Lurkers, register with SgForums and ask ur H2 Chem qns here!
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Originally posted by IronAge:
cool homework forum
Hi ultimaonline, regarding the discussion above, my thoughts are since N withdraws electron density away from O, wouldn't that cause O to have a even larger partial negative charge?
Hi IronAge, welcome to the forum. Are you a JC2 student this year? Or a private tutor? Or an MOE school teacher? Honest reply please.N doesn't withdraw electron density from O, but it just donates less to O than either H or R would. Can you see how that results in a less polar O-H bond? Or do you need me to explain further?
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Fyi (to everyone reading this), this forum is intended to help students, not private tutors. But nonetheless (and here's the irony), I'm far more willing to help out private tutors with their Chemistry questions when they honestly state they're private tutors when asked, compared to private tutors who dishonestly masquerade as school students when asking for help with Chemistry questions (and I can discern from your posts whether you're a student or not).
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Ultima, all these years when u take a level,
Dyou do all qns in Paper 3 or u choose 4 to do?
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Originally posted by Flying grenade:
Ultima, all these years when u take a level,
Dyou do all qns in Paper 3 or u choose 4 to do?
Depends on my mood *shrugs*.I can choose to do just 3 questions (eg. if 2 of the questions bore me), and I can still get my A grade. Or I can choose to do all 5 questions (and complete within the time allocated), and I still get my A grade.
But if you students do just 3 questions, you won't have enough marks to get A grade. If you students try to do all 5 questions, you won't have enough time to complete the paper.
So you should focus on yourself, and stop asking about me.
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Hi
Can we differentiate HOC6H4CH2NH2 and H2NC6H4CH2OH with acidified KMnO4(aq)?
How about Benzamide and ethanamide with Br2, FeBr3 + heat? thanks
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Originally posted by Mrworry:
Hi
Can we differentiate HOC6H4CH2NH2 and H2NC6H4CH2OH with LiAlH4 in dry ether? thanks.
First of all, both your analytes' formulae have a typo error, if there are 2 substituents on the benzene ring, it's -C6H4-, not -C6H5-.No, you can't. No group in either molecule can react with LiAlH4. You have to use K2Cr2O7 (but not KMnO4), heat, and observe for color change, which will occur for only 1 of the analytes.
For an alternative method, if both analytes are present in ppt form, only 1 will dissolve with NaOH(aq).
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I noticed my mistake (about using LiAlH4) and edited the question. Why with K2Cr2O7, but not KMnO4 though ? How about the 2nd part? tq
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Originally posted by Mrworry:
I noticed my mistake (about using LiAlH4) and edited the question. Why with K2Cr2O7, but not KMnO4 though ? How about Benzamide and ethanamide with Br2, FeBr3 + heat? thanks tq
Write out the products generated when you use K2Cr2O7, versus for KMnO4, then you'll see why.Yes, that might work, but the yield will not be high due to the deactivating amide group (which withdraws electrons by both induction and resonance). Although this isn't a synthesis question, but if the yield is minimal, any observation wouldn't be clearly visible, making this a poor method to distinguish between the two analytes.
An alternative and better method would be to use LiAlH4, followed by alkaline aqueous iodine.
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http://papers.gceguide.com/A%20Levels/Chemistry%20(9701)/9701_w13_qp_43.pdf
http://papers.gceguide.com/A%20Levels/Chemistry%20(9701)/9701_w13_ms_43.pdf
http://papers.gceguide.com/A%20Levels/Chemistry%20(9701)/9701_w13_er.pdf
In Q6 part f(i) , i dont understand 2 things.
First of all, i dont understand why G and H are not similar to F...if we rotate both the left and right hand side of the molecule, shouldnt we get F.
Secondly, how does rotating right hand side of J by 60o clockwise make ot identical to F as suggested by the examiner report. And why is Chirality of Left hand carbon of both G and H opposite to that of F.
After all this, how to solve f(iii) ie drawing the fourth optical isomers.
Is this related to R,S configurations or is this something else...as to where can i read more about this?
If you could explain this i detail, i would be grateful..thanks.
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Originally posted by Light5:
http://papers.gceguide.com/A%20Levels/Chemistry%20(9701)/9701_w13_qp_43.pdf
http://papers.gceguide.com/A%20Levels/Chemistry%20(9701)/9701_w13_ms_43.pdf
http://papers.gceguide.com/A%20Levels/Chemistry%20(9701)/9701_w13_er.pdf
In Q6 part f(i) , i dont understand 2 things.
First of all, i dont understand why G and H are not similar to F...if we rotate both the left and right hand side of the molecule, shouldnt we get F.
Secondly, how does rotating right hand side of J by 60o clockwise make ot identical to F as suggested by the examiner report. And why is Chirality of Left hand carbon of both G and H opposite to that of F.
After all this, how to solve f(iii) ie drawing the fourth optical isomers.
Is this related to R,S configurations or is this something else...as to where can i read more about this?
If you could explain this i detail, i would be grateful..thanks.
This is the toughest question in the paper for 2 reasons : because it's slightly beyond the A level syllabus, and also because it's extremely difficult to visualize (solely using imagination) the rotation of 3D structures.Going beyond the syllabus, ie. at Uni level, the correct way to approach such questions is to assign priorities to the 4 different groups based on the Cahn–Ingold–Prelog priority rules, and concordantly label each of the 2 chiral C atoms present as either R (Rectus) or S (Sinister). The 4 optical stereoisomers would then be (2R,3R), (2R,3S), (2S,3R) and (2S,3S), where (2R,3R) & (2S,3S), and (2R,3S) & (2S,3R), are the 2 sets of enantiomers, while (2R,3R) & (2S,3R), (2R,3R) & (2R,3S), (2S,3S) & (2S,3R), and (2S,3S) & (2R,3S), are the 4 sets of diastereomers.
Alternatively, or simultaneously, if you can obtain (or make for yourself) a 3D 'ball-&-stick' model of the molecule, and rotate each end, it'll help you to better visualize and solve this question. Of course, you won't be allowed to bring in such tools or toys in the A level exams, so you'll have to settle with using your hands : orientate your fingers (using both hands) in a tetrahedral geometry, with 1 finger towards you (the wedge), 1 finger away from you (the hash or dash), and 2 fingers (1 from each hand) neither towards or away (the normal bonds), then slowly rotate your hands. While simultaneously mentally keeping track of which finger representing which group of atoms, *and* the priority of each group (if you're assigning R vs S configuration to help you answer this question). Not an easy task, unfortunately.
You could say this question is unfair to be asked at A levels. But it's a bell-curve anyway, so since it's unfair for everyone, it's arguably still fair. Such questions are intended for only the top few % of the A level cohort to be able to answer correctly, otherwise too many people would score A grade. Regardless of where you're ranked in the cohort, exam-smart students would always skip such time-consuming questions first, complete the rest of the paper, then go back to such 'skipped' questions last. It'll be penny-wise-pound-foolish to waste 15 min struggling on this 3 marks question, then find out at the end of the exam that you only have 3 min left with 15 marks of much easier questions left unanswered.
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Lets assume that i know how to assign R and S priorities
To assign these, the lowest priority group has to be oriented away from us..in this case, the lowest priority group is the Hydrogen atom but the problem is that on the Left Hand Carbon atom for each of the structures, the H-atom is shown to lie on the plane rather than pointing out or into the plane...so assuming that we know how to assign R and S, how can we get past this issue?
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Originally posted by Light5:
Lets assume that i know how to assign R and S priorities
To assign these, the lowest priority group has to be oriented away from us..in this case, the lowest priority group is the Hydrogen atom but the problem is that on the Left Hand Carbon atom for each of the structures, the H-atom is shown to lie on the plane rather than pointing out or into the plane...so assuming that we know how to assign R and S, how can we get past this issue?
By visualizing the structure vividly in 3D space using your imagination, than slowly rotating (not just the single sigma bond, but move the entire structure) it in your mind, until the lowest priority substituent, is pointing away from yourself. And try using your fingers (orientated in a tetrahedral geometry, eg. left hand 3 fingers in a trigonal pyramidal tripod stand, right hand's forefinger representing lowest priority facing up, before rotating it till it's away from you) to help.Also see these videos :
https://www.youtube.com/watch?v=auphHXYK8tE
https://www.youtube.com/watch?v=kFD6hzLseVs -
What are the causes of amphoterism?
Tried to research online, but to no avail ! The closest relevant info i can find is 'Amphoterism depends on the oxidation states of the oxide' from wiki! But still, this piece of info doesnt help much!
Prior to researching i only know Al2O3 is amphoteric,
but after researching now i know Al(OH)3 , Zn (OH)2, Be(OH)2 are also amphoteric species.
I encountered a qn asking if SrO is amphoteric.
We only learnt and memorised that Al2O3 is amphoteric. So how can we determine(or how can we extrapolate the Only info that we know and memorised Al2O3 is amphoteric) which species is amphoteric when a qn ask?
I know Be2+ and Al3+ have high charge density, but i dk how to link this fact to the 2 ions being able to react with acid and bases
Is the reason for amphoterism high charge density? If it is, then i just memorise which ions have high charge density then amphoteric species?
Thanks for helping.
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Originally posted by Flying grenade:
What are the causes of amphoterism?
Tried to research online, but to no avail ! The closest relevant info i can find is 'Amphoterism depends on the oxidation states of the oxide' from wiki! But still, this piece of info doesnt help much!
Prior to researching i only know Al2O3 is amphoteric,
but after researching now i know Al(OH)3 , Zn (OH)2, Be(OH)2 are also amphoteric species.
I encountered a qn asking if SrO is amphoteric.
We only learnt and memorised that AL2O3 is amphoteric. So how can we determine which species is amphoteric is a qn asked?
A level shortcut : covalent = acidic, ionic = basic, hence partial covalent partial ionic = amphoteric.BedokFunland JC (ie University) level : to understand identify and understand why & how a species is amphoteric, and why the A level shortcut works (it's not an explanation, but merely an indirect indicator), you first need to understand why & how an acidic species is acidic, and why & how a basic species is basic, and you *must* draw out the curved-arrow electron-flow reaction mechanisms. It's too much detail to go into on the forums, and my BedokFunland JC students can ask me to teach them these mechanisms, while the rest of you can go ask your school teacher or private tutor if interested.
So do you think strontium oxide qualifies to be amphoteric? Btw, amphiprotic is a subset of amphoteric, but not vice-versa. Don't confuse these terms. Go ask your school teacher or private tutor if interested.
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Originally posted by Light5:
http://papers.gceguide.com/A%20Levels/Chemistry%20(9701)/9701_w13_qp_43.pdf
http://papers.gceguide.com/A%20Levels/Chemistry%20(9701)/9701_w13_ms_43.pdf
http://papers.gceguide.com/A%20Levels/Chemistry%20(9701)/9701_w13_er.pdf
In Q6 part f(i) , i dont understand 2 things.
First of all, i dont understand why G and H are not similar to F...if we rotate both the left and right hand side of the molecule, shouldnt we get F.
Secondly, how does rotating right hand side of J by 60o clockwise make ot identical to F as suggested by the examiner report. And why is Chirality of Left hand carbon of both G and H opposite to that of F.
After all this, how to solve f(iii) ie drawing the fourth optical isomers.
Is this related to R,S configurations or is this something else...as to where can i read more about this?
If you could explain this i detail, i would be grateful..thanks.
Hi Light5! Possible for u to upload a screenshot of the qn in discussion on top of the link of the source of the paper in the future ?
Some of us can't access/view the link
Would be awesome if u could upload a pic of the qn in discussion plus the link source of the content!
Thanks (:
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Originally posted by UltimaOnline:
A level shortcut : covalent = acidic, ionic = basic, hence partial covalent partial ionic = amphoteric.BedokFunland JC (ie University) level : to understand identify and understand why & how a species is amphoteric, and why the A level shortcut works (it's not an explanation, but merely an indirect indicator), you first need to understand why & how an acidic species is acidic, and why & how a basic species is basic, and you *must* draw out the curved-arrow electron-flow reaction mechanisms. It's too much detail to go into on the forums, and my BedokFunland JC students can ask me to teach them these mechanisms, while the rest of you can go ask your school teacher or private tutor if interested.
So do you think strontium oxide qualifies to be amphoteric? Btw, amphiprotic is a subset of amphoteric, but not vice-versa. Don't confuse these terms. Go ask your school teacher or private tutor if interested.
Thanks for helping !
amphoteric compound is a molecule or ion that can React both as an acid or a base
An amphiprotic molecule (or ion) can either donate or accept a proton, thus Acting either as an acid or a base.
For jc students : examples of the aforementioned can be found in the wikipedia page
SrO cannot be amphoteric. Charge density of Sr2+ not large enough to polarise Oatom's e- cloud to cause the SrO bond to become partially ionic partially covalent
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Oops
Extracted this from my notes which in turn was extracted from kim seng chan's book
Cause for amphoteric
High charge density high polarising power, polarises water molecules and weakens O-H bonds in water molecules, enabling water to protonate forming H3O+, causing the solution to be acidic, enabling it to react with strong bases
Explain why BeO is amphoteric whereas the other group 2 oxides are basic
Due to high charge density of Be ion, there is covalent character in Beryllium oxide. Be atom is highly electron deficient, giving it acidic characteristic, hence enabling it to react with strong bases such as NaOH. Presence of O2- allows BeO to react with acid to form salt and water
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Originally posted by Flying grenade:
Hi Light5! Possible for u to upload a screenshot of the qn in discussion on top of the link of the source of the paper in the future ?
Some of us can't access/view the link
Would be awesome if u could upload a pic of the qn in discussion plus the link source of the content!
Thanks (:
With the filename given, you can look up the same Cambridge paper yourself from other similar websites, eg. XtremePapers.com -
Originally posted by UltimaOnline:
N doesn't withdraw electron density from O, but it just donates less to O than either H or R would. Can you see how that results in a less polar O-H bond? Or do you need me to explain further?
Hmm i try,
O is more electronegative than N. O withdraws some e- density from N. Magnitude of delta -ve on O decreases. The O-H bond in hydroyxlamine is less polar compared to O-H bond in water. Hence strength of H bond between hydroxylamine molecules is weaker than strength of Hbond between water molecules.
Hence water has a higher B.P. than hydroxylamine even though hydroxylamine can from more Hbonds per molecule than water.
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Can help check or improve?
In any case, Thanks for your help and explanations in helping me solve chem qns !
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Originally posted by Flying grenade:
Hmm i try,
O is more electronegative than N. O withdraws some e- density from N. Magnitude of delta -ve on O decreases. The O-H bond in hydroyxlamine is less polar compared to O-H bond in water. Hence strength of H bond between hydroxylamine molecules is weaker than strength of Hbond between water molecules.
Hence water has a higher B.P. than hydroxylamine even though hydroxylamine can from more Hbonds per molecule than water.
Indeed, the O-H bond in hydroxylamine is less polar, compared to in H2O or in alcohols, due to the electronegative N atom. So far so good. But here's where your school cocked up : the boiling point of hydroxylamine should actually be higher, not lower, than H2O, as number of H bonds formed is more important than strength of H bonds. Simply look at their melting points, and you'll see I'm right. However, in practice, hydroxylamine never gets to reach its theoretical boiling point (ie. above 100 deg C), because it readily decomposes (at 58 deg C) into water, ammonia, and either dinitrogen monoxide or nitrogen gas (depending on mechanism pathway and reaction conditions). -
Page 160 cs toh advanced guide
'Maximum buffer capacity' section
Is it that for an acidic buffer or alkaline buffer, theoretically, the buffer can have the pH of the solution remains(almost) unchanged, when acid or base of concentration of up to 10 times the concentration of the weak acid / weak base, added to the buffer solution?
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Originally posted by Flying grenade:
'Maximum buffer capacity' section
Is it that for an acidic buffer or alkaline buffer, theoretically, the buffer can have the pH of the solution remains(almost) unchanged, when acid or base of concentration of up to 10 times the concentration of the weak acid / weak base, added to the buffer solution?
Yes, mathematically, a solution can be considered a valid buffer as long as the ratio of molarities of conjugate acid-base pairs lies within 1:10 and 10:1, ie. pH = pKa (+1 or - 1) -
Originally posted by UltimaOnline:
Indeed, the O-H bond in hydroxylamine is less polar, compared to in H2O or in alcohols, due to the electronegative N atom. So far so good. But here's where your school cocked up : the boiling point of hydroxylamine should actually be higher, not lower, than H2O, as number of H bonds formed is more important than strength of H bonds. Simply look at their melting points, and you'll see I'm right. However, in practice, hydroxylamine never gets to reach its theoretical boiling point (ie. above 100 deg C), because it readily decomposes (at 58 deg C) into water, ammonia, and either dinitrogen monoxide or nitrogen gas (depending on mechanism pathway and reaction conditions).Wow!!! Okay damn cool!!! Thanks for these info, really appreciate it (: