Lurkers, register with SgForums and ask ur H2 Chem qns here!
-
Originally posted by Flying grenade:
how is C-H(in HCFCs) more reactive than C-Cl or C-F(in CFCs) ???
BE(kjmol-1) C-F 486 , C-H 410, C-Cl 340
https://www.dropbox.com/s/jod3jnsesmr9c7m/20160923_170437-1.jpg?dl=0
The 'greater reactivity' described by CS Toh in this context isn't about bond strength. It's about H atoms being more readily reacted away (eg. combusted) by other environmental agents before reaching the stratosphere, compared to Cl atoms being less reactive in this other ways. Upon reaching the stratosphere, of course Cl atoms would be more free radical reactive and damaging to the ozone layer.Note that despite what CS Toh wrote, HCFCs, while less damaging than CFCs, are still damaging and thus still being phased out (like CFCs).
-
ahh i see ! understand.
what determines reactivity ? like e.g. combustion reactions
Ea, thermodynamics(energetics), stability of final pdts, kinetics, temp ?
isn't BE one of the factors that affects reactivity, and indeed require less energy(ea) to break a c-cl compared to c-h?
in the troposphere(below stratosphere ) less frs reactions
in the stratosphere, higher probability cl radicals would be formed compared to H•
-
Originally posted by Flying grenade:
ahh i see ! understand.
what determines reactivity ? like e.g. combustion reactions
Ea, thermodynamics(energetics), stability of final pdts, kinetics, temp ?
isn't BE one of the factors that affects reactivity, and indeed require less energy(ea) to break a c-cl compared to c-h?
in the troposphere(below stratosphere ) less frs reactions
in the stratosphere, higher probability cl radicals would be formed compared to H•
Yes, that's indeed 1 of the factors. Based on what CS Toh wrote, common sense will tell you that some other factors (beyond A levels) outweigh this factor. Chemistry is also about common sense. -
But if theres high oxidation state means it cant be oxidised further... and also high C:O ratio means harder to oxidise...
-
Originally posted by 8truthseeker8:
But if theres high oxidation state means it cant be oxidised further... and also high C:O ratio means harder to oxidise...
I said high O to C ratio, not C to O.The OS change is from +3 (for both C atoms) in ethanedioic acid, to +4 in carbon dioxide.
-
Yes, either way, that means the C is bonded to many O and cant be oxidised further... nitpicking on such small details wont do any good
-
which is the major product from
elimination from unsymmetric halogenoalkanes such as 2-bromobutane?
http://www.chemguide.co.uk/mechanisms/elim/elimunsym.html
didnt address it
is the major product
-
but-1-ene
-
cis-but-2-ene
-
or trans-but-2-ene?
-
-
ok, made easy organic book pg 77 and 78 have address this. understand alr.
but is cis but-2-ene or trans but-2-ene the major product?
-
eh in page 78 he wrote trans isomer is more stable than the cis isomer due to reduced steric hindrance
but he didnt comment/write on either the cis or trans would be the major product
so for the case of but-2-ene, is the trans isomer the major pdt? due to this reason of more stable? anymore reasons?
for other cases might be different? different geometric (cis-trans)isomers for different alkenes molecules have different major/minor products for diff reasons?
-
Originally posted by Flying grenade:
eh in page 78 he wrote trans isomer is more stable than the cis isomer due to reduced steric hindrance
but he didnt comment/write on either the cis or trans would be the major product
so for the case of but-2-ene, is the trans isomer the major pdt? due to this reason of more stable? anymore reasons?
for other cases might be different? different geometric (cis-trans)isomers for different alkenes molecules have different major/minor products for diff reasons?
First of all, be sure you don't confuse regiochemistry (Zaitsev vs Hofmann) with stereochemistry (cis vs trans, or E vs Z), and thermodynamic vs kinetic products.You can go read up on regiochemistry and thermodynamic vs kinetic products on your own if you like (don't ask me about it).
In terms of stereochemistry, whether the cis/trans or E/Z product is favored, depends on whether the elimination reaction (eg. dehydrohalogenation) occurs via E1 or E2 mechanism (go read up on the factors which influence E1 vs E2 on your own if you like, don't ask me about it).
When E1 occurs, the major product is the thermodynamic or more stable product (ie. trans geometic isomer).
When E2 occurs, there is only 1 product, which is determined by the stereochemistry of the reactant undergoing the elimination.
As Wikipedia states : "In the E2 mechanism, the two leaving groups need to be antiperiplanar, because an antiperiplanar transition state has staggered conformation with lower energy than a synperiplanar transition state which is in eclipsed conformation with higher energy ; consequently the reaction mechanism involving staggered conformation is favored for E2 reactions."
Concordantly and consequently, the elimination product of the E2 mechanism is already pre-determined by the stereochemistry of the reactant, and depending on case-by-case, will be *either* the cis or trans geometric isomer *only*.
If you get a mixture of cis and trans products, it's either because E1 occurred, or a mixture of E1 and E2 occurred.
Note that in reality (ie. at Uni level, in the chemical industry), a mixture of E1, E2, SN1, SN2 all occurs simultaneously and competitively, to generate a mixture of all possible products.
FlyingGrenade, you're the one who opened this can of worms by asking this question, your own fault blame yourself, don't expect me to clean up the worms after you. You'll probably end up with more questions than before, after reading this post above, but don't ask me any more questions on these matters.
For the sensible pragmatic A level H2 Chem student, just focus on your within H2 syllabus material and ignore whatever goes beyond (ie. for the Singapore H2 Chem exam, don't try to bring in Zaitsev or Hofmann or E1 or E2 or cis or trans or E or Z , unless specifically asked by the exam question).
For the already-getting-A-grades-for-Prelims enthusiastic A level student taking H3 or Olympiad Chem intending to study Medicine or Chemistry in the Uni, you can go read up on your own about these matters, and/or go ask your school teacher or private tutor.
Moving on...
-
https://en.m.wikipedia.org/wiki/Acetophenone
did the publisher copy the qn wrongly?
Should be A,B instead of A,B,C?
1-Phenylethan-1-one
cannot be oxidised by hot kmmo4(in acidic or alkaline conditions ) right? because there's no benzylic H?
-
Originally posted by UltimaOnline:
First of all, be sure you don't confuse regiochemistry (Zaitsev vs Hofmann) with stereochemistry (cis vs trans, or E vs Z), and thermodynamic vs kinetic products.You can go read up on regiochemistry and thermodynamic vs kinetic products on your own if you like (don't ask me about it).
In terms of stereochemistry, whether the cis/trans or E/Z product is favored, depends on whether the elimination reaction (eg. dehydrohalogenation) occurs via E1 or E2 mechanism (go read up on the factors which influence E1 vs E2 on your own if you like, don't ask me about it).
When E1 occurs, the major product is the thermodynamic or more stable product (ie. trans geometic isomer).
When E2 occurs, there is only 1 product, which is determined by the stereochemistry of the reactant undergoing the elimination.
As Wikipedia states : "In the E2 mechanism, the two leaving groups need to be antiperiplanar, because an antiperiplanar transition state has staggered conformation with lower energy than a synperiplanar transition state which is in eclipsed conformation with higher energy ; consequently the reaction mechanism involving staggered conformation is favored for E2 reactions."
Concordantly and consequently, the elimination product of the E2 mechanism is already pre-determined by the stereochemistry of the reactant, and depending on case-by-case, will be *either* the cis or trans geometric isomer *only*.
If you get a mixture of cis and trans products, it's either because E1 occurred, or a mixture of E1 and E2 occurred.
Note that in reality (ie. at Uni level, in the chemical industry), a mixture of E1, E2, SN1, SN2 all occurs simultaneously and competitively, to generate a mixture of all possible products.
FlyingGrenade, you're the one who opened this can of worms by asking this question, your own fault blame yourself, don't expect me to clean up the worms after you. You'll probably end up with more questions than before, after reading this post above, but don't ask me any more questions on these matters.
For the sensible pragmatic A level H2 Chem student, just focus on your within H2 syllabus material and ignore whatever goes beyond (ie. for the Singapore H2 Chem exam, don't try to bring in Zaitsev or Hofmann or E1 or E2 or cis or trans or E or Z , unless specifically asked by the exam question).
For the already-getting-A-grades-for-Prelims enthusiastic A level student taking H3 or Olympiad Chem intending to study Medicine or Chemistry in the Uni, you can go read up on your own about these matters, and/or go ask your school teacher or private tutor.
Moving on...
ok thanks god for your effort in typing and explaining.
no i dont blame you or myself for asking this qn. i think this chemistry is important. i never expect u to clean stuff up for me.
Thank you for taking effort to answer my(not within syllabus) conceptual, but important chemistry questions, within reasonable, introductory, basic limits , using legit chemistry, despite a toll on your patience, effort and willingness to share.
it helps. thank you
-
Originally posted by Flying grenade:
https://en.m.wikipedia.org/wiki/Acetophenone
did the publisher copy the qn wrongly?
Should be A,B,D instead of A,B,C?
1-Phenylethan-1-one
cannot be oxidised by hot kmmo4(in acidic or alkaline conditions ) right? because there's no benzylic H?
The benzylic H atom requirement for side-chain oxidation is specifically for hydrocarbon level. Since the side-chain is already semi-oxidized to the carbonyl level, KMnO4 can indeed further oxidize the carbonyl side-chain to a benzoic acid. -
Originally posted by UltimaOnline:
The benzylic H atom requirement for side-chain oxidation is specifically for hydrocarbon level. Since the side-chain is already semi-oxidized to the carbonyl level, KMnO4 can indeed further oxidize the carbonyl side-chain to a benzoic acid.Omg , i see. ok thanks !
-
Phenols can't be oxidised to benzoic acid because don't have a hydrogen atom on the carbon atom to which the OH group is attached ah?
-
Originally posted by Flying grenade:
Phenols can't be oxidised to benzoic acid because don't have a hydrogen atom on the carbon atom to which the OH group is attached ah?
Yeah, phenols don't even have a benzylic C atom, that's why. -
phenol oxidise to quinone?
http://www.mhhe.com/physsci/chemistry/carey/student/olc/ch24reactionphenols.html
- In general, phenols are more easily oxidized than simple alcohols.
-
Phenols are rather easily oxidized despite the absence of a hydrogen atom on the hydroxyl bearing carbon. Among the colored products from the oxidation of phenol by chromic acid is the dicarbonyl compound para-benzoquinone
https://www2.chemistry.msu.edu/faculty/reusch/virttxtjml/alcohol2.htm
-
https://en.m.wikipedia.org/wiki/Phenols#Oxidation
-
Originally posted by Flying grenade:
Phenols are rather easily oxidized despite the absence of a hydrogen atom on the hydroxyl bearing carbon. Among the colored products from the oxidation of phenol by chromic acid is the dicarbonyl compound para-benzoquinone
https://www2.chemistry.msu.edu/faculty/reusch/virttxtjml/alcohol2.htm
Indeed, when this question came out for a top JC Prelim paper a couple of years ago, only my BedokFunland JC students were able to draw the oxidation product.Why? I'll give you a hint : unlike Singapore JC H2 Chem students, only my BedokFunland JC students (together with H3 Chem and Olympiad Chem students), understand both the meaning of resonance contributors vs resonance hybrid, as well as my BedokFunland JC formula for calculating Oxidation State.
-
Q1: DHS/2015/P1/Q14. While acid strength (indicated by pKa) increases from HCl to HI, but due to the leveling effect, in aqueous solutions they are all strong acids which dissociate completely and hence result in the same pH.
HF only dissociates partially in aqueous solutions.
-
ground state doesn't only apply for atoms. it can be for other particles as well.
https://socratic.org/questions/how-do-you-determine-ground-state-electron-configurations-for-ions
-
what does this mean Ionic product" is always intended to be "initial before precipitation" ?
-
Originally posted by Flying grenade:
what does this mean Ionic product" is always intended to be "initial before precipitation" ?
Figure out for yourself. -
[Chemistry fact]
http://chemistry.stackexchange.com/questions/44488/stability-of-mno4-in-basic-vs-acidic-conditions
MnO4- is more stable in alkaline medium, compared to in acidic medium