BedokFunland JC's A Level H2 Chemistry Qns (Part 2)
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2013 A Level P2 Qn
Sushii asked :
i got a chem qn that sort of came out for p2 ... what determines basicity of amines? steric hindrance or the electron donating effect of alkyl groups?
Steric hinderance explains why a particular species, eg. LDA, might behave as a base instead of a nucleophile. But this question didn't ask why benzylamine is a base rather than a nucleophile, it asked what was the relative basicity of phenylamine vs benzylamine vs ethylamine.
Comparing benzylamine vs ethylamine, ethylamine is both a stronger base and a stronger nucleophile, because the lone pair on the N atom is more available for dative bond donation, in turn because :
1) ethylamine has 2 electron-donating by induction sp3 C atoms (as an ethyl group) bonded to the N atom, while benzylamine only has 1 sp3 C atom bonded to the N atom.
2) benzylamine's sp2 C atoms are electron-withdrawing by induction, as its orbitals have a higher % of s orbital character.
Students who blindly copied-and-pasted (mostly without understanding) their school notes on "overlap of p orbitals of N atom with pi orbitals of benzene ring" get zero marks here, because resonance is not relevant when comparing ethylamine versus benzylamine (ie. resonance delocalizaton of the N atom's lone pair is only relevant when comparing phenylamine versus the other two species).Sushiii replied :
wow thanks that cleared up quite a bit of misunderstanding i had learnt! (altho i answered it wrongly -_- )
No worries, as few H2 students could answer this part of the question correctly anyway.Although JC students generally said this year's 2013 paper was easy, sometimes it's because the students don't even realize that some tricky questions are tricky.
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Posted by Prototype445
Ahh one more question that I find rather peculiar
Saturated solutions Ksp=IP will not have any precipitation right? I saw one prelim paper that said it will precipitate out (it was a MCQ question)The typical A level H2 Chem question will ask, "what is the aqueous molarity of this species required to initiate precipitation?", I instruct my BedokFunland JC students to write in the 'A' level exam, "the aqueous molarity of this species must be greater than xx.xx mol/dm3 in order for precipitation to occur."
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Posted by iamsylar
For the question regarding benzylamine and ethylamine I argued that benzylamine was more basic cause I saw the benzene ring as electron rich, and hence the C6H5CH2- alkyl group attached to the nitrogen atom in benzylamine would have a higher electron density around the nitrogen atom which makes the lone pair more available for protonation? I see arguments are that the benzene ring causes the lone pair to partially delocalise into the benzene ring but isn't that inductive effect hindered by the carbon atom between the lone pair and the benzene ring?Nice try (with the idea of the electron rich benzene ring attracting protons); proton substitution (via electrophilic aromatic substitution) of the benzene ring does indeed occur, but only under strongly acidic conditions, and not proton addition (which is the definition of basicity), as such would disrupt its aromaticity.
If you read my previous posts, I've already explained why the benzene ring actually reduces electron density on the N atom in benzylamine by induction (due to greater % s orbital character of benzene's sp2 C atoms). Majority of H2 students, who (blindly followed school notes and) mentioned "delocalization of the lone pair into the benzene ring due to overlap of p orbitals... etc", ie. resonance, got it wrong, as resonance decreasing basicity applies only to phenylamine, not benzylamine.
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Posted by Pizzateddy
For paper 2 the bromine substituting onto the phenol with a ether group, is it supposed to be mono substitution or tri substitution?Some ppl argued that it should be tri sub(at 3,5,6) if I rmb correctly becos the ether is activating but since ethers wasn't taught in jc syllabus will Cambridge penalize?
Ah yes, this was one of the problematic questions for this year's Paper 2 (since 2011 when Cambridge began their sadism spree, there has always been several problematic / ambiguous / debatable questions for each H2 Chem A level exam; watch out for tomorrow's Paper 1!).
Ethers are indeed activating, because they donate electrons by resonance, though not as strongly activating as the phenolic group.
Therefore, one could argue for both answers : mono-substitution, or tri-substitution.
This is truly Cambridge's fault (for setting a question in which there is arguably more than one correct answer), should the Mark Scheme not accept both answers.
(To exacerbate this matter, the exact same problem arose twice in the question : halogenation and nitration).
As such, there is no way of saying for sure, that assuming Cambridge will only accept 1 of the answers, whether Cambridge is looking for mono-substitution or tri-substitution.
However, in such scenarios, the exam-smart candidate, will have to play the Criminal Minds psychological profiler game, and get into the minds of the Cambridge question setter. From the look of the question, specifically the structure of the reactant molecule, Cambridge was probably testing the H2 Chem student on directing effects (ie. ortho / para / meta), as you will notice that the (other) ortho and para positions relative to the phenolic group (ie. the single strongest activator) were blocked.
And because both the halogen and nitro substituents are deactivating, it could be argued that Cambridge was testing the candidate to see if he/she knows, should only mono-substitution occur, where would it occur?
Therefore, in conclusion, it is more likely, that (assuming Cambridge only accepts 1 answer; though it would be fairer if Cambridge, perhaps in retrospect after marking of exam scripts begin, hopefully revise their mark scheme to accept both answers), Cambridge was seeking mono-substitution (on the correct ortho position) rather than tri-substitution, as the single best correct answer.
Conclusion and Final Advice :
For tomorrow's Paper 1, if Cambridge again dishes out such frustrating annoyances, be exam-smart and play the Criminal Minds game and profile the Cambridge question setter's psyche. -
BedokFunland JC's 2013 H2 Chemistry Paper 1 Answers :
Q1 - B
Q2 - B
Q3 - D
Q4 - C
Q5 - A
Q6 - C
Q7 - A
Q8 - B
Q9 - D
Q10 - C
Q11 - D
Q12 - B (NO2 is a radical which damages the ozone layer, but thinning of ozone layer *reduces* global warming, not increases it! It's true that NO2 is a greenhouse gas, but it's not one of the most common or one of the most damaging greenhouse gases. Acid rain from both HNO3 and H2SO4 (for which NO2 indirectly catalyzes the formation of) is probably a greater concern than global warming, in the context of NO2. NO2 catalyzes oxidation of SO2 to SO3, which undergoes hydrolysis to form H2SO4.)
Q13 - D
Q14 - D
Q15 - B
Q16 - D
Q17 - D
Q18 - A
Q19 - C
Q20 - C ( X and Z are solely sp2-sp2 overlap, while W and Y have partial and full pi bonding respectively (for this resonance contributor anyway; the resonance hybrid actually has partial pi bonding for all the bonds) ; since there's no X and Z option, so yes the best answer is C - all bonds are sp2-sp2 overlap ... + some unhybridized p orbital overlap actually)
Q21 - C
Q22 - D
Q23 - B or D (debatable but more arguably B ; since temperature, duration, extent of hydrolysis not specified)
Q24 - B
Q25 - B
Q26 - B
Q27 - C
Q28 - C
Q29 - A
Q30 - D
Q31 - A
Q32 - D
Q33 - A
Q34 - B (tricky : the OS of Grp II metal does not change, but oxygen undergoes disproportionation)
Q35 - D
Q36 - B
Q37 - A
Q38 - B
Q39 - C
Q40 - A (the mechanism involved for this condensation reaction, is nucleophilic addition-elimination). -
A BedokFunland JC H2 Chemistry Qn
Work out the electronic configurations of :
a) Fe, Fe+, Fe2+, Fe3+
b) In, In+, In2+, In3+
Hence explain why compounds of In3+ do not have colour, unlike compounds of Fe3+.
Solution :
Fe : [Ar] 4s2 3d6
Fe+ : [Ar] 4s1 3d6
Fe2+ : [Ar] 4s0 3d6
Fe3+ : [Ar] 4s0 3d5In = [Ar] 4d10 5s2 5p1
In+ = [Ar] 4d10 5s2 5p0
In2+ = [Ar] 4d10 5s1 5p0
In3+ = [Ar] 4d10 5s0 5p0As shown above, In3+ in it's compounds have a fully filled d subshell, hence no d-d* electron transitions are possible, and therefore its compounds do not have colour.
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Originally posted by Blanklorh:
erm for qn 36, i worked out the elec config of In to be 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10 5p1
to my understanding, partially filled d orbital shld give colour no? then ionic salt of In(III) ought to be able to have a d-d transition isnt it? hence the colour?
Good thoughtful effort, you did indeed successfully work out the electron configuration.But unfortunately, you're slightly off target when it comes to ionizing, as electrons are always removed from outer electron shells first :
In = [Ar] 4d10 5s2 5p1
In+ = [Ar] 4d10 5s2 5p0
In2+ = [Ar] 4d10 5s1 5p0
In3+ = [Ar] 4d10 5s0 5p0
Hence In3+ in it's ionic salts have a fully filled d subshell, hence no d-d electron transitions, hence it's ionic salts are not coloured.
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2013 A level Qn
A Group II metal forms an oxide with the following formula : MO2
How many electrons are present in the anion?
Answer :
18 electrons in the peroxide anion.
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A BedokFunland JC H2 Chemistry Question & Answer
Originally posted by hoay:When considering the electron pairs repulsion in a molecule, why does a lone pair of elctrons repel more strongly than a bonding pair???
For A level purposes, it should be regarded and imagined that a lone pair of electrons is positioned 'horizontally' (ie. both electrons equally close to the atom's nucleus) and thus occupies more space that is in closer proximity to the nucleus, and hence experiences greater repulsions with the other electron pairs around the atom, compared to bond pairs which are positioned 'vertically' away from the nucleus (ie. the dot closer to one atom, and the cross nearer to the other atom bonded with; in fact, this is how Cambridge prefers A level students to draw their dot-and-cross diagrams, as opposed to how some schools erroneously teach), and thus the bond pair's repulsion is 'shared' between (repelling) electrons of both atoms, and hence lesser in magnitude, compared to the repulsion that occurs between lone pairs and other electron pairs of the atom.As an illustrative analogy, imagine you're on the MRT train.
The passengers standing upright holding onto (ie. bonded to) the overhead support handles represent the bond pairs, while the passengers lying down on the floor, not bonded to the overhead support handles, represent the lone pairs.
Obviously, the passengers lying down on the train floor (ie. the lone pairs), would occupy more floor space, annoying and repelling other passengers to a greater extent, compared to the passengers standing upright (ie. the bond pairs).
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A BedokFunland JC Original H2 Chemistry Qn & Soln
An organic molecule contains 1 benzene ring, 2 non-aromatic rings, 2 double bonds, and 2 triple bonds. What is the value of x in its molecular formula C15 Hx N2 O2 S F Cl Br I ?
Solution :
Halogen atoms are terminal, similar to hydrogen atoms, and thus may be considered as H atoms.
As far as degree of unsaturation is concerned, the molecular formula now stands as C15 H(x+4) N2 O2 S.
The presence, addition or subtraction of oxygen and sulfur atoms does not affect the molecule's degree of unsaturation, due to the bivalency of O and S.
As far as degree of unsaturation is concerned, the molecular formula now stands as C15 H(x+4) N2.
Without affecting the molecule's degree of unsaturation, each nitrogen atom that is inserted between a C atom and a H atom, also carries with it an additional H atom, due to the trivalency of N.
As far as degree of unsaturation is concerned, the molecular formula now stands as C15 H(x+4-2) = C15 H(x+2).
From the data given in the question,
1 benzene ring + 2 non-aromatic rings + 2 double bonds + 2 triple bonds
= 4 + 2 + 2 + 4 = 12 degrees of unsaturation.Since a completely saturated, acyclic hydrocarbon with 15 C atoms is expected to have the molecular formula C15 H32, hence the same molecule with 12 degrees of unsaturation is expected to have the molecular formula C15 H(32-24) = C15 H8.
Based on our work above, C15 H(x+2) = C15 H8
This implies x + 2 = 8 , hence x = 6.
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To fully understand oxidation of aromatic species, necessarily involves the full mechanism (including the MnO4- species and multiple intermediates), which is not just beyond H2 & H3 A levels, but is also beyond much of Uni level Chemistry as well (in which you will have to choose to focus on your specialties and subspecialties; this holds true for all sciences and all disciplines).
As such, even your school JC teachers, and even some of the Cambridge examiners, themselves would not know the mechanism for this reaction (*particularly* for this aromatic oxidation reaction, which is significantly more complicated than say, oxidation of alkenes, alcohols, aldehydes, etc).
Bottomline : as far as A levels H2 Chemistry is concerned, Cambridge only requires you to appreciate that the alkyl side chain (containing at least 1 benzylic H atom) of a benzene ring, can be oxidized by heating under reflux with acidified KMnO4 to generate benzoic acid, and other by-products (usually including CO2).
If the benzene ring's side chain contains a C=C group, you have to treat it as an alkene and oxidize it accordingly to generate the expected products on the other side of the C=C goup. Any remaining C atoms between the benzylic C atom and the double bond, you may assume to be oxidized to CO2.
Last but not least,hot alkaline oxidation and its difference between hot acidified oxidation by KMnO4
For A level purposes, temperature together with pH matters only for oxidation of alkenes, ie. cold alkaline KMnO4 generates vicinal diol, while hot acidified KMnO4 results in oxidative cleavage.Other than alkenes, the H2 syllabus ignores any differences between acidified vs alkaline KMnO4 (but "heat under reflux" must always be specified), other than (obviously!) the required protonation or deprotonation of the oxidation products.
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A BedokFunland JC H2 Chemistry qn
Cyanuric acid, a weak triprotic acid, has the empirical formula CHNO, and a molar mass of 129.1g, and exists as an equilibrium mixture of two tautomers. Draw their structures.
Solution :
From the molar mass and empirical formula, we work out the molecular formula as C3H3N3O3.
From the molecular formula, we work out the degree of unsaturation as 4.
4 degrees of unsaturation could imply a benzene ring, but since there are only 3 C atoms present per molecule, and there are 3 N atoms available, we modify the benzene ring by alternating the C and N atoms within the ring, ie. obtaining an aromatic triazine ring.
Note that the 3 phenolic OH groups (for the 1st tautomer) or the 3 NH groups (for the 2nd tautomer) are responsible for the triproticity of cyanuric acid, as the conjugate base(s) are stabilized by having the negative charge(s) delocalized by resonance and dispersed by induction over the multiple electronegative, electron-withdrawing O and N atoms.
The molecule exists as an equilibrium mixture of two tautomers :
Credits : image above owned by Wikipedia @ http://en.wikipedia.org/wiki/Cyanuric_acid
Bonus Qn :
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Originally posted by a mugger:
how about the formation of H20,still a little edgy for me, so all Ch2Ch3 becomes 2 c02 but why does ch2ch2ch3 becomes ch3cooh?
I can't understand because my school notes did not clearly mention the rationale behind this concept,but only shows the reaction and products only.
To fully understand the reaction, you have to appreciate its mechanism. Google "mechanism for oxidative cleavage of alkenes".For A level purposes, it will suffice to note the products based on the cleavage of the C=C group, and apply this accordingly to any alkene.
Ethene undergoes oxidative cleavage by KMnO4 to generate 2 molecules of carbonic(IV) acid, each of which exists in equilibrium with, and hence can decompose into CO2 and H2O, with favourable positive entropy change.
A molecule with the group R=CH2CH2=R will undergo oxidative cleavage by KMnO4 to generate 2 molecules of R=O (in which the R groups may or may not be the same), and ethandial (OS of C increases from -2 to +1), which is further oxidized to ethandioic acid (OS of C increases from +1 to +3), which is further oxidized to (2 molecules of) carbonic(IV) acid (OS of C increases from +3 to +4), which exists in equilibrium with, and hence can decompose into CO2 and H2O (OS of C remains unchanged, from to +4 to +4), with favourable positive entropy change.
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Another BedokFunland JC Original H2 Chemistry Qn
Q1. Fehling's solution usually gives a brick red precipitate with aliphatic aldehdyes, in which Cu(II) is reduced to Cu(I). What is the one special aldehyde, that is able to reduce Cu(II) all the way to Cu, generating a copper mirror?
Q2. Fehling's solution and Tollen's reagent usually do not react with carboxylic acids. What is the one special carboxylic acid, that both Fehling's and Tollen's are able to react with, to generate positive results (ie. copper mirror and silver mirror observed)?
Q3. Suggest why the unusual reactions described in (Q1 & 2) above, are able to occur.
Solution :
Q1. Methanal.
Q2. Methanoic acid.
Q3. Methanal can be very readily oxidized to methanoic acid, which can be very readily oxidized to carbonic(IV) acid, which exists in equilibrium with, and hence can decompose into, carbon dioxide and water.
As predicted by Le Chatelier's principle, the continuous shifting of the position of equilibrium to the right (as the result of gaseous cabon dioxide leaving the aqueous solution), as well as the associated thermodynamically favourable positive entropy change (from aqueous carbonic(IV) acid to gaseous carbon dioxide), together results in the reaction series of the oxidation of methanal to methanoic acid to carbonic(IV) acid and the subsequent decomposition into carbon dioxide and water, having very positive oxidation potentials, which consequently cumulates into these redox reactions having thermodynamically favourable, positive redox cell potentials and negative Gibbs free energy change values.
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Originally posted by ArJoe:
Why does copper react with concentrated nitric acid although copper is less reactive than hydrogen? why can hydrogen react with iron although it is less reactive (reduction of iron oxide by hydrogen)?
Regarding hydrogen reacting with iron :
First of all, you probably mean hydrogen reacting with compounds or oxides of iron (rather than iron itself). At room temperature and at standard molarities or partial pressures, the reaction will occur at a negligible rate. So at O levels, it's correct to say "iron is less reactive than hydrogen", at standard conditions.
However, when sufficient heat energy is supplied to overcome the Ea barrier, together with higher molarities or partial pressures of gaseous hydrogen, then the reaction may proceed (depending on the exact redox reaction). For industrial and pragmatic purposes, a transition metal catalyst is almost always employed to increase the rate of the redox reaction when employing gaseous hydrogen.
Moreover, if the reaction you're referring to, is the reduction of iron(III) oxide to iron(II) oxide, then note that the standard reduction potential of Fe3+ to Fe2+ is +0.77V, and thus the reaction with gaseous hydrogen is indeed expected to proceed under standard conditions. If the reaction you're referring to are the reduction of other metal oxides with hydrogen, then be aware that as long as the overall redox potential isn't too negative, some reaction can still occur with heat and at higher than standard molarities, it's only a matter of the % yield at equilibrium.
Regarding concentrated nitric acid :
This is because both the cation H+ and the anion NO3- are strong oxidizing agents, and when both are present in high molarities, can work together to oxidize even relatively unreactive metals (ie. with low or slightly negative oxidation potentials).
The NO3- in nitric(V) acid (ie. with nitrogen having the maximum OS of +5) can be reduced to species such as HNO2 (OS of +3) or NH4+ (OS of -3).
For an even more powerful oxidizing agent, which can oxidize even gold and platinum (which are even less reactive than copper), see Aqua regia :
http://en.wikipedia.org/wiki/Aqua_regia
The chemistry of Aqua regia involves both redox as well as coordination complex reactions, and can be used by Cambridge to craft an excellent and challenging H2 Chemistry A level exam qn.
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"After the Rainbow" is a CSB video safety message focusing on preventing accidents in high school chemistry labs.
http://www.youtube.com/watch?v=g6vR0BdRCNY
BedokFunland JC's Warning : The worst risks of methanol CH3OH, is not its flammability, but it's toxicity. Methanol poisoning (whether through the skin or ingested) can cause permanent blindless and multiple organ failure, leading to a very painful death.
Methanol : http://en.wikipedia.org/wiki/Methanol#Toxicity
Even more toxic and terrifying, is hydrofluoric acid HF. It's legendary toxicity (if you get a single drop of hydrofluoric acid on your finger, you may need to amputate your entire arm immediately or risk dying a horrifying death) is not due to its (weak) acidity, but due to its biochemically disruptive properties.
Hydrfluoric acid : http://en.wikipedia.org/wiki/Hydrofluoric_acid#Health_and_safety
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A BedokFunland JC Recommendation for H2 Chemistry Qn.
When liquid mercury is added to concentrated nitric(V) acid, the mixture appears to immediately 'boil', with large bubbles observed. Write equations to explain the observation.
Watch YouTube video : http://www.youtube.com/watch?v=x1bkkGb1k7k
Credits : TAOFLEDERMAUS
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