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Hello, may i ask a question? It's from A level 2007 Paper 3 Question 3 d)i) Suggest the strucutres of the three carbonyl compounds, and the ratio in which they might be produced.
I've gotten methanal and propanone but I don't know why ethanal is produced too. And how do we deduce the ratio?
Thanks!
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Originally posted by p0tat0:
Hello, may i ask a question? It's from A level 2007 Paper 3 Question 3 d)i) Suggest the strucutres of the three carbonyl compounds, and the ratio in which they might be produced.
I've gotten methanal and propanone but I don't know why ethanal is produced too. And how do we deduce the ratio?
Thanks!
In the original experiment, with calcium ethanoate (4 C atoms) reactant, the only products were propanone (3 C atoms) and carbonate (1 C atom). This gives you a clue that it's about balancing C atoms, ie. 1 C atom used to form the carbonate product, the remaining available C atoms used to form carbonyl compounds.In the new experiment, the reactants are 1:1 molar ratio of calcium methanoate (2 C atoms) and calcium ethanoate (4 C atoms). Since Cambridge stated that 3 different carbonyl compounds are produced, you must be able to deduce that the mechanism involves permutations from the available reactants (ie. 1 R group of the carbonyl may be from 1 reactant, while the other side R group of the carbonyl may be from the other reactant).
Since both are equally available in a 1:1 ratio, in addition to methanal (from calcium methanoate alone) and propanone (from calcium ethanonate alone), it is also possible to obtain ethanal (from both calcium methanoate and calcium ethanonate, providing H and CH3 R groups respectively).
As to the mathematical ratio, it's simple permutational probability. There's only 1 chance to obtain methanal (ie. using methanonate x methanoate) and also only 1 chance to obtain propanone (ie. using propanonate x propanoate), but there are 2 chances to obtain ethanal (ie. using methanonate x propanonate and propanonate x methanoate).
Since both reactants are equally available in a 1:1 molar ratio, it's like tossing a coin twice, and the probability are as follows :
Heads x Heads = 0.5 x 0.5 = 0.25
Heads x Tails = 0.5 x 0.5 = 0.25
Tails x Heads = 0.5 x 0.5 = 0.25
Tails x Tails = 0.5 x 0.5 = 0.25In other words, the ratio of Heads x Heads : Heads x Tails or Tails x Heads : Tails x Tails, is 0.25 : 0.25 + 0.25 : 0.25, ie. 1:2:1
Which is why methanal : ethanal : propanone is produced in the ratio of 1:2:1.
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Sgforums should make an app, i keep forgetting to check back. Thanks for reply, just on the side, but Pb has large ionic radius and oxidation number max +4 which then causes the magnitude of charge density to be small right? also it doesnt have any d orbitals vacant :O so how you extrapolate that it can form complexes ?
and so the answer for the TYS question is cus Pb forms a new complex and the conc of Pb2+ falls , then by lcp equi shifts to form more ions ?
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Thank you so much!! :) Your explanation is really detailed, i finally undestand it after so long TuT
May i ask another question which is from A level 2008 paper 3? Question 2 e)iii) use these data to calculate the H:C ratio in alkene D, and hence suggest its molecular formula.
I've gotten the empirical formula which is C2H5 but i don't know how to get the molecular formula through calculation.
And another question from the same paper question 4b. Write equations, including state symbols for these two reactions and explain the observation.
The answer key used Al2Cl6 for its equation but can we use AlCl3 instead?
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Hi, I have a question from VJC Prelim 2016 P3 Q2bii,
Why do we have to take 75 / 2 ?
Is it because there's an equal chance of the molecule getting attacked, so a racemic mixture is formed??
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Originally posted by theophilus:
Sgforums should make an app, i keep forgetting to check back. Thanks for reply, just on the side, but Pb has large ionic radius and oxidation number max +4 which then causes the magnitude of charge density to be small right? also it doesnt have any d orbitals vacant :O so how you extrapolate that it can form complexes ?
and so the answer for the TYS question is cus Pb forms a new complex and the conc of Pb2+ falls , then by lcp equi shifts to form more ions ?
For Pb, the 5d orbitals may be filled, but the 6d orbitals are vacant and energetically accessible.The charge density of many metal ions lower in the periodic table are higher than you might think, vis-Ã -vis their high proton number, due to the d-block contraction effect and other factors (Uni level Chemistry).
In addition, quite independent from charge density, different ligands have different affinities with different metal ions, ie. hard vs soft Lewis acids and bases (Uni level Chemistry).
Correct, the shifting of equilibrium position (including any ligand substitution) is as predicted by Le Chatelier's principle (you're not allowed to use acronyms for A level exams).
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Originally posted by p0tat0:
Thank you so much!! :) Your explanation is really detailed, i finally undestand it after so long TuT
May i ask another question which is from A level 2008 paper 3? Question 2 e)iii) use these data to calculate the H:C ratio in alkene D, and hence suggest its molecular formula.
I've gotten the empirical formula which is C2H5 but i don't know how to get the molecular formula through calculation.
And another question from the same paper question 4b. Write equations, including state symbols for these two reactions and explain the observation.
The answer key used Al2Cl6 for its equation but can we use AlCl3 instead?
No prob :)The molecular formula is almost always 2 x empirical formula. Check using degree of unsaturation (you can google it out yourself and learn it if your school teacher or private tutor didn't teach you).
The molar mass is specified in the question, hence your answer must be in concordance. In other questions, if molar mass or some other qualifying data isn't specified, then either AlCl3 or Al2Cl6 may be accepted (although you should take into consideration that different temperatures and pressures favor the monomer vs the dimer, you can google this out yourself for more info).
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Originally posted by supercat:
Hi, I have a question from VJC Prelim 2016 P3 Q2bii,
Why do we have to take 75 / 2 ?
Is it because there's an equal chance of the molecule getting attacked, so a racemic mixture is formed??
Yes, correct. This is concordant with the kinetics data that point to SN1 instead of SN2. -
ohh... so for alkane the degree of unsaturation is zero but C:H ratio (2:5) will give a magntiude of 1/2 using the formula for degree of unsaturation. Is that why i need to multiple by 2 to get C4H10 which will give me magntiude of zero?
Thanks a lot! :)
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ignore my previous question. i got it. Sorry and Thanks !
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Originally posted by p0tat0:
ohh... so for alkane the degree of unsaturation is zero but C:H ratio (2:5) will give a magntiude of 1/2 using the formula for degree of unsaturation. Is that why i need to multiple by 2 to get C4H10 which will give me magntiude of zero?
Thanks a lot! :)
Yes correct, degree of unsaturation should be zero for a fully saturated molecule (eg. alkane). Use degree of unsaturation to check your molecular formula. -
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i read wrongly, A to B,
should be A from B . by the time i realised and corrected my mistake , it was a little too late, u were alr typing liao, so i missed 2 chances , to correct before u see the qn, or to correct when u see my qn but haven't type
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Hi, I have questions from HCl prelim 2014 P1.
Q12) How to determine rate constant? My only guess is k = ln2/half life but that isn't the right answer.
Q26) Why won't option D give CHI3? Shouldn't structures of methyl ketones give CHI3?
Thank you!
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Originally posted by supercat:
Hi, I have questions from HCl prelim 2014 P1.
Q12) How to determine rate constant? My only guess is k = ln2/half life but that isn't the right answer.
Q26) Why won't option D give CHI3? Shouldn't structures of methyl ketones give CHI3?
Thank you!
Q12) Use ln2/(40min) = 1.7329x10^-2Q26) Esters won't give a positive iodoform result.
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Hi UltimaOnline,
Here are some DHS MCQs on which I would like seek your help:
DHS14/P1/12 [Ans: D]
Remarks: I do not understand the titration curve. My understanding is as follows - the first unlabelled equivalence point between A and B signifies the end of neutralisation between maleic acid and sodium hydroxide, and at the point, only -O2CHCHCO2- is present. However, why is the second equivalence point is at C, (when an additional 10cm³ of maleic acid is added from the first equivalence point), instead of an addition 30cm³ of maleic acid (given that there is 30cm³ of -O2CHCHCO2- at first equivalence pt)? Can I confirm that at D, the only compound present is HO2CHCHCO2-? So where is “mixture” to resist the pH change? If there were an option further down the graph from D, I would have picked that, but D doesn’t seem correct to me.
DHS14/P1/20
Remarks: I understand that elimination is out because it wouldn’t yield a product which rotates plane-polarised light. I also understand that both SN1 (since an optically pure enantiomer of 2-bromobutane was used since an optically pure enantiomer of 2-bromobutane was used) and SN2 mechanisms will yield an optically active product. Am I supposed to draw any inference from the reduction of the angle of rotation or is that just a red herring?
DHS14/P1/25
Remarks: I can understand that C is not the answer because it cannot be deduced from the information above. However, how do I know whether C-16O bond is stronger than C-18O bond?
DHS14/P1/28 [Ans: C]
Remarks: I’m rather confused here. Do the arrows in the choices signify a successive addition of the compounds without any discarding of the aqueous layer? Why do I add water in the first and last steps and how do I go about ascertaining the sequence of the addition?
Once again, thank you very much for your help! :)
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Originally posted by gohby:
Hi UltimaOnline,
Here are some DHS MCQs on which I would like seek your help:
DHS14/P1/12 [Ans: D]
Remarks: I do not understand the titration curve. My understanding is as follows - the first unlabelled equivalence point between A and B signifies the end of neutralisation between maleic acid and sodium hydroxide, and at the point, only -O2CHCHCO2- is present. However, why is the second equivalence point is at C, (when an additional 10cm³ of maleic acid is added from the first equivalence point), instead of an addition 30cm³ of maleic acid (given that there is 30cm³ of -O2CHCHCO2- at first equivalence pt)? Can I confirm that at D, the only compound present is HO2CHCHCO2-? So where is “mixture� to resist the pH change? If there were an option further down the graph from D, I would have picked that, but D doesn’t seem correct to me.
DHS14/P1/20
Remarks: I understand that elimination is out because it wouldn’t yield a product which rotates plane-polarised light. I also understand that both SN1 (since an optically pure enantiomer of 2-bromobutane was used since an optically pure enantiomer of 2-bromobutane was used) and SN2 mechanisms will yield an optically active product. Am I supposed to draw any inference from the reduction of the angle of rotation or is that just a red herring?
DHS14/P1/25
Remarks: I can understand that C is not the answer because it cannot be deduced from the information above. However, how do I know whether C-16O bond is stronger than C-18O bond?
DHS14/P1/28 [Ans: C]
Remarks: I’m rather confused here. Do the arrows in the choices signify a successive addition of the compounds without any discarding of the aqueous layer? Why do I add water in the first and last steps and how do I go about ascertaining the sequence of the addition?
Once again, thank you very much for your help! :)
Hi Gohby,DHS14/P1/12. Since you're adding a weak acid to a strong acid, hence maximum buffer capacity is double equivalence point instead of half equivalence point. Being a diprotic acid, there are of course, 2 maximum buffer capacities. The 1st being equal molarities of HA- and A2- (which occurs at [10+20]/2=15 cm3), and the 2nd being equal molarities of H2A and HA- (which occurs at 40 cm3). Clearly, the more effective buffer occurs at equal molarities of H2A and HA- (ie. weak acid and amphiprotic species) rather than equal molarities of HA- and A2- (ie. amphiprotic and strong base), as illustrated by the larger horizontal region of the titration curve. Besides, 15cm3 isn't even an option in this MCQ.
DHS14/P1/20. If it were SN1 only, and the reaction was complete (as implied), then the product would be optically inactive. If it were SN2 only, and the reaction was complete (as implied), then the product would rotate plane polarized light by 13.5 degrees in either direction. Hence a mixture of both SN1 and SN2 must have occurred.
DHS14/P1/25. Option D is correct because the reactant is cyclic, hence there is only 1 hydrolysis product, not 2. Addressing your specific query : heavier isotopes form stronger bonds. This is Uni level Chemistry, and for A level purposes, Cambridge can still tempt the students with beyond-syllabus MCQ choices, to test if the student is sufficiently proficient to recognize the within-syllabus correct MCQ answer, and disregard the other options.
DHS14/P1/28. Problem is alcohol, being part non-polar and part protic polar, is miscible with both water and alkyl halide. Addding [water and excess conc HCl] serves to protonate the alcohol, which then migrates into the aqueous layer. Discard the aqueous layer by decantation. Add excess Na2CO3(aq) to remove excess HCl(aq) in the form of CO2(g) and NaCl(aq). Add S2O3 2-(aq) to reduce all the I2(s/aq/alc) to I-(aq), forming S4O6 2-(aq). Add water to remove all ionic impurities, then carry out decantation to leave behind the alkyl halide.
No prob, Ghoby! :)
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Hi, could I ask a question about ionic equilibria?
In JJC 2014 prelim mcq Q10,
Maximum separation occurs when IP = Ksp of Fe(OH)2 right? With the [Fe] being 0.05, I can calculate [OH-] but I keep getting pH of about 7. May I ask where did I go wrong??
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Originally posted by supercat:
Hi, could I ask a question about ionic equilibria?
In JJC 2014 prelim mcq Q10,
Maximum separation occurs when IP = Ksp of Fe(OH)2 right? With the [Fe] being 0.05, I can calculate [OH-] but I keep getting pH of about 7. May I ask where did I go wrong??
A researcher accidentally mixed 50.0 cm3 of 0.100 mol dm–3 MgSO4 solution with 50.0 cm3 of 0.100 mol dm–3 FeSO4 solution in the laboratory. It is suggested that the metal ions can be separated through selective precipitation by adding solid sodium hydroxide. At 298 K, the solubility product of Mg(OH)2 is 1.8x10^–11 mol3 dm–9, and that of Fe(OH)2 is 4.1x10–15 mol3 dm–9. What would be the pH of the resultant solution when maximum separation has occurred?Your error is assuming maximum separation occurs when Qsp = Ksp of Fe(OH)2, when it should be Mg(OH)2, ie. the more soluble ionic compound. You have to add as much OH- as possible to precipitate all the less soluble Fe(OH)2, without precipitating out any of the Mg(OH)2. Hence at that point, Qsp = Ksp of Mg(OH)2, not Fe(OH)2.
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Originally posted by UltimaOnline:
A researcher accidentally mixed 50.0 cm3 of 0.100 mol dm–3 MgSO4 solution with 50.0 cm3 of 0.100 mol dm–3 FeSO4 solution in the laboratory. It is suggested that the metal ions can be separated through selective precipitation by adding solid sodium hydroxide. At 298 K, the solubility product of Mg(OH)2 is 1.8x10^–11 mol3 dm–9, and that of Fe(OH)2 is 4.1x10–15 mol3 dm–9. What would be the pH of the resultant solution when maximum separation has occurred?Your error is assuming maximum separation occurs when Qsp = Ksp of Fe(OH)2, when it should be Mg(OH)2, ie. the more soluble ionic compound. You have to add as much OH- as possible to precipitate all the less soluble Fe(OH)2, without precipitating out any of the Mg(OH)2. Hence at that point, Qsp = Ksp of Mg(OH)2, not Fe(OH)2.
Oh ok I understand, thank you!
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Hi, I have a question from MI 2014 prelim P1 Q24
Why are there only 2 optical isomers for R? Is it due to internal plane of symmetry? I counted a total of 4 chiral carbons in R. Did I do additional counting??
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Originally posted by supercat:
Hi, I have a question from MI 2014 prelim P1 Q24
Why are there only 2 optical isomers for R? Is it due to internal plane of symmetry? I counted a total of 4 chiral carbons in R. Did I do additional counting??
There are only 2 chiral C atoms. But due to an internal plane of symmetry, the R,S and S,R stereoisomers are the meso isomer and thus optically inactive, and hence the only 2 enantiomeric optical isomers are S,S and R,R.