it doesn't make sense if purple,(high conc mno4-) (implies kmno4 is inside the conical flask, and is being titrated onto) -> colorless(all mno4- reacted) -> one extra drop MnO4-(pale pink)
Originally posted by UltimaOnline:
No. Becoz in titration, no matter how many drops (and each drop isn't exactly 1cm3 in the first place) you use to reach endpoint, the moles of titrant used will always technically be either slightly before, or slightly after, the equivalence point anyway.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.
i see! thanks god !
this is sibei confusing.
in the first place, small concentration of MnO4- is pale purple right? (pale pink not accepted?)
Mn2+ is pale pink in color.
if Mn2+ is in low concentrations, almost colorless, very faint pink
also, don't have titration whereby we add Mn2+ ions. only MnO4-.
i cant think of any scenarios whereby add mno4- into a colorless soln, and the color change is purple->colorless-> pale pink/purple(with extra drop of MnO4-)
except u have dilute soln of KMnO4. add lots of water, turn colorless. add one drop, color of soln perhaps turn pale pink
i got it alr Ultima. no need address this liao. thanks for your previous replies.
the only answer for 2013/3/2cii is just first permanent pale pink coloration when 1 excess drop of KMnO4 is added, thats all.
Decolorisation of purple to colorless is wrong.
once u add kmno4 into soln while stirring, purple immediately decolorise
Originally posted by Flying grenade:N11/p1/26
is it possible for 1,2-butandiol to exist(and with only one chiral centre )
oxidised to 1-hydroxybut-2-one or 2-hydroxybutan-1-al
oxidised to butan-1,2-dione ? or is it too unstable
sorry my blunder.
butan-1,2-dione doesn't exist.
only 2-oxobutan-1-al (aldehyde and ketone in the same molecule)
can tertiary alcohol undergo elimination to give alkene?
both this sources say can , but nvr show the mechanism of the final alkene cpd , i dk and cannot visualise
http://chem.libretexts.org/Core/Organic_Chemistry/Hydrocarbons/Alkenes/Synthesis_of_Alkenes/Alkenes_by_Dehydration_of_Alcohols#Secondary_and_tertiary_alcohols_dehydrate_through_the_E1_mechanism
http://www.masterorganicchemistry.com/2015/04/16/elimination-reactions-of-alcohols/
e.g. N07/1/27
was thinking can (CH3)3COH undergo elimination rxn to give alkene?
ok realized it can. form (CH3)2C=CH2
Originally posted by Flying grenade:can tertiary alcohol undergo elimination to give alkene?
both this sources say can , but nvr show the mechanism of the final alkene cpd , i dk and cannot visualise
http://chem.libretexts.org/Core/Organic_Chemistry/Hydrocarbons/Alkenes/Synthesis_of_Alkenes/Alkenes_by_Dehydration_of_Alcohols#Secondary_and_tertiary_alcohols_dehydrate_through_the_E1_mechanism
http://www.masterorganicchemistry.com/2015/04/16/elimination-reactions-of-alcohols/
Figure out the mechanism for yourself, or go ask your school teacher or private tutor.
2014 p1 qn 22
important theory qn
for alkaline hydrolysis of amide, let's say for example, there are 4 amide linkages in a cpd. is it 4 moles of NaOH will react with the cpd?
or is it 8 mol of NaOH? Angeline Tan and Jasmine Wong writes NaOH will react with the COOH formed from the alkaline hydrolysis of amides.
is it straight away will from COO- liao instead of COOH then COO- , as suggested by Angeline Tan and Jasmine Wong?
i havent encountered this in all books and notes so far, except in the solution written by A&J. seems like they are the first few who thought of this
Originally posted by Flying grenade:2014 p1 qn 22
important theory qn
for alkaline hydrolysis of amide, let's say for example, there are 4 amide linkages in a cpd. is it 4 moles of NaOH will react with the cpd?
or is it 8 mol of NaOH? Angeline and jasmine writes NaOH will react with the COOH formed from the alkaline hydrolysis of amides.
is it straight away will from COO- liao instead of COOH then COO- , as suggested by Angeline and Jasmine?
i havent encountered this in all books and notes so far, except in the solution written by A&J. seems like they are the first few who thought of this
Wah seh, intimate first name basis leh, machiam they ur ex-girlfriends liddat...
Instead of spoon-feeding you the answer (and I don't like it when you openly criticize/point out the author's typos on the forum, even though in your mind you think you're doing them a favor by pointing out their typos ; or worse, when you ask *me* to publicly confirm who's right who's wrong, eg. Chan Kim Seng vs George Chong vs CS Toh etc), so I'll leave it to yourself to solve this problem, quite easily done, by drawing out the hydrolysis mechanism (try drawing both acidic vs alkaline mechanisms), and you'll get your answer.
And don't post the answer here on the forum (as you are wont to do). Anyone else interested can go draw out the mechanisms for themselves to arrive at the answer.
idk if its typo
i am asking becos i dunno the chemistry concept, not becos i am questioning their typo
im not and never my intention to criticise authors. my words never show that. i am asking qn. but if other people interpret in that way, let them be. i never write anything offending, and my words not against law
Originally posted by Flying grenade:idk if its typo
i am asking becos i dunno the chemistry concept, not becos i am questioning their typo
im not and never my intention to criticise authors. my words never show that. i am asking qn. but if other people interpret in that way, let them be. i never write anything offending, and my words not against law
Regardless, like everyone else who asks me Chem qns here on the forum, I'd rather you focus on asking about TYS or Prelim Papers qns, instead of comparing notes and books and narrowly zooming in on nitty-gritty details and OCDing over who's right who's wrong (which you can actually settle yourself by cross referencing and applying common, ok, chemistry sense).
OCD ppl like me like to learn true knowledge, after learning many wrong concepts/knowledge/things all my life(i dont blame anyone) , I am Tired of learning false and wrong concepts, i only desire to learn the Truth.
i am curious, i desire to understand the Sciences, i work extremely hard cross reference excessively, think, analyse, and then if truly in need, seek help and the Truth from one of the rare and true teachers, Ultima.
self discovery and learning and independent learning is supremely important (must put in effort think critically yourself first before asking) , with the guidance from extremely desired,sought after and Trustworthy, knowledgeable, credible educators like BFJC
yes i truely thank all the authors who write these books to help me gain knowledge ,and i love all these books they wrote, they are great books i treasure them
Love the Sciences , work hard to learn the Truth(extremely important)
Originally posted by Flying grenade:OCD ppl like me like to learn true knowledge, after learning many wrong concepts/knowledge/things all my life(i dont blame anyone) , I am Tired of learning false and wrong concepts, i only desire to learn the Truth.
i work extremely hard cross reference excessively, think, analyse, and then if truly in need, seek help and the Truth from one of the rare and true teachers, Ultima.
self discovery and learning and independent learning is supremely important (must put in effort think critically yourself first before asking) , with the guidance from extremely Trustworthy, knowledgeable, credible educators like BFJC
yes i truely thank all the authors who write these books to help me gain knowledge ,and i love all these books they wrote, they are great books i treasure them
Love the Sciences , work hard to learn the Truth
PS. The message above is directed at all Singapore JC students taking the 2016 A levels. Have fun! ;Þ
Yes, i settle for myself, for the alkaline hydrolysis of Amides, one mol of OH- will react with 1 mol of amide, forming the Carboxylate immediately.
Correct?
i drew the mechanism with help of made easy org book page 365( Thank you, George Chong), and the help of Google, and i realise the Carboxylate is formed(i.e. not carboxylic acid form then Carboxylate) , in alkaline hydrolysis.
Originally posted by Flying grenade:Yes, i settle for myself, for the alkaline hydrolysis of Amides, one mol of OH- will react with 1 mol of amide, forming the Carboxylate immediately.
Correct?
i drew the mechanism with help of made easy org book page 365( Thank you, George Chong), and the help of Google, and i realise the Carboxylate is formed(i.e. not carboxylic acid form then Carboxylate) , in alkaline hydrolysis.
Originally posted by UltimaOnline:
It's all well and good you passed your Prelims, but remember you're up against the best of the entire Singapore cohort of all 22 JCs. From now on, imagine yourself taken out of your school and thrust into RJC. That's who you're up against in the bell-curve. *evil laugh*PS. The message above is directed at all Singapore JC students taking the 2016 A levels. Have fun! ;Þ
yes , all of us students taking the 2016 A levels know we are up against everyone else. i never shy away from competition. i acknowledge, embrace and make the best out of it.
Originally posted by UltimaOnline:
Eh hello I told u liao, whatever conclusion you get, don't post on the forum! Don't spoil the fun for others! Encourage them to draw out the mechanism for themselves to get the answer for themselves! Therefore, I will neither confirm nor deny your finding.
okay thanks ultimaaaa!!
i was thinking about this
any reaction with MnO4-
for example, MnO4- + 8H+ 5Fe2+ -> Mn2+ + 5Fe3+ +4H2O
will always form pale pink Mn2+ ions
in particular the 2013 p3 qn2cii
add KMnO4 into colorless soln of so2(aq)
2MnO4- + 5SO2 +2H2O -> 2Mn2+ + 5SO42- + 4H+
purple MnO4- is immediately decolorised as it is the limiting reagent, and it reacts with SO2(aq) . Pale Pink Mn2+ is already formed.
the solution should already have pale pink color already.
it's the little excess amount of MnO4- (due to the excess drop) after the end point that gives the stronger, Pale Pink colour.
of course when MnO4- is in large excess, solution turns purple.
Originally posted by Flying grenade:i was thinking about this
any reaction with MnO4-
for example, MnO4- + 8H+ 5Fe2+ -> Mn2+ + 5Fe3+ +4H2O
will always form pale pink Mn2+ ions
in particular the 2013 p3 qn2cii
add KMnO4 into colorless soln of so2(aq)
2MnO4- + 5SO2 +2H2O -> 2Mn2+ + 5SO42- + 4H+
purple MnO4- is immediately decolorised as it is the limiting reagent, and it reacts with SO2(aq) . Pale Pink Mn2+ is already formed.
the solution should already have pale pink color already.
it's the little excess amount of MnO4- (due to the excess drop) after the end point that gives the stronger, Pale Pink colour.
of course when MnO4- is in large excess, solution turns purple.
U.O. : Though the word "delocalized" should be used in the context of resonance, while "dispersed" should be used in the context of induction. Here, only induction applies, not resonance (since both O atoms are sp3 hybridized).
when i want to explain why carboxylate ion or phenoxide ion is stable, relative to say, ethoxide,
must we say 'hence -ve formal charge is more dispersed'
as in , e.g. : phenoxide ion is more stable than ethoxide ion because the negative formal charge on O atom of phenoxide is delocalised by resonance into the benzene ring, (hence -ve formal charge is more dispersed ), hence more stable, hence phenol is more acidic than alcohol
the words in bracket need write? or can straight away write more stable after saying electron delocalisation
my school trains us to always mention about (+ve or -ve) charge dispersal when discussing stability of cations e.g. (in phenylammonium)/anions(phenoxide)
2014 qn
The best answer (expected by Cambridge) is indeed an alkyne. Just that Cambridge will probably (for A level purposes) accept a diene as well.
energy released in forming triple bond + 2 single bond greater than 2 double bond is it?
i was stuck here for a while, dk which is acceptable
then i thought of there will always be a mixture % of products
Originally posted by Flying grenade:U.O. : Though the word "delocalized" should be used in the context of resonance, while "dispersed" should be used in the context of induction. Here, only induction applies, not resonance (since both O atoms are sp3 hybridized).
when i want to explain why carboxylate ion or phenoxide ion is stable, relative to say, ethoxide,
must we say 'hence -ve formal charge is more dispersed'
as in , e.g. : phenoxide ion is more stable than ethoxide ion because the negative formal charge on O atom of phenoxide is delocalised by resonance into the benzene ring, (hence -ve formal charge is more dispersed ), hence more stable, hence phenol is more acidic than alcohol
the words in bracket need write? or can straight away write more stable after saying electron delocalisation
my school trains us to always mention about (+ve or -ve) charge dispersal when discussing stability of cations e.g. (in phenylammonium)/anions(phenoxide)