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

last SG · 2010-07-26T23:40:35+00:00

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Q.1 The eletrolysis of impure copper is carried out using CuSO4 as electrolyte, copper being pure Cu, while anode as Impure Cu.

The three observations are :

A) Anode dissolves

B) The mass of cathode increases

C) the blue color of solution fades.

Is the third observation correct??

Q.2 What impurities are present in impure Copper?

Q.3 What if the electrolyte is Copper nitrate?

Q.4 What will be the products left after this electrolysis is carried out? I mean the salts which are left behind.

Q1. Incorrect. Any changes to [Cu2+] are negligible, as far as 'O' and 'A' levels are concerned.

Q2. Other metal ions, both more electropositive and less electropositive, than copper.

Q3. No difference, as sulfate(VI) and nitrate(V) anions both have their heteroatoms already in their most positive OS possible, and cannot be oxidized further at the anode (to which they are electrostatically attracted).

Q4. Depends on the specific impurities present. Less electropositive metal ions are reduced at the cathode, while more electropositive metal ions remained in the aqueous oxidized state.

Impurities in this context, are essentially :

other metals which are less electropositive than Cu

and

other metals which are more electropositive than Cu

The former are mostly already in their reduced state (eg. Ag, Au, etc), and simply drop to the bottom as anodic sludge.

The latter are mostly already in their oxidized state (eg. Fe2+, Al3+, etc), and simply dissolve into the electrolyte solution.

Notice that both groups do not take part in any redox reaction at the anode, and therefore do not deprive the Cu (at the impure anode) from being oxidized at the anode.

Hence, at the anode, Cu is being oxidized to Cu2+, for every 2 moles of electrons transferred.

At the cathode, only Cu2+ is being reduced to Cu (as more electropositive metals, eg. Fe2+, Al3+, etc) prefer to remain in their oxidized, aqueous state.

Hence, at the cathode, Cu2+ is being reduced to Cu, for every 2 moles of electrons transferred.

Therefore, the [Cu2+] in the electrolyte solution remains almost unchanged (barring negligible factors such as evaporation of solvent, negligible oxidation of impurities closely similar to Cu in electropositivity, etc; all of which are not the concern of the 'O' or 'A' levels).

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Another BedokFunland JC H2 Chemistry Qn

In organic chemistry reactions, nucleophilic substitution competes with elimination. If the elimination products are desired over the nucleophilic substitution products, what two factors can we use to promote the reaction towards elimination, and away from nucleophilic substitution?

Answer

Solvent and Temperature.

Using an ethanol solvation shell promotes the reaction towards elimination, while using an aqueous solvation (ie. hydration) shell, promotes the reaction towards nucleophilic substitution.

This is because the ethanol solvation shell poses greater steric hinderance (compared to a hydration shell), consequently forcing the OH- ion to function as a Bronsted-Lowry base, rather than as a nucleophile.

Because elimination reactions involve a positive entropy change, hence as predicted by the formula for Gibbs free energy, utilizing a higher reaction temperature, promotes elimination (ie. the Gibbs free energy change or delta G value, becomes more negative and the reaction more thermodynamically feasible) to a greater degree, than it does nucleophilic sustitution (which involve relatively negligible entropy change, and whose enthalpy change could be either endothermic or exothermic).

Note that nucleophilic substitution still requires a high temperature (as the breaking of covalent bonds involve a high activation energy barrier), and hence heating (usually under reflux) in both nucleophilic substitution and elimination is still necessary; it's just that the higher the temperature, the greater the tendency towards elimination because of the positive entropy change.

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Another BedokFunland JC Question

What is the difference between "peptide" and "amide"? When asked to label a -CONH- group, which word should be written?

What does the term "secondary structure" of a protein refer to?

Answers :

Both words refer to the -CONH- group, but peptide refers to this group within proteins, while amide refers to this group (everywhere else) outside proteins.

To be more specific, peptide groups are -CONH- generated from the nucleophilic acyl substitution / addition-elimination / condensation reaction, between the alpha carboxylic acid COOH group of one (alpha) amino acid, with the alpha amine NH2 group of another (alpha) amino acid, within a protein.

Amide groups refer to all other types of -CONH- groups, most usually those in synthetic molecules, eg. drugs and polymers.

Special mention : when (in the theoretical event of) a -CONH- group is generated from the reaction between one amino acid's R group's COOH group, and another amino acid's R group's NH2 group, (or one alpha group reacting with one R group), the consequent -CONH- group is still regarded as an amide group, rather than a peptide group (peptide groups are generated strictly from alpha COOH and alpha NH2 groups).

On a related note, when asked what the "secondary structure" of proteins refers to, Cambridge requires the H2 Chem student to specify, "A protein's secondary structure refers to the alpha-helix or beta-pleated sheet structure of a polypeptide, generated as a consequece of the hydrogen bonding between peptide groups (and not the R groups) of amino acids in a protein."

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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

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

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.

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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).

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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 3d5

In = [Ar] 4d10 5s2 5p1
In+ = [Ar] 4d10 5s2 5p0
In2+ = [Ar] 4d10 5s1 5p0
In3+ = [Ar] 4d10 5s0 5p0

As 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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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

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 :

Draw the mechanism for the tautomerization between the two species.

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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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