https://www.dropbox.com/s/9ejng7bnfmwzjsz/png%3Bbase64709b07a18ab9e405.png?dl=0
for the lone pair on the N atom in pyridine ring, it's not delocalised by resonance right? because if it is, then the lone pair cannot be datively bonded to Co alr right?
is the answer B (cos co used 2 e- to bond with 2 O atoms)?
if the lone pair on N atom in this particular molecule in this particular qn, cannot delocalise by resonance, does the other 5 C atoms in that ring have partial double bond/have resonance?
Originally posted by Flying grenade:https://www.dropbox.com/s/9ejng7bnfmwzjsz/png%3Bbase64709b07a18ab9e405.png?dl=0
for the lone pair on the N atom in pyridine ring, it's not delocalised by resonance right? because if it is, then the lone pair cannot be datively bonded to Co alr right?
is the answer B (cos co used 2 e- to bond with 2 O atoms)?
if the lone pair on N atom in this particular molecule in this particular qn, cannot delocalise by resonance, does the other 5 C atoms in that ring have partial double bond/have resonance?
The pyridine ring is of course aromatic, with all bonds within the ring having partial double-bond character in the resonance hybrid. The lone pair is not needed (and not wanted!) for aromaticity here, hence it resides in a sp2 hybridized orbital instead of unhybridized p orbital (ala phenylamine), is hence available to accept a proton, to attack an electrophile, or function as a ligand. Consequently and concordantly, pyridine is more (both Lewis and Bronsted-Lowry) basic and nucleophilic, as compared to phenylamine.
in pyridine by itself, is the lone pair on N needed and wanted for aromaticity?
Originally posted by Flying grenade:in pyridine by itself, is the lone pair on N needed and wanted for aromaticity?
Originally posted by UltimaOnline:
Since the OS of the Co atom here is (-3) + (+5) = (+2), hence electron-configuration is [Ar] 4s0 3d7.The pyridine ring is of course aromatic, with all bonds within the ring having partial double-bond character in the resonance hybrid. The lone pair is not needed (and not wanted!) for aromaticity here, hence it resides in a sp2 hybridized orbital instead of unhybridized p orbital (ala phenylamine), is hence available to accept a proton, to attack an electrophile, or function as a ligand. Consequently and concordantly, pyridine is more (both Lewis and Bronsted-Lowry) basic and nucleophilic, as compared to phenylamine.
oh -3 because three dative bonds to co, +5 because co connected to five more electronegative atoms.
its like when calculating o.s. , treat the lp from N as donated and then (an e-) attracted back to it at the same time, lol. but its great, this is how chemistry works
Originally posted by Flying grenade:oh -3 because three dative bonds to co, +5 because co connected five more electronegative atoms.
its like when calculating o.s. , treat the lp from N as donated and then (an e-) attracted back to it at the same time, lol. but its great, this is how chemistry works
Originally posted by UltimaOnline:
Didn't u understand what I wrote? Obviously not (both u and the lone pair).
ok yes i got it alr ! lp on N in pyridine resides in sp2 hybridised orbital. not needed in pyridine's aromaticity ! hence pyridine more basic and nucleophilic than phenylamine !
Thanks, god !
Originally posted by UltimaOnline:
Good that u understand this (both the calculation and how chemistry works).
Thanks, God!
https://www.dropbox.com/s/837vt8arsov7rgr/20160913_084805.jpg?dl=0
https://www.dropbox.com/s/53zyvau1gruxljg/20160913_084812.jpg?dl=0
can this be considered tautomerisation?
in the first step, the pi bond is broken , and there is a intramolecular transfer of a proton
side chain oxidative cleavage of Ethylbenzene yields benzoic acid, CO2 + H2O?
i thought it's benzoic acid and ethanoic acid?
i don't get how ethanedioic acid is produced, which further oxidises to CO2 and H2O
for the oxidative cleavage, isn't only one side of the Ethyl have a new cooh attached, it is both side, to form ethanedioic acid?
when Al lose 3 e- through ionic bonding to form Al3+ , it has tripositive formal charge
Al in Alcl3 is e- deficient, having 6 e- around it , 3 bp, 0lp. Al in group 13, no formal charge
in Al2Cl6, is the formal charge of Al -1 , since it has 4 bp around it? but Al2Cl6 is indeed a neutral molecule
for the acidic hydrolysis of nitrile to form carboxylic acid, i was penalised for using Conc h2so4, HUR, instead of dilute h2so4, heat under reflux
is it liable to be penalised for reagents and conditions that can work, albeit unnecessary?
e.g. a rxn can procede with conc acid, or another rxn can proceed without heat, e.g. for the electrophillic addn of Br2 (aq) to phenol, or E add of hbr to alkene, can happen at room temp, but if a student write heat, will still be penalised becos unsure the rxn need input (heat) energy or not?
interesting that N1 and N4 forms a covalent bond with Zn
https://www.dropbox.com/s/egqpp5rzfk1b4hg/20160913_144056-1.jpg?dl=0
then it's like Zn as a metal didn't really get to donate(as in ionic bonding) it's two 4s electrons, it's merely sharing
also, Zn's electronic configuration in its ground state is alr [Ar] 3d10 4s2
how does Zn still have vacant, energetically accessible orbitals to accept dative bond from N2 and N4?
2011 p3 qn 2 iv
need to calculate pH at end point for the titration curve
cstoh's and sean chua's didn't indicate the pH of the two end points of this diprotic acid
is it look at the marks and evaluate whether cambridge want us to indicate pH at end point on the titration curve?
Originally posted by Flying grenade:https://www.dropbox.com/s/837vt8arsov7rgr/20160913_084805.jpg?dl=0
https://www.dropbox.com/s/53zyvau1gruxljg/20160913_084812.jpg?dl=0
can this be considered tautomerisation?
in the first step, the pi bond is broken , and there is a intramolecular transfer of a proton
Obviously not lah! First step is Bronsted-Lowry acid-base deprotonation, followed by nucleophilic addition. The product has 2x the no. of C atoms as the reactant, so how can they be structural isomers (ie. tautomers) of each other?!? Sheeeeesshh!!!
-_________-"
Originally posted by Flying grenade:side chain oxidative cleavage of Ethylbenzene yields benzoic acid, CO2 + H2O?
i thought it's benzoic acid and ethanoic acid?
i don't get how ethanedioic acid is produced, which further oxidises to CO2 and H2O
for the oxidative cleavage, isn't only one side of the Ethyl have a new cooh attached, it is both side, to form ethanedioic acid?
Originally posted by Flying grenade:when Al lose 3 e- through ionic bonding to form Al3+ , it has tripositive formal charge
Al in Alcl3 is e- deficient, having 6 e- around it , 3 bp, 0lp. Al in group 13, no formal charge
in Al2Cl6, is the formal charge of Al -1 , since it has 4 bp around it? but Al2Cl6 is indeed a neutral molecule
Originally posted by Flying grenade:for the acidic hydrolysis of nitrile to form carboxylic acid, i was penalised for using Conc h2so4, HUR, instead of dilute h2so4, heat under reflux
is it liable to be penalised for reagents and conditions that can work, albeit unnecessary?
e.g. a rxn can procede with conc acid, or another rxn can proceed without heat, e.g. for the electrophillic addn of Br2 (aq) to phenol, or E add of hbr to alkene, can happen at room temp, but if a student write heat, will still be penalised becos unsure the rxn need input (heat) energy or not?
Yes, u'll be penalized if u write heat when rtp will suffice, coz it tells Cambridge that u dunno ur Chem reagents & conditions, and no boss will hire u as a chemist coz u're wasting his electricity and money and resources to do unnecessary nonsense.
Originally posted by Flying grenade:interesting that N1 and N4 forms a covalent bond with Zn
https://www.dropbox.com/s/egqpp5rzfk1b4hg/20160913_144056-1.jpg?dl=0
then it's like Zn as a metal didn't really get to donate(as in ionic bonding) it's two 4s electrons, it's merely sharing
also, Zn's electronic configuration in its ground state is alr [Ar] 3d10 4s2
how does Zn still have vacant, energetically accessible orbitals to accept dative bond from N2 and N4?
http://sgforums.com/forums/2297/topics/491191?page=10#posts-11184163
Originally posted by Flying grenade:2011 p3 qn 5 iv
need to calculate pH at end point for the titration curve
cstoh's and sean chua's didn't indicate the pH of the two end points of this diprotic acid
is it look at the marks and evaluate whether cambridge want us to indicate pH at end point on the titration curve?
thanks god for your replies.
Originally posted by UltimaOnline:
Read my reply to Gohby's question at the bottom of the page.http://sgforums.com/forums/2297/topics/491191?page=10#posts-11184163
in this scenario, is the e config of zn [Ar] 3d10?
Originally posted by UltimaOnline:
If in doubt whether it's required, just do it.
ok got it. for 1st end point, use amphiprotic formula
for 2nd end pt, use kb , find poh then ph
retweet throwback
http://sgforums.com/forums/2297/topics/325645
written 2008
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?