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

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    • Memo asked :

      To a 1dm3 saturated solution of magnesium carbonate, 50.0 cm3 of 1.00moldm-3 aqueous magnesium chloride was added. Find the mass of the precipitate formed. Given : Ksp of Magnesium carbonate is 1.00 x 10^-5 mol2dm-6

      Yes, Tetrahedron has posted the correct solution (pun intended)

      Procedure :

      From the Ksp of MgCO3, find molarity of Mg2+ ions in the 1dm3 saturated solution (before MgCl2(aq) is added), hence find moles of Mg2+ and CO3 2- ions present then.

      Find the moles of Mg2+ ions present in the 50.0 cm3 of 1.00moldm-3 aqueous magnesium chloride.

      Based on the new volume of the solution after the MgCl2(aq) is added, find the new number of moles and hence new formal (ie. initial) molarities of both the Mg2+ and CO3 2- ions present (in the new solution), before precipitation occurs.

      Using algebra, let the moles of MgCO3 precipitated out be x. Hence the new equilibrium molarities of both Mg2+ and CO3 2- ions can be found (both terms will contain x, since the new initial molarities of both ions are known, the change will be minus x for both due to a 1 : 1 stiochiometry)

      Using the Ksp formula and value of MgCO3, and plugging in the new equilibrium molarities of both the Mg2+ and CO3 2- ions, solve for x.

      From moles of MgCO3, multiply by molar mass of MgCO3 to find the mass of the precipitate.

      Edited by UltimaOnline 06 Nov `12, 12:59AM
    • Q1)
      When ammonia is converted into nitric acid on a commercial scale, the following reactions can occur.
      In which reaction does the greatest change in oxidation number of the nitrogen occur?

      A 4NH3 + 5O2 → 4NO + 6H2O
      B 3NO2 + H2O → 2HNO3 + NO
      C 2NO + O2 → 2NO2
      D 4NH3 + 6NO → 5N2 + 6H2O

      Answer is A.

      Why is ethanoic acid a stronger acid in liquid ammonia than in aqueous solution?

      A Ammonia is a stronger base than water.
      B Ammonium ethanoate is completely ionised in aqueous solution.
      C Ammonium ethanoate is strongly acidic in aqueous solution.
      D Liquid ammonia is a more polar solvent than water.

      Answer is A.


      Comments on Q1. From -3 to +2, that's an increase of FIVE huge OS units.

      Comments on Q2. The strength of an acid is measured by how much % of it dissociates protons. In water, only a small % of the carboxylic acid molecules present will release / dissociate its protons. In ammonia, almost 100% of the carboxylic acid molecules present will release / dissociate its protons. Simply because ammonia is a hornier (ie. stronger) base than water, and will want to "shoot out its balls (ie. lone pair)" to grab the chiobu proton (remember, girls are positive, guys are negative), moreso than water... not because the balls (lone pair) of O aren't every bit as horny as the balls of N, mind you, but simply because they already kena hen-pecked by their own internal wife, ie. the protons in their own nucleus. Comparing O and N, since the shielding effect from inner shell electrons are the same, but because O has more protons than N, hence the lone pair (balls) are more tightly electrostatically held (groped) by the positively charged nucleus (the wife), and so are less available to flirt (shoot out balls) to accept a proton chiobu. Hence O atoms are less nucleophilic and basic, compared to N atoms.

      Edited by UltimaOnline 10 Nov `12, 12:06AM
    • KickMe asked :

      When the exam question asks to "Suggest the identity...." or "Suggest a reason.." or any other "suggest" questions, do we need to provide an answer that is backed up? For example, question 2 of 2011 Paper 3, 'A' levels.

      Also, is it correct to say that the CH3COOH group in HOOCCH2-COO- is electron withdrawing?

      To calculate pH at equivalence point of a weak acid that has 2 buffer regions, is it always correct to use the average of the pH at the two buffer regions?


      Yes, you should always elaborate to backup your "suggested" identity/reason, etc.

      If you can, always specify by induction or resonance (or hyperconjugation, but that's not required at H2 levels) :
      electron-donating by induction
      electron-withdrawing by induction
      electron-donating by resonance
      electron-withdrawing by resonance

      For instance, the aliphatic OH group (eg. in a 2-hydroxycarboxylic acid) is only electron-withdrawing by induction, and does not donate by resonance. But the aromatic or phenolic OH group is much more strongly electron-donating by resonance, compared to it's electron-withdrawing by induction effects.

      In the 2011 exam qn, although you could say that the other COOH group is electron-withdrawing (by induction only, not by resonance), this is a weaker explanation for why the Ka1 value is so large. A better answer, would be to draw the diagram to show how the conjugate base (uninegative) anion is effectively stabilized by intramolecular hydrogen bonding : this hydrogen bond is particularly strong, because the (boyfriend) O atom has a negative formal charge (and is thus more electron-rich) compared to a typical partial negative charge. In fact, the boyfriend O atom is even richer (ie. electron-rich) compared to the (covalently bonded) husband O atom, which makes this adulterous affair (hydrogen bond) all the more stronger and more stable. The more stable the conjugate base, the stronger the acid.

      If a solution contains only an amphiprotic species (ie. a species which can accept a proton or donate a proton, eg. HCO3-, H2PO4-, HPO4 2-, etc), then yes, the formula for pH of this solution is pH = 1/2 (pKa1 + pKa2).

      But the species must be reasonably amphiprotic, and not just in theory. For instance, in hydrazine, the first Kb value is fairly large (similar to ammonia, which is well known as a fairly effective base), but the 2nd Kb value is sooo small (approx 1 x 10^-16, which is even smaller than the [H+] or [OH-] from the auto-dissociaton/ionization of water!), that hydrazine should be more correctly regarded as a monoprotic, rather than diprotic base.

      In which case, treating the salt N2H5+ as only acidic rather than amphiprotic, and using the Ka formula to calculate pH, would give you a more accurate pH value, compared to using the amphiprotic formula.

      Why is hydrazine effectively only monoprotic, when there are two N atoms with their balls (lone pair) available? Two reasons.

      1 : Girls only like Guys, Girls don't like Girls. They're not Lesbo.
      Once protonated, we get the conjugate acid N2H5+, which is (obviously) a Girl. A proton H+ is also (obviously) a Girl. There would be intercationic or internuclei repulsion (ie. catfight), and thus the activation energy barrier Ea, required to cruelly force the N2H5+ Girl and the H+ Girl onto each other, would be rather high. Write this in the exam as, "Due to inter-cationic repulsion between N2H5+ and H+, the Ea for N2H5+ to accept a proton is high, and thus the Kb2 value is lower than expected."

      2 : The protonated N atom with the positive formal charge, is now strongly electron-withdrawing by induction (remember : an atom with a positive formal charge is much more strongly electron-withdrawing by induction, compared to if it had no formal charge; the formal charge factor even outweighs or is more important than the electronegativity factor), sucking away (by induction) the electron density of the remaning N atom's balls (lone pair), making the N atom fail to perform as a man (ie. Bronsted-Lowry base). Write this in the exam as, "Due to the electron-withdrawing by induction effect of the adjacent positive formal charged (protonated) N atom, the lone pair on the (unprotonated) N atom is now less available to accept a proton, and thus the Kb2 value of N2H4 (or the Kb value of N2H5+) is lower than expected."

      Edited by UltimaOnline 10 Nov `12, 11:49AM
    • KickMe asked :

      Regarding the more accurate way of calculating pH at equivalence point, for example, the first equivalence point of HOOCCH2COOH,

      I have 2 options:

      1) -OOCCH2COOH + H2O --> -OOCCH2COO- + H3O+
      2) -OOCCH2COOH + H2O --> HOOCCH2COOH + OH-

      How then do I know which one is the correct one?

      Here (and in point of fact, for most cases), this (ie. considering only either the Ka or the Kb alone, instead of considering both simultaneously vis-a-vis the amphiprotic formula) would be the less accurate way of determining pH.

      As long as the Ka (proton dissociation) and Kb (base hydrolysis) of the amphiprotic species are comparably large (and in a majority of cases they will indeed be), then the amphiprotic formula will be more accurate : pH = 1/2 (pKa1 + pKa2)

      But in rare cases, such as in the conjugate acid of hydrazine, ie. N2H5+, where the Ka value of this (theoretically) amphiprotic species is much larger than the Kb value, or vice-versa, then you should (obviously) use whichever value is larger.

      Consider the diprotic acid H2A.

      If Ka of HA- (ie. Ka2 of H2A) is >>> than Kb of HA- (ie. Kb2 of A2-), then regard HA- as only acidic and use the Ka of HA- (ie. Ka2 of H2A) to determine pH.

      If Kb of HA- (ie. Kb2 of A2-) is >>> than Ka of HA- (ie. Ka2 of H2A), then regard HA- as only basic and use the Kb of HA- (ie. Kb2 of A2-) to determine pH.

      But as long as the Ka and Kb of that species (eg. HA-) are comparably large (ie. less than 1 million times magnitude difference from each other), then this species is indeed (for all practical considerations) truly an amphiprotic species, in which case the amphiprotic formula pH = 1/2 (pKa1 + pKa2) would obtain a more accurate pH answer, because it takes into consideration the fact that this amphiprotic species would actually undergo hydrolysis both ways : as a Bronsted-Lowry acid and also as a Bronsted-Lowry base, simultaneously.


      KickMe remains terribly confused at this point, and no, at this point he still doesn't see it. At least not yet.

      Now I see! We consider the value of the Ka and Kb to determine if it is effectively an acid or a base. And in the case of Ka/Kb<10^7, and Kb/Ka<10^7, we use pH=1/2(pKa1+pKa2)

      However, what if it is the 2nd equivalence point of a diprotic acid that I am talking about? I will only have the value of one pKa to use. Then I will not be able to successfully determine the pH at the 2nd equivalence point using 1/2(pK1 + pK2)

      For example, given HOOCCH2COOH,


      For the first equivalence point, since Ka is more than one million times larger than Kb, we treat it as an acid, hence use -OOCCH2COOH --> -OOCCH2COO- + H+

      For the second equivalence point, Ka/Kb=1000 hence use 1/2(pK1 + pK2) But theres not pK2 in this case.

      And when I attempt to use the approach of "treat it as acid or base", since Ka is larger than Kb, I treat it as an acid, which in this case is wrong as I am suppose to treat it as a base.

      Once the 1st equivalence point is reached, you have HOOCCH2COO- (which is the only species present), which is clearly an amphiprotic species, hence pH = 1/2 (pKa1 + pKa2).

      (The only time you only need to compare Ka of a species, with Kb of that same species (ie. not Ka of an acid with Kb of that acid's conjugate base), is if the amphiprotic formula does not appear to be mathematically viable, ie. if the species' acidic capacity far outweighs that of its basic capacity, or vice-versa.)

      Once the 2nd equivalence point is reached, you have the dinegative -OOCCH2COO- (which is the only species present), which is clearly only basic (and not acidic at all, since there are no protons available to dissociate, obviously) and therefore you have no choice but to use the Kb (and not Ka, which applies only to acidic species that have protons available to dissociate) of -OOCCH2COO-, which could be labeled as Kb1, to calculate the [OH-], hence pOH, hence pH.

      When we define the various Ka and Kb values such that Ka1 > Ka2, and Kb1 > Kb2, then :
      Ka1 x Kb2 = Kw and Ka2 x Kb1 = Kw

      For now, the KickMe saga finally comes to a conclusion :

      I understood the second equivalence point. The only doubt I have now is why HOOCCH2COO- is "clearly an amphiprotic species". By calculating the Ka and Kb, I found the ratio of Ka/Kb=50 million, shouldn't then I treat it as effectively an acid?

      pKa1 = 2.65 hence Ka1 = 2.2387 x 10^-3 hence Kb2 = 4.4669 x 10^-12
      pKa2=5.70 hence Ka2 = 1.9953 x 10^-6 hence Kb1 = 5.0118 x 10^-9

      For the amphiprotic species, Ka2 and Kb2 are relevant.
      Since Ka2 is only 446,669 times larger compared to Kb2, hence this is indeed an amphiprotic species, and the amphiprotic formula for pH applies with validity.

      Edited by UltimaOnline 10 Nov `12, 8:59PM
    • I thought that was the end of it, but apparently not yet...

      KickMe persisted doggedly :

      Why do you get Kb2 from pKa1 and Kb1 from pKa2?


      Because as long as we consistently define Ka1 > Ka2 > Ka3 etc, and Kb1 > Kb2 > Kb3 etc, then :

      Let H2A represent a weak molecular diprotic acid (eg. the conjugate acid of A2-)
      Ka1 is the 1st proton dissociation, H2A --> H+ + HA-
      Ka2 is the 2nd proton dissociation, HA- --> H+ + A2-


      Let A2- represent a diprotic base (eg. the dinegative conjugate base of H2A).
      Kb1 is the 1st base hydrolysis, A2- + H2O --> OH- + HA-
      Kb2 is the 2st base hydrolysis, HA- + H2O --> OH- + H2A

      By observing the relevant species involved, notice that :

      Ka1 x Kb2 = Kw, since both Ka1 and Kb2 deal with the same conjugate acid-base pair of H2A / HA-.
      Ka2 x Kb1 = Kw, since both Ka2 and Kb1 deal with the same conjugate acid-base pair of HA- / A2-.

      The amphiprotic species here is obviously HA-.

      If HA- were to function as a Bronsted-Lowry acid,
      then it is Ka2 that is relevant, since HA- --> H+ + A2-

      If HA- were to function as a Bronsted-Lowry base,
      then it is Kb2 that is relevant, since HA- + H2O --> OH- + H2A


      The above equations and formulae regarding the various proton dissociations or base hydrolyses, can always be used, regardless of the symbols used, eg. say if the acid was the conjugate acid (eg. H2B2+) of a weak molecular diprotic base (eg. B), and the amphiprotic species would then be HB+.

    • Drbiology asked :

      Do you think that alkyl and hydride shift is likely to be tested in H2 syllabus? There are far-reaching implications if they chose to do so. For instance we must consider the Friedel-craft acylation instead of alkylation due to instability of carbocation intermediate formed.

      Anyway, am I allowed to use reagents I learned in H3 for synthesis questions in paper 3? For instance PCC, DIBAL-H, etc. My school chem teacher told me not to do so, but am I allowed to?


      Regarding the application of reagents or knowledge beyond that which is usually taught within the H2 syllabus, in answering H2 A Level exams, I still advise the same exam-smart strategies :

      Whenever you see the possibility of equally appropriate alternative answers, go ahead and write them all. Of course, if the most appropriate answer (ie. what you suspect Cambridge is most likely looking for) can be found within the H2 syllabus (and this is most usually the case), then go ahead and write that H2 answer in first.

      Thereafter, if you feel it appropriate (eg. if you feel a better answer would be one that you've learnt outside the H2 syllabus), then go ahead and add that in as well.

      Despite what some JC teachers might warn you about, writing in alternative answers in the A Level exams is an exam smart thing to do, as long as you carefully qualify your answers, and as long as none of these answer contradict each other in any way (in which case the marker would think you're trying to cheat and of course you'll get zero marks for the question).

      ( Examples of this would be to write all 4 of the following non-contradictory answers, when asked to describe the type of reaction between an alcohol and a carboxylic acid or acyl halide, to generate an ester :
      esterification, condensation, nucleophilic acyl substitution, addition-elimination )

      Since the H2 Chem papers are increasingly testing beyond the basic H2 syllabus, this issue is all the more relevant and pertinent. Bottonlime : When faced with such questions, go ahead and write in both the H2 answer as well as the better H3 answer, with an appropriate statement so the Cambridger marker understands you, eg. "we could use this H2 method OR this H3 method to synthesize this product, but the H3 method has the following advantages...". Just be careful that your answers don't contradict each other in any way.

      Usually, if the desired answer by Cambridge, involves concepts beyond that which is typically taught in H2 syllabus, they will give the necessary info on the concept as part of the question, ie. a data-based question. This will likely be the case for concepts such as carbocation rearrangements via alkyl and hydride shifts.

      If so, all the more power to those of you who made the effort to study beyond the basic H2 syllabus. But it's still fair to those who didn't (ie. the typical within-syllabus H2 student doesn't lose out too severely to the H3 or Olympiad or BedokFunland JC student, because all the necessary data will likely be provided in the question, or the answer can be extrapolated from the basic syllabus with a bit of thinking, etc).

      Yes, carbocation rearrangement can be asked by Cambridge in a H2 exam, in the form of "Carbocation rearrangements can occur as follows... Clemmenson reduction involves the following... The acylium ion is resonance stabilized as follows... Suggest why Friedal-Crafts alkylation to generate propylbenzene using propyl chloride has a low yield, and suggest how this difficulty can be overcome" or something similar.

      My BedokFunland JC Challenge Qn titled "H2Chem_Clemmenson_Reduction" deals with this problem and gives the solution.


      Dr Biology replied :

      Very insightful, thank you Ultima! But I guess there is still the 'time' factor which could really prevent most able students from writing the most appropriate answers with extensive line of reasoning. This would, more often than not, limit the answers to the perceived accurate answer.

      No problem at all.

      And yes you're right, time constraints will always be a factor of being exam-smart, as well. In this regard, my advice to students is to always skip any and all questions that you suspect will be time consuming (eg. difficult equilibria calculation qns, or Deductive Elucidation qns, etc), and/or questions on topics that (for whatever reason) you're ill-prepared to handle (eg. particularly so for the Planning Qn of Paper 2 ; for instance in last week's paper, unless you happened to have prepared very well for this Kinetics clock reaction Planning topic, it would have been exam-smart of you to have skipped it to focus on the other questions more productively first).

      Edited by UltimaOnline 12 Nov `12, 12:18AM
    • Memo asked :

      Speaking of Deductive Elucidation qns which will be appearing in P3. Is it wise to skip that question which consists of 8/9/10m organic elucidation qn? and choose the other 4qn to do?


      Depends. Have a quick read and spend a minute or so on the Deductive Elucidation qn. If you feel confident you've a rough idea on the skeleton of the structure, and can flesh it out pretty quick, then go ahead and go right into the question. But if there are too many details that confuse you, or if after a couple of minutes into the question you're stuck, then skip it and proceed to the next question.

      Killer Deductive Elucidation qns do not just appear in Prelim papers. In the last few years, Cambridge has been setting increasingly sadistic questions in this regard as well, particularly on conditions that involve isomeric branching and chiral carbons.

      And as always, keep an open mind for cyclo-compounds and dienes that can be oxidatively cleaved to generate ethandial that's further oxidized to ethandioic acid that's further oxidized to carbonic(IV) acid, that exists in equilibrium with and hence can decompose into, carbon dioxide and water.
      Also beware that "hot, acidified, KMnO4(aq)" does more than just oxidation : it can also protonate any basic groups (eg. amines), and can also hydroyse groups such as esters, amides and nitriles. Similarly for "hot, alkaline...".

      Also note that "a gas that turned moist red litmus paper blue" may not necessarily be ammonia, but can be any amine (primary, secondary or tertiary), or diamine, or triamine, etc, with a low boiling point.

      For practice, here's a fairly difficult (not the most difficult possible, and in fact I modified this question to make it more sadistically enjoyable (for me, of course) for my BedokFunland JC students to pratice on), but Cambridge questions should be of comparable difficulty to the original version of this) Deductive Elucidation question :

      Mr Chong's website :

      Edited by UltimaOnline 12 Nov `12, 1:28AM
    • KickMe asked :

      Just realised that there isnt any instructions on the no. of s.f. to give in the A levels. How do we know how many s.f. we should give?


      For H2 Chem purposes :

      Rule #1
      Follow the question's data, particularly if the question instructs "Give your answer to an appropriate number of significant figures or decimal places". Whether the question's data gives to 1 significant figure or 10 decimal places, follow suit.

      Rule #2
      Rule #1 takes priority, but if Rule #1 is not relevant or specified, then all intermediate workings should be given to 5 significant figures, final answer should also be given to 5 significant figures first, then rounded off and presented to 3 significant figures.

      Some JC teachers instruct their students to work with 3 significant figures all the way, which is risky and not advisable.

      Edited by UltimaOnline 13 Nov `12, 12:10AM
    • aleoca333 asked :

      Whats the exact difference between resonance and aromaticity?

      2. In the imidazole ring, why is it that one of N can protonate while the other one cant?


      The Prelim paper answer says "Lone pair not available to H+ ions, as lone pair is involved is involved in  delocalisation of 4n+2 pi electrons around heterocylic ring, which results in stability due to ring displaying aromaticity". But i dont get the answer..

      How are 4n+2 pi electrons when there are only 2 double bonds? Isnt the other N involved in delocalization as well?[/QUOTE]

      For H2 Chem purposes, you needn't concern yourself with Huckel's rule and the definition of aromaticity. You only need to appreciate that aromaticity is a special or extreme case of resonance delocalization of electrons [B][U]completely[/U][/B] around a ring (as opposed to say, resonance in a acyclic species such as a carboxylate ion), which due to such extensive delocalization (completely around a ring) affords an unusual magnitude of stability.

      The difference in stability and thus thermodyanic energy, between the resonance (whether aromatic or not) model and the non-resonance model (eg. benzene versus cyclohexa-1,3,5-triene), which can be experimentally determined, eg. by comparing their enthalpies of hydrogenation or combustion, is known as the resonance stabilization energy.

      Your Prelim Paper question is unlikely to be asked by Cambridge in a H2 Chem paper, because not only is Huckel's rule or the definition or aromaticity not within the H2 syllabus, but specifically, imidazole is particularly tricky in that of the 2 N atoms present, only the amine's lone pair is part of the aromatic system (since it's participation is required to complete the delocalization completely around the ring), while the imine's lone pair is available to accept a proton without disrupting aromaticity. This is one way (for another explanation involving resonance but not involving aromaticity, see my BedokFunland JC question H2Chem_Acid-Base_Equilibria_of_Histidine) to understand why in the imidazole ring, the imine is more basic compared to the amine (which is unusual because aliphatic amines are usually more basic compared to aliphatic imines and nitriles, due to the lone pair of imines occupying an sp2 hybridized orbital versus an sp3 hybridized orbital for the amine, and an sp hybridized orbital for the nitrile (Cambridge may ask "Suggest why nitriles are not basic, unlike amines"); since s orbitals are located closer to the nucleus compared to p orbitals, hence the greater the % s orbital character, the more tightly held the electrons will be by the positively charged nucleus, hence the less available the lone pair will be to accept a proton).

      3. For CH3COOH, is the bond angle between the two Cs 109.5 or 120 degree? Is the bond stronger or weaker than average C-C bonds?

      The answer is :

      109.5 deg for the sp3 hybridized C atom
      120 deg for the sp2 hybridized C atom
      (The exam-smart candidate will then label both atoms by drawing out the full displayed structural formula, and also label the bond angles on the diagram of the structural formula.)

      If Cambridge asks you if this C-C bond is stronger or weaker, they will specifically point out another C-C bond in the same molecule or another molecule (but with diagram given). The key to answering such questions is % s orbital character of the sigma bond.

      Hence, a sp3-sp3 sigma bond has a lower % s character and is thus longer and weaker, compared to a sp2-sp3 sigma bond, which in turn has a lower % s character and is thus longer and weaker, compared to a sp2-sp2 sigma bond, and so on.

    • Why ethanolic AgNO3 is able to react with alkyl halides (ie. the test to differentiate between an alkyl chloride vs bromide vs iodide).

      Due to the very low Ksp of silver halides, as well as the the Lewis basic capacity of the (electron-rich, partial negatively charged) halogens, the halogens are able to coordinate (shoot out a dative bond) to the Ag+ ion, and subsequently cleave away from the alkyl group.

      As part of either the transition state and/or intermediate (depending on whether the alkyl halide was primary, secondary or tertiary), two alternative boyfriends would volunteer to emotionally stabilize the transition state / intermediate carbocation girl (who upon loss of the halogen ex-boyfriend, now she lacks a stable octet and has an emotionally vacant orbital that needs a guy to fill her up), and these two horny / helpful guy nucleophiles are : the NO3- ion and the ethanol.

      NO3- has the advantage of being electron-richer (which girl doesn't love a rich boyfriend), but ethanol has the gangsterism advantage of larger numbers (by definition of being the solvent).

      So while the inorganic product is the silver halide ppt (of which the time taken to form this ppt, will reveal to us about the required activation energy which in turn will help us to deduce, together with the colour of the ppt, which test substance was the alkyl chloride vs bromide vs iodide), the organic product is a mixture of the love-childs of the two different couples : the alkyl nitrate and the ether (the immediate product is the protonated ether, which readily loses its proton to the ethanol solvent and/or the nitrate ion). The position of equilibrium eventually shifts to the side of the ether, since NO3- is a much better leaving group compared to RO-, leaving the major organic product as the ether.

      Added : One more important point (that Cambridge can test you on) regarding this precipitation reaction :

      Since ionic lattice formation enthalpy is strongly exothermic, the formation of the silver halide ppt, provides the required activation energy to cleave the C-X covalent bond, thermodynamically driving this reaction to completion (ie. a highly negative and feasible Gibbs free energy, translating into a product favoured high Kc value).

      Edited by UltimaOnline 14 Nov `12, 12:48PM
    • KickMe asked :

      The answer is: S in S2Cl2 has energetically accessible, vacant 3d orbitals to accept lone pair of electrons from oxygen in water and undergo hydrolysis. But... why not Cl? Cl also has 3d orbitals. Similar situation goes for CCl4, although C does not have 3d orbitals, Cl has!


      The S atoms are partial positively charged electrophile girls (due to Cl being a lot more electronegative than S), to whom the horny (partial negatively charged O atoms in) water nucleophile guys would want to shoot out their balls (lone pairs) to make love (chemically react) with. Hence hydrolysis occurs.

      Similarly for why CCl4 undergoes hydrolysis, but very slowly (much higher Ea required). For a look at why SiCl4 undergoes hydrolysis much more readily (much lower Ea) compared to CCl4, see Rod Beavon's :


      Edited by UltimaOnline 14 Nov `12, 12:58PM
    • 2012 H2 Chem Paper 3 Q5.

      Answers :

      The uninegative (mono-anion) conjugate base of cis-butenedioic acid (maleic acid) is indeed stabilized by intramolecular hydrogen bonding, while no such stabilization for trans-butenedioic acid (fumaric acid) exists.


      c) L is the cyclic carboxylic acid anhydride form of the alkenedioic acid =CH-(C=O)-O-(C=O)-CH=, where the two ends are joined in a ring. L is attacked by NH3 nucleophile at one of the two electrophilic acyl sites to generate M : H2N-(C=O)-CH=CH-COOH

      d) N is the keto-dicarboxylic acid. Because the ketone group is on a beta C with respect to one of the COOH groups, decarboxylation of the beta-keto-dicarboxylic acid can occur. Hence Q is CO2. Upon decarboxylation (ie. removal of CO2), the enolic-carboxylic acid tautomerizes to a keto-carboxylic acid. Hence P is H3C-(C=O)-COOH



      Edited by UltimaOnline 15 Nov `12, 11:47AM
    • Qn 3's mechanisms on N2O5 :


      Qn 5's mechanisms on Malic acid :

    • Question :

      3-methylpentane can undergo reaction with chlorine to form monosubstituted compounds that are optically active. How many possible isomers, including stereoisomers, can be formed in the reaction?


      Answer :

      8 mono-halogenated isomers may be generated, which are :


    • MCQ 2007 "A" levels
      A solid compound Z dissolved readily in water to give a weakly alkaline solution. On evaporation of the water, Z was recovered unchanged.

      What could Z be?



      Solution :

      This is a bit of an unfair question by Cambridge, because of the subjective "weakly" adjective. Both B and C are arguably correct, but when faced with two possible answers, you gotta choose the better answer.

      If option B was simply ethanoic acid, the question would be more problematic. But here, due to the strong electron-withdrawing by induction effect of the 3 Cl atoms, the strength of trichloroethanoic acid is increased by a tremendous magnitude.

      Compare the pKa values :
      http://en.wikipedia.org/wiki/Phenol vs http://en.wikipedia.org/wiki/Trichloroacetic_acid

      When the pKa value of an acid is very low, the acid is very strong, which means that the conjugate base does not undergo hydrolysis to any significant extent.

      If you've understood everything I said so far, the best answer to this MCQ should now be obvious to you.

    •  A 100cm^3 solution contains 0.2 mol/dm^3 MgCl2 and 0.1 mol/dm^3 CuCl2. A solution of sodium hydroxide is added to the mixture. Mg(OH)2 starts precipitating when 40 cm^3 of sodium hydroxide has been added.

      Ksp of Mg(OH)2 is 6.3 x 10^-10, Ksp of Cu(OH)2 is 2.2 x 10^-20

      What is the concentration, in mold/dm^3, of Cu^2+ in the solution when Mg(OH)2 just precipitates?

      Solution :

      This question has several red herrings.

      From Ksp of Mg(OH)2, and new molarity of Mg2+, find molarity of OH- at that point. Plug this molarity into the Ksp of Cu(OH)2, find molarity of Cu2+.

      Answer is 4.99 x 10^-12 mol/dm^3

    • 1. Which of the following statements contains one mole of the stated particle?
      A Molecules in 19.0 g of fluorine gas.
      B Electrons in 24.0 dm3 of hydrogen gas at room temperature and pressure.
      C Neutrons in 1.00 g of helium gas.
      D Protons in 2.02 g of neon gas.

      2. An experiment is conducted to investigate the kinetics of reaction between bromopropane and 0.1 mol dm-3 sodium hydroxide.

      The rate equation is as follows:

      Rate = k [bromopropane] [OH-]

      The half-life of bromopropane in one of the experiments is t minutes.

      What is the new half-life (in minutes) of bromopropane when the concentration of bromopropane is doubled and concentration of sodium hydroxide is reduced to 0.01 mol dm-3?

      A 0.05t
      B 0.1t
      C 5t
      D 10t

      3.Consider the following equilibrium system:

      H2 (g) + I2 (g) 2HI (g) H = +53 kJ mol-1

      Which of the following change is incorrect?

      A Numerical value of Kp is not equal to Kc at 25 C.

      B Increasing the mass of H2 will not cause the equilibrium constant to increase.

      C Increasing temperature increases the rate constant and equilibrium constant.

      D Rate of forward reaction is equal to rate of backward reaction when equilibrium is reached.

      4. The solubility product of iron(II) carbonate is 2.1  10–11 while that of silver carbonate is 8.1  10–12 at 25°C.

      Which of the following statements is true?

      A Addition of silver nitrate increases the solubility of silver carbonate.

      B Addition of sulfuric acid to a solution containing iron(II) carbonate increases the solubility product of iron(II) carbonate.

      C Iron(II) carbonate precipitates first when sodium carbonate is added to a solution containing equal concentrations of iron(II) and silver ions.

      D The solubility of iron(II) carbonate is higher than the solubility of silver carbonate.

      5. Liquid E has an H of vapourisation of 10.0 kJ mol1 and a boiling point of
      266 K.

      What is the S of condensation of vapour E?

      A –26.6 J mol1 K1

      B –37.6 J mol1 K1

      C +26.6 J mol1 K1

      D +37.6 J mol1 K1

      6. 2-methylpropane can react with bromine in the presence of sunlight to give two monosubstituted halogenoalkanes, 1-bromo-2-methylpropane and 2-bromo-2-methylpropane.
      Given the relative rates of abstracting H atoms are:
      Type of H atom primary secondary tertiary
      Relative rate of abstraction 1 4 6

      What is the expected ratio of 1-bromo-2-methylpropane to 2-bromo-2-methylpropane formed?

      A 9 : 1 C 6 : 1

      B 3 : 2 D 1 : 1

      7. In which sequences are the molecules quoted in order of decreasing boiling points?

      1 CH3(CH2)3CH3, (CH3)2CHCH2CH3, CH3C(CH3)2CH3

      2 AlBr3, AlCl3, AlF3

      3 SO2, SiO2, CO2

      8. A cell consisting of a V2+ (aq), V3+ (aq) | Pt (s) half-cell and a Au3+ (aq) | Au (s) half-cell is shown below using conventional notation.
      Pt (s) | V2+ (aq), V3+ (aq) || Au3+ (aq) | Au (s) Ecell = +1.76 V
      Which of the following statements is true?
      1 The mass of the Au electrode increase.
      2 The negative electrode is the Pt electrode.

      3 The standard electrode potential for Au3+ (aq) | Au (s) is +2.02 V.

      9. Oxytetracycline is a class of broad-spectrum antibiotics used to treat a variety of infections.

      Which of the following statements about oxytetracycline is correct?
      1 One mole of oxytetracycline reacts with three moles of thionyl chloride.
      2 One mole of oxytetracycline reacts with two moles of hot sodium hydroxide to liberate one mole of ammonia gas.

      3 One mole of oxytetracycline reacts with six moles of ethanoyl chloride.

      10. The mass percentage of magnesium in a mixture of magnesium chloride and
      magnesium nitrate was found to be 21.25%. What mass of magnesium chloride
      is present in 100 g of the mixture?

      A 47 g
      B 51 g
      C 53 g
      D 56 g

      11. Methane was burned in an incorrectly adjusted burner. The methane was converted into a
      mixture of carbon dioxide and carbon monoxide in the ratio of 98:2, together with water
      What will be the volume of oxygen consumed when y dm3
      of methane is burned?

      A 1.99y dm3
      C 0.995y dm3
      B 1.995y dm3
      D 0.99y dm3

      12. To identify an oxide of nitrogen, 0.10 mol of the oxide was mixed with 10 dm3
      of hydrogen gas and passed over a heated catalyst. At the end of the reaction, 0.4 dm3
      of hydrogen gas remained. The ammonia produced required 125 cm3 of 1.6 mol dm-3
      HCl for neutralisation. All gasoues volumes were measured at room temperature and pressure.

      What is the formula of the oxide of nitrogen?

      A NO C NO2
      B N2O D N2O4

      13. Which atom has the highest ratio of unpaired electrons to paired electrons in its ground

      A boron C nitrogen

      B carbon D oxygen

      14. Use of the Data Booklet is relevant to this question.

      Based on bond energies listed in the Data Booklet, what are the possible products of the
      following reaction?

      •CH3 + CH3CH2Cl ?

      A CH4 + CH3CHCl C CH3CH2CH3 + Cl•
      B CH3CH2CH2• + HCl D CH3CH2CH2Cl + H•

      15. Which of the following quantities is equal to the Avogadro constant?

      1 The number of oxygen atoms in 49.9 g of allactite, Mn7(AsO4)2(OH)8, of molar mas
      798 g mol^-1

      2 The number of aqueous chloride ions in a solution containing 0.5 mol of the comple

      3 The number of ions in 168 g of Reinecke’s salt, NH4[Cr(NH3)2(SCN)4], of molar mas
      336 g mol^-1

      16. Which of the following is hydrogen bonded in the liquid state?

      1 CH3NH2

      2 CH3CHO

      3 CH2F2

      17. Which of the following mixture produce ND3 gas upon heating?
      [D = an isotope of hydrogen]

      1 CaO (s) and ND4Cl (s)

      2 CH3CN and NaOD in D2O

      3 CH3CONH2 and NaOD in D2O


      (Partial) Solutions :

      Q1. From sample mass of neon gas, find moles of neon gas. Then find moles of protons (1mol of Ne contains 10mol of protons).

      Q2. Changing the molarity of a reactant does not alter its half-life. But changing the molarity of other reactants, will.

      Q3. Kp and Kc have the same mathematical value, under standard conditions. Alternatively, this MCQ can be solved by elimination.

      Q4. From Ksp, find molar solubility. Fe(OH)2 will have a smaller molar solubility.
      Note : usually for such questions, you can find the actual value of [OH-] at the point when the less soluble compound precipitates out. Plugging this molarity into the Ksp for the other (more soluble) compound, you can obtain the required molarity of the cation, which will be found to be mathematically larger than the existing molarity at this point.

      Q5. At boiling point (or melting point, for that matter), the system is at equilibrium and delta G = 0.

      Q6. Combine the two factors mathematically (by multiplying them together)
      - number of H atoms subtitutable
      - stability of alkyl radical intermediate

      Q7. For the first case, branching decreases the surface area available for van der Waals attractions.
      For the second case, the greater the number of electrons present, and the greater the molecular size, hence the more polarizable the electron charge clouds, hence the greater the magnitude of dipoles and partial charges, hence the stronger the electrostatic attractions, therefore the stronger the van der Waals.

      Q8. From the cell notation, Au is the cathode, Pt is the anode. Cell potential = Reduction potential @ Cathode + Oxidation potential @ Anode.

      Q9. The amide N atom and the enolic OH groups do not react with SOCl2.
      1 mole of OH- is required for (base-promoted) hydrolysis of the terminal / primary amide, generating NH3. Another 1 mole of OH- is required to substitute away the -NR2 group, generating NR2H (where R = CH3).

      Q10. Use algebra (only 1 algebraic variable required, since it's out of 100%) to solve.

      Q11. Write two separate equations, one for complete combustion, one for incomplete combustion. Add up total O2 required.

      Q12. From data, you can find the ratio of N to O, and also the actual moles of the oxide. Hence solve for x and y (in NxOy).

      Q13. For N, each p orbital is only singly filled and hence unpaired.

      Q14. Weaker bonds are more readily broken than stronger bonds, and stronger bonds are more readily formed than weaker bonds. This is why you cannot iodinate an alkane via free radical substitution.

      Q15. Avogadro constant = 1 mole.

      Q16. CH3NH2 can both accept and donate H bonds, while CH3CHO and CH2F2 can only accept H bonds.

      Q17. It's all about the mechanism.

      1 - CaO (s) and ND4Cl (s). D+ proton transferred from ND4+ to O2-, generating ND3 and OD-.

      2 CH3CN and NaOD in D2O. OD- attacks C, pi bond shifts to N, N grabs D+ proton; repeat until you obtain a geminal triol (which dehydrates into a carboxylic acid) and ND2-, which grabs a D+ proton from the carboxylic acid to generate ND3.

      3 CH3CONH2 and NaOD in D2O. OD- attacks C, pi bond shifts up, reforms carbonyl group, forming RCOOD while NH2 is eliminated as NH2-, which grabs a D+ proton from RCOOD to generate RCOO- and NH2D.

    • KickMe asked :

      Why is it that C2H6 has a greater entropy standard molar entropy that C2H4? Why does the stronger id-id interactions in C2H6 not result in a smaller entropy than C2H4?

      Answer :

      Regarding the van der Waals attractions, interestingly ethane does indeed have a slightly higher boiling point, but ethene has a slightly higher melting point (because melting point has to do with symmetry / stackability as well).

      As long as the state (eg. gaseous versus gaseous) is the same, for entropy we do not look at the strength of the intermolecular attractions, but rather we consider the total number of permutations and combinations possible.

      The more atoms are present, the more ways to permutate and combine, in terms of atom rotations, bond rotations, bond stretchings, bond bendings, etc. Hence entropy is greater for the molecule with greater number of atoms.

    • Originally posted by hoay:

      NaCl is not hydrolyzed by water instaed water just hydrates it. AlCl3 is hydrolyzed by water producing acidic solution. CH3Cl in hydrolzsed by NaOH (aq) and by water to form alcohol... Hydrolysis is the reaction with water or NaOH(aq)?? Please calrify and define hydrolysis.

      Technically, hydrolysis refers to the chemical reaction with water (don't worry about the "lysis" part, as either water or the other reactant will inevitably be broken up or "lysed" when the reaction occurs).

      However, because water is a weaker nucleophile, therefore to save time and money, we use a stronger nucleophile OH- in nucleophilic substitution reactions involving water, as the richer the electron-rich species, the stronger its nucleophilic strength, the lower the Ea required, the faster the rate of reaction.

      Furthermore, the final intended product is the same. For instance :

      R-X reacts with water to generate R-OH2+, which loses a proton to X-, hence generating R-OH and H-X as the final products. Higher Ea, slower reaction.

      R-X reacts with Na+OH- to generate R-OH, with Na+ and X- as counter ions, generating R-OH and Na+X- as the final products. Lower Ea, faster reaction.

      On a related note, notice that physical & inorganic chemists are lazier compared to organic chemists in some ways (eg. physical/inorganic chemists' quick dative bond arrowhead versus organic chemists' full curved-arrow mechanism), but are stricter in other ways (eg. physical/inorganic chemists say "hydration refers to physical interaction with water, while hydrolysis refers to chemical reaction with water"; but organic chemists say, "since dehydration means removing water from alcohol to alkene, then let's just call adding water to alkene as hydration; no need to specify hydrolysis rather than hydration"). 

    • EmpireCity asked :

      AlCl3 reacts with AlH4 and (CH3)3N to give (CH3)3NAlH3.
      Which statement about (CH3)3NAlH3 is correct?
      A It contains hydrogen bonding
      B It is dimeric
      C The Al atom is electron deficient
      D The bonds around Al atom are tetrahedrally arranged.

      Why is B untrue when the product has dative bonding?

      Mechanism :
      Hydride ion (H-) transfers from AlH4- to AlCl3, generating AlH3 (a Lewis acid or 'electrophile'), which is then available for trimethylamine (CH3)3N (a Lewis base or 'nucleophile') to attack, forming the Lewis acid-base product of (CH3)3NAlH3 with AlCl3H- as a byproduct.

      This (arguably) isn't considered 'dimerization' because an H- ion is lost from the two 'monomers' when forming the 'dimer' product. (It's a bit of an ambiguous area because you can also argue that polymerization can be either addition polymerization or condensation polymerization). So it's still a rather lousy option to include in this weak question.

    • EmpireCity asked :

      How is an ionic bond formed between magnesium and nitrogen?

      Answer :

      Metals like Mg like to be oxidized (ie. they have positive oxidation potential), as they are electropositive.
      Non-metals like N like to be reduced (ie. they have positive reduction potential), as they are electronegative.

      Mg loses 2 electrons to form Mg2+
      3Mg loses 6 electrons to form 3Mg2+

      N gains 3 electrons to form N3-
      2N gains 6 electrons to form 2N3-

      The nitride anions are electron-rich Guys.
      The magnesium cations are electron-deficient Girls (cationic ---> catgirls).

      Guys and Girls are both sexually highly charged (ie. high charge density),
      and are naturally sexually (ie. electrostatically) attracted to each other.

      The ratio of 3 Girls to 2 Guys (while a recipe for disaster in human terms) here is actually balanced in terms of charges, because these trinegative N3- guys have a huge sexual appetite, being able to take on 3 normal girls at once, while the 3 Mg2+ girls have a slightly smaller (but still twice as horny as a normal girl, ie. unipositive cation) sexual appetite, being able to take on 2 normal guys at once. Hence the final ratio of 3 Mg2+ girls being together with 2 N3- guys, is just right, electrostatically speaking).

      Once the orgy begins, the ferocious sexual attraction (ie. ionic bond) is too strong and it's hard to separate them. But initially, the guys and girls must overcome their initial shyness, and there's where you need to provide the enticing spark of activation energy, ie. you need to provide a lot of heat energy before N2(g) and Mg(s) will be willing to react together to generate Mg3N2(s).


      EmpireCity said : "N normally forms 3 bonds instead of two."

      You're confusing ionic bonds with covalent bonds. The "3 bonds" thingie only applies to covalent bonds.

      EmpireCity said : "I thought the N atoms with two bonds will form covalent bond with Mg."

      The difference between electronegativities of Mg and N are too large to form covalent bonds, hence ionic bonds occur instead.

      Mg2+ has sufficiently high charge density to form covalent bonds with elements that are less electronegative, eg. carbon. Grignard reagents are considered to have covalent bonds rather than ionic bonds, ie. C--Mg--X (where -- represents a covalent bond, and X presents a halogen other than F).


      You're referring to the recent TYS question where 4 N atoms forms bonds with the central Mg atom.

      The key to solving that question is recognizing formal charges :

      Notice that 2 of the N atoms have no formal charges (ie. 3 bond 1 lone), while the other 2 N atoms have a unipositive formal charge (ie. 4 bond, 0 lone).

      Hence, you are expected to deduce that the unipositive formal charges on the N atoms, are the consequence of the N atoms donating dative bonds.

      The remaining 2 N atoms have normal (ie. non-dative) bonds with the Mg atom. Since (as I just said in my previous post) the electronegativity difference between N and Mg is rather large, hence the bond is considered ionic bonding rather than covalent bonding.


      To show all charges (on the TYS diagram),

      For ionic bonding between N and Mg, it's N- (ionic bond) Mg2+ (ionic bond) N-
      (the N atoms have a uninegative formal charge, because it has 2 lone pairs, 2 bond pairs)

      For dative bonding between N and Mg,

      it's N+ (dative bond) Mg (dative bond) N+     (if you show formal charges after dative bonds are formed)


      it's N: (dative bond) Mg2+ (dative bond) :N     (if you show formal charges before dative bonds are formed)


      Edited by UltimaOnline 26 Nov `12, 12:34PM
    • The rate of metabolic removal of paracetamol from the human body is a first order reaction with k = 0.26 / hour. How much time will have elapsed, between taking a paracetamol pill, until only 25% remains in the body?

      Solution :

      [Final] / [Initial] = ( 1 / 2 ) ^ n            where n = number of half lives
      1 / 4 = ( 1 / 2 ) ^ n
      n = 2

      Since half-life = ln2 / k = 2.666 hours

      Hence time elapsed = (number of half lives) x (duration of half life) = 2 x 2.666 = 5.332 hours

    • The mass percentage of magnesium in a mixture of magnesium chloride and
      magnesium nitrate was found to be 21.25%. What mass of magnesium chloride
      is present in 100 g of the mixture?

      Solution :

      % by mass of Mg in MgCl2 = 2.5498 x 10^-1
      % by mass of Mg in Mg(NO3)2 = 1.6386 x 10^-1

      Since % by mass of Mg in mixture = 21.25%, hence :
      21.25 / 100 = X (2.5498x10^-1) + (1-X) (1.6386x10^-1)
      X = 53.38 g

      Edited by UltimaOnline 26 Nov `12, 6:54PM
    • Rate = k [Br2] [NO]^2

      With 3 mol/dm3 of NO, time taken for the molarity of Br2 to decrease to 1/8 its original molarity, is 15 hours. If 6 mol/dm3 of NO is used, what's the time taken for the molarity of Br2 to decrease to 1/8 its original molarity?

      Solution :

      Since 15 hours elapsed to reduce molarity from 1k to 1/8k,

      [Final] / [Initial]   =   (1/2) ^ n
      1/8 = (1/2) ^ n
      n = 3

      Since 3 (old) half-lives = 15 hours, hence 1 (old) half-life = 5 hours.

      Since molarity of the second-order reactant is doubled, the rate of reaction increases by (change in molarity) ^ (order of reactant) = 2 ^ 2 = 4 times.

      Hence (new) half-life of the other reactant is the (old) half-life decreased by 4 times, ie. 5 / 4 = 1.25 hours.

      Since molarity of Br2 is to be reduced from 1k to 1/8k,

      [Final] / [Initial]   =   (1/2) ^ n
      1/8 = (1/2) ^ n
      n = 3

      Since (new) half-live = 1.25 hours, hence time elapsed to decrease molarity of Br2 from 1k to 1/8k = 3 x 1.25 = 3.75 hours.

      Edited by UltimaOnline 27 Nov `12, 12:03PM
    • 2012 H2 Chemistry Paper 1 (Answers)

      Q1. C
      Q2. C
      Q3. D
      Q4. B
      Q5. C
      Q6. A
      Q7. D
      Q8. C
      Q9. A
      Q10. D
      Q11. D
      Q12. C
      Q13. B
      Q14. A
      Q15. A
      Q16. D
      Q17. D
      Q18. C
      Q19. C
      Q20. C
      Q21. D
      Q22. B
      Q23. B
      Q24. D
      Q25. C
      Q26. A
      Q27. C
      Q28. B
      Q29. B
      Q30. A
      Q31. B
      Q32. D
      Q33. A
      Q34. B
      Q35. B
      Q36. B
      Q37. D
      Q38. A
      Q39. B
      Q40. B

      Edited by UltimaOnline 28 Nov `12, 2:59PM
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