Originally posted by Flying grenade:in this scenario, is the e config of zn [Ar] 3d10?
YJC 2015 JC2 block test
https://www.dropbox.com/s/c030p3wzet45wbq/20160913_174110-1.jpg?dl=0
answer : https://www.dropbox.com/s/58htemwef1jluiv/20160913_145520-1.jpg?dl=0
option 3
methyl benzene and 4-bromo methyl benzene indeed have different boiling points arising from different strength of IMF , why it cannot be separated via distillation?
is it becos the two molecules will cling together strongly as similar Mr and hence Id-Id attns? mixture is miscible? or similar B.p. ? but one has Pd-Pd , the other only has idid, B.P. couldn't be similar right?
Originally posted by Flying grenade:YJC 2016 prelim
https://www.dropbox.com/s/j3ilvq87qk5grym/20160913_172208-1.jpg?dl=0
Answer given is In step 2, PCl5 will be hydrolysed in the presence of water
Heat is required for step 4
consider this qn
https://www.dropbox.com/s/g4mhqly1zxjp0sw/20160913_171845-1.jpg?dl=0
answer : https://www.dropbox.com/s/ftwnj586zg1x5ij/20160913_172024.jpg?dl=0
i dun understand option A
for option B, in anhydrous conditions, i know HCl(g) will be generated when PCl5 is reacted with R-OH.
in this scenario, in aqueous conditions, this rxn is flawed / cannot happen right?
I suspect u still won't understand this, so u better go ask ur school teacher or private tutor. Move on.
Originally posted by Flying grenade:YJC 2015 JC2 block test
https://www.dropbox.com/s/c030p3wzet45wbq/20160913_174110-1.jpg?dl=0
answer : https://www.dropbox.com/s/58htemwef1jluiv/20160913_145520-1.jpg?dl=0
option 3
methyl benzene and 4-bromo methyl benzene indeed have different boiling points arising from different strength of IMF , why it cannot be separated via distillation?
is it becos the two molecules will cling together strongly as similar Mr and hence Id-Id attns? mixture is miscible? or similar B.p. ? but one has Pd-Pd , the other only has idid, B.P. couldn't be similar right?
Originally posted by Flying grenade:YJC 2016 prelim
https://www.dropbox.com/s/j3ilvq87qk5grym/20160913_172208-1.jpg?dl=0
Answer given is In step 2, PCl5 will be hydrolysed in the presence of water
Heat is required for step 4
consider this qn
https://www.dropbox.com/s/g4mhqly1zxjp0sw/20160913_171845-1.jpg?dl=0
answer : https://www.dropbox.com/s/ftwnj586zg1x5ij/20160913_172024.jpg?dl=0
i dun understand option A
for option B, in anhydrous conditions, i know HCl(g) will be generated when PCl5 is reacted with R-OH.
in this scenario, in aqueous conditions, this rxn is flawed / cannot happen right?
Originally posted by UltimaOnline:
Correct, PCl5 (amongst other reagents) must be used only under anhydrous conditions.
means that MCQ qn epic fail right? as it writes B is the answer, but it's not. then all 4 options wrong
Originally posted by Flying grenade:means that MCQ qn epic fail right? as it writes B is the answer, but it's not. then all 4 options wrong
thanks god !!! you are god !!!! you are god
https://www.dropbox.com/s/t89auq5k0ztegoo/20160913_190411-1.jpg?dl=0
2011 /p3 / qn 2cii
answer shown above
If we students did get molecule F correct, most probably we will get the suggested structures in 2cii correct, from clever-guess
but how to use chemistry explain and how do we deduce/how do we know using legit chemistry?
so the mechanism, the NH2 and HCl attack the delta plus C as O is highly electronegative withdraws electrons away from C, pi bond shift up form lp in O ?
Originally posted by Flying grenade:https://www.dropbox.com/s/t89auq5k0ztegoo/20160913_190411-1.jpg?dl=0
2011 /p3 / qn 2cii
answer shown above
If we students did get molecule F correct, most probably we will get the suggested structures in 2cii correct, from clever-guess
but how to use chemistry explain and how do we deduce/how do we know using legit chemistry?
so the mechanism, the NH2 and HCl attack the delta plus C as O is highly electronegative withdraws electrons away from C, pi bond shift up form lp in O ?
Draw out the full mechanism, then upload it for me to check. Or ask ur school teacher or private tutor to explain to u the mechanism.
Originally posted by UltimaOnline:
But more pertinently, in this qn, the answer already specified it's becoz they need to product to have lower, not higher, boiling point, compared to the reactant.
but is this qn quite Lame, because the 4-bromomethylbenzene still can be separated from methylbenzene, both products can still be seperated , no matter which molecule(with higher or lower bp) ends up in the distillate collection beaker or remains in the round bottom flask right??
Originally posted by Flying grenade:but is this qn quite Lame, because the 4-bromomethylbenzene still can be separated from methylbenzene, both products can still be seperated , no matter which molecule(with higher or lower bp) ends up in the distillate collection beaker or remains in the round bottom flask right??
Originally posted by UltimaOnline:
Notice Cambridge didn't ask u to explain how u get the structure, u're just required to draw the structure, coz Cambridge knows students dunno how to explain.Draw out the full mechanism, then upload it for me to check. Or ask ur school teacher or private tutor to explain to u the mechanism.
ok got it. i think majority of students who dont know how or why to get the structures and the mechanisms got it correct by clever and/or lucky guess, without the firm, confirm and correct idea using chemistry
Originally posted by UltimaOnline:
Welcome to YJC.
im sure the Students in YJ love and appreciate their school.
anyway, Every Singapore JCs are awesome !
Originally posted by Flying grenade:ok got it. i think majority of students who dont know how or why to get the structures and the mechanisms got it correct by clever and/or lucky guess, without the firm, confirm and correct idea using chemistry
so do we always treat directly adjacent chiral carbons in a ring , as only capable to form 2^1=2 optical isomers?
Originally posted by Flying grenade:so do we always treat directly adjacent chiral carbons in a ring , as only capable to form 2^1=2 optical isomers?
2016 YJC JC2 Block test
https://www.dropbox.com/s/68y2bmpevqboy3g/20160913_203034-1.jpg?dl=0
is this plausible?
gaseous SOCl2 bubbled into a soln containing molecules with hydroxy group. the pH of resulting soln no doubt will be lower, with or without any molecule ,as long as one bubble socl2 into any soln
but is this reaction viable? does Vit A(in this qn) , gets its hydroxy group substituted with a Cl?
if we want to cause nucleophillic sub of a halogen onto Vit A containing the OH group, we would carry out in anhydrous condition , with Vit A as a solid right?
no doubt only a very small percentage of Vit A , will dissolve in water, as the molecule is majority Non-polar and hydrophobic
so perhaps it does undergoes nu sub for the halogen when Vit A is put in water and SOCl2 in bubbled into this soln of water containing Vit A , after substituting for halogen most likely still Solid, as molecule is largely non polar
some Continual Assessment test
https://www.dropbox.com/s/y6y6ajf0pzvaptv/20160913_204221-1.jpg?dl=0
for what type of reaction is the reaction in III, cher say it's disproportionation/redox
can i say it's disproportionation redox, acid base reaction?
Aldehydes can undergo rxn with NaOH(aq) just like that by itself without heat? do we need to know that aldehydes can ketone can react with strong bases?
https://www.dropbox.com/s/hzuqucmgqglddfi/20160913_204209-1.jpg?dl=0
option 3, fine , it's correct, 9 chiral centres after reaction with LiAlH4
have five sets of two-adjacent chiral carbons in a ring, total of 7 such carbons in this case
which set of two-adjacent chiral carbons can only produce 2 optical isomers?
how many stereo optical isomers are there in this molecule?
Originally posted by Flying grenade:2016 YJC JC2 Block test
https://www.dropbox.com/s/68y2bmpevqboy3g/20160913_203034-1.jpg?dl=0
is this plausible?
gaseous SOCl2 bubbled into a soln containing molecules with hydroxy group. the pH of resulting soln no doubt will be lower, with or without any molecule ,as long as one bubble socl2 into any soln
but is this reaction viable? does Vit A(in this qn) , gets its hydroxy group substituted with a Cl?
if we want to cause nucleophillic sub of a halogen onto Vit A containing the OH group, we would carry out in anhydrous condition , with Vit A as a solid right?
no doubt only a very small percentage of Vit A , will dissolve in water, as the molecule is majority Non-polar and hydrophobic
so perhaps it does undergoes nu sub for the halogen when Vit A is put in water and SOCl2 in bubbled into this soln of water containing Vit A , after substituting for halogen most likely still Solid, as molecule is largely non polar
U coconaden, anhydrous doesn't mean no solvent! The reactant is solvated (not solid) and thus can exists in a solution, just not with water (ie. not aqueous solution, a non-aqueous solution lah)!
The solvent has to be inert (ie. no protic OH or NH groups), so that it won't use up the PCl5 or SOCl2 reagent.
The YJC teacher's error : if it's an aprotic solvent which won't react with PCl5 or SOCl2, then neither will [H+] in solution increase, and hence there'll be no change in pH.
But for A level purposes, Cambridge may willingly commit the same 'error' as ur YJC teacher (ie. they may set this same qn), coz they don't expect A level H2 students to think deeply enough to recognize this as an error.
Originally posted by Flying grenade:some Continual Assessment test
https://www.dropbox.com/s/y6y6ajf0pzvaptv/20160913_204221-1.jpg?dl=0
for what type of reaction is the reaction in III, cher say it's disproportionation/redox
can i say it's disproportionation redox, acid base reaction?
Aldehydes can undergo rxn with NaOH(aq) just like that by itself without heat? do we need to know that aldehydes can ketone can react with strong bases?
https://www.dropbox.com/s/hzuqucmgqglddfi/20160913_204209-1.jpg?dl=0
option 3, fine , it's correct, 9 chiral centres after reaction with LiAlH4
have five sets of two-adjacent chiral carbons in a ring, total of 7 such carbons in this case
which set of two-adjacent chiral carbons can only produce 2 optical isomers?
how many stereo optical isomers are there in this molecule?
Due to the bicyclic rings, 4 chiral C atoms are 'merged' into 2 chiral C atoms, as far as calculation of stereoisomers are concerned.
However, because this is considered beyond A levels, hence in the A level exam, especially if it's MCQ, you may have to consider either possibility as correct (ie. Cambridge may think A level students aren't smart enough to understand or recognize the problematic bicyclic ring circumstance, and so the 'correct' answer may actually be the 'wrong' one; so if both possibilities appear as different options, then u jia lat liao).
If it's a P2 or P3 qn, the exam-smart BedokFunland JC student will give both possible answers, with qualifications and explanations.
Besides, this question (and hopefully Cambridge's questions) only asked for how many chiral C atoms, instead of asking how many stereoisomers are possible.
Originally posted by UltimaOnline:
Both u and YJC teacher made an error here (urs way more coconaden and unforgivable!).U coconaden, anhydrous doesn't mean no solvent! The reactant is solvated (not solid) and thus can exists in a solution, just not with water (ie. not aqueous solution, a non-aqueous solution lah)!
The solvent has to be inert (ie. no protic OH or NH groups), so that it won't use up the PCl5 or SOCl2 reagent.
The YJC teacher's error : if it's an aprotic solvent which won't react with PCl5 or SOCl2, then neither will [H+] in solution increase, and hence there'll be no change in pH.
But for A level purposes, Cambridge may willingly commit the same 'error' as ur YJC teacher (ie. they may set this same qn), coz they don't expect A level H2 students to think deeply enough to recognize this as an error.
great thanks. now that you wrote what anhydrous means, ppl reading this will finally know
so if the soln is inert, after pcl5 or socl2 reacted with -OH, the HCl will escape the reaction mixture as HCl(g)?
Originally posted by UltimaOnline:
Both u and YJC teacher made an error here (urs way more coconaden and unforgivable!).U coconaden, anhydrous doesn't mean no solvent! The reactant is solvated (not solid) and thus can exists in a solution, just not with water (ie. not aqueous solution, a non-aqueous solution lah)!
The solvent has to be inert (ie. no protic OH or NH groups), so that it won't use up the PCl5 or SOCl2 reagent.
The YJC teacher's error : if it's an aprotic solvent which won't react with PCl5 or SOCl2, then neither will [H+] in solution increase, and hence there'll be no change in pH.
But for A level purposes, Cambridge may willingly commit the same 'error' as ur YJC teacher (ie. they may set this same qn), coz they don't expect A level H2 students to think deeply enough to recognize this as an error.
ok dafuq, means we do tick the option as Correct in the exam, that anhydrous socl2 will result in a decrease in pH of the soln , even though we know that it won't, in a inert soln(e.g. ccl4)? force yourself get confused like this, not healthy
and is it like Since we know socl2 or pcl5 must react in anhydrous condition, if Cambridge present the question similar to above, do u think they expect us to think that the solvent must be inert?
However, because this is considered beyond A levels, hence in the A level exam, especially if it's MCQ, you may have to consider either possibility as correct (ie. Cambridge may think A level students aren't smart enough to understand or recognize the problematic bicyclic ring circumstance, and so the 'correct' answer may actually be the 'wrong' one; so if both possibilities appear as different options, then u jia lat liao).
wow, i hope they want us to select the correct Chemistry option, if this unfortunate event happens. i will take my chances and gamble or give or take that 1 mark away and shade the correct Chemistry answer
Originally posted by Flying grenade:great thanks. now that you wrote what anhydrous means, ppl reading this will finally know
so if the soln is inert, after pcl5 or socl2 reacted with -OH, the HCl will escape the reaction mixture as HCl(g)?