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

last SG · 2010-07-26T23:40:35+00:00

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Institute / Employer / Corporate Sponsor : "Eh Mr Singaporean Scientist, you better faster give me results on the research I directed you to do, so I can faster get my copyright patent and sell my billion dollar product soon! Don't waste my money researching some useless frontier stuff for that Nobel prize nonsense hor! Or else I'm gonna withdraw your team's funding, or even fire your ass!"

Singaporean Scientist : "I can't afford to get fired, with the costs of living rising in Singapore so steeply year after year, and with all the foreign talent competition coming in year after year. I need to feed my family, that's getting my priorities right. Besides, even if I had tens of millions of dollars of my own money (which I don't) and years of my own time to spend on frontier research to advance mankind's scientific understanding worthy of a Nobel prize, even in the slim chance I succeed (after spending many millions of dollars and many years of my life), there's no way I can easily convert such knowledge (which will be made freely available to all mankind, unlike products and technologies which can be copyrighted) into cash, and if anything, other MNC corporations and pharmaceutical giants are the ones who will eventually greedily profit from my years of Nobel prize winning research, and I get nothing! What the hell! It just doesn't make sense for me to do this. No point. Not worth it. If there's anything being a Singaporean has taught me, it's to be realistic, pragmatic and down-to-earth, prioritizing security and stability over all else. We're certainly as intelligent as scientists from any other country, but we'll leave the Nobel prize winning work to them (scientists who in some countries and institutions are better funded to pursue frontier research, and whose wealthy employers & sponsors passionately believe in contributing to mankind over copyrighting profit-making patents). Not worth it for us."

UltimaOnline SG

Hi Gohby,

Q1. Yes you're right. But in addition, even before LCP has a chance to participate, the fact is the question didn't specify the molarity of the KCl solution. In theory, if significantly higher than 1M (ie. concentrated), Cl- would be oxidized to Cl2 at the anode immediately. If significantly lower than 1M (ie. dilute), then H2O would be oxidized instead. Data Booklet redox potentials are only for standard molarities, ie. all species 1M on both LHS and RHS. In practice (for almost all aqueous solutions, unless super dilute or super concentrated), you'll almost always get a mixture of both O2 and Cl2 generated at the anode.

Q2. You're right for option 3. For option 2, Cr3+ has a high cationic charge density, and thus the hexaaquated complex ion will, just like Al3+(aq), lose protons to the solution, making it acidic, thus protonating carbonate ion, generating carbonic acid, which exists in equilibrium with, and hence can decompose into CO2(g) and H2O(l) with thermodynamically favourable positive entropy change, and also as predicted by Le Chatelier's principle.

You're welcome :)

Hi UltimaOnline,

I have some questions on electrochem:

1: VJC 2012 Prelim P1/Q8

Remarks: Depending on the answer scheme provided by different parties, the answer is either B or C. I would like to confirm if the answer is C. Here are my thoughts: comparing Eox potentials, water is oxidised preferentially to Cl- at the anode. Hence, H+ ions move towards Z, thus turning the colour of the litmus red. However, I would think after a few minutes, when sufficient oxygen has been evolved, owing to LCP, the oxidation potential of water would have been lower than -1.36V, thus chloride ions in the KCl starts to oxidise instead, thus turning the litmus white. Am I right?

2. Which of the following compounds do not exist under aqueous conditions?

1 Cobalt(II) chloride

2 Chromium(III) carbonate

3 Iron(III) iodide

Answer: 2 and 3.

Remarks: For 3 is it because the Ecell for the redox reaction between Fe3+ and that of I- is positive (+0.23V)? For 2 chromium (III) cannot undergo a redox reaction with water as the Ecell is negative - so why can’t it exist? What about 1?

Thank you! :)

UltimaOnline SG

Hi UltimaOnline,

I have a couple of questions on inorganic chemistry:

Which of the following is true?

BF4- is able to act as a ligand.

Aqueous bromine and aqueous chlorine can be distinguished by adding iron (II) sulphate, followed by aqueous sodium hydroxide to separate volumes of aqueous bromine and aqueous chlorine.

When AgNO3(aq) is mixed with FeCl2(aq), a grey precipitate is formed, which does not dissolve in NH3(aq).

Answer: 1, 2 and 3

Remarks: For 1, how is BF4- ion able to act as a ligand despite it not having lone pair of electrons? For 2, what happens? I suppose there could be a redox reaction between iron (II) and bromide/chloride ions, but how is this a way to distinguish between the halogens? For 3, what is the chemistry involved in this reaction? Isn’t AgCl of cream colour, which dissolves in ammonia?

2. [DHS Prelim 2012]

S is a transition element. The 3d sub–shell of S in the compound K[S(C2O4)2(NH3)2] contains 3 electrons. How many unpaired electrons does S contain when it is in the elemental state?

A: 2 B: 3 C: 4 D: 5

Answer: C

Remarks: I thought the answer is 6: from the compound, I gather that the charge on S is 3+. Hence, during the ionisation, I would think that 1 4s electron and 2 3d electrons will be taken away. As electronic configuration of S is [Ar]3d54s1 prior to ionisation, wouldn’t there be 6 unpaired electrons in the elemental state of S?

3. http://img.photobucket.com/albums/v700/gohby/Chemistry/trichloro_zpsb0anyptf.jpg

Answer: B

Remarks: I’d like to confirm if my understanding is correct. Firstly, sodium chloride and aqueous bromine will be formed. When I add excess concentrated sodium hydroxide, bromine will disproportionate (assuming that the reaction is carried at temperatures above 15 degrees) to Br- and BrO3-. I am assuming BrO3- to be colourless. As trichloromethane is denser than water it will form a colourless layer below all the ions.

Thank you! :)

Hi Gohby,

Q1. BF4- cannot act as a ligand, but it can function as a source of a F- ligand. (Qn is phrased ambiguously, so qn's fault). If molarities are less than standard molarities, only Cl2, but not Br2, is strong enough an oxidizing agent to oxidize Fe2+ to Fe3+, which you can confirm by the colour of the Fe(OH)3(s). The grey ppt isn't AgCl (which is white, btw), but Ag(s).

Q2. Ignoring the dative bonds from the ligands, since the 3d subshell of the metal ion contains 3 electrons, and it's OS is +3, thus the metal ion's 4s orbital is vacant. Hence reducing the metal ion to it's elemental atomic form, we add 3 electrons. 2 of these electrons will first go into 4s (since we're working backwards ; when ionizing from atom to cation, always remove 4s electrons before adding to 3d), and the 3rd will go into the 3d subshell. This gives a total of 4 electrons (all unpaired) in the 3d subshell.

Q3. You're correct. But just to add, the actual mechanism for all 3 halogens (chlorine bromine iodine), is the halogen first disproportionates into the halide ion and the halonate(I) ion. The halonate(I) ion then further disproportionates into the halide ion and the halonate(V) ion. For chlorine, the disproportionation of chlorate(I) ion occurs only upon heating at 70 deg C. For bromine and iodine, this occurs readily at room temperature (they require much lower Ea for the disproportionation redox reaction). Bromate(I) ion can only be isolated at 0 deg C, while iodate(I) ion can never be isolated at any temperature. Fluorine behaves very differently with NaOH(aq) because it is the king of electronegativity. Although the reaction of fluorine with NaOH(aq) is not in the H2 syllabus (and thus H2 student's needn't know or memorize what happens), but Cambridge may still come up with "suggest suggest" type of challenging questions for this beyond-syllabus reaction.

Welcome :)

UltimaOnline SG

Hi UltimaOnline,

Many thanks once again for your cogent explanation. :)

For 1, why isn’t Br2 able to oxidise Fe2+ to Fe3+ since the Ecell is +0.30V?

For 2, I don’t understand why wouldn’t the electronic configuration of the element be [Ar]3d⁵4s¹ instead of [Ar]3d⁴4s² since a half-filled 3d subshell would afford extra stability. If it were [Ar]3d⁵4s¹ then there would be 6 unpaired electrons.

5: What is the trend of the pH of aqueous Group II chloride going down the group? Are they all neutral?

6: Which of the following gives the best description of the reactions of Group II metals and their compounds?

All Group II oxides undergo neutralisation with hot acids to give a salt and water.

Beryllium hydroxide is amphoteric due to the high charge density of the Be²+ ion.

Remarks: The answer is 1 (I agree), but what’s wrong with choice 2? Is it because it does not give the best description of the Group II reactions (since it only talks about beryllium as opposed to Group II hydroxides in general), or is it because choice 2 should be “Beryllium hydroxide is amphoteric because it reacts with acids and bases” → but that begets the question as to why does it even react with acids and bases in the first place.

7: X is a mixture of two compounds. When X is treated with an excess of dilute hydrochloric acid, a colour gas is evolved and some, but not all of the mixture dissolves. Which one of the following mixtures could be X?

A Ba(NO3)2 and Ca(OH)2

B Ag2SO4 and CaCO3

C CaCO3 and MgSO4

D Ca(OH)2 and MgCO3

Remarks: Since not all of the mixture dissolves and silver chloride is insoluble, B is the only option - but what is the colour gas that is evolved? They aren’t referring to carbon dioxide right?

8: ACJC Prelim 2012 P1 Q17

Remarks: I agree with the answer B, but why is D not a possible answer? (I reckon the position of elements down the group would be X -> Z -> Y)

Hi Gohby,

Q1. Under standard molarities, yes (and H2 students must memorize this) both Cl2 and Br2 can oxidize Fe2+ to Fe3+. But since the answer given stated this option was correct (of course, the student can't see the mark scheme, but based on option elimination : A is 1+2+3, B is 1+2, C is 2+3, D is 1 only), therefore it implies the molarities are less than 1M for the halogens (ie. a diluted solution such that only Cl2, but not Br2, is able to oxidize Fe2+ to Fe3+). Even under standard molarities, there may be an observable difference between the amount of Fe(OH)2 vs Fe(OH)3 ppts generated using Cl2 vs Br2. But you're right, this question is problematic.

Q2. Usually for such qns, you'll not be expected to state the electron configuration as 4s1 3d5 unless the element is specifically stated as Cr. Because in most other such MCQs, the element may not be either Cr or Cu (where a tweak is needed). However in this particular MCQ, your argument that the qn implies it is Cr, and therefore having electron configuration 4s1 3d5 (instead of 4s2 3d4) is fair, and as such Cambridge (in a P2 or P3 qn) would accept either answer. However, since this MCQ did not have an option "6 unpaired electrons", therefore the DHS teacher didn't want the student to tweak the electron configuration. But you're right, this question is problematic.

Q5. They tend towards neutral down the Group. Near the top of the group, they are acidic, due to high cationic charge densities, their hexaaquated complex ions have weakened (less polarized) O-H bonds and tend to dissociate protons, making the solution acidic. The counter ion Cl- has low anionic charge density, and so its effect on pH is negligible.

Q6. [Edited 25 Nov 2015] : For the oxide and the hydroxide of any metal to behave as a Bronsted-Lowry base, is obvious, as both the oxide and hydroxide anions have a propensity to accept protons to lower their anionic charge densities to stabilize themselves. Both the oxide and the hydroxide of Be are equally acidic, because the high cationic charge density of Be enables both the oxide and hydroxide to accept 2 *additional* moles of OH-(aq) (the oxide will first undergo hydrolysis to generate 2 OH-(aq), which are already present in the hydroxide ; subsequently for every mole of Be2+, 4 moles of OH-(aq) ligands complex with the Be2+ Lewis acid to generate the ionic tetrahydroxoberyllate(II) coordination complex), thus behaving as a (Lewis, not Bronsted-Lowry) acid.

The reason why I earlier suggested replacing the word "hydroxide" with "oxide" in your statement about amphotericity, is simply because the A Level H2 Chemistry syllabus focuses on the acid-base-amphoteric nature of the oxides, rather than hydroxides. So yes, you're right that both the oxide and the hydroxide of Be are equally amphoteric, but for A levels, it's the oxide that's tested rather than hydroxide, so if it were a P2 or P3 qn, I would still recommend you to write "oxide" rather than "hydroxide". Since we're left with P1, yes indeed, you (all students taking the A levels) should be aware that both the oxide and the hydroxide of Be are equally amphoteric.

Q7. Replace the typo word "color" with "colorless", and you're good to go.

Q8. As you say, D is only *possible*, it is not *confirmed*. And therefore, D is wrong. D implies "confirmed", not "possible".

UltimaOnline SG

How to choose whether Cu2+ will be reduced to Cu+ or Cu in electrochemistry questions?

i.e reaction between KI (aq) and Cu(SO4)2 (aq)?

The standard reduction potential of Cu2+ to Cu is +0.34V, which is more positive compared to the standard reduction potential of Cu2+ to Cu+, which is +0.15V. Hence, always reduce Cu2+ to Cu (especially for electrochemistry questions, since regardless of whether the standard cell potential using Cu2+ to Cu+ reduction half-equation is feasible or not, the standard cell potential using Cu2+ to Cu reduction half-equation will always be more positive and hence more thermodynamically feasible, ie. Gibbs free energy more negative), *unless* otherwise indicated / hinted by the question (eg. in inorganic deductive-elucidation questions), or when it's for Fehling's and Tollen's, in which H2 Chem students are required to memorize and be able to write out the balanced redox equation in which the aldehyde* is oxidized to carboxylate ion, and Cu2+ is reduced to Cu+ in brick-red / brownish-red ppt of Cu2O(s)*.

Update : Turns out MrWorry was asking about the special case reaction to generate copper(I) iodide by mixing an aqueous solution of sodium or potassium iodide and a soluble copper(II) salt such copper sulfate.

Cu²⁺ + 2I⁻ → CuI₂

The CuI₂ immediately decomposes to iodine and insoluble copper(I) iodide, releasing I₂.

2 CuI₂ → 2 CuI + I₂

Under standard conditions, the cell potential (ie. reduction potential @ cathode + oxidation potential at anode) for this reaction is negative, ie. Gibbs free energy is positive and hence thermodynamically infeasible. So why does this reaction occur to generate copper(I) iodide?

As it turns out, as a consequence of the CuI(s) precipitation, [Cu+] decreases, shifting the position of equilibrium to the RHS as predicted by Le Chatelier's principle, making the reduction potential of Cu2+ to Cu+ more positive. Concordantly, the cell potential under these non-standard conditions (where CuI(s) precipitates out), becomes positive, and the reaction becomes thermodynamically feasible.

Further Update :

Hi Ultima, pertaining to the above concern, how do we reconcile the possibility of the precipitation of insoluble CuI also lowering [I-] and thus shifting the position of eqm of I2/I- reduction equation to the right hence making it more positive and in that sense still make the cell potential for the reaction negative?

Thanks so much!

Excellent point, Sugarfortress. This *appears* to be irreconcilable at A levels, given the stoichiometry and magnitudes of the relevant oxidation and reduction potentials. A easy workaround would be to suggest that excess I- be required (again, non-standard molarities). However, as it turns out, there's actually a more important reason that enables the redox reaction to proceed to generate CuI(s) even without excess I-. Which is, the reduction potential of Cu2+(aq) to Cu+(s) precipitate is actually far more positive than given in the Data Booklet, which is for aqueous species, ie. Cu2+(aq) to Cu+(aq). So much so that as CuI(s) precipitates out, the magnitude of the decrease in oxidation potential of I- to I2, is still far outweighed by the magnitude of increase in reduction potential of Cu2+(aq) to Cu+(aq) to Cu+(s). As such, admittedly my earlier explanation is simplified and incomplete, but would still suffice for A levels, should Cambridge ever ask this question.

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* The following are exceptions (not taught in Singapore JCs to H2 Chem students) for Fehling's and Tollen's :

Methanal is (the only aldehyde to be) able to reduce the Cu2+ in Fehling's solution to either Cu+ or Cu, and hence (depending on factors such as molarity) you may get either a red ppt of CuO(s), or a copper mirror of Cu(s), or both.

Methanoic acid is (the only carboxylic acid to be) able to react and give a positive result with both Fehling's and Tollen's. This is because the oxidation potential of methanoic acid to carbonic(IV) acid is large and positive, concordant to the observation that H2CO3(aq) exists in equilibrium with, and hence can decompose into, CO2(g) and H2O(l) with thermodynamically favorable positive entropy change, and also as predicted by Le Chatelier's principle since CO2(g) leaves the reaction mixture, pulling the position of equilibrium over to the RHS.

Ketones do not react with Fehling's and Tollen's. Alpha-(pri)hydroxy-ketones are (the only ketones to be) able to react and give a positive result with both Fehling's and Tollen's.

(Bonus challenge : Draw the curved-arrow electron-flow mechanism in which the alpha-(pri)hydroxy-ketone prototropically tautomerizes first to an ene-diol, then further prototropically tautomerizes to a alpha-(sec)hydroxy-aldehyde ; the redox reaction with Fehling's and Tollen's pulls the position of equilibrium to the RHS, as predicted by Le Chatelier's principle).

Benzaldehyde and its derivatives only give a positive result for Tollen's, but not Fehling's. There are 2 reasons for this :

Firstly, the more alkaline Fehling's solution results in benzaldehyde undergoing the Cannizzaro disproportionation reaction instead (which occurs for aldehydes unable to undergo the aldol reaction due to lack of an enolizable acidic alpha proton, such as in benzaldehye ; however in other aldehydes (ie. those with an enolizable acidic alpha proton), the redox reaction with the oxidizing (and tartarate and citrate ligand coordination complexed to prevent Cu(OH)2 precipitation in Fehling's and Benedict's respectively) Cu2+ ion takes thermodynamic (but not kinetic, as heating is required) priority over Aldol reactions, and as predicted by Le Chatelier's principle since the redox reaction is largely irreversible, while the Aldol reactions involve reversible steps. The Aldol reaction may still occur as a minor side-reaction, but does not affect the observable generation of the red Cu2O ppt).

Secondly, using my BedokFunland JC formula "Oxidation State = Formal Charge + Electronegativity consideration", notice that while the oxidation state (OS) of the aldehyde C atom in aliphatic aldehydes is +1, the OS of the benzaldehyde C atom in the original resonance contributor is +1, while in all other resonance contributors (with pi electrons within the benzene ring withdrawn out towards the electron-withdrawing carbonyl oxygen) is 0, and thus the OS of the benzaldehyde C atom in the resonance hybrid is much closer to 0 than +1. As such, it requires a greater magnitude of thermodynamic Gibbs free energy, and thus a stronger oxidizing agent (Tollen's is a stronger oxidizing agent compared to Fehling's, as the standard reduction potential of Ag+ to Ag is +0.80V, which is more positive compared to the standard reduction potential of Cu2+ to Cu, which is +0.34V, even as the redox potentials are tweaked in the presence of the ammonia and tartarate ligands in Tollen's and Fehling's respectively) to oxidize benzaldehyde (OS of carbonyl C atom = 0) to benzoic acid (OS of acyl C atom = +3).

UltimaOnline SG

TYS 2010/P1/33

why is option 3 correct? Like how will osmosis increase entropy?

Analogy :

2 rooms, linked by only 1 door (the partially permeable membrane).

Room A only got girls (solute).

Room B only got guys (solvent).

Both rooms (ie. the system) have low entropy.

But now, if some guys from room B with higher guy (solvent) potential crosses the door (partially permeable membrane) to room A with the lower guy (solvent) potential, there will be in increase in disorderliness in room A (coz now got mix of guys and girls), and thus there is a positive entropy change in both room A and the overall system.

UltimaOnline SG

TYS2014/P3/5

Why is H2+Ni @150degrees able to reduce the benzene ring but not the amine and carboxylic acid?

Is there any decarboxylation reaction in the syllabus we need to know? Like some sodalime reaction to decarboylate a carboxylic acid

H2+Ni @150degrees is able to reduce the benzene ring *as well as* the nitrile group, but not the carboxylic acid.

First of all, although hydrogenation/reduction of the benzene ring is not in the H2 syllabus, but Cambridge *did* give you the Mr of both Z and Omega, which will enable you to deduce the hydrogenation/reduction of the benzene ring.

Next, in the H2 syllabus you ought to have memorized that heating with H2 gas over Ni catalyst at 140 deg C can reduce nitrile to amine (CS Toh pg 244). And also in the H2 syllabus you ought to have memorized that under no circumstances can H2 gas reduce carboxylic acid to aldehyde or primary alcohol. LiAlH4 (in dry ether) is required for that.

Finally, both hydrogenation of the benzene ring to cyclohexane and hydrogenation of nitrile to amine is easier to achieve using H2 gas, compared to reduction of carboxylic acid using H2 gas (which cannot be done), because the former 2 are simpler addition reactions (ie. adding H2(g) to reactant to get the product), while the latter involves not just addition of H, but also requires elimination of an O atom, which requires chemical reagents (to carry away the O) in a solvated environment (ie. not just gaseous H2).

Decarboxylations aren't in the H2 syllabus, but Cambridge can (and have, in recent years) give you sufficient info in the question to deduce that decarboxylation has occurred.

Two notable decarboxylation reactions that are useful for A level students to be familiar with, are the use of soda lime, and heating beta-keto-carboxylic acids (Singapore-Cambridge H2 Chemistry A Level 2012 Paper 3 Qn 5d).

As a challenge for H2 Chem students, try drawing out the decarboxylation mechanisms for both reactions described above.

Additional decarboxylation reactions that Cambridge can use in data-based questions, are the Hunsdiecker reaction and the Kochi reaction.

UltimaOnline SG

TYS 2014/P3/4(d)(i)

Is it possible for the answer to be acid base reaction of the phenol in tyrosine and the compound in (i)? And also if there is acid base reaction, will an ionic bond be formed since there will be an anion and a cation?

TYS 2014/P3/4(b)(i)

Can use LiAlH4 to reduce followed by protonation?

TYS2010/P2/5

Is imine group considered and amino acid? also what is the name of the bond formed? Like can is be considered an amide or peptide bond?

TYS 2014/P3/4(d)(i)

No, "acid-base" reaction is not accepted, because the question is asking for an "interaction" not "reaction" and furthermore the question defines the "interactions" as that responsible for the tertiary structure of the protein.

Yes, the required answer is ionic interaction with the aspartate ion (deprotonated conjugate base form at pH 7).

TYS 2014/P3/4(b)(i)

Cannot, coz -CN would be reduced to -NH2.

TYS2010/P2/5

No, an imino carboxylic acid is an imino carboxylic acid, not an amino carboxylic acid. And no, imino carboxylic acids do not form peptide bonds or groups, as an imine group is significantly less nucleophilic compared to an amine group, due to the lone pair residing in a sp2 hybridized orbital, which has greater % s-orbital character, compared to the sp3 hybridized orbital of an amine group (assuming lone pair isn't delocalized by resonance).

Note however, that for this question, Cambridge is loosely (and obsoletely and arguably erroneously) including a secondary amine group in the definition of an "imine group". A secondary amine group, is after all, still an amine group (ie. lone pair residing in an sp3 hybridized orbital, assuming lone pair isn't delocalized by resonance). As such, it is sufficiently nucleophilic to generate the peptide group, and so, yes in this context (as per Cambridge's loose, obsolete and erroneous definition of an imino acid), the so-called imino acid does form the peptide group and peptide bond, in the protein.

In biochemistry contexts, real or actual alpha-imino acids (ie. properly defined as carboxylic acids with the C=N imine group on the alpha C atom, rather than Cambridge's secondary amine group trying to pass of as an imine group) involved in protein synthesis, are usually metabolic intermediates converted from one particular proteinogenic amino acid, catalyzed by a specific enzyme, and subsequently converted into another particular proteinogenic amino acid, catalyzed by another specific enzyme.

UltimaOnline SG

TYS 2013/P1/30

why is option D considered an amino acid when there is no proper amino can carbonyl group? (if I’m not wrong it is an imine group? reference to TYS2010/P2/5)

Also, since the question ask for hydrogen bond to be formed, then ALL the options would be able to form hydrogen bond because that is a characteristic of the secondary structure of protein, then is there something wrong with this question?

And to cause a “bend” in the alpha helix would mean that the R group is bulky am i right? then from option A B and C, only C would cause the bend or some form of distortion in the alpha helix right?

What is the difference between the Alpha helix and the Beta pleated sheet? like why would one occur instead of the other?

TYS 2013/P1/28

why is the formation of disulfide bridge considered an oxidation reaction? Like why is condensation wrong? (am i right to say that H2 is evolved in the formation of the disulfide bridge)

TYS 2013/P1/30

As explained previously, proline is more correctly considered an alpha-amino acid with a secondary amine group, rather than an alpha-imino acid (since it doesn't have a true C=N imine group).

You must be able to interpret the Cambridge phrasing correctly. The question is asking for you to verify the capacity for both ends (ie. the amine end and the carboxylic acid end) of each amino acid (in the 4 options) to each form a peptide group that are both capable of both accepting H bonds (ie. by the -C=O peptide group) and donating H bonds (ie. from the -N-H peptide group).

Thus, the answer is proline (option D), because upon nucleophilic acyl substitution and condensation to generate the peptide group, the secondary amine has lost it's only H atom (eliminated as H2O, enzyme catalyzed in biosynthesis), and as such the peptide group generated at this end, does not contain a H atom, and thus is incapable of donating a H bond.

The standard usual hydrogen bond for secondary structure of proteins, involve the -C=O group acting as hydrogen bond acceptor, and -N-H peptide group acting as hydrogen bond donor.

Consequently, the proline amino acid residue's peptide group's inability to donate a H bond, together with the steric hindrance effect caused by the exceptional conformational rigidity of proline (in turn due to the distinctive cyclic structure of proline's R group side chain), results in proline acting as a structural disruptor in the middle of secondary structures such as alpha helices and beta sheets.

As to your loaded question on why some sections of a polypeptide chain, preferentially form alpha helices while others form beta sheets, this goes beyond both H2 Chemistry and H2 Biology. Of course it has to do with the primary structure (of which the secondary, tertiary and quaternary structures build upon), or the sequence of amino acids in the section of the polypeptide chain, and the configuration of different R groups of the different amino acid resides exerting a compound effect on whether the alpha helix or beta sheet would be more thermodynamically favorable.

Certain amino acid residues favor forming the alpha helix (ie. high alpha-helical propensities), while other amino acid residues favor forming the beta sheet (ie. low alpha-helical propensities), depending on the amino acid's R group. Check out the following table on amino acid alpha-helical propensities :

https://en.wikipedia.org/wiki/Alpha_helix#Table_of_standard_amino_acid_alpha-helical_propensities

UltimaOnline SG

TYS 2013/P1/28

why is the formation of disulfide bridge considered an oxidation reaction? Like why is condensation wrong? (am i right to say that H2 is evolved in the formation of the disulfide bridge)

There are 2 reasons why this reaction is more properly labeled as a redox reaction, rather than a condensation reaction.

1stly, because sulfur is more electronegative than H, the OS of the S atoms become more negative when adding H atoms to break the disulfide bond or bridge into 2 separate thiol groups of cysteine amino acid, hence this is a reduction reaction. And when the disulfide bond or bridge is generated from the removal of H atoms from 2 separate thiol groups of cysteine amino acid, the OS of the S atoms become more positive, hence this is an oxidation reaction.

(Note : the actual numerical OS values of the S atoms before and after the disulfide bond forming / breaking redox reaction, depends on which electronegativity system you ascribe to : some electronegativity systems regard S to be more electronegative than C, while other electronegativity systems regard C to be more electronegative than S. The different electronegativity systems, named after the different chemists using different experimental criteria and data, are Pauling's, Mulliken's, Allred–Rochow's, Sanderson's, and Allen's.)

2ndly, a condensation reaction (in organic chem, not physical chem) refers to the actual elimination of 1 small molecule to join 2 molecules together. You do *not* eliminate an actual H2 molecule when forming the disulfide bond or bridge. Rather, the mechanism (whether enzyme-catalyzed in biosynthesis, or laboratory / industrial synthesis) involves eliminating 2 H+ ions and 2 electrons (instead of a H2 molecule), and hence this does not strictly qualify as a condensation reaction.

UltimaOnline SG

Delta G (Gibbs free energy) = 0

when there is equilibrium or state change like from H2O(l) <=> H2O(g),

but from H2O(s) to liquid the entropy will change so is the Gibbs free energy of that reaction not 0?

At 273K (ie. the melting / freezing point of water), the kinetic rate and thermodynamic feasibility of both the freezing and melting processes are the same, hence Gibbs free energy change for both the forward (eg. melting) and backward (eg. freezing) reactions are both zero.

It's true that the exact values for entropy change and (even) enthalpy change of a particular reaction depends on its temperature; but for A level purposes, in order to simplify calculations, you have no choice but to make the assumption (and Cambridge has indeed asked before in past A level papers, "State the assumptions in carrying out your calculations") that the variation in entropy and/or enthalpy across the temperature range (involved in the question) is negligible, when plugging in the delta S and delta H values into your delta G formula.

UltimaOnline SG

[Malaysia] - Man falls into factory’s sulfuric acid tank!

https://sg.news.yahoo.com/man-falls-factory-sulphuric-acid-073452179.html

YouTube videos of a simulated acid burn wound :
https://www.youtube.com/watch?v=vq0wPfQyEzw
https://www.youtube.com/watch?v=F47mjNVi6K4

UltimaOnline SG

General Conceptual Questions:

(1) Is there resonance in CH3CO3- anion? (I'm guessing no, but I don't know why.)

(2) Why is it a "reflux" for KCN(alc) reactions?

A Level Questions:

2013P3Q2(b)[ii]

The answer key says "NH3 acts as a reducing agent as it reduced HNO2 to N2, itself being oxidised to N2." It's a bit confusing to me. Is the thing being reduced N as well? Is this a disproportionation reaction?

2013P3Q2(d)[ii]

"Benzylamine does not decolourise reddish-brown bromine water and does not form white ppt."

Why does it not decolourise reddish-brown bromine water? Is it not a ring activated by a NH2 group that can have Br groups attached to it?

2013P3Q3(c)[ii]

The answer said that "KMnO4 reacts with SO2 to form colourless products" then proceeded to say "a permanent pale pink colur is seen at the end-point". Is it colourless or pale pink? (Saw both colourless and pale pink in a lot of other Qs too, very confused on this front.)

2014P3Q5(c)[i]

Why is H2O not oxidised instead? The answer placed E value of Ni2+/Ni instead, but E value of H2O (-0.83) is more negative than Ni2+/Ni (-0.25)?

2014P3Q5(d)

"silver will drop to the base as anode sludge."

What is anode sludge and how does it work? Why does it not stay at the anode? (This is very confusing for me because even though it's a alloy with many components oxidising away gradually, I don't see how it will just fall off? *confused face* *brings out hot tea to calm self* Sorry if the question is kinda dumb, haha.)

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Thanks :)

General Conceptual Questions:

Q1. No, and there are 2 ways of looking at why not. 1stly, in a peroxycarboxylate ion (conjugate base), the central O atom is sp3 hybridized (2 lone pairs + 2 bond pairs), hence no unhybridized p orbital available to overlap sideways, required for resonance delocalization of p or pi electrons to occur.

2ndly, if you try to delocalize the -ve formal charge on the terminal O atom, by using lone pair form bond pair with the central O atom, the central O atom will have it's octet configuration 'kena violated' (recall period 2 elements do not have vacant, energetically accessible 3d orbitals to accommodate an expanded octet).

Q2. Yes, you need to know how to draw, label and describe the use of the reflux apparatus (when it came out in the Singapore A levels a few years ago, most Singapore JC students couldn't do this question). There are 2 main types of reflux setups that can be tested at A levels : simple reflux, and reflux with distillation.

Heating is required for most organic chemistry reactions (other than those whose reactants or products are vulnerable to thermal decomposition or undesired side reactions, then room temperature or warming is required instead of heating), as most organic chemistry reactions involve breaking of covalent bonds and hence involve high Ea.

However, heating alone without reflux (ie. the use of a condenser) causes problems : organic reactants and solvents, being simple covalent molecular with weak intermolecular van der Waals are volatitle (ie. low boiling point). As such, either the solvent, or one or more of reactant(s), may vaporize and escape, instead of remaining in the reaction mixture, which sabotages the reaction by preventing it from continuing and/or reaching equilibrium yield for the product.

Hence heating under reflux (ie. with the use of a condenser) has 2 main purposes :
1) Ensuring the solvent does not vaporize away and cause the reaction vessel to boil dry, stopping the reaction. In addition, as a particular solvent has its own fixed boiling point, hence by careful choice of solvent, you can control the temperature within a very narrow range, to allow the reaction will proceed at the temperature ideal for that particular reaction. The constant boiling action also serves to continuously mix the solution, ensuring the reactants can continue to effectively mix and collide and thus reach equilibrium yield for the product.
2) Ensuring that the organic, volatile reactants (1 or more of which may have boiling points below the solvent and products) do not vaporize away and cause the reaction to slow down and stop, which prevents reaching the maximum equilibrium yield possible.

UltimaOnline SG

General Conceptual Questions:

(1) Is there resonance in CH3CO3- anion? (I'm guessing no, but I don't know why.)

(2) Why is it a "reflux" for KCN(alc) reactions?

A Level Questions:

2013P3Q2(b)[ii]

2013P3Q2(d)[ii]

"Benzylamine does not decolourise reddish-brown bromine water and does not form white ppt."

Why does it not decolourise reddish-brown bromine water? Is it not a ring activated by a NH2 group that can have Br groups attached to it?

2013P3Q3(c)[ii]

2014P3Q5(c)[i]

Why is H2O not oxidised instead? The answer placed E value of Ni2+/Ni instead, but E value of H2O (-0.83) is more negative than Ni2+/Ni (-0.25)?

2014P3Q5(d)

"silver will drop to the base as anode sludge."

_________________________________________________

Thanks :)

A Level Questions :

2013P3Q2(b)[ii] : This is not a disproportionation reaction, but a symproportionation or comproportionation reaction. It does not change the fact that NH3 is being oxidized and hence is the reducing agent, and that HNO2 is being reduced and hence is the oxidizing agent.

2013P3Q2(d)[ii] : This molecule isn't phenylamine, it is benzylamine. The NH2 group cannot donate electrons by resonance into the benzene ring, as it is inhibited or blocked by the sp3 C atom, and hence resonance delocalizaton of the N atom's lone pair to the benzene ring sp2 C atom (remember that resonance requires sideways overlap of unhybridized p orbitals) cannot occur without 'violating' the octet of the sp3 C atom.

2013P3Q3(c)[ii] : Purple MnO4- is reduced to Mn2+, which is a very pale pink colour, which in diluted solutions appear colorless. The end-point is indicated by the first sighting of a permanent pink colour (even after swirling the conical flask). This "permanent pink colour" is *not* due to the very pale pink colour of Mn2+, but instead is due to the last drop (from the burette) of the purple KMnO4 which indicates the end-point has been reached, and that the last drop is an 'extra' drop of KMnO4 in excess which doesn't react away (since the analyte has already been fully oxidized), hence the "permanent pink colour".

And if you're wondering "but won't that mean the volume reading won't be accurate, since there's an 'extra' drop in excess?", then 2 things : 1stly, that single extra drop is relatively inconsequential and negligible, compared to the other inevitable experimental limitations : limitations of measuring instruments, uncertainty values, human errors, systematic errors, and random errors. 2ndly, remember that the end-point only approximates equivalence point. We can only claim the value obtained to be the end-point, which we can only hope (with minimal experimental error) to be sufficiently close to the true equivalence point. Hence to ensure reliability and reproducibility, you have to repeat the titration several times until you obtain concordant titres.

2014P3Q5(c)[i] : Eh walau you. -_-". -0.83V is the standard reduction potential of H2O to H2 gas. Since the standard reduction potential of Cu2+ to Cu is +0.34V (ie. reduction potential more positive than -0.83V the standard reduction potential of H2O to H2), obviously Cu2+ wins and is hence reduced to Cu at the cathode (instead of H2O being reduced to H2 gas). If you wanna compare the standard oxidation potential of Ni to Ni2+ which is +0.25V, to the standard oxidation potential of H2O to O2 which is -1.23V, then obviously Ni wins (ie. oxidation potential more positive) and is hence oxidized to Ni2+ at the anode (instead of H2O being oxidized to O2 gas). And you can use +0.83V only if you want to oxidize H2 gas (in alkaline solution) to H2O, but wherefore cometh the H2 gas (and the required OH-(aq) ions) in this question?

2014P3Q5(d) : "Anode sludge" is the term given for the gooey disguisting stuff (ie. sludge) that is deposited below the anode. Why does it fall off and drop down? Because metals are insoluble in water *and* denser than water. Imagine you have a cake with many small stones embedded in it. As the cake is being eaten away by ants, won't the stones fall? As the Cu atoms in the block of impure Cu ore rock (which is the anode) is being oxidized to *aqueous* Cu2+ (ie. cations are soluble (unlike solid metals in atomic state) thanks to ion - permanent dipole interactions can dissolve and become aqueous, making the block of impure Cu ore rock which is the anode appear to dissolve away), the less reactive (ie. less electropositive and hence harder to oxidize) metals such as silver, remaining as the unoxidized solid metal, and solid metals being insoluble in water and denser than water, naturally drop down from the impure copper block of ore rock (ie. the anode), which together with any unreactive (organic or inorganic) material left over from the ore rock (eg. sand, soil, etc), becomes the gooey disgusting mess below the anode, which we call "anode sludge".

UltimaOnline SG

Using Data Booklet redox potentials, show that reaction between CuSO4 (aq) and KI (aq) to generate CuI(s) is unlikely to occur. Suggest why the reaction does in fact occur.

With regards to mrworry's qn on the reaction between CuSO4 and KI,

1. E cell= -0.39V < 0, hence, according to the data booklet this reaction is not feasible.

2. However, Cu2+ gets reduced to form Cu+. Cu+ is precipitated to form CuI. By LCP, position of equilibrium shifts to the right. Therefore, E Cu2+/Cu+ is more positive than +0.15 as stated in data booklet. So this reaction is feasibl.

Excellent, Tjyj. Your answer should get full marks from Cambridge, but to play safe, also do mention that "as a consequence of the CuI(s) precipitation, [Cu+] decreases" then your Le Chatelier statement, and finally the conclusion that "hence the cell potential under these non-standard conditions (where CuI(s) precipitates out), becomes positive, and the reaction becomes thermodynamically feasible."

UltimaOnline SG

A buffer solution is to be made using 1.00 mol dm–3 ethanoic acid, CH3CO2H, and
1.00 mol dm–3 sodium ethanoate, CH3CO2Na.
Calculate to the nearest 1 cm3 the volumes of each solution that would be required to
make 100 cm3 of a buffer solution with pH 5.50.
Clearly show all steps in your working.

Ka (CH3CO2H) = 1.79 × 10–5 mol dm–3

Answer.

Putting pH and pKa in the Handerson-Hassalbach Equation we will get [A]/[HA] = 0.753..... Now to calulate the volume i want to use c = n / v; c = 1 M but no clue where to find the number of moles ??

Let the volumes (in dm3) of A- and HA be V and (0.1-V) respectively. Then (V dm3 x 1M) / ((0.1-V) dm3 x 1M) = (753 / 1000). Solve for V, then convert from dm3 to cm3, round off to nearest 1cm3.

UltimaOnline SG

Hi UltimaOnline,

I have further questions on electrochem:

I. The diagram below shows the sodium-nickel(II) chloride battery, which is a high power, high capacity cell suitable for electric traction applications.

The electrolyte used is molten sodium aluminium chloride, which has a melting point of 157°C. In the reaction, nickel(II) chloride is reduced to nickel. Which of the following statements is incorrect?

A: Terminal E is the negative terminal.

Question: How do I know which is the positive and negative terminal?

II: In the construction of pacemakers for the heart, a tiny magnesium electrode can be used to create an electrical cell with the inhaled oxygen. The relevant half‐equations are as shown:

Mg2+ + 2e− ⇌ Mg (Equilibrium 1)

½ O2 + 2H+ + 2e− ⇌ H2O (Equilibrium 2)

In the body, a potential of 3.20V is usually obtained. What is the best explanation for this e.m.f.?

A The small size of the magnesium electrode

B The low concentration of Mg2+ surrounding the magnesium electrode

C The high resistance of the body fluids surrounding the electrodes

D The physiological pH of between 7 and 8 of the body fluid surrounding the electrodes

Answer: D

Remarks: Since the Ecell according to the Data Booklet for this reaction +3.61V, and in the body the usual emf is 3.20V, this would mean either the [Mg2+] is greater in reality, or that there or lower [H+] or [Oxygen] in reality. I can infer that D is correct because of the low [H+] in a pH environment between 7 to 8, but how do I know the other choices are wrong/not good explanations?

III: http://img.photobucket.com/albums/v700/gohby/Chemistry/sludge_zpsimhrdrbj.jpg

Answer: D

Remarks: Comparing the Ered values, I can gather that Cu2+ will be reduced to Cu at the cathode.

(i) What is meant by “Ag and Fe impurities”? If they refer to metallic compounds, wouldn’t it be unfavourable for the metallic ions to be oxidised (since it is at the anode) even further?

(ii) If “Ag and Fe impurities” refers to the metals itself, how do I know that Cu2+ and Ag+ will be formed, since the Eox (Cu/Cu2+) and Eox (Ag/Ag+) are negative? And how does Ag “fall off” the electrode and form the sludge?

IV: http://img.photobucket.com/albums/v700/gohby/Chemistry/doublelec_zpshzwpp2ja.jpg

Answer: A

Remarks:

(i) If it is an open-ended question, will I know if the product at S would be Fe2+ (instead of Fe)? I think it can be both (so long as the Ecell is positive) but I can’t be sure.

(ii) At electrode Q, comparing the Ered potentials between (Cu2+/Cu) and (H+/H2), wouldn’t Cu2+ be preferentially reduced compared to the latter - so why would hydrogen gas be the products formed at Q?

Thank you! :)

Hi Gohby,

QI. Oxidation potential of Na to Na+ is more positive than oxidation potential of Ni to Ni2+. Hence Na is oxidized, so Terminal E is the anode.

QII. For electrochem electrodes, size doesn't matter. [Mg2+] should be larger, not smaller. If resistance is higher, voltage is higher, not lower.

QIII. Impurities could refer to either (reduced or unoxidized) metallic or (oxidized) ionic state. At the anode, both Fe and Cu are oxidized, while Ag remains unoxidized and hence metallic and hence falls off into the anode sludge. At the cathode, Cu2+ is reduced, while Fe2+ remains in aqueous state. Regarding the sludge, this is what I wrote in another thread :

"Anode sludge" is the term given for the gooey disguisting stuff (ie. sludge) that is deposited below the anode. Why does it fall off and drop down? Because metals are insoluble in water *and* denser than water. Imagine you have a cake with many small stones embedded in it. As the cake is being eaten away by ants, won't the stones fall? As the Cu atoms in the block of impure Cu ore rock (which is the anode) is being oxidized to *aqueous* Cu2+ (ie. cations are soluble (unlike solid metals in atomic state) thanks to ion - permanent dipole interactions can dissolve and become aqueous, making the block of impure Cu ore rock which is the anode appear to dissolve away), the less reactive (ie. less electropositive and hence harder to oxidize) metals such as silver, remaining as the unoxidized solid metal, and solid metals being insoluble in water and denser than water, naturally drop down from the impure copper block of ore rock (ie. the anode), which together with any unreactive (organic or inorganic) material left over from the ore rock (eg. sand, soil, etc), becomes the gooey disgusting mess below the anode, which we call "anode sludge".

QIV. Under aqueous conditions, you cannot reduce Fe2+ to Fe. You'll need molten iron for that. But since electrolysis is expensive, the blast furnace method is utilized instead. You cannot use standard redox potentials, because in aqueous soluions, molarity of water is much larger than 1M (molarity of pure water is 55.555M, and Cambridge may ask A level students to show this via calculations), hence position of equilibrium shifts to RHS, hence reduction potential of H2O becomes more positive than stated in Data Booklet.

At cathode Q, H+ is reduced to H2 gas instead of Cu2+ being reduced to Cu, because the electrolyte doesn't have any Cu2+ to begin with. Sure, Cu2+ is being generated at anode P, but at the start of the electrolysis (for which the qn is based on), the molarity of H+ far outweighs the molarity of Cu2+ (which is technically zero M at initial).

No prob :)

UltimaOnline SG

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