for delocalisation of electrons to occur, the adjacent atom with respect to the atom containing the unhybridised p orbital with a lone pair of e- must have an empty unhybridised p orbital?
Hi, I have a question regarding 2014 AJC Prelim, paper 2 Qn 5.
Regarding option C, when 4 methylbenzoic acid reacts with Br2(aq), why won't C8H8O2Br2 form? The positions of CH3 (2,4 directing) and COOH (1,3 directing) will support bromination right?
SORRY CORRECTION
R GROUP NH2 is more basic than Alpha NH2 group in the FIRST PLACE, and hence, will definitely be protonated first, followed by the alpha NH2 group.
i gt good smart qn haha
lp of e- of N atoms of R group of Arginine are delocalised by resonance. since the extra stability conferred by resonance,
how would we know if the NH2 R group would be less willing in accepting a proton(as in due to delocalisation by resonance), and hence less basic, compared to the alpha NH2 group? (or more basic, due to the rule that Rgroup NH2 is more basic than Alpha NH2 group )
true case : imine group more basic. reasons described below.
misinterpreted case : R group NH2 less basic due to resonance.
https://en.m.wikipedia.org/wiki/Arginine
this is the issue that was bugging my mind just now, but couldn't gather my thoughts fast enough.
the two N atoms of the R group of Arginine has partial double bond due to delocalisation of electrons by/due to resonance.
will the / how do we know if, either of the R group N atoms, is more stable, and hence, may or may not readily accept proton compared to Alpha N atom of Alpha NH2 group, and hence may or may not be more basic than the alpha NH2 group,
or the R group NH3+ may or may not be more acidic than alpha group NH3+ as the tendency to be protonated is more or less
Originally posted by Flying grenade:for delocalisation of electrons to occur, the adjacent atom with respect to the atom containing the unhybridised p orbital with a lone pair of e- must have an empty unhybridised p orbital?
Originally posted by supercat:Hi, I have a question regarding 2014 AJC Prelim, paper 2 Qn 5.
Regarding option C, when 4 methylbenzoic acid reacts with Br2(aq), why won't C8H8O2Br2 form? The positions of CH3 (2,4 directing) and COOH (1,3 directing) will support bromination right?
Even without the electron-withdrawing COOH group, methylbenzene still isn't sufficiently activated to be able to attack Br2(aq). You need anhydrous Br2 with AlBr3 or FeBr3 Lewis acid catalyst for that.
And if free radical substitution of the methyl group is desired, exposure to UV light is required.
Originally posted by Flying grenade:SORRY CORRECTION
R GROUP NH2 is more basic than Alpha NH2 group in the FIRST PLACE, and hence, will definitely be protonated first, followed by the alpha NH2 group.
i gt good smart qn haha
lp of e- of N atoms of R group of Arginine are delocalised by resonance. since the extra stability conferred by resonance,
how would we know if the NH2 R group would be more willing in accepting a proton, and hence more basic, compared to the alpha NH2 group?
https://en.m.wikipedia.org/wiki/Arginine
this is the issue that was bugging my mind just now, but couldn't gather my thoughts fast enough.
the two N atoms of the R group of Arginine has partial double bond due to delocalisation of electrons by/due to resonance.
will the / how do we know if, either of the R group N atoms, is more stable, and hence, may or may not readily accept proton compared to Alpha N atom of Alpha NH2 group, and hence may or may not be more basic than the alpha NH2 group,
or the R group NH3+ may or may not be more acidic than alpha group NH3+ as the tendency to be protonated is more or less
Your confusion lies in thinking the lone pair of the R group N atom is delocalized *before* protonation, and hence less available to accept a proton. Nonsense.
First of all, only the R group amine's lone pair is slightly delocalized by resonance before protonation, but the R group imine's lone pair isn't, and it is the R group imine which is protonated (not the amine). Thereupon, the R group amine's lone pair then becomes extensively delocalized, in order to allow the positive formal charge on the conjugate acid to be effectively delocalized, across both N atoms.
which substituent group has *strongly* activating or deactivating capabilities? xD
Originally posted by UltimaOnline:
Because the benzene ring of para-methylbenzoic acid isn't sufficiently activated to attack the Br2(aq), since CH3 is only electron-donating by induction and hence weakly activating, while COOH is electron-withdrawing by both induction and resonance, and is hence moderately deactivating.Even without the electron-withdrawing COOH group, methylbenzene still isn't sufficiently activated to be able to attack Br2(aq). You need anhydrous Br2 with AlBr3 or FeBr3 Lewis acid catalyst for that.
And if free radical substitution of the methyl group is desired, exposure to UV light is required.
lmao, we learn Br2(l) with anhydrous FeBr3 for electrophilic substitution of Br to methylbenzene
small little difference in details , these details, small and big
but we as students and educators try to be correct to the fullest extent as possible so we always aspire to learn the correct stuff
Originally posted by UltimaOnline:
Obviously the R group imine is more basic than the alpha amine, especially in Arginine. Because now there are 2 reasons, not just 1. Induction (electron-donating by induction R group) and resonance (the conjugate acid is stabilized by having its positive formal charge delocalized by resonance over 2 N atoms).Your confusion lies in thinking the lone pair of the R group N atom is delocalized *before* protonation, and hence less available to accept a proton. Nonsense.
First of all, only the R group amine's lone pair is slightly delocalized by resonance before protonation, but the R group imine's lone pair isn't, and it is the R group imine which is protonated (not the amine). Thereupon, the R group amine's lone pair then becomes extensively delocalized, in order to allow the positive formal charge on the conjugate acid to be effectively delocalized, across both N atoms.
OMG , thanks for clarifying, elucidating, and correcting my severe wrong thinking for this . Thank you.
Originally posted by Flying grenade:lmao, we learn Br2(l) with anhydrous FeBr3 for electrophilic substitution of Br to methylbenzene
small little difference in details , these details, small and big
but we as students and educators try to be correct to the fullest extent as possible so we always aspire to learn the correct stuff
Originally posted by Flying grenade:which substituent group has *strongly* activating or deactivating capabilities? xD
Originally posted by UltimaOnline:
That's also correct and equivalent. Both reagents obviously (if you don't know why, you're screwed) have to be anhydrous. No point if only 1 reagent is anhydrous when you're gonna mix them anyway, is there? Common error by Singapore JC students is to write "Br2(aq) with anhydrous FeBr3", zero marks.
ahh yes yes, thanks for pointing that out !!! it is indeed a common error by both students and teachers.
Now, the people who have read these posts, know better, the correct reagents.
from the back of my head, an example of a strongly (ring) deactivating group is Cl , strongly (ring) activating group is OH or O-
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Thank you very much, I understood your explanation. For the same question (2014 AJC Prelim, paper 2 Qn 5), for option B, there won't be bromination then? Since it is COOCH3, I infer that it is also electron withdrawing but still not sufficient for bromination.
For 2014 AJC Prelim, paper 3 Qn 2ciii, may I ask how is T formed? What type of reaction could possibly occur to form the alkene by-product? At first I thought that it could be SN1 mechanism since it is secondary alkyl halide. But it will not be able to form an alkene, especially with the removal of O, since O is needed for attacking the alkyl halide.
For 2014 AJC Prelim, paper 3 Qn3d, I had trouble deducing Geraniol. I could deduce N,P,Q and R, but I wonder why Geraniol doesn't have 2 terminal alkene. Since heating geraniol with excess KMnO4 will give N and P, and we see that 2 carbon has been removed, so why is there no 2 terminal alkene present? 2 CO2 could be removed. Instead, the product is a diol. Are diols colourless gases?
Originally posted by supercat:Thank you very much, I understood your explanation. For the same question (2014 AJC Prelim, paper 2 Qn 5), for option B, there won't be bromination then? Since it is COOCH3, I infer that it is also electron withdrawing but still not sufficient for bromination.
For 2014 AJC Prelim, paper 3 Qn 2ciii, may I ask how is T formed? What type of reaction could possibly occur to form the alkene by-product? At first I thought that it could be SN1 mechanism since it is secondary alkyl halide. But it will not be able to form an alkene, especially with the removal of O, since O is needed for attacking the alkyl halide.
For 2014 AJC Prelim, paper 3 Qn3d, I had trouble deducing Geraniol. I could deduce N,P,Q and R, but I wonder why Geraniol doesn't have 2 terminal alkene. Since heating geraniol with excess KMnO4 will give N and P, and we see that 2 carbon has been removed, so why is there no 2 terminal alkene present? 2 CO2 could be removed. Instead, the product is a diol. Are diols colourless gases?
Originally posted by supercat:Thank you very much, I understood your explanation. For the same question (2014 AJC Prelim, paper 2 Qn 5), for option B, there won't be bromination then? Since it is COOCH3, I infer that it is also electron withdrawing but still not sufficient for bromination.
For 2014 AJC Prelim, paper 3 Qn 2ciii, may I ask how is T formed? What type of reaction could possibly occur to form the alkene by-product? At first I thought that it could be SN1 mechanism since it is secondary alkyl halide. But it will not be able to form an alkene, especially with the removal of O, since O is needed for attacking the alkyl halide.
For 2014 AJC Prelim, paper 3 Qn3d, I had trouble deducing Geraniol. I could deduce N,P,Q and R, but I wonder why Geraniol doesn't have 2 terminal alkene. Since heating geraniol with excess KMnO4 will give N and P, and we see that 2 carbon has been removed, so why is there no 2 terminal alkene present? 2 CO2 could be removed. Instead, the product is a diol. Are diols colourless gases?
Hi supercat, do you mind uploading the picture of the questions? Thanks !
Originally posted by supercat:Thank you very much, I understood your explanation. For the same question (2014 AJC Prelim, paper 2 Qn 5), for option B, there won't be bromination then? Since it is COOCH3, I infer that it is also electron withdrawing but still not sufficient for bromination.
For 2014 AJC Prelim, paper 3 Qn 2ciii, may I ask how is T formed? What type of reaction could possibly occur to form the alkene by-product? At first I thought that it could be SN1 mechanism since it is secondary alkyl halide. But it will not be able to form an alkene, especially with the removal of O, since O is needed for attacking the alkyl halide.
For 2014 AJC Prelim, paper 3 Qn3d, I had trouble deducing Geraniol. I could deduce N,P,Q and R, but I wonder why Geraniol doesn't have 2 terminal alkene. Since heating geraniol with excess KMnO4 will give N and P, and we see that 2 carbon has been removed, so why is there no 2 terminal alkene present? 2 CO2 could be removed. Instead, the product is a diol. Are diols colourless gases?
2014 AJC Prelim, paper 2 Qn 5 : Correct, ester group is electron-withdrawing by both induction and resonance, hence the benzene ring is deactivated, no halogenation via electrophilic aromatic substitution occurs.
2014 AJC Prelim, paper 3 Qn 2ciii : E2 reaction occurs in competition with SN2. While the Singapore A level H2 Chem syllabus simplifies the matter of whether nucleophilic aliphatic substitution (ie. SN1 vs SN2) or elimination (ie. E1 vs E2) occurs as simply a matter of solvent (ie. aqueous vs ethanolic), but in practice (ie. real-life and Uni level Chem), all 4 compete against each other, and there are other factors involved in the 4-cornered fight : SN1 vs SN2 vs E1 vs E2.
While E1 and E2 mechanisms are not required for the Singapore A level H2 Chem syllabus, but nonetheless as evidenced by the AJC Prelim paper (and all Singapore JC Prelim papers and significantly tougher recent Singapore-Cambridge A level H2 Chem papers since 2010), it pays for students gunning for A grade to have a deeper understanding of Chemistry beyond the basic A level H2 syllabus. Which is the advantage that H3 Chem, Olympiad Chem, and BedokFunland JC Chem students have over H2 Chem only students.
2014 AJC Prelim, paper 3 Qn3d : You're right that the 2 missing C atoms are eliminated in the form of CO2(g). But you're incorrect that 2 terminal alkene groups must therefore be present in geraniol. Because if that were the case, how would heating geraniol with acidified KMnO4 yield N and P and CO2(g)? The given formula of geraniol precludes the acidic hydrolysis of ester or amide group as a possible explanation.
Concordantly, the student is expected and required to deduce and elucidate that the functional group in question (that results in 2 C atoms eliminated as CO2) is either the =CHCH2OH group, or the =CHCH= group, either of which is oxidized by KMnO4 with oxidative cleavage into ethandial, which is further oxidized to ethandioic acid, which is further oxidized to 2 moles of carbonic(IV) acid, which exists in equilibrium with, and hence decomposes into, 2 moles of CO2(g) + H2O(l), with thermodynamically favorable positive entropy change, and with the position of equilibrium shifting to the RHS as predicted by Le Chatelier's principle (since CO2(g) leaves the aqueous reaction mixture).
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Hi Flying grenade, prelim papers are available on the internet, I googled them online.
http://score-in-chemistry.weebly.com/2014-jc-prelim-papers-and-solutions-from-internet.html
^ AJC 2014 prelim
For E1 and E2, I searched them online. So I take it that nucleophilic sub occurs together with elimination? And SN1, E1 favours sec/tertiary compounds while SN2, E2 favours pri/sec compounds? In this case, then I see nucleophilic elimination "similar" to normal elimiation reactions (dehydration) in the sense that a C=C bond is formed? From my understanding, alcoholic solvents tends to promote elimination while (aq) tends to promote substitution. Does acidic/alkaline conditions play a role too?
Btw, why are methanoic acid and ethanedioic acid so special? They are the only 2 unique carboxylic acid that can be further oxidised to CO2 and H2O. Are there any reasons for this? Is it because of their structure?
Originally posted by supercat:Hi Flying grenade, prelim papers are available on the internet, I googled them online.
http://score-in-chemistry.weebly.com/2014-jc-prelim-papers-and-solutions-from-internet.html
^ AJC 2014 prelim
For E1 and E2, I searched them online. So I take it that nucleophilic sub occurs together with elimination? And SN1, E1 favours sec/tertiary compounds while SN2, E2 favours pri/sec compounds? In this case, then I see nucleophilic elimination "similar" to normal elimiation reactions (dehydration) in the sense that a C=C bond is formed? From my understanding, alcoholic solvents tends to promote elimination while (aq) tends to promote substitution. Does acidic/alkaline conditions play a role too?
Btw, why are methanoic acid and ethanedioic acid so special? They are the only 2 unique carboxylic acid that can be further oxidised to CO2 and H2O. Are there any reasons for this? Is it because of their structure?
Because of the high O to C ratio for these molecules, allowing ease of further oxidation, and hence more positive oxidation potentials.
Hi supercat, thanks for the link
can individual amino acids (residues ) lys and ser undergo neutralisation?
what is the R group interaction between lys and ser in a protein? is it ionic interaction as a result of neutralisation between the R groups? or is it H bonding? or is it indeed ionic interactions, but for reason idk
can neutralisation between R groups occur within the protein?