Originally posted by Flying grenade:N13/1/20
can resonance occur in the straight chain of the molecule(and therefore throughout the whole molecule including the benzene ring) ? the region indicated by "z" in the qn
if yes, then that molecule shown in the qn, is one of the many resonance contributors?
why is the carbon with unipositive formal charge have a hybridisation of sp3?
https://www.dropbox.com/s/j7sbvcugdwn6ahn/20160907_031503-1.jpg?dl=0
N09/1/37
Originally posted by Flying grenade:why is the carbon with unipositive formal charge have a hybridisation of sp3?
https://www.dropbox.com/s/j7sbvcugdwn6ahn/20160907_031503-1.jpg?dl=0
N09/1/37
Originally posted by UltimaOnline:
It doesn't.
ok , cs toh wrong for that. so that C with uni+ve formal charge is sp2 hybridised right?
but for the other carbon, is sp3.
from cs toh advanced guide, another version of carbocation intermediate formed is the cyclic bromonium ion(pg 306). in that case, both C is sp3 hybridised
Originally posted by Flying grenade:ok , cs toh wrong for that. so that C with uni+ve formal charge is sp2 hybridised right?
but for the other carbon, is sp3.
from cs toh advanced guide, another version of carbocation intermediate formed is the cyclic bromonium ion(pg 306). in that case, both C is sp3 hybridised
2013 paper 2 qn 5bi
i dont understand how to get the primary structure of cys-tyr-ile-gln-asn-cys-pro-leu-gly
why cannot(or can?) the other way round , gly-leu-pro-cys-asn-gln-ile-tyr-cys ?
i understand by convention, N terminal is on the left, C terminal is on the right, but i can't tell/dk how use this convention here
part bii)
both phenol(at tyr there) and amine (at cys there) can be deprotonated and protonated at pH 7?
Originally posted by Flying grenade:2013 paper 2 qn 5bi
i dont understand how to get the primary structure of cys-tyr-ile-gln-asn-cys-pro-leu-gly
why cannot(or can?) the other way round , gly-leu-pro-cys-asn-gln-ile-tyr-cys ?
i understand by convention, N terminal is on the left, C terminal is on the right, but i can't tell/dk how use this convention here
part bii)
both phenol(at tyr there) and amine (at cys there) can be deprotonated and protonated at pH 7?
This is known as post-translational modification of proteins (H2 Biology). Since NH3 isn't an amino acid, the C terminus remains as Gly, and hence the only correct acceptable answer is to begin with the correct N terminus, ie. Cys.
Yes, Cambridge will accept either phenol or amine as the answer. Don't be confused : phenol and amine exist in equilibrium with phenoxide and ammonium respectively, it doesn't mean at pH 7, phenol is completely deprotonated (it's not), or that amine is completely protonated (it's not). That's why Ka and Kb values exist, to describe the position of equilibrium and extent of hydrolysis for acidic and basic groups.
OHHH. i see, thanks Ultima !!!
yea, I just realised the free NH3 group
Originally posted by UltimaOnline:
The trickiness of this question (other than the disulfide bond / linkage / bridge), is that the last amino acid residue gly's COOH terminus has been acylated (ie. nucleophilic acyl substitution, addition-elimination, condensation) with NH3 to generate a terminal amide, which Cambridge deliberately employed to confuse students as to whether or not to begin with Gly or Cys as the N terminus.This is known as post-translational modification of proteins (H2 Biology). Since NH3 isn't an amino acid, the C terminus remains as Gly, and hence the only correct acceptable answer is to begin with the correct N terminus, ie. Cys.
Yes, Cambridge will accept either phenol or amine as the answer. Don't be confused : phenol and amine exist in equilibrium with phenoxide and ammonium respectively, it doesn't mean at pH 7, phenol is completely deprotonated (it's not), or that amine is completely protonated (it's not). That's why Ka and Kb values exist, to describe the position of equilibrium and extent of hydrolysis for acidic and basic groups.
nu sub of NH3 to COOH group to form amide is not in the H2 syllabus right?
yield super low?
for formation of amide, acyl halide + amine , carboxylic acid anhydride+amine , better yield?
Originally posted by Flying grenade:nu sub of NH3 to COOH group to form amide is not in the H2 syllabus right?
yield super low?
for formation of amide, acyl halide + amine , carboxylic acid anhydride+amine , better yield?
for the titration using KMno4 , for example, KMno4 with SO2 soln in 2013/p3/2cii , for detection of end point, is the observation from purple to colorless after adding KMno4 dropwise, better than, observing the first permanent pink color after adding one excess drop of KMno4 to a colorless soln?
Originally posted by Flying grenade:for the titration using KMno4 , for example, KMno4 with SO2 soln in 2013/p3/2cii , for detection of end point, is the observation from purple to colorless after adding KMno4 dropwise, better than, observing the first permanent pink color after adding one excess drop of KMno4 to a colorless soln?
[1 mark] ~
Until the first permanent pink colour
OR
Until colour change from colourless to pink
[1 mark] ~ Repeat until two titres are concordant within 0.1 cm3
whut, for titrations using KMnO4, end point detected when purple soln turns colorless, not accurate? ??
Originally posted by Flying grenade:whut, for titrations using KMnO4, end point detected when purple soln turns colorless, not accurate? ??
in practical sessions, i remembered i didn't titrate using kmno4 from burette, always is some other titration scenarios/cases, which always only involves adding 5 drops of kmno4 or something? i can't remember): and have alr thrown away all my practical worksheets . it always state kmno4 decolorises only. no have from colorless to pink
of the many questions i did before,ive only seen purple turn colorless, this is the 1st time(2013 p3) i encounter a answer booklet says kmno4 colorless to pink :/
Originally posted by Flying grenade:in practical sessions, i remembered i didn't titrate using kmno4 from burette, always is some other titration scenarios/cases, which always only involves adding 5 drops of kmno4 or something? i can't remember): and have alr thrown away all my practical worksheets . it always state kmno4 decolorises only. no have from colorless to pink
of the many questions i did before,ive only seen purple turn colorless, this is the 1st time(2013 p3) i encounter a answer booklet says kmno4 colorless to pink :/
.
Originally posted by UltimaOnline:
Purple to colorless first, then colorless to pink.
yes i know that..
but my doubt is why must drop 1 more kmno4 to turn it from colorless to pink
wouldn't when kmno4 decolorises , i.e. purple to colorless, suffice, for detection of end pt
furthermore, 1 drop is approx 1cm3, so which method more accurate
Originally posted by Flying grenade:yes i know that..
but my doubt is why must drop 1 more kmno4 to turn it from colorless to pink
wouldn't when kmno4 decolorises , i.e. purple to colorless, suffice, for detection of end pt
OMG ULTIMA SIBEI SMART
THANKS BOSS
zai kia
err.. so do we take Volume of KMnO4 to turn soln into a permanent pink color, minus 1cm3 ? since that 1 extra drop is in excess
Originally posted by Flying grenade:err.. so do we take Volume of KMnO4 to turn soln into a permanent pink color, minus 1cm3 ? since that 1 extra drop is in excess
So if you minus 1 drop, your moles of titrant used will be slightly before equivalence point. If you don't minus 1 drop, your moles of titrant used will be slightly after equivalence point. Since either way it'll be slightly inaccurate (can't be helped), so it makes more sense to just use the titre volume to reach endpoint, without further complicating errors by minusing away 1 drop.
Now you understand why we call this 'endpoint' instead of 'equivalence point'? Endpoint merely approximates equivalence point.
wait.
the answer provided by sean chua(SAP publisher) is wrong .
2013/p3/2cii
KMnO4 is added from burette.
solution is colorless. end point is detected when one extra drop of KMnO4 turns the solution a permanent pink color soln.