BedokFunland JC's A Level H2 Chemistry Qns (Part 2)
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Originally posted by Flying grenade:
Hi Ultima, hope u could examine this
qn : 2004 A level chemistry
https://www.dropbox.com/s/e5rkrmko3uhrac0/20160720_163024-1.jpg?dl=0 , part 2cii
my answer and teacher's answer
https://www.dropbox.com/s/0jqju117dpcdesm/20160720_163753-1.jpg?dl=0
my doubt : E°cell = 0.00V , can reaction still proceed/ thermodynamically favourable?
qn say 'two reagents mixed together'
can VO2+ form?
is it VO2+ can form because its aqueous medium, where H2O and H+ are present uh?
In this case, you're wrong, but your teacher isn't fully correct either.Vanadium's maximum OS is +5, not +3.
Based on standard oxidation potentials of V 3+, compared to the standard reduction potential of Cu+2 to Cu, you need only consider oxidation of V 3+ to VO 2+ (ie. OS of V = +4), and to nothing further (eg. to VO3 - or VO2 +) under standard conditions. But of course, the exam question can be more sadistic and replace V 3+ with V, and Cu2+ with a stronger oxidizing agent, then you need to consider each and of the possible oxidation products for vanadium all the way to +5, including memorizing (if asked by the question) all the color changes.
Minor point : Your teacher used the 'minus' version of the cell potential formula, but the 'plus' version is actually more meaningful, chemistry-wise (ie. for students & teachers who understand their chemistry more deeply).
Cell potential = Reduction potential @ Cathode + Oxidation potential @ Anode = (+0.34 V) + (-0.34V) = 0.00 V
If the calculated (standard or otherwise) cell potential is 0 V, it means the system with standard molarities on both sides of the equation would be at equilibrium (ie. Qc = Kc; it just so happens that in rare cases like this case, also that Kc = 1).
If the system is already at equilibrium, no net visible change occurs (since rate of forward reaction = rate of backward reaction, but be careful : k(forward) = k(backward) only if Kc = 1, a rare case). But note that at initial, you only mixed standard molarities (ie. 1M) of V 3+ and Cu 2+ together, which means Qc is not yet = Kc. Hence a net visible change will indeed occur, until Qc = Kc (which happens to be Kc = 1 in this reaction). But (your teacher didn't mention this) the reaction will be significantly slower (exactly how much slower, depends on kinetics and activation energy for the forward reaction), compared to if the calculated (standard or otherwise) cell potential was a lot more positive. This is because of the intimate relationship between electrochemistry and equilibria and kinetics.
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Originally posted by ArJoe:
Hi, for 2015 tjc paper 3 qn 3i, on why aspirin is insoluble in water, my answer differs fromthe answer scheme.
I said that the formation of predominantly instantaneous dipole induced dipole interactions due to large non polar benzene ring between aspirin and h2o molecules releases insufficient energy to overcome hydrogen bonds between h2o and idid interactions between aspirin. Dissolution is energetically unfavourable.
is my answer acceptable?
No, your answer is not acceptable.1st error : between polar and non-polar molecules, it's not instantaneous dipole - induced dipole London dispersion van der Waals interactions, it's permanent dipole - induced dipole Debye van der Waals interactions.
2nd error : you didn't mention that the COOH and OCO groups of aspirin can and will participate in limited hydrogen bonding with the water solvent, but the solubility of aspirin as a whole is still limited by the weaker permanent dipole - induced dipole Debye van der Waals interactions of the non-polar hydrophobic benzene ring with water solvent.
Lastly, and this is partly the question's fault : if you say "dissolution is energetically unfavourable", it implies aspirin is completely insoluble (eg. like SiO2 or diamond, etc), but aspirin is actually slightly soluble and does dissolve slightly in water under standard conditions. Which is why it's better to state and explain why "the solubility of aspirin is hence limited" in water.
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Originally posted by gohby:
Hi UltimaOnline, Sorry to trouble you again, but here are some questions which I have
NJC/2012/P1/Q22
Why can’t B be an answer, given that acyl chlorides can be made by reacting thionyl chlorides with carboxylic acid?
NJC/2012/P1/Q25
I understand that C is the answer (http://www.chemguide.co.uk/organicprops/haloalkanes/making.html), but why can’t the other options be prepared with distillation?
NJC/2012/P1/Q27
I think there are 3 ways which the elimination can take place given that there are 3 B hydrogens. Applying Saytzeff’s Rule, I can only eliminate the option where the hydrogen is abstracted from the methyl substituent. The other options (where the hydrogen is abstracted from the CH2, and where the hydrogen is abstracted from the cyclohexane substituent) seem equally substituted to me. So what would be the major product of the reaction? Further, wouldn’t either of the products contain 2 double bonds, thereby making the total number of stereoisomers 4?
Once again, thank you for your help. :)
Hi Gohby, no prob :)NJC/2012/P1/Q22 : B is incorrect because the amine group present is even more nucleophilic than the OH of COOH, and hence would react even more readily with the thionyl halide electrophile, in this case the primary amine would undergo 2 consecutive nucleophilic substitutions with the thionyl bromide to generate the N=S=O group, eliminating 2 molecules of H-Br. This is beyond the A level H2 syllabus, and so isn't completely fair, but nonetheless Cambridge *could* ask this question as a challenging A grade question, to determine if the candidate understands the concepts of relative nucleophilicities and electrophilicities, and choose the best possible option among the 4 presented.
Cambridge has done indeed so many times in the past, especially in recent years, deliberately presenting 2 likely options : 1 option going beyond the A level H2 syllabus so candidates cannot be sure if it's correct, the other option within the A level H2 syllabus so competent candidates should be sure if it's correct or wrong, and therefore choose accordingly.
NJC/2012/P1/Q25 : Options A & D can be carried out at room temperature, and hence doesn't require this setup (and therefore should not be used). B requires heating under reflux. The setup shown is heating with distillation to separate out the desired product which has a lower boiling point compared to the reactants.
NJC/2012/P1/Q27 : This reaction involves competition between the Zaitsev aka Saytzeff elimination, and the Hofmann elimination. While the Bronsted-Lowry base employed is the sterically unhindered OH- ion, nonetheless the cyclohexane ring does pose some degree of steric hindrance, but in this case the Zaitsev product should ultimately still dominate (and conveniently also because the Hofmann elimination isn't taught in the H2 syllabus).
The major product which is the Zaitsev product, is therefore the product that has the double bond with the cyclohexane ring, as this product would be a more substituted internal alkene (with no H atoms bonded to its sp2 C atoms), compared to the alternative product that has the double bond with the alkyl side chain (with 1 H atom bonded to its sp2 C atoms).
The number of stereoisomers of the major product is therefore 2 (ie. as a result of the pre-existing alkene double bond in the longer side chain), instead of 4 (cis-cis, cis-trans, trans-cis, trans-trans).
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Originally posted by Flying grenade:
is it
CxHyOH + [x + (y+1)/4 -1] O2 -> xCO2 + (y+1)/2 H2O ?
Almost correct, but your coefficient of O2 is slightly off. Here's the correct working to get the correct coefficient of O2.Qn : Write the balanced equation for the complete combustion of CxHyOH.
2k + 1 = 2x + (y+1)/2
2k = 2x + (y+1)/2 - 1
k = (x + (y+1)/4 - 1/2) -
Originally posted by senga:
Hi. New poster. Here couldn't create a new thread so I have to ask here. So I've decided to retake my A levels this year and I am only starting to study for it now due to certain circumstances. I know it is quite late to study for it now but I really want to give it a shot. Not expecting stellar results, just want to get grades good enough to enter an engineering course in NTU or NUS. Preferably NUS due to personal reasons but I'm fine with NTU as well. (Government calling for more students to pursue engineering isn't very comforting for me because I'm scared that demand for the courses will go up and the competition would be stiff for the limited slots.)
Currently I am starting on chemistry first because I have some notes on it. I don't remember much so I thought starting with topical would be better. Doing a topical TYS which I have from my jc days. Feel that the answers are abit lackluster. Not sure whether I should continue using it to study up each of the topics. Can't provide a picture now. I know that ultimaonline recommend CS Toh books. Wonder if it is suitable for someone like me who is starting from close to zero.
I'm taking PCME as my subject combination and I will be taking paper 5 for the sciences.
As for other subjects, I'm not sure how to study for it. I'm just using a topical TYS for physics which I think has detailed explanation. Don't know how to start for math, econ and gp. I need some advice as well as recommendation for which books I should get to study for A levels. It's hard to explain my situation in detail. Hope that someone could give me some guidance. Thanks.
Yup, just use CS Toh's Study Guide and memorize the entire book.And buy at least 2 different H2 Chem TYS available from Popular, coz different TYS give different answers (since Cambridge has a legal agreement with Sg Govt to keep the actual Sg A level mark schemes a secret), so you should cross reference across both answer booklets. If for questions that give totally different answers (and there will be many), and if you're not sure which answer is better or more correct, post about it here, and I'll advise you.
Don't bother doing Prelim papers, coz at this late stage when you've not yet started preparing for H2 Chem, doing Prelim papers will only demoralize you and be counterproductive. Better to focus on doing only Sg TYS papers.
Originally posted by senga:So I should start with a topical TYS and get a yearly TYS for cross referencing of answer, right? My concern for the answer sheet is about the answering method, to ensure i get the marks. The current Topical TYS I'm using is from 2003-2012. No problem with that right? How about the CS Toh practice book?
No, don't use Topical. Go buy 2 or 3 different TYS, all yearly. Go to a big Popular bookstore, eg. Bras Basah, which carry a wider variety of stock. If you get topical, harder to cross reference across all 2 or 3 different TYS. And at this late stage, no point practicing Topical. You should practice yearly format across all topics.In addition, if you have Android handphone with internet data access and/or can read Google Play ebooks on your phone or PC, and you can use your parents' or your own credit or debit cards, I strongly recommend you also buy CS Toh's TYS solutions (only soft copy available, but not pdf formal, you need to login to access the Google server each time you want to read CS Toh's ebook), to cross reference against the other 2 or 3 hardcopy TYS you're gonna buy from Popular.
Trust me and use this approach : I myself have bought and use 4 different TYS solutions every year, and have noted that answers vary widely across the different TYSes (just as they would across different Singapore JCs for Prelim papers and TYS papers, since every H2 Chem teacher in every JC has a different opinion of what the best answer should be). Your best bet is to cross reference the different answers across the TYSes, pick 2 of the best answers, and combine their points to write the best possible answer that would secure your marks in the 2016 A levels.
Lastly, be sure to post here on this forum, whenever you come across any such different answers and you're not sure which is the best answer. There are hundreds of JC students who lurk on the forums here, and they'll also benefit from your questions here. I'll advise all of you on which are the best answers to write for your A levels.
Purchase CS Toh's H2 Chem TYS ebooks on Google Play here :
http://www.post-1.com/step-by-step/?a=book_al_soln_chem
Originally posted by senga:So I should read through the study guide once then start doing the yearly TYS for my study approach. For paper 5, just need to diligently practise the planning questions from paper 2 right? thanks for the advice. Will post my queries here.
No, don't read the Study Guide, then do the yearly TYS. Rather, use the yearly TYS directly now to start learning & revising as you go from question to question, based on the different solutions from different publishers. And for each question, also cross reference with CS Toh's A level Study Guide (ie. essential H2 Chem syllabus lecture notes you need to memorize). Learn and Revise as you go through each question in the TYS, that's the only feasible approach now at this late stage, only 3 months away from A levels.Yes, no worries about Planning and Paper 5. For Paper 5, just follow the given instructions carefully. For Planning Qn (Paper 2), skip it first. Complete the rest of the Paper, then go back to it after you're done with the rest of the Paper, and write whatever you can.
If you *really* want to prepare for Planning Qn (ie. if you need A grade for H2 Chem because of your desired Uni course), you should buy and study Chan Kim Seng & Jeanne Tan's H2 Chem Planning Book, available from Popular Bookstore (the larger outlets, not the smaller outlets) :
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This challenging Cambridge exam question would ruin most Singapore H2 Chem students if it came out for Singapore A levels.
To solve the 1st part of the Cambridge question, use the formula
ΔS surroundings = - ΔH / T
(Cambridge may, in a challenging Singapore A level H2 Chem exam question, require you to qualitatively explain this formula, ie.
1. an exothermic reaction (ie. negative ΔH) generates heat which results in an increase in the disorderliness and hence entropy of the surroundings
2. the more negative the value of ΔH, ie. the more exothermic the reaction, the more positive the entropy increase of the surroundings
3. the same amount of energy released into the surroundings will make a bigger difference at a lower ambient temperature, hence the division by T)To solve the 2nd part of the Cambridge question, use
ΔS total = ΔS system + ΔS surroundings
To solve the 3rd part of the Cambridge question, note that :
ΔS total = ΔS system + ΔS surroundings, hence
ΔS total = ΔS system + (- ΔH / T ), hence
Multiplying through by -T, we get :
- T x ΔS total = ΔH - T x ΔS system
ie. Δ Gibbs free energy = ΔH - T x ΔS system
ie. Hence, when ΔS total = 0 (as specified by the exam question), which means Δ Gibbs free energy = 0, then
0 = ΔH - T x ΔS system, hence
ΔH = T x ΔS system, hence
T = ΔH / ΔS system, which will enable you to solve the 3rd part of the Cambridge question.
To solve the 4th part of the Cambridge question, you need to understand that the reaction will only proceed if the total entropy change of the universe is zero or positive, ie. Gibbs free energy change = zero or negative. Hence to ensure the reaction proceeds, ie. for the reaction to be thermodynamically feasible, you need the temperature to be above the value calculated when you used (as instructed by the question) total entropy change = zero.
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Originally posted by Flying grenade:
why stability I- >Cl- > Br- > F- ?
I- less electronegative, and anionic charge density is smaller?
The 2 factors are opposing, you have to decide (based on experimental evidence), which factor outweighs which factor. Indiscriminately writing in both factors in the A level exam in a shotgun approach, will be shooting yourself in the foot, and you'll get zero marks (because Cambridge will be thinking, "KNN! This bugger trying to cheat by giving both opposing factors, expecting us Cambridge examiners to choose the correct answer for him! PCB!" -
Originally posted by Nikkilyx:
Hi, first timer here! i am facing some problems with chem equilibrium/thermochemistry... the front page of my lecture notes states that we are to 'show understanding that the position of equilibrium is dependent on the standard Gibbs free energy change of reaction' but nowhere in the lecture notes mention anything about it... can you please explain how they are related?
Welcome, and feel free to post your H2 Chem qns here. Nonetheless, do note that :Tis the Age of the Internet
Google be thy Sword
Wikipedia be thy Shield
- BedokFunland JCFor general Chem conceptual qns like these, there are many websites on the internet that you're encouraged to check out on your own first (you know what they say about Singapore education system spoonfeeding students with notes, until most Sg students don't know or aren't used to taking the initiative to research out and learn for themselves in self-directed lifelong learning).
The best type of questions to ask here, is when you have doubts or disagreements with TYS or Prelim answers, and seek clarification or a 2nd opinion. That's the best way I can help you guys here.
The relevant formula is ΔG = - R T ln Kc
Mathematically, JC students are expected to know that ln 1 = 0. So if position of equilibrium lies on the LHS, Kc < 1, then ΔG > 0, hence the forward reaction is not thermodynamically feasible (under standard conditions and molarities), the backward reaction is.
Conversely, if position of equilibrium lies on the RHS, Kc > 1, then ΔG < 0, hence the forward reaction is thermodynamically feasible (under standard conditions and molarities), the backward reaction isn't.
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Originally posted by Nikkilyx:
thanks for your reply. sorry to bother you so late. i saw this formula on my brother's old uni notes but not on my lecture notes... My lecture notes also mention that 'a quantitative treatment is not required', which i assume we have to explain without numbers/ formulas... and this is where i am having problems with... this is my explanation: when temperature increased for an exothermic reaction, TdeltaS also increases. Hence deltaG becomes more negative since deltaG = deltaH - TdeltaS and deltaH is negative. Hence the reaction occurs more readily and more products are formed. Since K = [Products]/[Reactants], and [Products] increases while [Reactants] decreases, K increases correspondingly, thus the equilibrium shifts to the right... do you think my explanation is correct?
Your explanation is incomplete, because you failed to specify if entropy change was positive or negative. Otherwise it's ok.Just because 'a quantitative treatment is not required', does NOT mean you "*have* to explain without numbers/ formulas", and it also does NOT mean that Cambridge won't accept your explanation using mathematical formulae, as long as the formulae is valid, correct and relevant, even if the formulae isn't taught within the syllabus.
Lastly, to 'show understanding that the position of equilibrium is dependent on the standard Gibbs free energy change of reaction' as specified in the syllabus, all you need to do is simply state in words, that when the position of equilibrium lies on the LHS, Gibbs free energy change (ΔG) is positive and cell potential (E°cell) is negative ; conversely and concordantly when position of equilibrium lies on the RHS, Gibbs free energy change (ΔG) is negative and cell potential (E°cell) is positive; no further explanation in either words or formulae is required.
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Originally posted by Flying grenade:
what flows through the salt bridge? is it electrons and spectator ions ?
Counter spectator ions only. -
Originally posted by iSean:
CIE AS Level M/J 2016: 9701/22/M/J/16
There is no Marking Scheme for this Answer until Results Day on 11 August.
How do you do (i) and (iii)?
IE of … Y log 10 Difference by log Difference by I.E
fith 6542 3.82 +0.15 +2820
sixth 9362 3.97 +0.07 + 1656
seveth 11018 4.04 +0.49 + +22588
eighth 33606 4.53
I think my answer will be either Na/Mg/Cl
(iii) Big jump between sixth and seventh I.E.
So is it Sulfur?
(i) Group 17 or VII. The magnitude of increase from 7th to 8th ionization energy is significantly greater than preceding increases, because the 8th electron removed (ie. 8th ionization energy used) is from an inner electron shell closer to the nucleus, which is thus more tightly held electrostatically by the positively charged nucleus, as predicted by Coulomb's law, and also because of less shielding effect.
(iii) Correct, it is sulfur : 1s2 2s2 2p6 3s2 3p4
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Originally posted by Flying grenade:
cs toh writes Amides are slowly hydrolysed by heating under reflux with Aqueous alkali or acid
my cher say reagents and conditions is 6moldm-3 , prolonged heating
so in Cambridge exam write conc acid/alkali or aq?
Aha! You've shown you've a serious misconception right there! One that almost all Singapore JC students (and maybe even some JC teachers) have!Can you figure out what your misconception is before I reveal it?
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Originally posted by Flying grenade:
when asked to 'give equation for the reaction'
e.g. pre u paper 2014 p3 qn 5b ii
can we draw the molecules, instead of giving
C6H5OH + 3Br2 → C6H2Br3OH + 3HBr? will it be marked correct?
Depends on context. Even in contexts where drawing out the structures and/or mechanisms is acceptable by Cambridge as an alternative to the written equation, you'll need to be careful in the stoichiometry and balancing the reaction correctly. If unsure of acceptability of context, always play safe and give both (ie. draw out structures and/or mechanism if it helps you, *and* write out the balanced equation), as long as you know your chemistry well and have good time management.In the context of your particular question, if all you mean is drawing out the benzene ring, yes it's acceptable to do so (instead of just writing the formula), particularly since in this case phenol is tribrominated (on all 3 of the ortho and para positions) and hence in this case there are no positional isomer ambiguity problems (eg. if it was only monobrominated or dibrominated, then you would have a mixture of different positional isomers, in which case it would be more correct to write the condensed structural formula C6H4BrOH instead of drawing it out).
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Originally posted by Flying grenade:
2015 TPJC p1 qn 19. dont understand the soln and dk how get the answer.
a thought of mine is this , why can't [Pt(NH3)6]4+.4Cl- then?
Your formula has 4 Cl- counter ions, while the question specified 2 Cl- counter ions (you gotta know how to interpret the question correctly, even if they don't phrase it as well as I would if I were to set the question).
Originally posted by Flying grenade:just realised cstoh book dont have zaitsev's rule
does zaitsev's rule apply for E1 and E2 ?
Yes, but depending on the sterics of both the Bronsted-Lowry base and acid, either the Zaitsev product or the Hofmann product would dominate as the major product. -
Originally posted by Nikkilyx:
On the same topic about ionisation energy, i have this question that i need some clarifications.
Element Z is the successive element from potassium across the period. Compare and contrast the first ionisation energy of potassium and element Z (calcium). Support your answers with reasons.
At first, my answer was:
K - 1s2 2s2 2p6 3s2 3p6 4s1; Z (Calcium) - 1s2 2s2 2p6 3s2 3p6 4s2
Z < K as valence electron to be removed from Z is paired while that from K is unpaired, thus less energy is required due to inter-electronic repulsion.
However, after checking the Data Booklet, I realised that the first ionisation energy of calcium is higher than potassium, so I reconstructed my answer. My question is why, unlike sulfur and phosphorus, there is no inter-electronic repulsion that causes the first ionisation energy of calcium to be lower?
Chemistry is a microcosm of Real Life. When there are opposing factors, in some cases the 1st factor wins, in some cases the 2nd factor wins.Considering K vs Ca, the factor of increased nuclear charge outweighs the factor of inter-electron repulsion, consequently ionization enthalpy for Ca is more endothermic than for K. Considering P vs S, the factor of inter-electron repulsion outweighs the factor of increased nuclear charge, consequently the ionization enthalpy for S is less endothermic than for P.
In Chemistry, including for A levels, you can only show your understanding of both (or all) opposing factors, then carry out experiments to determine which factors outweigh which. To discuss this deeper at higher levels (beyond A levels), yes it's possible to show with more involved calculations why this factor outweighs that factor at this point in the periodic table, but these would not be relevant or useful for A level purposes, and would only confuse the A level student and is therefore counterproductive.
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Originally posted by Metanoia:
Hi all,
seeking some clarification on what consitutites a repeating unit during condensation polymerisation (O-Level Pure Chem)
The monomer below is able to polymerise on its own
HO-X-COOH
The question requires the student to represent the polymer using "two repeating units"
Will the correct answer be
1) -O-X-COO-X-CO-
Or
2) -O-X-COO-X-COO-X-COO-X-CO-
I thought it would be the first answer, but my tutee tells me that her school expects the 2nd option.
Opinions and inputs welcomed.
Your student's school teacher is wrong here. Because the monomer (undergoing condensation polymerization) here is a hydroxycarboxylic acid, so for 2 repeat units, you just require 2 R groups (usually illustrated by a shaded-box at O levels) in the section of the polymer. Exactly as you've drawn.The reason for your student's school teacher's error, is because in some other instances of condensation polymerization, 2 different monomers are required, eg. a diamine and a dicarboxylic acid in Nylon 66. In such instances, 2 repeat units must include 4 R goups and 0.5+3+0.5 amide linkages, ie. -- diamine -- dicarboxylic acid -- diamine -- dicarboxylic acid --.
Thus, perhaps by misguided instruction from the HOD, many secondary school teachers just apply this requirement blindly across the board, including for condensation polymers that only involve a single monomer, eg. hydroxycarboxylic acid.
Long story short : you're correct, Metanoia. Your student's school teacher is wrong.
Originally posted by Metanoia:Thanks for making it much clearer.
I agree that there is a chance that the teacher was using the 'blanket rule' which caters to two different condensation mononers.
It was actually from a past year prelim question, and the teacher insisted that answer key was correct to draw option 2.
I've asked the student to clarfy again with the teacher.
Sometimes they can get quite distraught with conflicting expectations.
Oh no worries (about your student being distraught over conflicting answers). just tell your students what I tell my students, "For your school exams, write your answers the way your school teachers taught you. For your O/A level exams, write your answers the way I taught you. Because I personally get A grade every year, so that's Cambridge saying you should trust me over your school teachers." -
Originally posted by ArJoe:
Hi, qn1: while hcl reacts readily with alkenes under normal conditions, hcn does not. Explain using chemical bonding?
i am not very sure but i think it has something got to do with extent of orbital overlap. C-hdond have greater extent of orbital overlap, stronger bond, require more energy to break, in order for the alkene to be protonated??Qn2: explain why cr(nh3)6 3+ is coloured.
as i explain, should i state in cr3+, the 5 d orbitals have the same energy level….
or should i state in cr(nh3)6 3+, the 5 d orbitals….. which phrasing is correct?qn 3: why doea cr3+ on reaction with na2co3 gives co2?
is it because cr(h2o)63+ have high charge density, high polarising power, undergoes cation hydrolysis by polarising oh bond in h2o, causing h atom to ionise to form h+. Since solution is acidic it undergoes acid base neutralisation with na2co3 to form co2Are my answers correct? Thanks!
Q1. The C-H sigma bond is strong due to sp-s overlap, but more importantly, HCN is a weak acid because the conjugate base is significantly unstable due to the relatively low electronegativity of C atom that bears the negative formal charge (it is only the sp hybridization that just barely allows the C atom to bear the negative formal charge). The weakness of HCN as an acid results in high Ea for the alkene to act as a Bronsted-Lowry base to abstract the proton from HCN, hence the relative inertness.Q2. You can go straight into "In the presence of the 6 NH3 ligands, the d orbitals become non-degenerate... etc etc".
Q3. Yes, that's correct. But in addition to a textual explanation, you also need to write out the relevant balanced equations to obtain full marks.
Originally posted by Flying grenade:C-H sigma bond of alkene is formed by sp2-s overlap
compared to what/which bond? bond between HCl? bond between HCN (sp-s) overlap?
does hybridisation apply for HCl and other molecules, or only for Carbon/nitrogen bonded molecules?
what does breaking the C-H bond got to do with the electrophilic addition of HCl or HCN to alkene? shouldn't it be breaking the pi bond between C atoms in the alkene?
damn stress now
thanks Ultima for explaining
Compared to most other C-H bonds which are 2sp2-1s or 2sp3-1s overlap. H-Cl bond is even weaker because it is 3sp3-1s overlap.Orbital hybridization applies for all molecules, eg. 3sp3d2 hybridization for octahedral centers (eg. PCl6- ion).
If you draw out the mechanism (for the reaction that does not readily occur due to high Ea required), you'll see that you're breaking the 2p-2p pi bond of the alkene (ie. the weak Bronsted-Lowry base), as well as the 2sp-1s sigma bond of the H-CN (ie. the weak Bronsted-Lowry acid).
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Originally posted by iSean:
CIE W14 Q5.
Just wondering for this reaction besides Na, how does vanillin react for Br?
Why it react on C-5 and not C-3?As vanillin is named clockwised and not anticlockwised.
Also you mind explaining the other reactions that I'd highlighted? Especially RCOCl and Fe3+ thanks :)
Halogenation via electrophilic aromatic substitution.As long as you correctly substitute the incoming electrophile onto an available ortho or para position relative to the phenolic group, Cambridge will accept.
Diazotization and diazonium coupling is not in the Singapore H2 syllabus. If you're interested, go read up on it yourself.
Acyl halides are electrophiles attacked by the most nucleophilic group, ie. amine on L-Dopa, and phenol in Vanillin.
Of course, depending on the solvent, the alpha-amino-carboxylic acid of L-Dopa may exist as zwitterionic form, in which case the NH3+ is no longer nucleophilic and cannot attack the acyl halide electrophile. Note that acyl halides must be used in anhydrous form, which implies the other reagent (ie. L-Dopa) should not be used in aqueous solvent.
Fe3+ (in neutral solution, can't be too acidic or too alkaline, interested students can ponder over why, and check your answer with your school teacher or private tutor) coordinates & complexes with phenolic ligands to generate a violet coloured coordination complex.
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Originally posted by iSean:
Also we didn't learn the terms Para (1,4-) and Ortho (1,2-) but since you said it was phenolic, so it should be Electron Donating Group? How do we determine it as the phenolic group as the first place as Vanillin has an Acetyl group and Aldehyde group bonded to the Benzene ring with the hydroxyl group.
Is not like students can determine when 2 different natures of groups are bonded to the same ring and the Data Booklet doesn't even help a bit.... When considering bonding at C2, C4 or C3. And we have to determine where will be our C1 at the first place...
(I'd just finished whole Organic Chemistry in 2 week time... So I'm still relative new to this)
Whether the student can determine the ortho-para-meta directing nature of each substituent, is up to the student to learn about this as you/he/she wills. Different JCs, different teachers, different private tutors, teach to differing extents. Depending on your own willingness and choice, you can read up about this on your own, even if your school teacher chooses not to teach it.The phenolic OH group is electron-donating by resonance moreso than it is electron-withdrawing by induction, hence each phenolic OH group is activating and ortho-para directing. The alkyl group (sp3 C atoms) in L-Dopa is electron-donating by induction (relative to the sp2 C atoms of the benzene ring), and is hence a weak activator and an ortho-para director. The benzaldehyde group in Vanillin is electron-withdrawing by both induction (due to a significant partial positive charge on the carbonyl sp2 C atom as a result of both induction and resonance) and resonance (due to the sideways overlap of the unhybridized p orbitals of sp2 C and O atoms of the carbonyl C=O group), and is hence deactivating and meta directing.
Hence, the strongest activator (for both L-Dopa and Vanillin) being the phenolic OH group, wins and gets to direct the incoming electrophile. Because L-Dopa contains 2 phenolic OH groups, Cambridge will accept any of the 3 possible answers, although the Mark Scheme answer is the major product due to least steric hindrance.
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Originally posted by MightyBiscuits:
Hi UltimaOnline, just a little concept clarification. Is there a buffer region for weak acid and weak base titration? Thanks
Technically, yes. Buffer regions still occur when you've close to equal molarities of HA and A-, as well as close to equal molarities of B and BH+ (note that the actual maximum buffer capacities of the solution, however, is no longer based simplistically on having equal molarities of both members of each conjugate acid-base pair, since the other conjugate acid-base pair also exerts its influence on pH and buffering capacity; exact calculations here are significantly beyond the scope of A levels).In addition, there is an additional (albeit weaker) buffer region (unique to weak acid - weak base titrations) that spans from the 1st buffer region (ie. near half-equivalence point) all the way to the 2nd buffer region (ie. near double-equivalence point), including the titration equivalence point, since both cation and anion of the salt at equivalence point, are themselves a weak acid and a weak base.
If Cambridge were to ask Singapore students to calculate the pH at equivalence point of a weak acid - weak base titration, 99.99% of Singapore H2 Chem students (and even many JC teachers and private tutors) would die. But if you're interested to challenge yourself, try solving such a question. Hint : you'll need to solve simultaneous equations which takes into account the simultaneous hydrolysis of both cation and anion of the salt.
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Originally posted by Flying grenade:
okay i see. for cyclo alkanes, only substitutents exhibit cis trans only. for cycloalkenes, the whole molecule cis or trans about the double bond.
Of course, it's possible for cycloalkenes to have geometric isomerism for both the double bond as well as substituents. In which case (for A levels), u just need so specify whether the cis-trans isomerism refers to the substituents or the double bond, eg. "This molecule is cis in regard to the double bond, but trans in regard to the 2 Cl atoms on the ring". Cambridge won't require you to name such complicated molecules with more than 1 set of cis-trans isomerism present simultaneously. -
Originally posted by LavXuan:
(Planning qns) In starch-iodine titration, why must starch indicator only be added near the titration end point and not that the start of titration?
If starch solution is added at the start of the titration, the insoluble dark blue colored triiodide - starch clathrate precipitate formed will cause a time delay in releasing the triiodide ion (to be reduced by the thiosulfate ion), which may cause your titer volume and hence your titration results to be inaccurate. In other words, you may already have added enough thiosulfate solution from the burette to reach end point (which approximates equivalence point), but the dark blue color may remain for several seconds longer, fooling you into adding another couple of drops of thiosulfate solution from the burette, ruining the accuracy of your titration results.This is much less of a problem if the starch solution is added only near the end of the titration, as there will be a much smaller amount of triiodide ions left to be released from the decomposition of the much smaller amount of dark blue colored triiodide - starch clathrate precipitate, which will take much less time (almost instantaneous) and hence poses little risk of error to your titer volume and titration results.
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Originally posted by Flying grenade:
there's no methanal generated anywhere in the mechanism is it?
e.g. george chong organic book page 83
is it that the alkene is oxidised to ethanedioic acid then h2co3 then further to CO2 and H2O?
The CH2 side of a terminal alkene is oxidized to methanal which is further oxidized to methanoic acid which is further oxidized to carbonic acid which decomposes to CO2 and H2O, provided you heat under reflux with MNO4-/H+. -
Originally posted by ACA-Milkshake:
2011 TYS Paper 1
Qns 8: How do we know that the conditions are cold such that Cl2 can disproportonate in NaOH? Is it because as long as the reaction takes place in rtp, we assume that the conditions are cold?
2011 TYS Paper 2
Qns 4c(iii): Do we use the bond energies found in 4c(ii) to answer? Or the bond energies are not relevant in this qns?
Qns 6a(ii): Based on reaction 1 alone, shouldn't the functional group be an acyl chloride since it can react with silver nitrate immediately to produce a white ppt? So why is Halogenoalkane the answer when they have to undergo additional steps after adding silver nitrate?
Q8. Yes.Q4ciii. The stability of CCl4 over SiCl4 is *DESPITE* the stronger Si-Cl bonds (as a challenging A grade question, Cambridge can ask you to suggest at least 2 different possible explanations for this anomaly of C-Cl bond being weaker than Si-Cl bond despite C and Si being in period 2 and 3 respectively ; I expect my BedokFunland JC students to be able to rise up to Cambridge's challenge!).
So the bond enthalpies are not irrelevant, they are meant for Cambridge to (using A level exam questions) educate the student who understands what's going on, "Oh yah hor, how come Si-Cl bond stronger, yet SiCl4 kena hydrolyzed more readily leh? Chemistry damn interesting sia! I love Chem!"
The reasons for the stability of CCl4 over SiCl4 is covered in the basic H2 syllabus (although Singapore JCs usually teach only 1 simplified reason, while I teach my BedokFunland JC students (at least those who ask me) 3 different reasons for this).
Q6aii). Acyl chloride will give a white ppt at rtp. The qn specified warming is necessary, thus alkyl chloride.
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Originally posted by Flying grenade:
pg 112 , same book as above
lone pair of O resides in a sp3 hybridised orbital, can undergo delocalisation? i thought it's only for lone pairs residing in unhybridised p orbitals
That's exactly right. Since the lone pair on the O atom in phenol *is* indeed delocalized by resonance into the benzene ring, obviously the lone pair doesn't occupy an sp3 hybridized orbital, geddit?