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

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    • Originally posted by Ng.keebin:

      Why does pKa value of hydrohalic acids decrease down the group?

      The strength of HX increases (ie. Ka increases or pKa decreases) down the halogens (Group VII or Group 17) for 2 main reasons : decreasing magnitudes of endothermic H-X proton dissociation enthalpies (quote relevant Data Booklet H-X bond enthalpies, and explain that the effectiveness of orbital overlap to form the H-X sigma bond decreases with increasing principal quantum number and the concordant increasing diffusiveness of electron orbitals), and increasing stabilities of X- halide ion conjugate base due to decreasing anionic charge densities in aqueous solutions.

    • A BedokFunland JC original H2 Chemistry challenge

      The new 2016 H2 Chemistry syllabus (9729) will test Singapore H2 Chemistry candidates more rigorously on their understanding (not just memorizing) of reaction mechanisms. Test your own understanding, familiarity and competency in proposing, suggesting and constructing curved-arrow electron-flow reaction mechanisms in the following chemical reaction.

      The self-disproportionation, self-decomposition redox reaction of hydrogen peroxide can occur through different pathways, depending on pH and the availability of a catalyst, as well as the specific type of catalyst. Based on your understanding of reaction mechanisms as required by the H2 Chemistry syllabus, suggest and draw 4 alternative curved-arrow electron-flow mechanisms for this reaction (for the sake of simplicity, without involving pH or catalyst ; such a reaction would still be thermodynamically feasible, but would be much slower due to high Ea).

      A free radical mechanism, given that the overall order of reaction is 2.

      An ionic mechanism, given that the overall order of reaction is 2.

      A free radical mechanism, given that the overall order of reaction is 1.

      An ionic mechanism, given that the overall order of reaction is 1.

      After you've drawn all 4 mechanisms as described above, suggest (and explain your reasoning) which of the 4 mechanisms would be the most thermodynamically feasible and hence the most likely mechanism for the reaction (without involving pH or any catalyst).

      Edited by UltimaOnline 03 Feb `16, 5:40PM
    • Originally posted by Ephemeral:

      1. What happens when CH3CO(CH2)2CH=C(CH3)2 reacts with HBR?


      2. A certain industrial cleaner and paint solvent was distilled to produce a single compound D. When D reacted with 2,4-DNPH, an orange ppt was produced. With alkaline aq I2, D gave a pale yellow ppt. D did not react either with warm acidifed KmnO4 or with aq Br2. Reduction of D with H2 over a catalyst produced equimolar mixture of 2 isomers E and F with molecular formula C4H10O. 

      Is D an aldehyde or ketone? Reduction of D with H2 over a catalyst shows that it is an aldehyde (my notes said that ketone cannot be reduced by H2) but no reaction with acidified KmnO4 shows that it is a ketone. 

      Q1. Electrophilic addition of HX unto the nucleophilic alkene occurs. As for determining the major product, Cambridge will *not* accept "because of Markovnikov's rule" as an explanation. You need to specify "Tertiary carbocations are more stable than secondary carbocations ; the more stable the intermediate, the lower the Ea required, hence the faster the rate of reaction, generating more of the product, which we label as the major product".

      As far as the H2 syllabus is concerned, carbonyl compounds do not react with hydrogen halides or hydrohalic acids, as the X- nucleophile has lower anionic charge densities and thus less nucleophilic, Bronsted-Lowry basic and Lewis basic, compared to say, the cyanide ion (written as CN-, but when drawing out the mechanism, Cambridge requires you to clearly indicate the uninegative formal charge on the C atom and not the N atom, or you'll be penalized), as C is not electronegative enough to stabilize a negative formal charge (the only reason why CN- can even exist as an intermediate is due to the high % s orbital character of the sp C in CN-).

      Beyond the H2 syllabus, the addition of HX to carbonyl compounds does occur, but the position of equilibrium lies mostly to the carbonyl side, as determined thermodynamically by favourable entropy change. This is similar to the position of equilibrium lying significantly more towards the carbonyl side than the geminol diol side, but even moreso.

      Q2. -__-" Don't trust your school notes. Gaseous hydrogen (with nickel or platinum catalyst) does indeed reduce aldehydes, ketones and even nitriles, but not carboxylic acids, esters and amides.

    • Originally posted by Ephemeral:

      What is the type of reaction in the second step when using Grignard reagent? 

      Why does the second step occur and why does it not stay as -- O^-MgX^+?

      Source : https://en.wikipedia.org/wiki/Grignard_reaction

      Bronsted-Lowry acid-base protonation or proton transfer reaction. In anhydrous solvents, it does stay as O- ionically bonded to +MgX (with some degree of covalent character). But because you (ie. whoever carries out the reaction) want to obtain the organic alcohol product (with a longer carbon chain), thus you would subsequently add water (the least toxic and cheapest Bronsted-Lowry acid available) to protonate the alkoxide (which is significantly Bronsted-Lowry basic because O- is relatively unstable due to relatively high anionic charge density as the negative formal charge is on the small O atom with small atomic radius being a period 2 element) to obtain your desired (conjugate acid) alcohol product now with a longer carbon chain (as Grignard reagents are carbon nucleophiles which undergo nucleophilic addition), together with (MgX)OH as a by-product (not side-product, don't confuse the two).

      Edited by UltimaOnline 07 Feb `16, 5:28PM
    • Originally posted by Ng.keebin:

      If benzene ring contains delocalised electrons, why can't it conduct electricity?

      Because unlike the giant covalent lattice structures of graphene (which stack together in layers to form graphite), benzene consists of simple, discrete covalent molecules, and hence can only delocalize electrons within a single small molecule of benzene (ie. electrons cannot jump from one benzene molecule to the next, since the molecules are separate or discrete), while in graphene (being a giant covalent lattice of benzene rings joined together, so to speak, see image below), the delocalized electrons can travel across the entire giant covalent lattice of the graphene, which is the medium that conducts the electricity (ie. flow of electrons).


      Image source : https://en.wikipedia.org/wiki/Graphene

      Edited by UltimaOnline 10 Feb `16, 10:50AM
    • Originally posted by gohby:

      Hello UltimaOnline,


      I have some questions at hand:




      With reference to the second and fourth row in the diagram, what’s the mathematical basis behind the linearising of the first and second order concentration against time graphs? In other words, why do I get a linear graph when I ln the concentration in a first order reaction and when I reciprocate the concentration in a second order reaction?


      Q2: ACJC 13/P1/Q5


      I find the phrasing of the choices confusing. If I accept the answer as D, can I take “the phosphorus atoms of both oxides” to mean some of the phosphorus atoms of both oxides? (Although this will also make C and B acceptable choices too.)


      Alternatively, if I take  “the phosphorus atoms of both oxides” to mean all of the phosphorus atoms of both oxides, there would have been no correct answer, given that there is a trigonal pyramidal P at P4O9?


      Q3: If I mix tetrachloromethane with methanol, what are the intermolecular forces of attraction formed (debye forces?) and destroyed (presumably id-id and H bond). Debye forces are weaker than pd-pd forces but how does that account for no heat being evolved? Likewise, when I mix trichloromethane (pd-pd) with propanone (pd-pd), why would heat be evolved?


      Thank you! :)

      Hi Gohby,

      Q1. The linearizing of molarity-against-time graphs to determine 2nd and 3rd orders by use of ln and reciprocal functions are beyond the H2 syllabus, and thus not required for students to know. Cambridge will give relevant guidance if these functions are to be used. Nonetheless, it's a useful bonus to teach your students about using these, and will still be acceptable by Cambridge (though they'll usually structure their questions such that the student will need to do other within-syllabus procedures and checks that will already be sufficient to determine the order of reaction, so this is just an extra bonus).

      To address your question more directly is in the realm of mathematics rather than chemistry, and thus if you would like to pursue a deeper explanation, you could start another thread requesting the assistance of the resident mathematics experts Eagle and WeeWS. The key to understanding the linearity of the molarity-against-time graphs after the ln and reciprocal functions are applied, are deriving the Integrated (ie. applying the Integration function of Calculus unto the rate laws) 1st and 2nd Order Rate Laws. Once these are derived using Calculus, then the graphical linearity becomes a natural and mathematically obvious expression of the Integrated Rate Laws. For further reading, see links below.


      One last point though, is that for the H2 syllabus, Cambridge will instead use another graphical linearizing method to hint to the student that the reaction is of 2nd order kinetics with respect to a particular reactant. Graphical linearizing can be achieved by squaring (ie. power raised to 2nd order) the molarity or partial pressure of the reactant, and plotting against initial rate (with x and y axes interchangeable, after all it's a linear graph). This is to test A level students to see if they are able to correctly interpret the linear graph to mean that the rate of reaction is *directly proportional* (hence linear) to the square (ie. raised to the power of 2) of the molarity or partial pressure of the reactant, and thus deduce the order of reaction is 2nd order with respect to that reactant. Specifically, see Singapore A levels 2010 P2 Q3 for such an example.

      Q2. The question's fault for not being specific. In which case you have to consider all the options together and see which interpretation makes the most sense (here, it is ALL atoms rather than SOME atoms). So D is still the best answer. The "tetrahedral arrangement" in option D refers to the electron geometry rather than molecular geometry. Most Singapore JC school teachers (and private tutors) unwisely teach their students 'shape' or 'geometry' to directly mean 'molecular geometry' or 'ionic geometry', when the correct pedagogical approach would be to teach students to first work out the 'electron geometry', and thereafter derive the concordant 'molecular geometry' or 'ionic geometry'.

      On a related note, most Singapore JC school teachers (and private tutors) also unwisely teach students to identify orbital hybridization (sp, sp2, sp3, etc) in terms of single vs double vs triple bonds, when the correct pedagogical approach would be to teach orbital hybridization in terms of electron geometries. In other words, do H2 Chem students truly understand the *purpose* and *nature* of orbital hybridization (ie. in relation to VSEPR theory, sigma & pi bonds, and lone pair residence), or do they blindly apply the 'single vs double vs triple bonds' rules of their school lecture notes without understanding?

      An effective way to test your students on their understanding on this, is to ask them to identify orbital hybridizations of non-carbon atoms in a variety of molecules, and going beyond just sp, sp2 and sp3 orbital hybridizations. For more able students, you can bring in resonance as well, and how it relates to and/or affects orbital hybridizations and vice-versa.

      Q4. Tetrachloromethane with methanol : since H bonding exists between methanol, and only van der Waals (a mixture of all 3 : Keesom or pd-pd, Debye or pd-id and London Dispersion or id-id) forces exist between tetrachloromethane with methanol, thus newer intermolecular attractions are weaker than original intermolecular attractions, and thus the solvation or mixing will be endothermic.

      Trichloromethane with propanone : slightly trickier, since both are polar aprotic molecules. In such a case, at A levels, students are not expected to state for certain (ie. without experimental data) whether the mixing will be exothermic or not, but A level students are only required to state that the reaction *may* or *could* be exothermic, because the newly formed intermolecular attractions *may* or *could* be slightly stronger than original intermolecular attractions (which for MCQs this understanding will usually suffice to determine the answer, especially when combined with elimination MCQ strategy).

      Not all van der Waals (ie. whether we're talking about Keesom or pd-pd, Debye or pd-id and London Dispersion or id-id) forces are equal in strength, before and after mixing of 2 species. The newly formed (Keesom or Debye or even London Dispersion) van der Waals forces could be weaker or stronger than the original van der Waals forces (even when considering the same specific type of van der Waals forces, eg. Keesom forces).

      Why? Consider the molecular geometries and dipoles present : trichloromethane (with longer C-Cl bonds and a shorter C-H bond ; the Cl atoms are strongly delta negative while the C atom is strongly delta positive) is tetrahedral while propanone is trigonal planar (with C being strongly delta positive and O being strongly delta negative due to both induction and resonance).

      Hence, the newly formed molecular interactions (mainly and most significantly Keesom forces, but all 3 types of van der Waals forces are of course present) are stronger, because the geometries and dipoles allow it : the strongly delta negative O of propanone forms strong Keesom attractions with the strongly delta positive C (with H posing negligible steric hindrance) of trichloromethane, while the strongly delta negative Cl atoms of propanone forms strong Keesom attractions with the strongly delta positive C of propanone.

      A level students are strongly advised to draw out both molecules to illustrate (and annotate) these intermolecular Keesom van der Waals forces, even if the Cambridge question doesn't specifically require so. This is because Cambridge will award marks as long as relevant content is written by the student, regardless of whether textual or graphical.

      Edited by UltimaOnline 11 Feb `16, 7:24AM
    • Here's a Chemistry Calculation Challenge for all of you 2016 Chemistry students (both JC1 & JC2, both H1 & H2).

      Approximately 7.98 g of an unknown metal oxide reacts with 3.667 dm3 of carbon monoxide to generate carbon dioxide and the solid metal (at conditions where the molar volume of a gas may be taken to be 24.45 dm3). 125.0 cm3 of 1.60 mol/dm3 of HCl (aq) was required to completely convert all of the solid metal into its chloride salt. Using mathematical calculations and chemical reasoning, identify the metal.

      Warning : You may *not* assume the oxidation state of the metal is necessarily the same in both its oxide and its chloride ; ie. the OS of the metal may or may not be the same, eg. V2O5 --> V --> VCl3.

      After attempting the above question, you may post your final answer (ie. identity of metal) here and I'll tell you if your answer is correct, but do not post the complete solution here (don't ruin the fun for others).

      Edited by UltimaOnline 15 Feb `16, 8:07PM
    • Originally posted by gohby:

      Hello UltimaOnline,

      I have some questions which I would like to clarify:

      1. HCI/08/P1/Q4


      For option B, both hydrazine and hydrogen peroxide form hydrogen bonds and dispersion forces. What makes the bp of hydrazine lower? Also, I was thinking that D could be the answer because the δ- on N is weaker than the δ- on O, thus it would be a weaker base.


      2. RVHS/13/P1/Q32


      For option 1, I can accept that magnesium has a melting point greater than 97.8. However, without memorising the actual melting point of magnesium, how would I know that its melting point is below 1083? Also, could you explain why option 2 is factually correct (only the electron for copper in the 4s subshell is delocalised right), and that such information can be deduced from the table?


      3. NJC/13/P1/Q31


      How can we be sure the enthalpy changes in 1 & 2 cannot be determined experimentally?


      Thank you! :)

      Yo Gohby,

      Q1. Yes, because N is less electronegative than O.

      Q2. Lousy qn. From the given data alone, all 3 options cannot be deduced. From data given + Data Booklet / Periodic Table (which is thus implied), options 1 & 2 can be deduced to be *probable*. 3 is wrong because it's too generalized, and Chemistry (being a microcosm of real life) always has exceptions to every rule. You can only deduce the melting point of Mg is *probably* somewhere in between, as (by using Data Booklet) the cationic charge density is also somewhere in between. For metallic bonding, the d electrons also participates, though to a much lesser extent than the s electrons.

      Q3. Because in adding water, you can't guarantee every water molecule added will be used for hydration instead of for solution. If you say do backwards and carry out dehydration instead (a more controllable process), that's applying Hess Law already *evil laughter from Hess*. C and H2 do not spontaneously (ie. delta G = +ve) form methane. You need to couple it with other thermodynamically favourable processes (eg. biochemical or industrial) to form methane through a series of steps. See https://en.wikipedia.org/wiki/Methane#Production

    • Originally posted by gohby:

      Hi UltimaOnline,

      Am I right to say that the bp of hydrazine is lower than that of hydrogen peroxide because of stronger pd-pd AND H-bonds (even though they both form 2 H bonds/molecule) because O is more electronegative than N, thus the aforementioned forces formed will experience stronger attraction?


      I agree with A being the "best" answer, but is C an incorrect answer, given that magnesium chloride is acidic and magnesium oxide is basic?

      Yes, though focus on the H bonding, not the van der Waals.

      Option C is wrong, because MgO while basic, isn't sufficiently soluble (due to highly endothermic lattice dissociation enthalpy) to generate a strong alkali (ie. high pH). Even option A is incomplete or partially incorrect, as aqueous alkalis at room temperature cannot dissolve SiO2 (due to an even higher endothermic lattice dissociation enthalpy, in turn due to SiO2 being a giant covalent lattice structure). Molten (not aqueous) alkalis, or (at the very least) concentrated alkalis at high temperatures, are required.

      Originally posted by gohby:

      Thanks UltimaOnline! :)

      How do I compare the basicity between hydrazine and hydrogen peroxide then. Although the lps on H2O2 belong to oxygen and are thus less available for donation, there are 4 electron pairs on hydrogen peroxide compared to 2 on hydrazine - doesn't that matter?

      How exactly does increasing the concentration of [OH]- increase the feasibility of the reaction with silicon dioxide?

      No prob Gohby :)

      No of lone pairs doesn't matter as much as electronegativity (afterall, no matter how many lone pairs you have, you'll still incur a unipositive formal charge upon donating a dative bond to the proton). Thus amines and hydrazines are more nucleophilic and more Bronsted-Lowry basic, compared to alcohols and peroxides.

      Molarity affects both kinetics and equilibria (which are themselves inter-connected, ie. Kc = kf / kb), and thus thermodynamic feasibility and favourability.

      Edited by UltimaOnline 21 Feb `16, 5:37AM
    • Originally posted by Light5:



      Please take a look at Q3 part b(ii).

      I dont understand how nitrate ion can act as ligand while BF3 cannot. In both NO3- and BF3, central atom(N and B respectively) have no lone pair of electrons which means by this logic, both NO3- and BF3 cannot act as ligands.

      By another logic, the O-atoms in NO3- have lone pairs to donate and so do the F-atoms in BF3 so both should be able to function as ligands.

      Why does NO3- act as ligand while not BF3?

      Thank you for explaining this.

      NO3- is anionic, electron-rich, and hence can function as a nucleophile, Bronsted-Lowry base, ligand, and Lewis base. The donor atom (donating the dative covalent bond) would be any of the O atoms with a uninegative formal charge (in the resonance hybrid, all 3 O atoms bear a 2/3 negative formal charge and are thus equivalent).

      BF3 is a neutral molecule, but because the B atom lacks a stable octet, it is in that sense electron-deficient (even though the B atom is formally neutral without any formal charge, but it does bear a significant partial positive charge, due to the electron-withdrawing by induction effects of the electronegative F atoms, though the lack of a stable octet is the most important consideration for its Lewis acidic behaviour), and thus BF3 can function as an Lewis acid (with the electrophilic B atom as the recipient atom accepting the dative covalent bond).

      F being the most electronegative element, has low propensity for donating dative bonds (due to the unipositive formal charge it would consequently incur). Which is why BF3 is certainly a lot more Lewis acidic (or electrophilic) than it is Lewis basic (or nucleophilic). Rather, it is the F- ion (ie. fluoride ion, with a uninegative formal & ionic charge), being electron-rich, that does indeed function as a strong Lewis base, Bronsted-Lowry base, nucleophile and ligand, with a strong (due to effective overlap of the F atom's 2nd electron shell orbitals) dative covalent bond (with significant ionic character) thusly formed.

      Edited by UltimaOnline 22 Feb `16, 3:09AM
    • Originally posted by Light5:



      In Q3(c) why isnt MnO4- formed when Mn+2 reacts with H2O2 despite the fact that Eo of H2O2/H2O is +1.77 and that for MnO4/Mn+2 is lower so theoretically this reaction is possible...so why is Mn+3 formed?

      In Q7 part b(ii) why is an ionic compound i.e NH2+Br- formed with CH3CH2Br and why isnt HBr formed along with -NHCH2CH3 if the reaction is nucleophilic substitution?



      In Q1(a) why cant volume/24000 method be used..and isnt PV=nRT defined only for ideal gases??

      In Q1(c) if we increase pressure, then volume reduces so assuming we use Mr=mRT/PV (as is used by examiner)..shouldnt PV remain constant meaning Mr value is unaffected...i understand how in Q1(b) the calculated value of Mr will be lower than actual value but even if we increase pressure to reduce volume to try to compensate for the incorrectly recorded temperature, shouldnt PV remain constant and so no effect on calculated value of Mr from (b)...please explain why my logic is wrong?

      Thank You

      Q3c) This Cambridge question is problematic, because actually you do *not* get Mn3+(aq) either. Any MnO4-(aq) generated from the oxidation of Mn2+ by H2O2, being a strong oxidizing agent, will immediately react with H2O2 to be reduced back to Mn2+ and/or MnO2 (depending on pH), so you don't get MnO4-(aq) as a stable product. But because Mn3+ is also *extremely* unstable and a strong oxidizing agent, any Mn3+(aq) generated from the oxidation of Mn2+ by H2O2, will also immediately react with H2O2 to be reduced back to Mn2+, and/or undergo hydrolysis and/or precipitate out (depending on pH) to any of the following significantly more thermodynamically and kinetically stable species : pentaaqua-monohydroxo-manganese(III) complex ion [Mn(OH)(H2O)5]2+(aq), and/or manganese(III) oxide-hydroxide MnO(OH)(s), and/or manganese(III) oxide Mn2O3(s), and/or manganese(II,III) oxide MnO.Mn2O3(s), and/or undergo disproportionation into Mn2+(aq) and MnO2(s).

      So this Cambridge question is problematic. If MnO4- is rejected as an answer, then Mn3+ should also be rejected. If Mn3+ is accepted (based simply on redox potentials), then MnO4- should also be accepted. Don't worry too much about this (every year in every Cambridge Chemistry paper, there's bound to be a couple of problematic questions, but since all candidates will struggle with such questions together, and since grades are assigned on a bell-curve anyway, so it's no real cause for worry).

      In Q7 part b(ii). The immediate product of an SN2 reaction (draw the mechanism and you'll see why) is actually the ionic salt (ie. ammonium halide), which exists in equilibrium with the Bronsted-Lowry proton-transferred species of amine + hydrogen halide (ie. the products you suggested). But since the question specified "the products are more soluble in water than in organic solvents", the question is testing if you understand that this refers to ionic species rather than molecular species (as stronger ion-permanent dipole interactions ensure aqueous solubility) and hence Cambridge will only accept the ionic salt as the required answer for this question.

      As an aside, HCl (and any hydrogen halide HX) could also undergo electrophilic addition with alkenes (though this reaction is more controlled in the gaseous state, because in the aqueous state water is a competing nucleophile), but since this question Cambridge specified the molecular formula of the desired product, hence the required answer does not involve this side reaction.

      Q1a). First of all, 24dm3 is a terribly inaccurate approximation itself obtained from PV=nRT (if you carry out the calculations, you'll find 24dm3 to be a 2 sig fig approximation, not even acceptable for 3 sig fig, so only use 24dm3 when the Cambridge question specifies you can use it), so your 24dm3 is even less accurate than using PV=nRT (to at least 3 sig fig). And yes, it's for ideal gases, which at A levels you have no choice but to use the assumption that the ideal gas formula is an acceptable approximation for the properties of the real gas (otherwise if you want greater accuracy, you'll have to use the Van der Waals equation, taught only at Uni level).

      The second and more relevant reason for this question, is that at room temperature and pressure, this species exists as a liquid (it only exists as a gas when you heat it beyond room temperature), which means that under the conditions this species exists as a gas (ie. *not* at rtp), you cannot use 24dm3 gaseous molar volume (which is a 2 sig fig approximation for rtp *only*), and hence you *must* use PV=nRT.

      Q1b(i&ii). LOL this is where "2 wrongs make a right". You would be right if you increased the pressure *and* you honestly and correctly used the newly increased pressure in your calculations. But Cambridge (evidently from their mark scheme) is implying that you are to cunningly & stealthily increase the pressure *but* you continue to dishonestly use the old lower pressure value in your calculations, ie. using a wrong pressure value to rectify a wrong temperature value, to obtain the (more) correct molar mass of the gas.

      Edited by UltimaOnline 23 Feb `16, 2:33PM
    • JC2 students, how long it takes you to solve the following deductive elucidation question on esters, is indicative of your grade in the A level exams. Time yourself to find out how prepared you are for the upcoming A levels.

      If you can correctly solve all of it within 10 minutes, you're an A grader.

      If you can correctly solve all of it within 20 minutes, you're a B or C grader.

      If you can correctly solve most of it within 30 minutes, you're a D or E grader.

      If you can correctly solve at least half of it but need more than 30 minutes, you're a S grader.

      If you still can't solve at least half of it even after spending more than 30 minutes, you're a U grader.


      Edited by UltimaOnline 25 Feb `16, 5:44AM
    • Originally posted by gohby:

      Hello UltimaOnline,


      How does L exhibit geometric isomerism?

      Thank you

      Hi Gohby,

      Cycloalkenes with at least 8 membered rings or larger, are able to exhibit geometric isomerism without excessively destabilizing ring strain caused by angle strain. If the ring was too small (eg. cyclohexane), only the cis isomer would exist but not the trans (otherwise ring strain caused by angle strain would be excessively destabilizing), and hence geometric isomerism doesn't exist for small cycloalkenes.

      BedokFunland JC Challenge Qn (for H2 Chem students aiming for the A grade) :

      Is the cis or trans geometric isomer more stable for cycloalkenes with a ring between 8 to 11 atoms? How about 12 atoms or larger? Explain your answer for each case.

      I won't post the answer here. Interested students can go ask your school teacher or private tutor.

    • Originally posted by gohby:

      Hi UltimaOnline,


      Apologies for digging out this post but I realised some of the discrepancies with sf arose because I did the approximation method to negate the dissociation of the acid when I shouldn't have done so. Hence I am relooking at other questions where there are discrepancies wrt the accuracy of the answers.

      For 3(ii) (Ka of acid = 5.9x10^-4M), here are my workings but I think I have missed out on an aspect which would explain the disparity between my answer and the suggested answer of 0.0937M. Could you see if my workings/assumptions are in order?

      For 4a would it be wrong if I suggest litmus as a suitable indicator? The answer suggested methyl orange but I doubt the working range of methyl orange coincides with the rapid pH change over the equivalence point.

      As for 4b, my workings are as follows:

      The answer suggested that the pH is 9.24 - is there something wrong with my workings?


      Thank you :)

      Hi Gohby,

      For Q3 your working is fine. You can check your assumption of mathematical validity by plugging in the values after you get your final answer. Is initial molarity >> change, such that equilibrium molarity can be approximated back to initial molarity to 3 sf? Since 0.0935 - (10^-5.9) = 9.3499 x 10^-2 = 0.0935 (to 3 sf), hence your approximation is mathematically valid.

      For Q4a, litmus changes colour at pH 7, since pH at equivalence point for this titration is approx 5 (based on graph), hence litmus won't be acceptable. But methyl orange (which changes color at pH 3.8) won't be acceptable either. A better indicator would be methyl red (which changes color at pH 5).

      For Q4b, first of all the question is wrong to say H2SO4 is a strong diprotic acid, the 1st proton dissociation is certainly strong, but the 2nd proton dissociation is slightly weak. But in this case, because NH3 is a significantly stronger Bronsted-Lowry base than SO4 2-, hence the final answer for this question remains the same (admittedly, correctly considering the strengths of the 1st vs 2nd proticities of H2SO4 is somewhat beyond the H2 syllabus, and Cambridge will give additional guidance in the question if it expects students to take this into consideration).

      I can't say for sure what's wrong with your working, but this is how I'd recommend students to approach this question. Since the approximation for H2SO4 as a strong diprotic acid is acceptable for this qn, the equation should be written as H+ + NH3 --> NH4+. Doing the ICF (Initial Change Final) table for the Bronsted-Lowry proton transfer reaction, the final moles are 0 (for H+, not H+ in soln, but H+ from H2SO4), 0.0010476 (for NH3) and 0.001 (for NH4+). Plugging these moles (no need to convert moles into molarities, since molarities/molarities in the same solution = moles/moles) into the Henderson-Hasselbalch equation, ie. pH = pKa + log ( [base] / [acid] ), the pH will be calculated to be 9.2429, ie. 9.24 (to so-called 3 sf, for to be precise 9.24 is to 2 sf, as due to the log function, only the digits behind the decimal point are counted as significant figures, but apparently this isn't taught in Singapore JCs, and Cambridge will accept the 9.24 (2 sig fig) answer, unless otherwise specified by the qn).

      Edited by UltimaOnline 25 Feb `16, 6:23PM
    • Originally posted by CKTR:



      Hi! May i ask how to solve this qns?

      Modified 2010 Alevel P3 Q3 (d)

      Alcohol J, CxHyOH, is a volatile fungal metabolite whose presence when detected in air can indicate hidden fungal attack on the timbers of a house.

      When 0.10cm^3 of liquid J was dissolved in an inert solvent and an excess of sodium metal added, 10.9cm^3 of hydrogen gas (measured at 298 k) was produced according to the following equation:

      CxHyOH+Na ---> CxHyONa+1/2 H2

      When a 0.10cm^3 of liquid J was combusted in an excess of oxygen in an enclosed vessel, the volume of gas (measured at 298K) was reduced by 54.4cm^3. The addition of an excess of NaOH caused a reduction in gas volume of 109cm^3 (measured at 298K)

      Use the data to calculate the values of x and y in the molecular formula for J, CxHyOH.


      Balance the equation in terms of x and y. Coefficients should be 2 and (2x + (y-1)/2) --> 2x and (y+1).

      If the alcohol was gaseous, based on the redox acid-base reaction between the monoprotic alcohol and electropositive sodium metal, determine the volume of gaseous alcohol (ie. if the liquid alcohol was present in the gaseous state) to be 21.8cm3. This is done out of convenience, to work in gaseous volumes instead of moles. If you prefer, you can convert everything to moles instead of working in gaseous volumes.

      Hence, based on CO2 generated, and based on the stoichiometry of the balanced equation, we have 21.8x = 109, hence x = 5.

      Reduction of 54.4 cm3 implies : [Used Up gaseous volume] - [Generated gaseous volume] = 54.4 cm3.

      Alternatively, reduction of 54.4 cm3 implies : -[Used Up gaseous volume] + [Generated gaseous volume]= -54.4 cm3.

      Used up gaseous volume refers to oxygen (since the alcohol was actually liquid, ie. zero gaseous volume), which is [ (21.8 / 2)(10 + (y-1)/2) ] cm3, based on the stoichiometry of the balanced equation.

      Generated gaseous volume refers to carbon dioxide, ie. 109 cm3.

      Hence y = 11.

    • Originally posted by CKTR:

       I have got another question. 

      A 0.0250 mol sample of an insoluble metal oxide is known either to exist as MO or M2O3, metal M is the metal. The sample was dissolved in 150cm^3 of 1.3mol dm^-3 hydrochloric acid. The resulting solution was made up to 500 cm^3 with distilled water. 25.0cm^3 of the solution then required 36.40 cm^3 of the 0.200 mol dm^-3 aqueous sodium hydroxide for neutralisation.

      (A) Calculate the no. of moles of HCl acid that reacted with the metal oxide. (I got 0.0494 mol)

      (B) Hence deduce the possible identity of the metal oxide (having trouble with this)

      Due to experimental limitations, the values have to be rounded off, ie. 0.0494 rounded off to 0.0500 and hence the metal oxide must have been MO (eg. MgO or CaO or BaO, etc), since 0.0250 mol of MO consists of 0.0250 mol of diprotic dinegative oxide anions O2-, which accepts 0.0500 mol of H+ in a Bronsted-Lowry acid-base proton transfer reaction to generate 0.0250 mol of water, ie. 2H+ + O2- --> H2O

    • Originally posted by LavXuan:

      Compound P has the molecular formula C8H9ClO. When P is reacted with PCl5, white fumes are evolved. Treatment of P with neutral iron (lll) chloride did not produce a rep purple coloration . p is optically active and when it is boiled with aq sodium hydroxide l, followed by acidification with dilute nitric acid, and addition of silver nitrate, a white precipitate is produced. When P is reflexes with aq sodium hydroxide, Q, C8H8O, which gives a reddish brown precipitate with Fehling's solution is obtained.


      Suggest the structural formula of P and Q, explaining clearly your reasoning.

      P is C6H5CH2CHClOH
      Q is C6H5CH2CHO

      Edited by UltimaOnline 27 Feb `16, 6:19PM
    • Originally posted by LavXuan:

      An organic compound A, C9H11Br, on treatment with hot aqueous potassium hydroxide gave a compound B, C9H12O. B responded to oxidation in three different ways. With acidified potassium dichromate it yielded C, C9H10O. With sodium hydroxide and iodine it yielded D, C8H7O2Na and a yellow ppt. With hot, acidic potassium manga are (vii) it yielded E, C7H6O2. 


      Identify compound A-E and explain the above reactions. 

      A is C6H5CH2CHBrCH3, and the rest follows accordingly

    • Originally posted by Ephemeral:

      1. Are peroxides or oxides more stable? If peroxides are unstable why do they form? Which Grp II metals form peroxides, and do they require varying degree of heat to form oxides? What is the general trend in forming oxides from peroxides?

      2. Pg 76 of George Cheong Inorg Chem: 'The single covalent bond between oxygen atoms in peroxide is relatively weaker due to repulsion.' What and where is the repulsion?

      3. Will all peroxides ultimately form oxides and if so, does that mean peroxide can be considered an 'intermediate'? (Pg 77 of George Cheong Inorg Chem)

      4. Pg 93 of George Cheong Inorg Chem: If it is considered a disproportionate reaction with reference to BaO2, then what about water? OS of O in H20 is -2 which is the same as Ba(OH)2. 

      Q1 & 2. Oxides are more stable in the solid state, stabilized by stronger ionic bonding (but oxides, unlike peroxides, are too unstable to exist in the aqueous state, due to too high an anionic charge density). Peroxides are generally considered more unstable than oxides, because of 2 reasons : the -1 OS of oxygen, which being electronegative, prefers an OS of -2 ; the inter-electron repulsion between the (lone pairs of the) partially (if organic peroxides) or formally (if inorganic peroxides) negatively charged atoms of the O-O peroxo group weakens the (unpolarized and hence also weaker) covalent O-O bond. Stability is relative, and even though peroxides are relatively less stable, they're still relatively stable enough to exist, and thus as a thermodynamic manifestation of entropy, anything that can exist (ie. be generated) by a valid reaction pathway will exist ; it's only a matter of relative kinetics, stabilities, equilibria and formation abundance, amounts or concentrations. Theoretically, any metal ion can ionically bond (with varying degrees of covalent character) with the peroxide ion if present (you can look up the redox mechanisms for formation of peroxides if interested), and hence all metal peroxides are possible. But if the metal cation has high cationic charge density, the peroxide anionic charge cloud becomes polarized and distorted to a greater extent, hence decomposing more readily (ie. at a lower temperature).

      Q3. If heated, yes. But you can dissolve the metal peroxide in water to generate hydrogen peroxide and the metal hydroxide (though the industrial manufacture of H2O2 uses more economical and controllable synthesis methods), or you can react the metal peroxide with carbon dioxide to generate oxygen gas and the metal carbonate, so you cannot say "all peroxides ultimately form oxides".

      Q4. Water (ie. neither the H or O atoms) is not itself oxidized or reduced. Rather, it is the peroxo O-O group (atoms) which undergoes disproportionation, from OS of -1, to OS of 0 (in O2 gas) and -2 (in oxide anion).

      Edited by UltimaOnline 29 Feb `16, 3:35PM
    • Originally posted by Ephemeral:

      What chemical tests can be carried out to differentiate between bromobenzene, (bromomethyl)benzene and ethanoyl bromide? Reagents provided are aq NaOH, NH3, HNO3, HCL, AgNO3 and distilled water.

      Differentiate via ease of hydrolysis, the rate of which can be measured by the rate of AgBr(s) precipitation. Adding AgNO3(aq) to all 3 test substances will suffice.

      Read CS Toh's A Level Study Guide : "Relative Strength of the C-Hal Bond" page under the chapter of "Halogen Derivatives", and also "Acyl Chlorides" page under the chapter on "Carboxylic Acids & Derivatives" .

      Alternatively, read the following pages on Jim Clark's A Level Chemistry website :




    • Originally posted by Flying grenade:

      How are free radicals electrically neutral??

      It's about formal charges. Consider the methyl radical. The C atom has 3 bond pairs and 1/2 a lone pair. That's 7 valence electrons in terms of a stable octet (ie. lacks a stable octet), and 4 valence electrons in terms of formal charge, and because C is in Group 14, hence the C atom has no formal charge, and hence the methyl radical is a neutral molecule without any ionic charge.

    • Originally posted by gohby:

      Hello UltimaOnline,


      I would like to enquire about the following questions:

      HCI 08/P1/Q31


      Is 2 be expected solely from diagram which shows that C forms 3 bonds in C60, thus delocalised electrons in the unhybridised 2p orbitals exists?


      How can I ascertain that 3 is not an answer, given that the write-up mentions that it was discovered in the produced formed when graphite was vapourised?


      HCI 08/P1/Q32


      Why is 2 not an answer? If we have the enthalpy change of atomisation of AlF3, then the need for the enthalpy change of formation of AlF3, enthalpy change of atomisation of aluminium and the enthalpy change of formation of fluorine would be obviated wouldn’t it?


      Thank you! :)


      Yo Gohby!

      Yes, the diagram is sufficient for students to interpret, deduce and infer this. But 'solely'? The more capable (ie. A grade) H2 Chem students will be expected to already be familiar with the various allotropes of C, including fullerenes (of which Buckminsterfullerene is the most well known) and their properties, even if not included in the basic H2 syllabus.

      Because Buckminsterfullerene exist as simple, discrete molecules, hence electrons can only delocalize within the individual molecule, and not across the individual molecules, and thus Buckminsterfullerene cannot conduct electricity (which refers to a function at the macro-scale, not micro-scale; ie. a solid piece of graphite, being a giant covalent lattice, held in your hands can conduct electricity, but a solid piece of Buckminsterfullerene, being a large no. of individual C60 molecules held together by van der Waals forces, held in your hands cannot conduct electricity).

      Then it wouldn't be called a Born-Haber cycle anymore, would it? (It would then be a generic enthalpy or energy cycle, but not Born-Haber cycle, which is a specific subset or type of enthalpy or energy cycle). The question specifies Born-Haber cycle. And also, specifically for A level H2 Chem purposes, atomization enthalpy is always applied to individual elements only. Thus Born-Haber cycles (especially for A levels) follows a fixed set of steps strictly. Clockwise : from the constituent elements, atomization of both constituent elements, IE for the metal, EA for the non-metal, lattice formation enthalpy, to get the compound in standard state. Anti-clockwise : from the from the constituent elements in standard state, formation enthalpy, to get the compound in standard state. So while you have a valid point about this being a lousy or debatable question, and unlikely to be asked by Cambridge (though Cambridge certainly has its fair share of lousy or debatable questions every year), nonetheless the best answer for this MCQ is still the one given by Hwa Chong.

      No prob, Gohby ;)

      Edited by UltimaOnline 03 Mar `16, 6:01AM
    • A BedokFunland JC H2 Chemistry Qn

      What is the pH of a solution containing 5.0 x 10-8 M of HCl?

      A) 7.30
      B) 6.82
      C) 6.89
      D) 7.00

      Edited by UltimaOnline 03 Mar `16, 5:56AM
    • Originally posted by Ng.keebin:

      do you need to heat the reaction moisture before testing for ammonia test or can you just test for it straight with a litmus paper after the reaction?

      If you plunge the litmus paper into the solution, a red-to-blue color change may be due to OH- ions present from other basic sources, making the solution alkaline. Warming the solution and allow any ammonia or volatile amine present to vaporize, reach and hydrolyze on the moist litmus paper held above the solution to effect a red-to-blue color change, is thus a more reliable indicator for the presence of ammonia or a volatile amine.

    • Originally posted by Ephemeral:

      My Qs come from George Cheong Organic Chem book:

      1. Must a tertiary alcohol be attached to 3 R groups? What if there is a double bond and hence there is only 2 R-groups?

      2. Is water slightly acidic at rtp? (Pg 187) Do we just take it as water is neutral at rtp when answering Qs (for acid-base equilibria)?

      3. (Pg 190) Why is organic solvent required to dissolve all 3 compounds? Why can't the experiment be carried out just as it is?

      4. I don't understand Example 7.7 on pg 191. What do the pKa values mean and how do we use them to solve the Q?

      5. Is formation of ester with excess alcohol and conc h2so4 at 140 degree celsius in syllabus? 

      6. Why is H3PO4 a weaker oxidising agent? What are the relative strength of some of the oxidising agents? (Pg 194) 

      7. (pg 196 fig.7-5) Why doesn't Et-O-H form protonated ethanol when it abstracts a H atom?

      Q1. Then the functional group is called enol, not tertiary or secondary alcohol.

      Q2. Yes, you can regard tap water to be of neutral pH, unless the question specifies otherwise. For instance, the qn may ask why rain water (even in non-polluted areas) is slightly acidic (you know why?). On pg 187, George Chong is explaining why the Ka of water is 1.8 x 10^-16, why the pKa of water is 15.74, and why the Kw of water is 14.0 ; which has nothing to do with your own question. Does this mean you don't understand George Chong's explanation and the entire page?

      Q3. The objective of the experiment itself is to use different solvents to separate out the different solutes (ie. via their different solute-solvent affinities). So how can you carry out the experiment (which is about using solvents) if you don't use the solvents? Do you even understand what the experiment is about?

      Q4. This qn is from Singapore-Cambridge A levels 2012 P3 Q5. The intramolecular hydrogen bonding in the uninegative conjugate base for the cis isomer, results in the Ka1 for the cis isomer to be larger than the Ka1 for the trans isomer (since the more stable the conjugate base, the stronger the acid), and also results in the Ka2 for the cis isomer to be smaller than the Ka2 for the trans isomer (since for the cis isomer, to dissociate the 2nd proton, you need to endothermically break or dissociate the intramolecular H bond *in addition* to the O-H covalent bond).

      Q5. Yes it is. Students aiming for A grade should also be familiar with the electron flow mechanism for both the forward (ie. esterification & nucleophilic acyl substitution & condensation & addition-elimination reaction) and backward (acidic and alkaline hydrolyses) reactions. And if you meant "ether" instead of "ester", Cambridge can also ask you to draw the mechanism to generate the ether, as a challenging A grade exam qn.

      Q6. Compared to H2SO4, H3PO4 is a weaker oxidizing agent for 2 reasons : S is more electronegative than P, and the OS of the heteroatom is +6 (in H2SO4) vs +5 (in H3PO4). Using the Data Booklet redox potentials (topic : electrochemistry), you can deduce the relative oxidizing and reducing strengths. For species not included in the Data Booklet, you'll have to apply your knowledge of H2 Chem (across topics) to make reasonable deductions.

      Q7. It does, of course. Protonated ethanol CH3CH2OH2+ is an electrophilic intermediate (remember that an atom with a positive formal charge is more strongly electron-withdrawing by induction compared to if it had no formal charge), attacked by a 2nd ethanol molecule, the nucleophile. As this is a primary alcohol, SN2 occurs instead of SN1 (George Chong missed out the curved arrow depicting the elimination of the H2O+ leaving group, draw it into your book yourself).

      Edited by UltimaOnline 13 Oct `17, 12:13AM
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