2017 H2 Chemistry JC1 & 2 students post your questions here
-
Hi UltimaOnline :P
Mind solving this question for me?
I'd provided the possible structures for dichorobenzene. -
Edited : my bad, but this is an idiotic qn!
-
Originally posted by UltimaOnline:
Edited : my bad, but this is an idiotic qn!Not really. I wanna see your opinions haha.
Because my answer is 3.Examiner Reports for this paper haven't came out. :P
-
Originally posted by iSean:
Not really. I wanna see your opinions haha.
Because my answer is 3.Examiner Reports for this paper haven't came out. :P
Edited : my bad, but this is an idiotic qn! -
Originally posted by UltimaOnline:
Edited : my bad, but this is an idiotic qn!Ease of questioning you to explain to me why is not 3.
Basically this isn't a question that most teachers really touch on, or just my college only.
Where most students can answer it either be 4, 5 or 6. -
Originally posted by iSean:
Ease of questioning you to explain to me why is not 3.
Basically this isn't a question that most teachers really touch on, or just my college only.
Where most students can answer it either be 4, 5 or 6.
Edited : my bad, but this is an idiotic qn! -
Originally posted by UltimaOnline:
Edited : my bad, but this is an idiotic qn!Well, I'd "solved" it. But not technically the answer wanted by Cambridge, which is B. Hence I seek your wisdom, @UltimaOnline.
Apparently they were looking for 4 structures.So I only can conclude that Structure 4. Is what they wanted.
As 5 and 6 correspond back to 2 and 3.
But under the resonant state does really 1 and 4 couldn't consider as one? -
Originally posted by iSean:
Well, I'd "solved" it. But not technically the answer wanted by Cambridge, which is B. Hence I seek your wisdom, @UltimaOnline.
Apparently they were looking for 4 structures.So I only can conclude that Structure 4. Is what they wanted.
As 5 and 6 correspond back to 2 and 3.
But under the resonant state does really 1 and 4 couldn't consider as one?
Ah yes, you're right. It's my bad, in failing to recognize it's a matter of 'camera angle', why 2 = 5 and 3 = 6.But it's also Cambridge's (and all Singapore JCs and Malaysian schools who set such questions) bad, for setting such an idiotic question. It's idiotic because cyclohex-1,3,5-triene doesn't and cannot exist, and so questions shouldn't be set on it. Also, calling it "Kekule structure" is erroneous, even if Mr Kekule himself got it wrong back then. Kekule structure = Lewis structure (originally, slight difference, but these days they're used interchangeably). Cyclohex-1,3,5-triene should be called "resonance contributor" of benzene, not "Kekule structure" of benzene.
To answer your question, in the real world, ie. resonance hybrid, 1 = 4. But in human minds, ie. resonance contributor, 1 not = 4. Your confusion proves my point, this is an idiotic question.
-
^_^
-
The Chemical Mystery of the Kim Jong Nam Assassination (and why 2 separate women assassins were required) :
https://sg.news.yahoo.com/toxic-mystery-behind-kim-jong-nams-assassination-045311729.html
-
Hi ultima,could you help me to solve this question?
I got negative values....
"0.7g of a hydrocarbon K was burnt completely in excess dry oxygen.2.2g of co2 is formed.100cm3 of K required 600cm of o2 gas,measured at same temp and pressure for complete combustion.Determine the molecular formula of K."
-
Originally posted by Carychidestar:
Hi ultima,could you help me to solve this question?
I got negative values....
"0.7g of a hydrocarbon K was burnt completely in excess dry oxygen.2.2g of co2 is formed.100cm3 of K required 600cm of o2 gas,measured at same temp and pressure for complete combustion.Determine the molecular formula of K."
Hi Carychidestar,(x + y/4) = 6
[ 0.7 / (12x + y) ] x = [ 2.2 / 44 ]
Solve the simultaneous equations.
-
Hi, I have a question, for N2014/P3/Q2(d),(e) why does the cyclic compound form an aliphatic compound after adding LiAlH4? I understand that the ketone group will be reduced to an alcohol but unsure about the ester functional group.... thanks
-
Originally posted by CKTR:
Hi, I have a question, for N2014/P3/Q2(d),(e) why does the cyclic compound form an aliphatic compound after adding LiAlH4? I understand that the ketone group will be reduced to an alcohol but unsure about the ether functional group.... thanks
LiAlH4 reduces ester group into 2 alcohols, at least one of which will be a primary alcohol (since it was obtained from the carboxylic acid part of the ester).No prob.
-
^_^
-
^_^
-
2 ang moh teenage slackers try out RJC life
-
hi
for Oxidation state questions, when do i use the O level formula and when do i draw out the structure and figure out the OS?
since sometimes both methods give different answers
-
Originally posted by glitter58:
hi
for Oxidation state questions, when do i use the O level formula and when do i draw out the structure and figure out the OS?
since sometimes both methods give different answers
Depends on the question and the molecule / ion involved, ie. simple vs complicated structure. If in doubt, draw out the structure to work out the OS of the individual atom in the individual resonance contributor (using my BedokFunland JC / University level formula or method for working out OS using structure), then give the more relevant answer (eg. for Organic Chem, in the oxidation of alcohols to carbonyl compounds to carboxylic acid, obviously the O level method, ie. an average OS per element, won't suffice), or if in doubt, give both answers (with explanation of each working). -
Hi UltimaOnline, is S in SO3 an expanded octet? Thank you! (:
-
Originally posted by keefay:
Hi UltimaOnline, is S in SO3 an expanded octet? Thank you! (:
Yo keefay,There are 7 major resonance contributors for SO3.
For A level H2 purposes, you can take it that the required resonance contributor (ie. the structure that you need to draw if asked in the A level exams) is the one without formal charges, ie. S has 6 bond pairs, 0 lone pairs. Hence in terms of a stable octet, the S atom (in this resonance contributor) has 12 valence electrons, ie. expanded octet.
But as a BedokFunland JC / Olympiad / H3 Chem student, I expect you pple to be cognizant of the other resonance contributors, in 3 of which the central dipositively formal charged S atom has exactly a stable octet with 1 doubly bonded no formal charged O atom and 2 singly bonded uninegatively formal charged O atoms, as well as another 3 resonance contributors in which the central unipositively formal charged S atom has an expanded octet with 2 doubly bonded no formal charged O atoms and 1 singly bonded uninegatively formal charged O atom.
In reality, the resonance contributor without formal charges, though usually taught in Singapore JCs, is actually the most minor of the resonance contributors, in spite of it having no formal charges (can you figure out why? My BedokFunland JC students can ask me during tuition, all other students can go ask their school teacher or private tutor).
Concordantly, all resonance contributors considered, you should be able to elucidate the resonance hybrid. But for A level H2 purposes, unless otherwise specified by the question, it will suffice to draw the simplest resonance contributor without formal charges.
Bonus BedokFunland JC challenge question : identify the orbital hybridization involved in all 4 atoms in the SO3 molecule, and hence describe the exact nature (ie. what type of overlap involving which specific orbitals) of each of the sigma and pi bonds (for the resonance contributors that have pi bonds).
No prob, keefay! :)
-
why is 1-ethylcyclohexanol a tertiary alcohol?
thanks!
-
Originally posted by glitter58:
why is 1-ethylcyclohexanol a tertiary alcohol?
thanks!
Because the alpha C atom is bonded to 3 other C atoms.https://www.google.com.sg/search?q=1-ethylcyclohexanol&source=lnms&tbm=isch
No prob!
-
another question, when butan-2-ol undergoes elimination, how do i know whether the major product is CH3CH2CH=CH2 or CH3CH=CHCH3 ?
-
Originally posted by glitter58:
another question, when butan-2-ol undergoes elimination, how do i know whether the major product is CH3CH2CH=CH2 or CH3CH=CHCH3 ?
There is the kinetic product, then there is the thermodynamic product. With sufficiently high temperatures, you can safely assume the major product is the thermodynamic product*.Internal alkenes are more thermodynamically stable compared to terminal alkenes*, and as such CH3CH=CHCH3 is expected to be the major product over CH3CH2CH=CH2.
(*Why? BedokFunland JC students can ask me during tuition, all other students can go ask their school teacher or private tutor.)