Lurkers, register with SgForums and ask ur H2 Chem qns here!
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Originally posted by Flying grenade:
Nitration of methyl benzene occurs more readily than Nitration of benzene , because ch3 group donates e- into the benzene ring, hence making it more nucleophilic ah?
Yes of course, but you didn't specify "electron donating" by which modality, induction versus resonance.In answering A level exam questions, you're advised to specify "electron donating by induction" or "electron donating by resonance" or "electron withdrawing by induction" or "electron withdrawing by resonance".
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Thank you
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Originally posted by UltimaOnline:
CS Toh's book is the basic H2 syllabus summary, so it may lack elaborative terms such as "steric hindrance" and "electronics". It's still a must-have book for all H2 students, but students intending to score A grade should also obtain further materials in addition (not in replacement) of CS Toh's book, see my BedokFunland JC's list of recommended books for H2 Chemistry :
http://infinity.usanethosting.com/Tuition/#Books_for_H2_ChemistryTertiary alkyl halides (such as in ur qn part d), undergo SN1, and primary alkyl halides undergo SN2, for 2 reasons : sterics and electronics.
Sterics and electronics are also the reasons why aldehydes are more electrophilic than ketones, so if you only have 1 mole of a nucleophile reacting with 1 mole of an electrophile with both aldehyde and ketone groups present, it is the aldehyde, not the ketone, which will be the electrophilic group accepting the nucleophile.
Electronics refers to the relative magnitudes and distribution of charges (both formal charges in resonance contributors and partial charges in resonance hybrids) as a consequence of either induction or resonance or hyperconjugation (the 1st 2 of which are more important for the H2 syllabus).
What about secondary alkyl halide?
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Originally posted by Flying grenade:
What about secondary alkyl halide?
Teo Chee Hean : "What do you think?"
Tis the Age of the Internet,
Google by thy Sword,
Wikipedia be thy Shield- BedokFunland JC
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Ultima are you proficient in literature ?
Bedok jc flair
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Originally posted by Flying grenade:
Ultima are you proficient in literature ?
Bedok jc flair
All great men are polymaths. To be precise, everyone can be a polymath (ie. great in the sense you maximize your potential) as long as you/he/she allows this. Being a polymath doesn't mean you know everything (that would be Wikipedia). Being a polymath means you allow yourself to explore, and give it your best efforts, in as many subjects, fields and domains as would interest you.Of course, modern society (in which the average poor man, not the rich elite, spend most of his time slaving at work to pay bills) doesn't naturally encourage self-exploration in multiple fields, so it's up to yourself to want to do so.
On a related note, if anyone reading this ever feels deeply conflicted by personal or religious issues, and would like advice or counseling from a spiritual (including supernatural*) but non-religious, non-dogmatic, non-judgmental, point of view or perspective, you may PM me.
*Actually everything is natural (including matters that most people would deem supernatural), it's only a matter of whether humanity and science on this planet has sufficiently evolved to be able to understand it or not.
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Hi ultima, may i enquire this,
Can bromoethene have p orbital overlapping between alkene c atoms and Br ,hence having partial double bond hence strengthening the R-X bond?
My tcher says so, he explained that there are lonepairs on Br, and there is pi electron cloud at the alkene c atoms , hence there is p orbital overlap
I was doubtful, becos there isn't any delocalisation by resonance , unlike chlorobenzene
Please advise , thank you !
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Originally posted by Flying grenade:
Can bromoethene have p orbital overlapping between alkene c atoms and Br ,hence having partial double bond hence strengthening the R-X bond?
Tcher says so
I was doubtful cos there isn't any delocalisation by resonance , like chlorobenzene
Please advise , thank you !
Writing (or typing) out in words is useless for questions like these. In the A level exam, you need to draw out the relevant resonance contributors and resonance hybrids for the vinyl halide to illustrate why the C-X bond in the resonance hybrid has partial double bond character, and thus vinyl halides (like aryl halides) are resistant to hydrolysis. -
Ah i see. Okay thank you so much ultima.
Page 212 of cs toh book, and also i have revised the drawing of bromoethene that I've learnt at BedokJc.
Thanks for your help again.
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Hi ultima, may i ask if i could enlist your help,
https://www.dropbox.com/s/jidtc1u05d4hbcp/20160331_194130.jpg?dl=0 , for this no link qn
Qn 1b)
Is this tutorial qn epic fail or is it a legitimate qn?
Im not sure how bronsted-lowry theory of acid and bases can apply to this qn, cant visualise any H+ donated or accepted
Thanks, Ultima
Update : is it H2O dissociates to form OH-and H+ , while NaH dissociates to form Na+ and H- , thereafter NaOH is formed from Na+ and OH- , while H2 is formed by H+ and H-
Since H- accepts the proton (H+) from NaH, is it that the proton donor i.e. bronsted-lowry acid is NaH?
Im also confused because i know amphiprotic water is both a weakacid and weak base,
So I'm deeply conflicted with the qn, not sure how do do this qn
Thanks for the help if possible, Ultima!
Update : typo error
I meant to type * Since H- accepts the proton (H+) from H2O, is it that the proton donor i.e. bronsted-lowry acid is H2O?
Instead of ''Since H- accepts the proton (H+) from NaH, is it that the proton donor i.e. bronsted-lowry acid is NaH?''
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Originally posted by Flying grenade:
Hi ultima, may i ask if i could enlist your help,
https://www.dropbox.com/s/jidtc1u05d4hbcp/20160331_194130.jpg?dl=0 , for this no link qn
Qn 1b)
Is this tutorial qn epic fail or is it a legitimate qn?
Im not sure how bronsted-lowry theory of acid and bases can apply to this qn, cant visualise any H+ donated or accepted
Thanks, Ultima
Update : is it H2O dissociates to form OH-and H+ , while NaH dissociates to form Na+ and H- , thereafter NaOH is formed from Na+ and OH- , while H2 is formed by H+ and H-
Since H- accepts the proton (H+) from NaH, is it that the proton donor i.e. bronsted-lowry acid is NaH?
Im also confused because i know amphiprotic water is both a weakacid and weak base,
So I'm deeply conflicted with the qn, not sure how do do this qn
Thanks for the help if possible, Ultima!
To properly understand this reaction, you *must* (so the sooner you accept this fact and work favorably with it, the better for your own sake) draw out the mechanism, which Cambridge can (and hopefully will, soon) ask H2 Chem students to do so, albeit as an A grade distinction challenging question to discriminate the elite Chem students from the rest of the cohort.Hydride ion, H-, from NaH (bear in mind that being ionic, it exists as Na+H- in the solid or solvated-in-inert-solvent state) is the Bronsted-Lowry base, which abstracts the proton H+, from the H2O molecule, which is the Bronsted-Lowry acid.
I expect all my BedokFunland JC students to be able to draw out the mechanism (usually only taught to H3 Chem and Olympiad Chem students, but if as a H2 Chem student, you can't draw out the mechanism, then you're not truly understanding this reaction, just blindly following 'rules' or 'patterns', such a tragedy for Singapore schools' H2 Chem students), otherwise (if you can't draw out simple mechanisms like these, then) you have no right to call yourself a BedokFunland JC student.
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T_T why for WeakAcid , at eqm, [HA] approx [HA]initial ?
Is it because HA only partially dissociates, then can just assume conc approx concentration initial?
Is it always the case?
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Originally posted by Flying grenade:
T_T why for WeakAcid , at eqm, [HA] approx [HA]initial ?
Is it because HA only partially dissociates, then can just assume conc approx concentration initial?
Is it always the case?
That's actually the case only for very weak acids or bases AND you start with a relatively large initial molarity of the weak acid or base, where hence the change in molarity is much smaller than initial molarity, and therefore the approximation is mathematically valid, otherwise you need to use graphing calculator to solve quadratic equation.For A level purposes, because Cambridge wants to leave the testing of mathematics to H2 Math instead of H2 Chem, and also because Cambridge is worried not all H2 Chem students are H2 Math students who can afford graphing calculators, therefore Cambridge has stated that they will always select acids or bases that are so weak, and/or start you off with a relatively large initial molarity of the weak acid or base, such that the approximation of equilibrium molarity back to initial molarity will always be mathematically valid, so no worries for A levels.
However, in some rare questions, eg. when the pH at equilibrium (eg. equivalence point) is given to you and you're tasked to solve for Ka or pKa or Kb or pKb, then because you need NOT solve quadratic equations, hence (by right) you have NO excuse to use approximation, then (by right) you will NOT be allowed to use approximation (again this is by right, but Cambridge tends to be lenient to H2 Chem students in this regard, and MAY accept approximation, but by right there's no valid reason to approximate, since there's no need to solve quadratic equation, so you're advised to give the correct answer without approximation).
Such questions have been asked in both Singapore JC Prelim paper questions, as well as by Cambridge many years ago, but not in recent years. Nonetheless, do be prepared.
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As always, thank you so much ultima for elaborating
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Ultima, may i ask for your help,
https://www.dropbox.com/s/ic7dycmt7rj35h4/20160331_234915.jpg?dl=0
Qn 7 a iii)
I can only think of common ion effect(CIE), CIE causes the dissociation of a salt to be reduced, if the solution already contains one of the ions. which i think might be wrong?Cos CIE only affects salt is it? Can it apply for Weak acid/bases ?
I thought of another
Or is it because the [CN-] increase with the addition of NACN , then the by le chatelier's principle, eqm shift back to the left, causing formation of [CN-] at the R.H.S of equation to be less ah?
Help ):
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Originally posted by Flying grenade:
Ultima, may i ask for your help,
https://www.dropbox.com/s/ic7dycmt7rj35h4/20160331_234915.jpg?dl=0
Qn 7 a iii)
I can only think of common ion effect(CIE), CIE causes the dissociation of a salt to be reduced, if the solution already contains one of the ions. which i think might be wrong?Cos CIE only affects salt is it? Can it apply for Weak acid/bases ?
I thought of another
Or is it because the [CN-] increase with the addition of NACN , then the by le chatelier's principle, eqm shift back to the left, causing formation of [CN-] at the R.H.S of equation to be less ah?
Help ):
A couple of things. Don't use acronyms in the A level exams. CIE more properly stands for "Cambridge International Exams" not "Common Ion Effect". Next, it is actually wrong to say "by Le Chatelier's principle" as commonly taught in Singapore JCs, the correct phrasing is "as predicted by Le Chatelier's principle", since Le Chatelier isn't some supernatural being who can magically shift the position of equilibrium in chemical reactions just by invoking his name.Also, and again despite what your school teacher may have said, Cambridge will never require, nor award marks, for the phrase "Le Chatelier's principle" in the A level exams. Otherwise, students who have no idea how to solve the question, other than knowing the obvious that it's an equilibria question, will just write the words "by Le Chatelier's principle" and hope to smoke out 1 mark out of nothing.
Common ion effects applies to all ionic equilibria, just as common species (including ions) effect applies to all equilibria reactions, as predicted by Le Chatelier's principle.
Correct Answer : When NaCN(s) is added to the solution, [CN-] increases, causing the position of equilibrium in the proton dissociation equation HCN(aq) (insert double half-arrow) H+(aq) + CN-(aq) to shift to the LHS. Concordantly, the extent or degree or percentage of HCN proton dissociation will decrease, and the pH of solution will become less acidic, compared to if no NaCN(s) was added.
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Thank you imba god help students
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Thank you for all the details, accurate phrasing and all
Thank u ultima
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Heard from my cher that the syllabus is changing the room temperature, no longer 25degrees C, changing to 27 or 28
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Originally posted by Flying grenade:
Heard from my cher that the syllabus is changing the room temperature, no longer 25degrees C, changing to 27 or 28
Yes, due to global warming. And the fact that this is April Fool's day. -
A BedokFunland JC H2 Chemistry Qn
Which molecule (phenylamine or pyridine) is a stronger Bronsted-Lowry base? Explain your reasoning.
Phenylamine is the stronger Bronsted-lowry base
the availiability of lone pair on phenylamine is greater than pyridine.
because N atom of phenylamine is sp3 hybridised, N atom of pyridine is sp2 hybridised.Steric hindrance by benzene bulky group, increasing bronsted-lowry basicity of phenylamine.
the magnitude of partial positive charge on N atom of phenylamine is larger due to benzene ring withdraws e- (from N) by resonance ? (idk)
the lone pair on pyridine is delocalised within the aromatic ring , decreasing it's basicity (and nucleophilicity?)
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Hi Ultima , may i ask you,
Steric hindrance by bulky groups (either on electrophile reducing electrophilicity, or on nucleophile reducing nucleophilicity and increasing Bronsted-Lowry basicity instead) around the electrophilic or nucleophilic atom.
i thought nucleophilicity and basicity are related ?
why steric hindance can reduce nucleophilicity and increase basicity?
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Originally posted by Flying grenade:
Hi Ultima , may i ask you,
Steric hindrance by bulky groups (either on electrophile reducing electrophilicity, or on nucleophile reducing nucleophilicity and increasing Bronsted-Lowry basicity instead) around the electrophilic or nucleophilic atom.
i thought nucleophilicity and basicity are related ?
why steric hindance can reduce nucleophilicity and increase basicity?
Teo Chee Hean : "What do you think?"I don't want to give away the answer to this particular question online. My BedokFunland JC students can ask me during tuition. Other students can go ask their school teacher or private tutor.
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Originally posted by Flying grenade:
A BedokFunland JC H2 Chemistry Qn
Which molecule (phenylamine or pyridine) is a stronger Bronsted-Lowry base? Explain your reasoning.
Phenylamine is the stronger Bronsted-lowry base
the availiability of lone pair on phenylamine is greater than pyridine.
because N atom of phenylamine is sp3 hybridised, N atom of pyridine is sp2 hybridised.Steric hindrance by benzene bulky group, increasing bronsted-lowry basicity of phenylamine.
the magnitude of partial positive charge on N atom of phenylamine is larger due to benzene ring withdraws e- (from N) by resonance ? (idk)
the lone pair on pyridine is delocalised within the aromatic ring , decreasing it's basicity (and nucleophilicity?)
Wrong, wrong and wrong.If this was a Singapore-Cambridge A level H2 Chemistry question, it would be an A grade distinction question, so it's expected or even intended that not all H2 Chem students will be able to correctly answer it.
Hints : there are 4 required marking points, each one leading to the next.
PS. Try asking your school teacher and see what he/she says! ;Þ
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Hi Ultima, may i ask you,
in continuation to this https://www.dropbox.com/s/obq34agw4who7ay/forum%20ss.PNG?dl=0 (pg 19 of this forum thread) ,would ring strain affect geometric isomerism?
i've googled, and found these images, with 3 C atoms [http://images.tutorvista.com/cms/images/44/example-of--cis-trans-isomerism.png] ,
5 C atoms [http://wps.prenhall.com/wps/media/objects/340/348272/Instructor_Resources/Chapter_05/Text_Images/FG05_20-17UN.JPG] ,
and 6 C atoms [http://mcat-review.org/cis-trans-ring.gif]
, that can exhibit geometric isomerism, hence unsure how *big* ring is needed to exhibit geometric isomerism? ( in regard to your advice *This ring is too small to have geometric isomerism about the alkene group.* )