly substitute the incoming electrophile onto an available ortho or para position relative to the phenolic group, Cambridge will accept.
Diazotization and diazonium coupling is not in the Singapore H2 syllabus. If you're interested, go read up on it yourself.
Acyl halides are electrophiles attacked by the most nucleophilic group, ie. amine on L-Dopa, and phenol in Vanillin.
Fe3+ (in neutral solution, can't be too acidic or too alkaline, interested students can ponder over why, and check your answer with your school teacher or private tutor) coordinates & complexes with phenolic ligands to generate a violet coloured coordination complex.
Also we didn't learn the terms Para (1,4-) and Ortho (1,2-) but since you said it was phenolic, so it should be Electron Donating Group? How do we determine it as the phenolic group as the first place as Vanillin has an Acetyl group and Aldehyde group bonded to the Benzene ring with the hydroxyl group.
Is not like students can determine when 2 different natures of groups are bonded to the same ring and the Data Booklet doesn't even help a bit.... When considering bonding at C2, C4 or C3. And we have to determine where will be our C1 at the first place...
(I'd just finished whole Organic Chemistry in 2 week time... So I'm still relative new to this)
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Also I just realized a problem with L-DOPA.
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Originally posted by iSean:
Also we didn't learn the terms Para (1,4-) and Ortho (1,2-) but since you said it was phenolic, so it should be Electron Donating Group? How do we determine it as the phenolic group as the first place as Vanillin has an Acetyl group and Aldehyde group bonded to the Benzene ring with the hydroxyl group.
Is not like students can determine when 2 different natures of groups are bonded to the same ring and the Data Booklet doesn't even help a bit.... When considering bonding at C2, C4 or C3. And we have to determine where will be our C1 at the first place...
(I'd just finished whole Organic Chemistry in 2 week time... So I'm still relative new to this)
Whether the student can determine the ortho-para-meta directing nature of each substituent, is up to the student to learn about this as you/he/she wills. Different JCs, different teachers, different private tutors, teach to differing extents. Depending on your own willingness and choice, you can read up about this on your own, even if your school teacher chooses not to teach it.The phenolic OH group is electron-donating by resonance moreso than it is electron-withdrawing by induction, hence each phenolic OH group is activating and ortho-para directing. The alkyl group (sp3 C atoms) in L-Dopa is electron-donating by induction (relative to the sp2 C atoms of the benzene ring), and is hence a weak activator and an ortho-para director. The benzaldehyde group in Vanillin is electron-withdrawing by both induction (due to a significant partial positive charge on the carbonyl sp2 C atom as a result of both induction and resonance) and resonance (due to the sideways overlap of the unhybridized p orbitals of sp2 C and O atoms of the carbonyl C=O group), and is hence deactivating and meta directing.
Hence, the strongest activator (for both L-Dopa and Vanillin) being the phenolic OH group, wins and gets to direct the incoming electrophile. Because L-Dopa contains 2 phenolic OH groups, Cambridge will accept any of the 3 possible answers, although the Mark Scheme answer is the major product due to least steric hindrance.
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Originally posted by iSean:
Also I just realized a problem with L-DOPA.
Depending on the solvent, the alpha-amino-carboxylic acid of L-Dopa may exist as zwitterionic form, in which case the NH3+ is no longer nucleophilic and cannot attack the acyl halide electrophile. Note that acyl halides must be used in anhydrous form, which implies the other reagent (ie. L-Dopa) should not be used in aqueous solvent. -
So basically the structures 1,2,4 I drawn for L-DOPA will all be possible then :) ?
Thanks for letting me know :D -
Originally posted by iSean:
So basically the structures 1,2,4 I drawn for L-DOPA will all be possible then :) ?
Thanks for letting me know :D
Yes, correct. No prob. -
(Planning qns) In starch-iodine titration, why must starch indicator only be added near the titration end point and not that the start of titration?
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Originally posted by LavXuan:
(Planning qns) In starch-iodine titration, why must starch indicator only be added near the titration end point and not that the start of titration?
If starch solution is added at the start of the titration, the insoluble dark blue colored triiodide - starch clathrate precipitate formed will cause a time delay in releasing the triiodide ion (to be reduced by the thiosulfate ion), which may cause your titer volume and hence your titration results to be inaccurate. In other words, you may already have added enough thiosulfate solution from the burette to reach end point (which approximates equivalence point), but the dark blue color may remain for several seconds longer, fooling you into adding another couple of drops of thiosulfate solution from the burette, ruining the accuracy of your titration results.This is much less of a problem if the starch solution is added only near the end of the titration, as there will be a much smaller amount of triiodide ions left to be released from the decomposition of the much smaller amount of dark blue colored triiodide - starch clathrate precipitate, which will take much less time (almost instantaneous) and hence poses little risk of error to your titer volume and titration results.
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Originally posted by UltimaOnline:
If starch solution is added at the start of the titration, the insoluble dark blue colored triiodide - starch clathrate precipitate formed will cause a time delay in releasing the triiodide ion (to be reduced by the thiosulfate ion), which may cause your titer volume and hence your titration results to be inaccurate. In other words, you may already have added enough thiosulfate solution from the burette to reach end point (which approximates equivalence point), but the dark blue color may remain for several seconds longer, fooling you into adding another couple of drops of thiosulfate solution from the burette, ruining the accuracy of your titration results.This is much less of a problem if the starch solution is added only near the end of the titration, as there will be a much smaller amount of triiodide ions left to be released from the decomposition of the much smaller amount of dark blue colored triiodide - starch clathrate precipitate, which will take much less time (almost instantaneous) and hence poses little risk of error to your titer volume and titration results.
Thank you for accurate and detailed explanation. my teacher took several lessons to cover it during J1(possibly to just stretch the time), and without all the specific molecules names, he just say starch-iodine complex, and without all the accurate explanations. Thank you Ultima
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2011 TYS Paper 1
Qns 8: How do we know that the conditions are cold such that Cl2 can disproportonate in NaOH? Is it because as long as the reaction takes place in rtp, we assume that the conditions are cold?
2011 TYS Paper 2
Qns 4c(iii): Do we use the bond energies found in 4c(ii) to answer? Or the bond energies are not relevant in this qns?
Qns 6a(ii): Based on reaction 1 alone, shouldn't the functional group be an acyl chloride since it can react with silver nitrate immediately to produce a white ppt? So why is Halogenoalkane the answer when they have to undergo additional steps after adding silver nitrate?
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Originally posted by ACA-Milkshake:
2011 TYS Paper 1
Qns 8: How do we know that the conditions are cold such that Cl2 can disproportonate in NaOH? Is it because as long as the reaction takes place in rtp, we assume that the conditions are cold?
2011 TYS Paper 2
Qns 4c(iii): Do we use the bond energies found in 4c(ii) to answer? Or the bond energies are not relevant in this qns?
Qns 6a(ii): Based on reaction 1 alone, shouldn't the functional group be an acyl chloride since it can react with silver nitrate immediately to produce a white ppt? So why is Halogenoalkane the answer when they have to undergo additional steps after adding silver nitrate?
Q8. Yes.Q4ciii. The stability of CCl4 over SiCl4 is *DESPITE* the stronger Si-Cl bonds (as a challenging A grade question, Cambridge can ask you to suggest at least 2 different possible explanations for this anomaly of C-Cl bond being weaker than Si-Cl bond despite C and Si being in period 2 and 3 respectively ; I expect my BedokFunland JC students to be able to rise up to Cambridge's challenge!).
So the bond enthalpies are not irrelevant, they are meant for Cambridge to (using A level exam questions) educate the student who understands what's going on, "Oh yah hor, how come Si-Cl bond stronger, yet SiCl4 kena hydrolyzed more readily leh? Chemistry damn interesting sia! I love Chem!"
The reasons for the stability of CCl4 over SiCl4 is covered in the basic H2 syllabus (although Singapore JCs usually teach only 1 simplified reason, while I teach my BedokFunland JC students (at least those who ask me) 3 different reasons for this).
Q6aii). Acyl chloride will give a white ppt at rtp. The qn specified warming is necessary, thus alkyl chloride.
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ultima can help to check these
even though Si-Cl bond is stronger than C-Cl bond, SiCl4 undergoes hydrolysis more readily than CCl4 because
1. Steric hindrance of Cl blocking the C atom, preventing water (nucleophile) from attacking it readily. Si is larger so it is more susceptible to attack by the nucleophile.
2. Lack of low-lying energetically accessible d orbitals of the C atom.
d orbitals creates the potential for addition-elimination hydrolysis mechanisms with lower activation energy).
accept dative bonds from water nucleophiles without needing to first cleave existing Si-Cl covalent bonds
3. The C-Cl bond is less polarised than the Si-Cl bond, because the difference in electronegativity between C and Cl is smaller than that between Si and Cl. ,and hence Si is more susceptible to nucleophillic attack compared to C? so what if Si-Cl bond is more polar ?
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http://sgforums.com/forums/2297/topics/392932?page=8
qn error Why is CCl4 inert to hydrolysis while SiCl4 is resistant to hydrolysis?
shld be SiCl4 is readily hydrolysed
1 : Because unlike C, Si has vacant energetically accessible 3d orbitals to use to expand their octet, and as such, is able to accept dative bonds from water nucleophiles without needing to first cleave existing Si-Cl covalent bonds, resulting in a lower activation energy required for the hydrolysis.
Reason #2 : The Si-Cl bonds are longer (but not weaker! See Data Booklet bond energies) compared to the C-Cl bonds. Accordingly, the partially (or delta) negatively charged Cl atoms in SiCl4 provide poorer shielding against the incoming water nucleophiles, compared to the Cl atoms in CCl4.
how does less steric hindrance provided by Cl atoms in SiCl4 , is related to bond energy?
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Why can NCl3 be hydrolyzed but CCl4 cannot?
http://chemistry.stackexchange.com/questions/38251/why-can-ncl3-be-hydrolyzed-but-ccl4-cannot
despite the N atom having no energetically accessible d-orbitals, similar to C atom
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Hi ultimaonline.
I recently did a question on why pka of peroxyethanoic acid(as shown in picture) is more than pka of ethanoic acid. my school answer is :The negative charge on oxygen in CH3CO2-
is very effectively dispersed through delocalisation within the ‒COO- π system. However, the anion CH3CO3- is not able to delocalise its electrons
over the C=O group. Hence, CH3CO2‒ is more stable than CH3CO3-.May I ask if it is correct to say that since the extra o atom in the peroxyethanoic acid has p orbital, resonance of pi electron cloud can also form in the ‒COOO- group in the conjugate base of peroxyethanoic acid , the lone pair of the extra o atom increases electron density of pi electron cloud, hence intensifies the negative charge of conjugate base , making it more unstable, extent of dissociation of peroxyethanoic acid is lesser, ka is smaller, pka is bigger. correct me if im wrong in my reasoning. Also, is the answer for your cadmium sulfide ppt qns u posted on your website is 5.12x10^(59) mol^(-5)dm^(15)? thank you.
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Originally posted by Rapidestlime:
Hi ultimaonline.
I recently did a question on why pka of peroxyethanoic acid(as shown in picture) is more than pka of ethanoic acid. my school answer is :The negative charge on oxygen in CH3CO2-
is very effectively dispersed through delocalisation within the ‒COO- π system. However, the anion CH3CO3- is not able to delocalise its electrons
over the C=O group. Hence, CH3CO2‒ is more stable than CH3CO3-.May I ask if it is correct to say that since the extra o atom in the peroxyethanoic acid has p orbital, resonance of pi electron cloud can also form in the ‒COOO- group in the conjugate base of peroxyethanoic acid , the lone pair of the extra o atom increases electron density of pi electron cloud, hence intensifies the negative charge of conjugate base , making it more unstable, extent of dissociation of peroxyethanoic acid is lesser, ka is smaller, pka is bigger. correct me if im wrong in my reasoning. thank you.
Hi Rapidestlime, to fully understand the answer to this question, why the negative formal charge on the terminal O atom in COOO- cannot be delocalized by resonance, you have to go beyond just simply the idea of sideways overlapping of p orbitals making resonance delocalization of pi electrons and hence delocalization of a formal charge possible.You actually have to draw out the resonance contributors in each case, ie. COO- vs COOO-, to explain why resonance delocalization of the negative formal charge is not possible in COOO-. Because unless it involves complete delocalization around a ring, ie. aromaticity, otherwise, lone pairs (in unhybridized p orbitals) cannot delocalize into lone pairs on other atoms by resonance, lone pairs can only at most delocalize into pi bond pairs.
In the case of COOO-, the central O atom's lone pair occupying a unhybridized p orbital can indeed delocalize by resonance as a pi bond pair with the acyl sp2 C atom, which can indeed accept the pi bond pair (which happens in the resonance stabilized COO- conjugate base) without having to expand its octet (which it can't do so, being in period 2), because the C=O pi bond can simultaneously delocalize into a lone pair on the O atom.
However, the lone pair on the terminal O atom bearing the negative formal charge in the COOO- conjugate base, *cannot* at the same time delocalize by resonance to form a pi bond with the central O atom in any resonance contributor (ie. regardless of whether the central O atom has delocalized it's lone pair unto the acyl C atom or not), because the central O atom is unable to accommodate an expanded octet configuration (forcing the central O atom to accept a pi bond from the terminal O- atom will result in the central O atom having either 2 lone pairs + 3 bond pairs in a resonance contributor, or 1 lone pair + 4 bond pairs in another resonance contributor, which cannot be done because period 2 elements only have 1 x 2s orbital and 3 x 2p orbitals, and do not have vacant, energetically accessible 3d orbitals to accommodate an expanded octet).
If (like all other H2 students) you've no experience with drawing resonance contributors and hybrids, the above will be tough for you to understand. It's best if you get your school teacher or private tutor to draw them out for you.
Of course, Cambridge will be delighted if you are able to do so (ie. draw out the resonance contributors and hybrids) as part of your explanation, but at the very least, giving your school's (ie. Singapore JCs') simpler answer will be sufficient to obtain the 1 mark for this question. No worries, when the question goes beyond the H2 syllabus and it's not fair to everyone*, it's still fair : it's a bell-curve afterall.
*Of course, H3 Chem, Olympiad Chem, and BedokFunland JC students will know how to draw out resonance contributors and hybrids. It's still fair because they've put in additional effort to understand Chem deeper. Why else join H3 Chem, Olympiad Chem, and BedokFunland JC?
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Originally posted by Rapidestlime:
Hi ultimaonline.
Also, is the answer for your cadmium sulfide ppt qns u posted on your website is 5.12x10^(59)mol^(-5)dm^(15)? thank you.
Very good, Rapidestlime. That is correct. -
RI 2015 Prelim
Paper 1
Q12. How do I infer option B?
Q36. Why is option 2 wrong? Cl is reduced frm 0 in cl2 to -1 in cl- and oxidised to +5 in cl03-, so technically it is reduced right?
Paper 2
Q4bi. Why is the rate determining step not the step involving the formation of nitronium ion? Generation of nitronium ion involves 2 neutral molecules while the next step involves a positively charged ion so isn't the reaction faster in second step?
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2015 HCI Prelim paper 1
Q25. Why is option D wrong? I mean to me both option C and D are correct...
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Originally posted by Ephemeral:
RI 2015 Prelim
Paper 1
Q12. How do I infer option B?
Q36. Why is option 2 wrong? Cl is reduced frm 0 in cl2 to -1 in cl- and oxidised to +5 in cl03-, so technically it is reduced right?
Paper 2
Q4bi. Why is the rate determining step not the step involving the formation of nitronium ion? Generation of nitronium ion involves 2 neutral molecules while the next step involves a positively charged ion so isn't the reaction faster in second step?
2015 RJC P1 Q12. Vol of vessel = 2dm3, hence 0.1mol = 0.05M. Concordantly, at 4 mins, molarity increases from 0.05 to 0.10.2015 RJC P1 Q36. Yes, but "reduced" implies "reduced only" while here Cl is "disproportionated", ie. "both reduced and oxidized". Blame RJC teacher if you disagree, but even Cambridge often 'chut' ambiguous stunts like these, and Singapore JC teachers often disagree on the best answer.
2015 RJC P2 Q4bi. Because even temporarily losing aromaticity is a hugely destabilizing, reluctant step.
2015 HCI P1 Q25. The MCQ asks for "best describes". NO2 isn't itself a greenhouse gas (that would be N2O, not NO2), it only contributes to the formation of tropospheric ozone which is a greenhouse gas (in contrast, the beneficial ozone layer refers to stratospheric ozone). Instead, NO2 pollutant is far more damaging to humans and the environment as a major contributor to acid rain.
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2012 TYS Paper 2
Qns 5b(i): The answer that my school gave was that "All molecules present in the same chirality, that us, only one enantiomer is present." But can I just say that the lactic acid contains chiral carbons that's why it causes the rotation of th eplane polarised light?
