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BedokFunland JC's A Level H2 Chemistry Qns (Part 2)

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  • Moderator
    H2 Chemistry @ BedokFunland JC (near VJC & TJC)
    UltimaOnline's Avatar
    15,154 posts since May '05
    • Originally posted by Ng.keebin:

      In grp ll, there is greater covalent character in mgcl2 than those below it but in group Vll, there is greater covalent character in silver astatide than in silver iodide though astatine is below iodine. Is it something about those cations in group ll having smaller radius and hence higher polarising power while for the anions under group Vll, larger anions are more polarisable...? 


      Because Group II or 2 ions are the cations with the polarizing power, hence the higher the cationic charge density and concordantly the greater the polarizing power up the Group, the higher the covalent character for the ionic compound.

      Conversely Group VII or 17 ions are the anions with the polarizable electron charge clouds, hence the lower the anionic charge density and concordantly the greater the polarizability down the Group, the higher the covalent character for the ionic compound.

    • Originally posted by Flying grenade:

      AlI3 has weaker Al-hal bond(since more covalent) compared to Alcl3?

      That, plus less effective overlap of hybridized orbitals to form the sigma bond due to more diffused valence orbitals of the halogen atom down the Group.

  • Flying grenade's Avatar
    1,243 posts since May '15
    • 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



      U coconaden. Some tertiary alcohols can kena dehydrated into alkenes, some cannot. It's *not* a matter of whether it's a pri or sec or tert alcohol, rather it's a matter of availability of beta protons to be abstracted by the Bronsted-Lowry base in the E1 or E2 mechanism.

      Figure out the mechanism for yourself, or go ask your school teacher or private tutor.

      Edited by UltimaOnline 09 Sep `16, 6:12AM
  • Moderator
    H2 Chemistry @ BedokFunland JC (near VJC & TJC)
    UltimaOnline's Avatar
    15,154 posts since May '05
    • 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

      "Angeline and jasmine writes..."

      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.

      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 

      Fair enough, I meant you "criticize/point out the author's typos", rather than criticize the authors themselves (I'll go edit my previous posts to be fair to you).

      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).

      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 

      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

      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.


      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.

      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.

      Edited by UltimaOnline 09 Sep `16, 4:42PM
    • Originally posted by nicolemantou:

      HCI 2014 paper 2
      Q4a) why doesnt the reagent dilute HNO3 need conc H2SO4 for electrophilic sub? Why dilute HNO3 dont react with compound C?

      If the reagent were conc HNO3 + conc H2SO4 (ie. strong nitrating agent), then all 4 molecules would be nitrated. The question deliberately specified dilute HNO3 (ie. weak nitrating agent), to see if you can tell which benzene rings are sufficiently activated to react with dilute HNO3 (ie. weak nitrating agent).

      Only D is sufficiently activated, because of the 2 electron-donating by resonance groups. If you're wondering why only mononitration occurs, it's because while OH and OR are both activating as they donate electrons by resonance, OH is a stronger activator than OR (why? I've already explained this somewhere among my 100+ questions on my BedokFunland JC website, go ask your school teacher or private tutor if you don't know), *and* NO2 group is strongly deactivating (ie. electron-withdrawing by *both* induction and resonance). Concordantly, while D is the only molecule sufficiently activated to undergo nitration with dilute HNO3 (ie. weak nitrating agent), but only mononitration occurs, ortho to the OH group.

      C is not sufficiently activated (actually it's not activated at all), because by induction, the vinyl side-chain's sp2 C atom (ie. not being an sp3 C atom) does not donate electrons by induction. By resonance, the vinyl side-chain is in equal measure electron-donating and electron-withdrawing (vis-à-vis the benzene ring), and is hence neither. Concordantly, C is neither an activated or deactivated benzene ring, and requires conc HNO3 + conc H2SO4 (ie. strong nitrating agent) to undergo nitration.

    • Originally posted by Flying grenade:

      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 require or expect 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.

    • Originally posted by gohby:

      Hi UltimaOnline, I have 2 questions here:


      Q1: In a question which requires “state what you would observe when A is converted to B” or “state the observations”, would a response of “white fumes are evolved” (instead of “white fumes of HCl (g) evolved”) suffice? I am of the opinion that given one cannot possibly observe the identity of the fumes evolved, a response of “white fumes are evolved” is sufficient and correct, and any further details as to the identity of the gas is superfluous (and wrong).


      Q2: NYJC/2012/P1/Q18 [Ans: D]

      When I mix A & B or A & C together, I will obtain AgCl (s), which will be the white ppt soluble in ammonia as shown in row 1. However, I do not understand the observations shown in row 2. Mixing A & C should produce iron (II) nitrate and silver chloride - so where is there no white ppt produced, as per row 1, and why is there a grey ppt (most likely it is iron) formed?

      Thank you, UltimaOnline :)

      Hi Gohby, no prob :)

      Q1. Yes, you're correct.

      Q2. Ag+ from AgCl(s) is reduced to Ag(s) a grey ppt insoluble in NH3(aq), and Fe2+(aq) is oxidized to Fe3+(aq).

      Edited by UltimaOnline 11 Sep `16, 4:40AM
    • Originally posted by gohby:

      Hi UltimaOnline - I've some further enquiries here:

      [YJC 2012 Prelim]


      I get that there are 5 dative bonds in this complex, but how does that point to the electronic configuration of the cobalt cation, given that in this case, I think it would have been the 1x4s, 3x4p and 1x4d orbitals which will hybridise to accept the lone pairs from the dative bonding?



      My thoughts: When I add iron (II) sulphate to bromine and chlorine, a redox reaction will occur and Fe3+ & chloride/bromide will be formed. When I then add sodium hydroxide to the mixture, iron (III) hydroxide will be formed. However, how can I distinguish between the solution containing chloride and that containing bromide? According to the E⦵ values, neither chloride nor bromide will be able to reduce iron (III) hydroxide to iron (II) hydroxide.


      NJC/2012/P1/Q15 [Ans: B]

      Upon heating with barium carbonate, the barium oxide formed will react vigorously with water - however, what is the gas evolved during the "effervescence" stated in the question? Unless the question used the term "effervescence" to describe the vigorous reaction - which would be an inaccurate characterisation, no?

      Thank you for your help. :)

      NJC Qn : this question has come up time and time again, because it is, as you suspect, a faulty question. While there is a slight difference in the color of solution (because color isn't just dependent on cation, but on ligand substitution equilibria as well), it is not sufficient to reliably distinguish between iron(III) chloride vs bromide. And while the differing magnitudes of thermodynamic feasibility translates into differing extents of equilibrium positions (and thus differing amounts of ppt obtained), this isn't an appropriate pedagogical assessment context for the H2 syllabus. In summary, this is a faulty question, but which by elimination still finds that option to be the most feasible (for MCQs, choose the 'most correct' answer, even if it isn't 100% correct).

      YJC Qn : for such questions in the context of the A level H2 syllabus, instead of considering the transition metal orbitals in which the dative bond electrons reside in, treat such questions as simply asking you to deduce the OS of the metal cation, and concordantly, the electron-configuration. In other words, disregard the dative bonds residing in the transition metal orbitals.

      NJC Qn : BaCO3 is significantly more stable than CaCO3, and at temperatures which decompose CaCO3, BaCO3 remains undecomposed and thus results in "effervescence" when mixed with HCl(aq).

      No prob, Gohby :)

    • Originally posted by [email protected]@ddy:

      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

      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.

      Originally posted by [email protected]@ddy:
      Would it be wise to skip doing paper 1 and leave it till when im done with all papers? Thats the advice Im getting which I feel is slightly dubious! And I should I bother doing the top tier JC prelims or stick to mid tier JC prelim papers! Thanks so much, the A level grind is real.

      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?


      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.

      Originally posted by [email protected]@ddy:

      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 ?

      Geometrical isomerism consists of cis–trans isomerism and E-Z isomerism. If there are only 2 R groups (need not be the same as each other), then cis–trans isomerism can be used. If there are 3 or 4 R groups, then E-Z isomerism needs to be used.

      Edited by UltimaOnline 11 Sep `16, 11:14PM
    • _

      Edited by UltimaOnline 13 Oct `17, 2:15AM
    • Originally posted by 8truthseeker8:

      Hi, can I ask two questions? I searched online for the answers but I cant seem to find a consensual answer and also the answers were not available.

      First, why are geminal diols unstable? I know they undergo dehydration but why?

      Second, why does Fehling's not react with aromatic aldehydes? I search online apparently its because of a step in the mechanism, but I cannot find the mechanism for oxidation using Fehling's anywhere online?

      Because of thermodynamically favorable positive entropy change, which shifts the position of equilibrium (ie. it's an equilibrium, you'll still get some germinal diols).

      As for why Tollen's can oxidize benzaldehydes while Fehling's can't, there are several reasons for this.

      For A level purposes, the more relevant concept is the more positive reduction potential of Ag+ to Ag, compared to Cu2+ to Cu+.

      At the same time, note that benzaldehyde is more resistant to oxidation compared to aliphatic aldehydes, as illustrated by the greater magnitude of oxidation state difference between the reactant (C atom's OS approximately 0.5 in the resonance hybrid for benzaldehyde, versus C atom's OS of +1 in aliphatic aldehydes) and the product (OS of +3 in the carboxylate ion).

      Oxidizing benzaldehydes are not a problem for stronger oxidizing agents K2Cr2O7 and KMnO4 (with more positive reduction potentials), which are (by virtue of higher oxidation states and charge densities of the Mn and Cr atoms) able to utilize a stronger ionic oxidation mechanism, as opposed to a weaker free radical oxidation mechanism for Fehling's and Tollen's, in which the more positive reduction potential for Ag+ to Ag over Cu2+ to Cu+ just sufficiently overcomes the larger thermodynamic barrier for the oxidation of benzaldehyde (as opposed to aliphatic aldehydes) via the weaker free radical mechanism (as opposed to the stronger ionic mechanism).

    • Originally posted by Flying grenade:


      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? 

      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.

      Originally posted by Flying grenade:

      in pyridine by itself, is the lone pair on N needed and wanted for aromaticity?

      Didn't u understand what I wrote? Obviously not (both u and the lone pair).

      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

      Good that u understand this (both the calculation and how chemistry works).

      Edited by UltimaOnline 13 Sep `16, 3:23AM
    • Originally posted by Flying grenade:

      YJC 2016 prelim


      Answer given is In step 2, PCl5 will be hydrolysed in the presence of water

      Heat is required for step 4


      consider this qn


      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? 

      Due to the restricted rotation of the sigma bonds within the bicyclic ring (if u dunno why the ring must be bicyclic, go ask ur school teacher or private tutor), it's impossible to have 2^2 optical stereoisomers (considering only the 2 chiral carbons); instead the configuration is fixed at either RS or SR or RR or SS (u can work out exactly which urself, using Cahn-Ingold-Prelog rules). And therefore its mirror image will be SR or RS or SS or RR (respectively). That's why the YJC answer considered both chiral carbons as just one chiral carbon, instead of 2, when working out no. of stereoisomers.

      I suspect u still won't understand this, so u better go ask ur school teacher or private tutor. Move on.

      Edited by UltimaOnline 13 Sep `16, 7:37PM
    • Originally posted by Flying grenade:

      2016 YJC JC2 Block test


      is this plausible?

      gaseous SOCl2 bubbled into a soln containing molecules with hydroxy group. the pH of resulting soln no doubt will be lower, with or without any molecule ,as long as one bubble socl2 into any soln

      but is this reaction viable? does Vit A(in this qn) , gets its hydroxy group substituted with a Cl? 

      if we want to cause nucleophillic sub of a halogen onto Vit A containing the OH group, we would carry out in anhydrous condition , with Vit A as a solid right?


      no doubt only a very small percentage of Vit A , will dissolve in water, as the molecule is majority Non-polar and hydrophobic

      so perhaps it does undergoes nu sub for the halogen when Vit A is put in water and SOCl2 in bubbled into this soln of water containing Vit A , after substituting for halogen most likely still Solid, as molecule is largely non polar 

      Both u and YJC teacher made an error here (urs way more coconaden and unforgivable!).

      U coconaden, anhydrous doesn't mean no solvent! The reactant is solvated (not solid) and thus can exists in a solution, just not with water (ie. not aqueous solution, a non-aqueous solution lah)!

      The solvent has to be inert (ie. no protic OH or NH groups), so that it won't use up the PCl5 or SOCl2 reagent.

      The YJC teacher's error : if it's an aprotic solvent which won't react with PCl5 or SOCl2, then neither will [H+] in solution increase, and hence there'll be no change in pH.

      But for A level purposes, Cambridge may willingly commit the same 'error' as ur YJC teacher (ie. they may set this same qn), coz they don't expect A level H2 students to think deeply enough to recognize this as an error.

    • Originally posted by Flying grenade:

      some  Continual Assessment test


      for what type of reaction is the reaction in III, cher say it's disproportionation/redox

      can i say it's disproportionation redox, acid base reaction?

      Aldehydes can undergo rxn with NaOH(aq) just like that by itself without heat? do we need to know that aldehydes can ketone can react with strong bases?


      option 3, fine , it's correct, 9 chiral centres after reaction with LiAlH4

      have five sets of two-adjacent chiral carbons in a ring, total of 7 such carbons in this case

      which set of two-adjacent chiral carbons can only produce 2 optical isomers?

      how many stereo optical isomers are there in this molecule?


      U coconaden! That's the Cannizzaro reaction lah! (pronounced machiam hokkien KAN-NI-ZAR-RO). Overall the only acceptable adjective is redox disproportionation (or disproportionation redox, Cambridge accepts either) reaction. It involves a hydride H- transfer, followed by a proton H+ transfer at the end. So to say it's a Bronsted-Lowry acid-base proton transfer, rather than more accurately a redox disproportionation reaction, is misleading and credit unworthy.

      Due to the bicyclic rings, 4 chiral C atoms are 'merged' into 2 chiral C atoms, as far as calculation of stereoisomers are concerned.

      However, because this is considered beyond A levels, hence in the A level exam, especially if it's MCQ, you may have to consider either possibility as correct (ie. Cambridge may think A level students aren't smart enough to understand or recognize the problematic bicyclic ring circumstance, and so the 'correct' answer may actually be the 'wrong' one; so if both possibilities appear as different options, then u jia lat liao).

      If it's a P2 or P3 qn, the exam-smart BedokFunland JC student will give both possible answers, with qualifications and explanations.

      Besides, this question (and hopefully Cambridge's questions) only asked for how many chiral C atoms, instead of asking how many stereoisomers are possible.

      Edited by UltimaOnline 14 Sep `16, 2:24AM
    • Originally posted by Flying grenade:

      ok dafuq, means we do tick the option as Correct in the exam, that anhydrous socl2 will result in a decrease in pH of the soln , even though we know that it won't, in a inert soln(e.g. ccl4)? force yourself get confused like this, not healthy

      and is it like Since we know socl2 or pcl5 must react in anhydrous condition, if Cambridge present the question similar to above, do u think they expect us to think that the solvent must be inert?

      wow, i hope they want us to select the correct Chemistry option, if this unfortunate event happens. i will take my chances and gamble or give or take that 1 mark away and shade the correct Chemistry answer

      Eh don't OCD and worry so much lah. It's a bell-curve afterall. If it's not fair to everyone, it's still fair. Be like me, any ambiguous qn like these, just make a relaxed, stress-free educated guess on what Cambridge wants (based on what Cambridge thinks A level students know) and make ur choice, or (if u're really good enough like me), skip the qn also still can get A grade. *evil laugh*

    • Ex-RJC students post their condolences on Facebook RJC Confessions, and hope to guide current RJC students stressed out with life.


    • Originally posted by Flying grenade:

      2009 p3 qn 1a bi) 

      given pka of glutamic acid is 4.3

      of course many of us know the COOH R group of glutamic acid will be deprotonated above pH of 4.3

      i know pH above 4.3 is alkaline relative to the pka of 4.3 of glutamic acid

      but i cant answer when my friend ask me why

      why a pH above the pka of an acidic(cooh) group will cause it to be deprotonated?

      no good definition of pka online

      how to answer the above question using chemistry concepts using chemistry terms? 

      For every acidic group (this includes the conjugate acids of basic groups), there will be a specific pH of the solution in which you've equal molarities of conjugate acid and conjugate base (for this particular acidic group), correct? (this is common sense).

      Chemists call this pH (which is unique for all acidic groups) the pKa value (for this particular acidic group).

      Hence, when pH starts getting more alkaline, you get higher molarity of conjugate base, lower molarity of conjugate acid. Hence, when pH starts getting more acidic, you get higher molarity of conjugate acid, lower molarity of conjugate base.

      If you and your friend still don't understand after reading the above, then you and your friend have no chemistry sense, no hope liao.

      And bear in mind this is about equilibria, which means no matter what the pH is, you'll always have some molarity of all possible conjugate acids and conjugate bases, eg. H3A2+, H2A+, HA, A-, etc, it's only a matter of correctly identifying the major species present at any given pH for A levels (at Uni levels, you need to identify the 1st most major species, the 2nd most major species, the 3rd, and so on, at any given pH).

      Edited by UltimaOnline 14 Sep `16, 6:22PM
    • Difference between inverse agonist and antagonist.



    • Originally posted by MightyBiscuits:

      Hi UltimaOnline, in George CHong's organic chem book pg 43, it gives an example of FRS in the methyl group in methylbenzene which makes me confused as I had always thought that FRS only occurs in alkanes. 

      If free radical substitution occurs can occur in alkene, how does it work?

      Also for George Chong's Organic chem book pg 45, I dont get how the ratio is formed. Thanks

      Alkyl side-chain of benzene is considered an 'alkane'.

      To determine product distribution of halogenation via free radical substitution, mathematically combine both factors :

      No. of H atoms substitutable x Stability of alkyl radical intermediate (also known as Kinetic Reactivity of Pri vs Sec vs Tert H atoms)

      Take the halogenation of butane for instance. In terms of "No. of H atoms substitutable", you'd expect 1-halobutane to be the major product (since 6 H atoms vs 4 H atoms). In terms of "Stability of alkyl radical intermediate", you'd expect 2-halobutane to be the major product (since the secondary alkyl radical intermediate to generate secondary alkyl halide 2-halobutane, is more stable than the primary alkyl radical intermediate to generate primary alkyl halide, 1-halobutane ; the more stable the intermediate, the lower the Ea, hence the faster the rate of reaction, resulting in the major product).

      Overall, to determine which factor outweighs which to result in the major product, you'd have to combine both factors mathematically (don't memorize the relative stabilities or reactivities, the exam question will provide it, as this factor depends on the specific halogen, as well as specific temperature).

    • Originally posted by senga:

      2011 TYS question 6 part c v and d ii.

      Which other reaction confirms W is aromatic? Explain your answer. 

      Why can't I use reaction 5 since reaction 5 proves that there is phenol functional group?

      Explain clearly why you have placed each of the functional groups in their particular positions.

      Why can't the chlorine atom be attached to the benzene ring? Why didn't they specify the adding of NaOH or KOH for reaction 1? Sorry if this question seemed too simple but I'm really confused over here. I thought NaOH/KOH is needed to substitute the chlorine no matter whether it is halogenoalkane or halogenoarene. 


      Edit. I think I get the answer. Reaction 3 specify that organic product produced has Mr of 138. Which means chlorine is removed. If reaction 3 is not there, can I assume that chlorine is attached to the ring instead of the alkyl group? The answer sheets say that reaction 1 shows that chlorine is attached to the alkyl group instead of the benzene ring, which doesn't make sense for me.


      Do I need to remember how to produce phenol from benzene?

      Reaction 3 is a better choice, as only alkyl benzenes can be oxidized by KMnO4 to hydroxybenzoic acid. Reaction 5 is a weaker reason, as it could imply an acyl halide side chain (although the degree of unsaturation won't fit, but most H2 students won't know how to check for this anyway).

      Without OH-(aq), only alkyl halides can be hydrolyzed upon warming (adding OH- would of course speed up the reaction). If the Cl atom was attached to the benzene ring as you suggest, then due to the C-Cl bond strengthened by having partial double bond character in the resonance hybrid, concordantly and consequently not only would OH-(aq) be required (now you understand why Cambridge didn't use OH-(aq)? now you understand the answer sheet?), but a much higher temperature is required (the question specified 'warm'), and besides, the nucleophilic aromatic substitution hydrolysis (via the elimination-addition mechanism involving a benzyne intermediate) of aryl halides to phenol is beyond the H2 A level syllabus.

      At University and professional levels, there are many different ways to generate phenol from benzene, but none of these are taught at A levels. In other words, within the H2 A level syllabus, there is no (within syllabus) method you can generate phenol from benzene. Of course, if you're aiming for a distinction A grade, it's always useful to expose yourself beyond the syllabus.

    • Originally posted by gohby:

      Hi UltimaOnline,


      I would like to seek your help with regard to the following questions:


      Q1: DHS/2015/P1/Q14


      What’s wrong with Choice B, given that acid strength increases from HCl to HI? Interestingly when I researched on the pH of these compounds in a fixed concentrations their pH are the same. (Source: http://www.aqion.de/site/191)


      Q2: DHS/2015/P1/Q34 [Ans B]


      Why is the answer not A? I think choice 3 is correct - vanadium oxide gets reduced before being oxidised. (Source: http://www.chemguide.co.uk/inorganic/transition/vanadium.html) Or am I missing out on something here?


      Q3: VJC/2015/P1/Q3 [Ans C]


      Which one of the following is a possible configuration of a stable M3+ ion in the ground



      Remarks: I do not understand this question. Are they asking for the configuration of M or M3+? If it’s the latter, then why is there a need to add the words “in the ground state”?


      Q4: VJC/2015/P1/Q14 [Ans C]

      Remarks: I thought the answer should have been B instead. This is because at B, since the concentration of lead (II) containing species is at its lowest, [Pb2+] should be at its lowest at V. I didn’t think C was the correct answer, because at any point between 0cm3 to Vcm3, although the ionic product was initially greater than the solubility product, the ionic product being equal to the solubility product at any point through the formation of the ppt.


      Many thanks, UltimaOnline for your help! :)

      Hi Gohby, no prob ;)

      Q1: DHS/2015/P1/Q14. While acid strength (indicated by pKa) increases from HCl to HI, but due to the leveling effect, in aqueous solutions they are all strong acids which dissociate completely and hence result in the same pH.

      Q2: DHS/2015/P1/Q34. You're right. Perhaps (lame, but) the DHS teacher question-setter was nitpicking that "likely" should be replaced with "definitely", since the maximum OS of V is +5 (based on electron configuration).

      Q3: VJC/2015/P1/Q3. The question is asking for the configuration of M3+. "Ground state" must always be specified, otherwise there would be many permutations and combinations possible for excited states.

      Q4: VJC/2015/P1/Q11 (it's Q11, not Q14). "Ionic product" is always intended to be "initial before precipitation". This is a somewhat unfair question, as the mathematics involved goes beyond A levels, but conceptually still examinable as a sadistic distinction question.

      The VJC teacher question-setter's thinking (and feeling rather proud of him/herself for setting such a tricky question, no doubt) is that in sufficiently large excess of halide ions, not only would all the precipitate dissolve, but even the molarity of aqueous Pb2+ ion, would eventually decrease to become even less than at point V, due to large excess of halide ions pulling the position of equilibrium over to the RHS to form the coordination complex ion, ie. [PbCl4]2- and [PbI4]2-, which are not considered as Pb2+(aq).

      Edited by UltimaOnline 25 Sep `16, 4:00PM
    • Originally posted by Flying grenade:

      eh in page 78 he wrote trans isomer is more stable than the cis isomer due to reduced steric hindrance

      but he didnt comment/write on either the cis or trans would be the major product 

      so for the case of but-2-ene, is the trans isomer the major pdt? due to this reason of more stable? anymore reasons?

      for other cases might be different? different geometric (cis-trans)isomers for different alkenes molecules have different major/minor products for diff reasons?

      First of all, be sure you don't confuse regiochemistry (Zaitsev vs Hofmann) with stereochemistry (cis vs trans, or E vs Z), and thermodynamic vs kinetic products.

      You can go read up on regiochemistry and thermodynamic vs kinetic products on your own if you like (don't ask me about it).

      In terms of stereochemistry, whether the cis/trans or E/Z product is favored, depends on whether the elimination reaction (eg. dehydrohalogenation) occurs via E1 or E2 mechanism (go read up on the factors which influence E1 vs E2 on your own if you like, don't ask me about it).

      When E1 occurs, the major product is the thermodynamic or more stable product (ie. trans geometic isomer).

      When E2 occurs, there is only 1 product, which is determined by the stereochemistry of the reactant undergoing the elimination.

      As Wikipedia states : "In the E2 mechanism, the two leaving groups need to be antiperiplanar, because an antiperiplanar transition state has staggered conformation with lower energy than a synperiplanar transition state which is in eclipsed conformation with higher energy ; consequently the reaction mechanism involving staggered conformation is favored for E2 reactions."

      Concordantly and consequently, the elimination product of the E2 mechanism is already pre-determined by the stereochemistry of the reactant, and depending on case-by-case, will be *either* the cis or trans geometric isomer *only*.

      If you get a mixture of cis and trans products, it's either because E1 occurred, or a mixture of E1 and E2 occurred.

      Note that in reality (ie. at Uni level, in the chemical industry), a mixture of E1, E2, SN1, SN2 all occurs simultaneously and competitively, to generate a mixture of all possible products.

      FlyingGrenade, you're the one who opened this can of worms by asking this question, your own fault blame yourself, don't expect me to clean up the worms after you. You'll probably end up with more questions than before, after reading this post above, but don't ask me any more questions on these matters.

      For the sensible pragmatic A level H2 Chem student, just focus on your within H2 syllabus material and ignore whatever goes beyond (ie. for the Singapore H2 Chem exam, don't try to bring in Zaitsev or Hofmann or E1 or E2 or cis or trans or E or Z , unless specifically asked by the exam question).

      For the already-getting-A-grades-for-Prelims enthusiastic A level student taking H3 or Olympiad Chem intending to study Medicine or Chemistry in the Uni, you can go read up on your own about these matters, and/or go ask your school teacher or private tutor.

      Moving on...

      Edited by UltimaOnline 25 Sep `16, 2:30PM
    • Originally posted by Flying grenade:




      did the publisher copy the qn wrongly?

      Should be A,B,D instead of A,B,C?


      cannot be oxidised by hot kmmo4(in acidic or alkaline conditions ) right? because there's no benzylic H?

      The benzylic H atom requirement for side-chain oxidation is specifically for hydrocarbon level. Since the side-chain is already semi-oxidized to the carbonyl level, KMnO4 can indeed further oxidize the carbonyl side-chain to a benzoic acid.

      Originally posted by Flying grenade:

      Phenols can't be oxidised to benzoic acid because don't have a hydrogen atom on the carbon atom to which the OH group is attached ah?

      Yeah, phenols don't even have a benzylic C atom, that's why.

      Edited by UltimaOnline 25 Sep `16, 3:15PM
    • Originally posted by Flying grenade:

      Phenols are rather easily oxidized despite the absence of a hydrogen atom on the hydroxyl bearing carbon. Among the colored products from the oxidation of phenol by chromic acid is the dicarbonyl compound para-benzoquinone


      Indeed, when this question came out for a top JC Prelim paper a couple of years ago, only my BedokFunland JC students were able to draw the oxidation product.

      Why? I'll give you a hint : unlike Singapore JC H2 Chem students, only my BedokFunland JC students (together with H3 Chem and Olympiad Chem students), understand both the meaning of resonance contributors vs resonance hybrid, as well as my BedokFunland JC formula for calculating Oxidation State.

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