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

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    H2 Chemistry @ BedokFunland JC (near VJC & TJC)
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    • Originally posted by Flying grenade:

      ok thanks god for ur patience, benevolence and still not giving up when i ask qns (high tolerance)

      2007 p3 qn 4c


      what does he mean by since reagents used are non chiral, the reactions involved are identical?


      That statement of CS Toh's, is an additional (and secondary) point : that because hot Al2O3 is a 'chirally blind' or non-stereospecific reagent, as opposed to say using the biological enzymes in your human body, hence whether you use enantiomer D or E, you end up with the same products F and G.

    • [Chemistry / Medicine] - In 1982, a group of 6 Americans became mysteriously paralyzed. Psychiatrists thought it was a neurological problem and neurologists thought it was psychiatrical problem.

      Eventually, they were diagnosed with Parkinson's disease, which is a crippling neurodegenerative disease affecting the elderly. But these were 6 relatively young Americans, including 2 young women aged 21 and 30 years old.

      After further medical and chemical investigation (including critical evidence turned up by mass spectrometry), it turned out that they had consumed an illegal synthetic opiod drug called MPPP (left in diagram below), which on its own would not have caused any problems.

      But if the chemistry synthesis was poorly carried out, the MPPP would be contaminated by a side-product (not by-product) MPTP (right in diagram below), which would cross the blood brain barrier and be metabolized by monoamine oxidase enzymes into MPP+, a powerful neurotoxin that would destroy all the dopamine-releasing neurons in the substantia nigra of the brain that controls bodily movement.

      Among the worst hit of the 6 victims was also the youngest : a 21 year old girl named Connie, who became permanently and near-completely paralyzed for the rest of her life.



      Edited by UltimaOnline 27 Sep `16, 3:34AM
    • Originally posted by Flying grenade:


      why first step higher Ea? both species are not same charged, no repulsion. step 1 involves breaking of a pi bond and forming a bond.

      step 2 involves breaking of a sigma bond



      Your points are valid, but at the same time, bear in mind that the Ea for the 1st step is somewhat higher than you would expect, due to electrostatic repulsions between the partial negatively charged O atoms between both reactants. The Ea for the 2nd step to cleave the C-O bond is somewhat lower than you would expect, because of electrostatic repulsions between the (partial and formal) negatively charged O atoms in the unstable intermediate.

      Finally, For MCQs, always choose the best answer. Options A & D can be ruled out for obvious reasons. Between C and B (both of which are not an accurate portrayal of the energy magnitudes involved), B is more likely, because in option C, the products are so thermodynamically unstable, that you would unlikely get much alkaline hydrolysis of the ester (despite the non-reversible step 3), and yet you should already be familiar with the fact that alkaline hydrolysis of esters occurs readily (ie. thermodynamically feasible) upon heating.

      Originally posted by Flying grenade:

      how do u know that hydrolysis of ester is an exo reaction,  the products of hydrolysis have a lower energy content of ester(reactant) ?


      The options given are inaccurate portrayals. Also, it's actually the final products (after proton transfer in step 3) that have less energy than the initial reactants.

      Fyi, thermodynamic energy / enthalpy and entropy wise, because esterification and hydrolysis are not redox reactions (the OS of the C atoms are exactly the same pre and post esterification, pre and post hydrolysis), and because there is no dramatic difference in steric strain or resonance stabilization energy in the reactants vs products, as well as number of reactants vs products, any difference in thermodynamic energy / enthalpy and entropy, won't be too dramatic.

      Edited by UltimaOnline 27 Sep `16, 8:47PM
    • Originally posted by Flying grenade:

      why 1. NaBD4 2. D2O

      is it followed by D2O ah?

      i dont get this thing ....

      Draw out the mechanisms for LiAlH4 on aldehyde, ketone, carboxylic acid, ester, amide, nitrile (which Cambridge *can *ask as a distinction A grade H2 Chem question) to properly understand why the reagents and conditions have to be written this way (ie. why 1st step must be anhydrous, why 2nd step must hydrolyze).

      If you cannot draw the mechanisms, go ask your school teacher or private tutor to teach you.

    • _

      Edited by UltimaOnline 13 Oct `17, 2:15AM
    • Originally posted by Flying grenade:

      tried to find out the reaction between S2O32- and H+, not much on the internet except this

      NaS2O3 (aq) + 2HCl (aq) <----> 2 NaCl (aq) + SO2 (g) + S (s) + H2O (l)


      Obviously because the disproportionation of S2O3 2- under acidic conditions is thermodynamically feasible, you coconaden.

    • Originally posted by Flying grenade:

      my question : why is the bond between two sulfur atoms in S2O32- a single bond? isn't forming a double bond more thermodynamically favourable?


      interestingly, the Resonance Hybrid in the wiki page shows S-S !!! why?? (i.e. no partial double bond between S atoms in the hybrid so no S=S in any contributors?

      however, some other google images show its possible for S=S (in resonance contributor) ?


      my teacher say

      consider S2O32- + H+ -> S + SO32-  --(1)

      more difficult to break double bond, require more energy

      experiment shows S forms relatively easy in (1) when S2O32- is reacted with acid  (rxn 1 proceeds relatively easy)

      hence should be S-S


      help reconcile this ultima

      No, it is more thermodynamically feasible for S to form single bonds. You asked this question before on the forum, and I already replied you on the forum with the answer. Don't waste my time. You keep forgetting stuff I already taught you, how to get A grade for A levels???

      S=S is a minor resonance contributor, S-S is a major resonance contribubtor (Cambridge can ask why). For the resonance hybrid, partial double bond character must be shown, or Cambridge will penalize you.

      If you can't figure out why, go ask your school teacher or private tutor. Move on.

    • Originally posted by gohby:

      Hi UltimaOnline,


      I would like to seek your help on some of the HCI questions here:


      HCI/2015/P1/28 [Ans A]


      Remarks: I think the answer is C instead. PCl5 reacts with the alcohol and COOH (and not the phenol) and it reacts with sulphuric due to the close proximity between the COOH and OH.


      HCI 2015/P1/25 [Ans C]


      Remarks: I’m not entirely sure why the answer is C, but here are my deductions:


      A is wrong because even though NO converts CO and unburnt hydrocarbons in the catalytic converter, the carbon dioxide formed is not harmless because it’s a greenhouse gas.

      B is wrong because it should be N2O.

      D is wrong because the reaction named is dissolution.


      I referred to wikipedia to evaluate choice C [https://en.wikipedia.org/wiki/Acid_rain]:


      In the gas phase sulfur dioxide is oxidized by reaction with the hydroxyl radical via an intermolecular reaction:

      SO2 + OH· → HOSO2·

      which is followed by:

      HOSO2· + O2 → HO2· + SO3

      In the presence of water, sulfur trioxide (SO3) is converted rapidly to sulfuric acid:

      SO3 (g) + H2O (l) → H2SO4 (aq)

      Nitrogen dioxide reacts with OH to form nitric acid:

      NO2 + OH· → HNO3


      However I’ve some disquiet about this choice:


      Firstly, such detailed knowledge and pathway of acid rain formation seems to be beyond the reach of the syllabus.

      Secondly, even if the student possesses knowledge on how exactly acid rain is formed, I wouldn’t say it catalyses the formation of acid rain from atmospheric sulphuric dioxide. This is because sulphuric acid is formed from sulphur dioxide, and the acid contributing to acid rain by nitrogen dioxide is nitric acid. Hence nitrogen dioxide doesn’t catalyse the formation of acid rain from sulphuric dioxide (unless it somehow forms sulphuric acid).


      HCI 2015/P1/18 [Ans C]


      I can accept the answer because that forms an integral part of the definition of transition element. However, what’s wrong with B as the answer?


      HCI 2015/P1/19 [Ans A]


      Is there a manner for students to make an intelligent guess/extrapolate from their knowledge in the syllabus to obtain the answer without understanding and appreciating their individual mechanisms? Don’t think the Grignard reagent in step 1 is part of the syllabus so I’m thinking there should be a more efficient way of solving this question.


      Many thanks Ultima :)

      No prob Gohby :)

      Q28. You're right, both are possible products, depending on whether conc H2SO4 is in excess or limiting.

      Q25. You're right (although D should be hydrolysis). For A level purposes, just use the simplified equations that NO2 oxidizes SO2 to SO3, which undergoes hydrolysis to generate H2SO4 (acid rain). The NO reacts with O2(g) to regenerate NO2, hence a catalyst.

      Q18. The other options are *properties* of transition metals, not definition.

      Q19. Even if the student isn't intimately familiar with the mechanism, but by inspection, should still be able to deduce whether addition, substitution, elimination, oxidation, reduction or hydrolysis has occurred. Also, the other options are inadmissible due to errors in steps 1, 2 or 3.

    • Hi Gohby,

      DHS/2015/P1/11. For option B, the graph obtained is determined by stoichiometry, not by kinetics. Regardless of order of reaction for H2O2, you'll get the same graph anyway. For option C, you can't use the half-life method because H2O2 and H+ are not in large excess. Even if they were, you would only be able to determine the order of reaction of I- (which generates I2, the molarity of which is represented by the y-axis). But you can still compare gradient of tangent at initial to determine the order of reaction wrt H2O2 using option C. For option D, since the graphs are linear, H2O2 must be first order.

      DHS/2015/P1/16. Adding H2SO4 protonates the NH3 ligands, consequently they can no longer function as ligands.

      VJC/2015/P1/8. The conditions within the battery in the human body cannot possibly be at standard conditions (ie. body temperature isn't standard *this fact in itself will suffice for this MCQ*; furthermore molarities of Li+ and I- have no reason to be standard, even if they began as standard molarities, which is silly because that doesn't serve its medical function, they will no longer be standard as the reaction proceeds). For option D, a quick calculation reveals more than 9g of Li is required for the battery to last 5 years.

      VJC/2015/P1/34. Option 2 is wrong because the OS of the alkyl group C atoms do not change, it's the OS of the carboxylic acid (acyl) C atoms which change. Option 3 is incorrect, because the redox potentials at the cathode and anode are determined by the voltage of the power supply, ie. this isn't a Galvanic-Voltaic cell, it's an electrolytic cell.

      No prob, Gohby :)

      Originally posted by gohby:

      Thank you UltimaOnline.


      I have some other questions here. I've penned down my thought process and doubts wrt some of the individual choices (pls pardon the lengthiness!). Thanks again.


      DHS/2015/P1/11 [Ans B]


      I’m not too sure what to make out of the choices but here are my workings:


      A: Order of reaction wrt H2O2 can be done by juxtaposing the gradient of [I-] taken at 0s (initial rate of rxn)

      B: the absorbance corresponds to the amount of I-(which doesn’t block out light) vs iodine (which blocks out light). But how do I know I can’t get the H2O2 order?

      C: No idea on how to analyse - firstly, the 2 graphs look significantly different for a diffenence in 0.02M of hydrogen peroxide. Secondly I tried to establish the half life of iodine (0->0.5->0.75…) by making a scaled drawing but got nowhere.

      D: I think the 2 graphs with differing [I-] is jus a red herring. I can note the rate of reaction at a particular [H2O2] at either of the graphs, double that [H2O2], and see to what extent the rate increases


      DHS/2015/P1/16 [Ans D]


      Regarding choice C, why does the addition of H+ in step 3 (to a mixture of [Cu(NH3)4]SO4 + Sn2+ + H2O) cause the formation of [Cu(H2O)6]2+?


      VJC/2015/P1/8 [Ans C]


      (i) I understand that the answer cannot be B because of the physiological environment (pH being between 7-8) in which the pacemaker is placed, but how exactly does that affect the Ecell, given that neither of the redox equations involve H+?

      (ii) Is D incorrect because it is the LiI crystal which is being exhausted so the choice should have referred to an amount of LiI instead of Li(s) instead?


      VJC/2015/P1/34 [Ans D]


      Is choice 2 wrong because RCO2Na is oxidised to RCOO- and H2O is reduced to H radical? Although at first blush I might have accepted choice 2 because the R-R and H2 were formed thereafter from RCOO- and the H radical. What does choice 3 mean when it suggests that the redox/reduction potential is least/most positive?

    • _

      Edited by UltimaOnline 13 Oct `17, 2:16AM
    • Originally posted by Flying grenade:

      ooh H2SO3 is a weak acid ?

      H2SO4 is a strong acid right

      Eh you jialat lah, your overly simplistic statement shows a lack of BedokFunland JC depth of understanding of acid-base equilibria.

      H2SO4 is a diprotic acid. The 1st proton dissociation is complete hence is technically strong (ie. stronger than hydroxonium ion). the 2nd proton dissociation occurs incompletely in aqueous solution, yet to a significant extent, hence HSO4- is technically weak (ie. weaker than hydroxonium ion) but is still considerably acidic (HSO4- is actually even more acidic than Al3+).

      H2SO3 is a diprotic acid. The 1st proton dissociation occurs incompletely in aqueous solution, yet to a significant extent, hence H2SO3 is technically weak (ie. weaker than hydroxonium ion) but is still considerably acidic (H2SO3 is actually even more acidic than Al3+). The 2nd proton dissociation is rather weak (ie. significantly weaker than hydroxonium ion), and simultaneously considering both the acidic and basic hydrolysis of the amphiprotic HSO3- ion, the pH of a solution containing only the HSO3- ion would be only approximately 4.5 (ie. HSO3- is actually even weaker an acid compared to Al3+).

      Edited by UltimaOnline 03 Oct `16, 10:17PM
    • Originally posted by Flying grenade:
      • Same physical properties except that they rotate plane polarised light in opposite directions
      • Same chemical properties except when another chiral molecule is involved.  <-- what does this mean?

      Biological reactions involve chiral enzymes acting on chiral substrates, hence only one enantiomer is the correct enantiomer to produce to correct effect. The other enantiomer can either be useless, or extremely toxic (eg. carcinogenic). This is an important consideration when manufacturing pharmaceutical medicinal drugs.

    • Originally posted by Flying grenade:

      the oxdn number of carbonyl C atom in benzaldehyde is between 0 and +1 ?


      how to determine the exact value for these type of cases? also the N atom in nitrobenzene is between +3 and +5?

      If resonance is involved, the actual value will be a *weighted* average of the values in the resonance contributors.

      You need to correctly deduce the OS value in the individual resonance contributors, before you can correctly elucidate the actual OS value in the resonance hybrid.

      You go draw out the resonance contributors of nitrobenzene, then see if you get " between +3 and +5" or not, you coconaden.

    • Originally posted by Flying grenade:




      how is it that Uand V yield the same products on ozonolysis

      W and X also


      U,V and  W,X all have diff structural formula 




      only ketone CH3CH2COCH3 generated from oxidation of X via ozonolysis can be distinguished only 

      First of all, if you use ozonolysis, you *must* specify "oxidative workup" or "reductive workup" or Cambridge will penalize you.

      Second (once you've chosen oxidative or reductive), then draw out all the products.

      If they're not the same, then they're not the same lah! (in which case, authors also can make typos ok! you urself know can liao!)

      If they're the same, then *that's why* they're the same lah! (in which case, who say different reactants cannot generate the same products after a reaction?)

      Edited by UltimaOnline 05 Oct `16, 10:28PM
    • Originally posted by supercat:

      Hi, may I ask A Level 2012 P3 q5 c?

      For this structure elucidation question,I derived that L has no aldehyde/ketone/alcohol/COOH groups. 


      M has to be an acidic salt, and alkene is present in M.


      I have problems deducing M, assuming my L is right.

      L is an acid anhydride right? But my L seems to be "cyclic" and I don't know how to add NH3 to it. From the question, it seems that the whole NH3 is added. How can NH3 be added when nothing is removed??

      Some of my thinking: L is similar to acyl chloride, so reaction may be similar. But to form salt...?


      Correct. Carboxylic acid anhydrides are electrophiles similar to acyl halides.

      Draw out (if you're not familiar, you can google it online and familiarize yourself with it, it's compulsory for H3 syllabus, and very useful / good-to-know for H2 syllabus) the addition-elimination mechanism for the nucleophilic acyl substitution, together with a subsequent Bronsted-Lowry acid-base proton transfer.

      Only by drawing out the mechanism of this reaction for yourself, will you be able to truly, properly and correctly understand this reaction.

      Originally posted by supercat:

      Is it like this? http://imgur.com/mdIfXtV

      Wrong. You need to reform the acyl group (ie. lone pair on negative formal charged O atom, shifts down to form a pi bond pair) and eliminate the RO leaving group as alkoxide RO-, before proton transfer occurs.

      Edited by UltimaOnline 07 Oct `16, 11:51PM
    • Originally posted by Flying grenade:

      why larger size is a weaker base? less electronegative, lp of e- more readily donated and available , wouldn't it more readily accept proton?


       i know larger size e- cloud more diffused, less effective overlap when forming covalent bonds, result in weaker covalent bonds

      I won't spoonfeed you (or anyone) on the forum with the answers. Go figure it out for yourself, and/or go ask your school teacher or private tutor.

      What you've described is about nucleophilicity, but Bronsted-Lowry basicity is a different matter altogether. A stronger nucleophile may or may not be a stronger Bronsted-Lowry base, case-by-case basis. Similarly, for ligand strength (which is also a separate matter from ligand field strength).

    • Originally posted by Flying grenade:


      what kind of molecule is V ?

      The reagents are in error. Sodium nitrite, not sodium nitrate, should be used.

      Protonating O=N--O- generates O=N--O-H, which can be further protonated into O=N--OH2+ which can decompose into H2O and O=N+, an electrophile to be attacked by the amine (ie. electrophilic substitution) to generate the final product of N,N-dimethylnitrosamine.

      Edited by UltimaOnline 09 Oct `16, 12:52AM
    • Originally posted by Flying grenade:

      alkaline hydrolysis of that two amide product all colorless

      idk color of phenylamine relative to what 

      Phenylamine in the absence of solvent is a colorless or pale yellow liquid.

      In aqueous solution (bear in mind you just carried out hydrolysis, be mindful of the situation dah dey!), phenylamine (ie. beyond the very small amount that is soluble or miscible in water), will appear as a emulsion or suspension or 'precipitate' (even if it's technically not solid at rtp).

      If the volume of phenylamine is comparable with the volume of water (ie. water is no longer a solvent, because solvent by definition must be in large excess), then eventually the 2 immiscible species will separate into 2 immiscible layers, based on differing density.

    • Originally posted by hoay:

      If we have a solution of S2O8-2 (+2.01V) ion and Ag+ (+0.80V) ion then which wil be discharged at cathode? Assuming both have 1 M solution.  

      An excellent trick question you can use to test your students!

      Although the standard reduction potential of peroxodisulfate(VI) aka peroxydisulfate(VI) to sulfate(VI) is more positive than the standard reduction potential of silver(I) to silver(0), and hence peroxodisulfate(VI) aka peroxydisulfate(VI) will seem to be more thermodynamically favored to be reduced to sulfate(VI) at the cathode, but that does NOT occur at all.

      (Note that the term "discharged" can only be used specifically if the species loses a charge, ie. Ag+ discharged to Ag, but when S2O8 2- is reduced to SO4 2-, the negative charge remains, so advise your students to write "oxidized at anode" and "reduced at cathode" instead of loosely (and often erroneously) misusing the term "discharge").

      The reason for this, is because the peroxodisulfate(VI) aka peroxydisulfate(VI) ion is ANIONIC, and hence electrostatically migrates to the ANODE (instead of cathode), but cannot be further oxidized (so some other anion present, or H2O, will have to be oxidized instead, exactly which, depends on their oxidation potentials).

      The Ag+ CATION will migrate to the CATHODE and be reduced to Ag(s) instead. So in this setup, to answer your question, Ag+ is reduced instead of the peroxodisulfate(VI) aka peroxydisulfate(VI).

    • Originally posted by gohby:

      Hi UltimaOnline,


      I have some questions here which I'd like to seek your clarification -


      Q1: Revisiting: HCI 2015/P1/18 [Ans C]


      18 Which statement correctly defines a transition element?

      A Transition elements exhibit more than one oxidation state in their compounds.

      B Transition elements form many coloured compounds.

      C Transition elements have partially filled d orbitals.

      D Transition elements or their compounds are widely used as catalysts.  


      Although the definition talks about partially filled d orbitals, isn’t it an inaccurate definition of transition elements because it should have been the ions of the transition elements which have partially filled d orbitals (or else by this definition suggested in C, scandium would be a transition element and copper would not be regarded as one).


      Q2: If I were to increase the concentration of the electrolyte, what will happen to the rate of electrolytic plating? On one hand there are sources to suggest that the rate of electrolysis will remain unchanged as the current remains unchanged, but this website (http://www.easychem.com.au/shipwrecks-and-salvage/3-electrolytic-cells/factors-that-affect-an-electrolysis-reaction) suggests that “Increased concentration of ions in electrolyte→ Increased current, and therefore increased rate of electrolysis”


      Q3: In suggesting why phathalic acid is more acidic than terephthalic acid, the answer scheme suggests that the acid anion of phthalic acid is more stable than that of terephthalic acid as intramolecular H bonds can exist due to the close proximity of O with the negative chareg and H atom of the COOH group.

      Ok, I don’t dispute the veracity of the explanation, but how does it afford an explanation to the question, since phthalic acid itself can also form intramolecular H bonds, which therefore increases its stability as well. Hence, I don’t see why the ability of phthalic acid anion to form intramolecular H bonds a good reason why it’s more acidic than terephthalic acid, given that the ability of the phthalic acid to form H bonds could well work against the formation of phthalic acid anion through its increased stability.


      Thank you UltimaOnline! :)

      Hi Gohby,

      Q1. Yes, you're correct, it's the qn's fault, but you still gotta choose the best option, which remains as C.

      Q2. No change in rate, the website author is erroneously lumping together galvanic-voltaic cells with electrolytic cells.

      Q3. You're right that H bonding is also possible before 1st proton dissociation, which would have the effect of weakening / decreasing Ka1.

      However, upon deprotonation, the COO- group (having a negative formal charge delocalized by resonance over both O atoms) is a far stronger hydrogen bond acceptor (compared to the neutral COOH group), thus more significantly stabilizing the amphiprotic conjugate base, with the overall effect of strengthening / increasing Ka1.

      No prob, Gohby! :)

    • Originally posted by supercat:

      Oh no ok let me re-draw everything out.

      For Qn 12 (same paper), could you please take a look at my working? Where did I go wrong? 


      I put B as my answer, because it's slightly more than 10cm3.

      Maleic acid, being a weak diprotic acid, can be represented by H2A. Note that the x-axis specifies volume of weak acid added, not strong base added.

      At 20cm3 of H2A added, you have 1st equivalence point, hence you only have HA-, which being amphiprotic, *can* function as a buffer.

      (Fyi, 2nd equivalence point to exactly fully deprotonate all of H2A, is at 10cm3 of H2A added against 20cm3 of NaOH.)

      However, at 40cm3 of H2A added (ie. 20cm3 excess H2A), you have equal moles hence equal molarities of HA- and H2A, hence you have *maximum* buffer capacity. Concordantly, D is the answer.

      Options A and B are obviously wrong, because you have a significant excess of the strong base NaOH, which means all of the 1st acidic protons have been removed, and most of the 2nd acidic proton as well. In fact, for option A, all of H2A is completely deprotonated to A2-, which means it is only basic, hence not a buffer at all). While for option B (which at 12cm3 of H2A against 20cm3 of OH-, leaving behind only a little bit of HA- left) is only effective at buffering against incoming strong acids, but almost totally useless at buffering against incoming strong bases. Thus options A and B are obviously wrong.

      Edited by UltimaOnline 16 Oct `16, 11:23PM
    • Originally posted by iSean:

      CIE M/J/16/Q4(a) Regarding drawing the structure, A and B are my suggested structure. And the most right is Googled Structure. 
      Are my Drawn Structure Correct? Also why does NH2 groups are in different planes? 
      Does it mean the drug is Tetrahedral, so it is not superimposed? 

      Additional Question, why the molecule is neutral? Despite consisting of C2O4 2- ion? 

      The google image's stereochemistry is a result of the cyclohexane adopting a non-planar conformation to maximize stability (each C atom is sp3, not sp2 like in benzene).

      The coordination geometry about the Pt2+ ion (the dipositive charge or +2 oxidation state of Pt, explains why the coordination complex is neutral, asked in another of your questions) has to be square planar, as predicted by the Jahn-Teller effect (Uni level Chemistry, for A level purposes, just accept and memorize the geometries, either tetrahedral or square planar, of the commonly encountered 4-coordinate complexes).

      And because each ligand is bidentate, hence only the cis-isomer is possible, not the trans isomer. Hence no cis-trans geometrical isomerism is possible.

      You might be creative and force a trans isomer by placing 1 ligand above and 1 ligand below the square planar complex, but it would be too thermodynamically unstable due to severe deviations from the ideal bond angles about the donor atoms vis-à-vis orbital hybridization concordant to VSEPR theory.

      Originally posted by iSean:

      Okay thanks for answering :) 
      But despite the NH2 group exists in on different planes, it joins coordinate bond as a square plannar? And not exists in Tetrahedral.... huh? 

      Due to Jahn-Teller effect, the coordination-complex has to be square planar, which means the N atom has to be co-planar or at least peri-planar with the other 4 donor atoms.

      Edited by UltimaOnline 24 Oct `16, 8:45PM
    • Originally posted by Flying grenade:

      AJC 2013 qn 2biv

      pic of qn here



      i thought steric hindrance is often applied for groups directly bonded to the atom of interest(in this case the carbonyl carbon) ?

      the bulky R groups connected to the carbon adjacent to the carbonyl carbon can also provide steric hindrance ? 

      Also it's the C atom connected to the carbonyl carbon in the ester( R-COO-R) that's changing, not the carbonyl carbon that's changing. the number of R groups connected to the adjacent carbon can also cause the carbonyl carbon to be less e- deficient? The more frequent cases we encounter is , the no. of R groups directly connected to the C atom of interest, or ethyl or methyl group

      (ok i do realise that benzene ring or long alkyl chains do provide steric hindrance for a molecule, overall)




      Steric hindrance applies to groups, not just individual atoms, obviously. If a huge bus is blocking the entrance of your school gate, do you expect only the student standing next to the bus to be blocked, or all students in the area kena blocked? U coconaden!

      Inductive effects fall rapidly with distance. Hence in this question, only steric hindrance is a correct reason, the inductive effect is insignificant. So fair enough, you're somewhat correct in your thinking here.

    • Originally posted by Flying grenade:

      for Histidine, 

      why Rgroup Basic group(which is more basic) has smaller pka , which in turn has bigger pkb value , than alpha Basic group ?

      The R group iminium is less acidic (ie. smaller Ka or larger pKa) than the alpha COOH, but more acidic (ie. larger Ka or smaller pKa) than the alpha NH3+.

      The R group imine is more basic (ie. larger Kb or smaller pKb) than the alpha COO-, but less basic (ie. smaller Kb or larger pKb) than the alpha NH2.

      Bonus : For a challenging A grade question, Cambridge can ask you to *explain* (ie. not just assign values without explanation) why is it the case that for Histidine (unusual among alpha amino acids), the R group imine is more basic (ie. larger Kb or smaller pKb) than both the alpha COO- and the R group amine, and why the R group imine is less basic (ie. smaller Kb or larger pKb) than the alpha amine. Or conversely, explain the relative acidities of the conjugate acids of these (abovementioned) groups and their Ka or pKa values.

      My BedokFunland JC students can check their answers with me during their next tuition session, while the rest of you can go ask your school teacher or your own private tutor. I won't reveal the answers here.

      Edited by UltimaOnline 25 Oct `16, 3:47PM
    • Originally posted by Flying grenade:

      ajc 2013 p1 qn 36

      why molecule 3,  https://en.m.wikipedia.org/wiki/Cyclooctatetraene doesn't exhibit aromaticity and is not a planar molecule, even though all the carbons are sp2 hybridised?

      Why should a molecule be planar just because all C atoms present are sp2 hybridized? Just because benzene is so, doesn't mean all molecules are so. That's a erroneous fallacious assumption on your part.


      Cyclooctatetraene isn't aromatic because in addition to being non-planar, the no. of pi electrons fail to satisfy Huckel's Rule.

      Originally posted by Flying grenade:

      Siala. quite difficult 


      cos we learn that sp2 hybridisation is trigonal planar

      Just because sp2 hybridisation is trigonal planar, doesn't mean a molecule with all C atoms being sp or sp2, has to be planar. Here's a simple example to illustrate this.



      Originally posted by Flying grenade:

      then how do we deduce if any given molecule is planar anot ?

      means we can't deduce if any given molecule is planar or not, even with the ability to deduce  hybridisation and given the non 3d projections(i .e. no dash wedge line shown) but given drawn chemical kekule/lewis structure ?

      unless we can somehow deduce, or have the chemical knowledge , by just looking at the non 3d structure of various molecules such as allenes and annulenes and cycloalkanes , they have to conform to some other conformations due to steric strain/angle strain/ring strain/vdw repulsions/torsional strain

      By looking at the entire structure case-by-case and applying your knowledge of different aspects of chemistry in combination.

      You seem to be whining for a super simple easy way to tell if a molecule is planar or not. Such simple easy ways are for Ordinary O Level Chemistry, not Advanced A Level Chemistry.

      Of course, Cambridge will use an appropriate range of difficulty in their questions. They may ask on super easy structures, or average difficulty intermediate structures, or slightly beyond A levels difficult structures (for A grade questions, only a few of such questions will appear in every paper, you shouldn't worry about such questions, it's a bell-curve, just do your best, and see if you're better than the rest).

      Edited by UltimaOnline 25 Oct `16, 9:24PM
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