Lurkers, register with SgForums and ask ur H2 Chem qns here!

last SG · 2015-08-19T22:54:00+00:00

Topic : atomic radius Hi! This is from singapore cambridge nov a level exam year 2010,Paper 1 qn 1Qn: which element has the largest atomic radius?BrKKrScSo...

UltimaOnline SG

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

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

Flying grenade SG

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.

UltimaOnline SG

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?

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.

Flying grenade SG

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.

Flying grenade SG

okay thanks ultimaaaa!!

Flying grenade SG

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.

UltimaOnline SG

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.

Yes, which is why Cambridge will also accept "pale purple" as an alternative to "pink" for endpoint.

Flying grenade SG

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)

Flying grenade SG

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

UltimaOnline SG

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'

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)

Don't need to add that line. If you want to add, then say "negative charge more dispersed", not "negative formal charge more dispersed", because formal charge is on an individual atom within an individual resonance contributor.

UltimaOnline SG

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

Whether there'll be a mixture of products, or only 1 product, depends on other conditions and factors (eg. solvent, catalyst, pH, temperature, pressure, etc) beyond the A level syllabus. And it's not always just about bond strength enthalpy change alone which determines the product(s), but other factors are also relevant, eg. ring strain, angle strain, steric strain, torsional strain, induction, resonance, reagent mechanism, entropy, etc.

All of which you needn't bother with, in the context of this A level question (ie. Cambridge doesn't expect or require you to think about all these factors). If (in the 2016 A level exam), you're not sure which product you'll get, be exam smart and give both products, with qualifications and explanations as to why each alternative product is possible or possibly more favored / stable.

Flying grenade SG

for the Sn1 mechanism and energy profile diagram, why is there energy given out from the first transition state (first hump) to the intermediate? (why the curve come down from the apex of first hump(when the bond just break right?) to the trough of the curve, intermediate ?)

from the intermediate to the second transition state, i understand have activation energy so there is energy required

Flying grenade SG

2014 p1 qn 29

the typical alkyl group only donates electrons by induction, which is a much weaker effect and insufficient for halogenation via electrophilic aromatic substitution.
understood

Interestingly for this particular question, the alkyl group actually withdraws electrons both by induction (due to higher % s character of the sp2 C atom of the alkene group),

ohhh ok i understood alr , sorry for posting, i got it after posting

as well as by resonance delocalization of electrons over to the electron-withdrawing COOH group as there is conjugation throughout the entire molecule from one end to the other.

understood

UltimaOnline SG

from the intermediate to the second transition state, i understand have activation energy so there is energy required

Because the intermediate is more stable than the transition state (which by definition are always the most unstable), obviously.

[email protected]@ddy SG

So 2 months out till A levels, any tips on an effecient study plan to stick to? Like should I just keep doing papers? is there value in doing topical? Cus I find myself constantly stuck in the B range ... and its damn difficult to hit that glorious A

UltimaOnline SG

Keep spamming Prelim papers. Focus more on your weakest topics, skip your strongest topics and easy questions. And track and increase your speed (often, students know how to do the questions, they just don't have the time to do them) until you're fast enough to finish Prelim papers (specifically top JCs) within the time allocated.

[email protected]@ddy SG

Oh and a chem question, how come adding HCL will protonate an alchol causing it to be more soluble in water? I thought alchohol is considered neutral?

UltimaOnline SG

Paper 1, up to you. No big deal either way. Yes, just keep spamming top tier JC Prelim papers.

Relative to more acidic HCl, alcohols are basic. Relative to more basic species (eg. NH2-), alcohols are acidic. As Einstein figured out, everything is relative, including morality.

[email protected]@ddy SG

Wow thanks that was deep, one last question how can an alkene exhibit geometrical isomerism with more than 2 different distinct R groups ??? ( E.g CH3, H, CH3CH2 ) like how do you tell ?

« Previous Page Next Page »