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

last SG · 2010-07-26T23:40:35+00:00

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CIE W14 Q5.

Just wondering for this reaction besides Na, how does vanillin react for Br?
Why it react on C-5 and not C-3?

As vanillin is named clockwised and not anticlockwised.

Also you mind explaining the other reactions that I'd highlighted? Especially RCOCl and Fe3+ thanks :)

Halogenation via electrophilic aromatic substitution.

As long as you correctly substitute the incoming electrophile onto an available ortho or para position relative to the phenolic group, Cambridge will accept.

Diazotization and diazonium coupling is not in the Singapore H2 syllabus. If you're interested, go read up on it yourself.

Acyl halides are electrophiles attacked by the most nucleophilic group, ie. amine on L-Dopa, and phenol in Vanillin.

Of course, depending on the solvent, the alpha-amino-carboxylic acid of L-Dopa may exist as zwitterionic form, in which case the NH3+ is no longer nucleophilic and cannot attack the acyl halide electrophile. Note that acyl halides must be used in anhydrous form, which implies the other reagent (ie. L-Dopa) should not be used in aqueous solvent.

Fe3+ (in neutral solution, can't be too acidic or too alkaline, interested students can ponder over why, and check your answer with your school teacher or private tutor) coordinates & complexes with phenolic ligands to generate a violet coloured coordination complex.

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Also we didn't learn the terms Para (1,4-) and Ortho (1,2-) but since you said it was phenolic, so it should be Electron Donating Group? How do we determine it as the phenolic group as the first place as Vanillin has an Acetyl group and Aldehyde group bonded to the Benzene ring with the hydroxyl group.

Is not like students can determine when 2 different natures of groups are bonded to the same ring and the Data Booklet doesn't even help a bit.... When considering bonding at C2, C4 or C3. And we have to determine where will be our C1 at the first place...

(I'd just finished whole Organic Chemistry in 2 week time... So I'm still relative new to this)

Whether the student can determine the ortho-para-meta directing nature of each substituent, is up to the student to learn about this as you/he/she wills. Different JCs, different teachers, different private tutors, teach to differing extents. Depending on your own willingness and choice, you can read up about this on your own, even if your school teacher chooses not to teach it.

The phenolic OH group is electron-donating by resonance moreso than it is electron-withdrawing by induction, hence each phenolic OH group is activating and ortho-para directing. The alkyl group (sp3 C atoms) in L-Dopa is electron-donating by induction (relative to the sp2 C atoms of the benzene ring), and is hence a weak activator and an ortho-para director. The benzaldehyde group in Vanillin is electron-withdrawing by both induction (due to a significant partial positive charge on the carbonyl sp2 C atom as a result of both induction and resonance) and resonance (due to the sideways overlap of the unhybridized p orbitals of sp2 C and O atoms of the carbonyl C=O group), and is hence deactivating and meta directing.

Hence, the strongest activator (for both L-Dopa and Vanillin) being the phenolic OH group, wins and gets to direct the incoming electrophile. Because L-Dopa contains 2 phenolic OH groups, Cambridge will accept any of the 3 possible answers, although the Mark Scheme answer is the major product due to least steric hindrance.

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Hi UltimaOnline, just a little concept clarification. Is there a buffer region for weak acid and weak base titration? Thanks

Technically, yes. Buffer regions still occur when you've close to equal molarities of HA and A-, as well as close to equal molarities of B and BH+ (note that the actual maximum buffer capacities of the solution, however, is no longer based simplistically on having equal molarities of both members of each conjugate acid-base pair, since the other conjugate acid-base pair also exerts its influence on pH and buffering capacity; exact calculations here are significantly beyond the scope of A levels).

In addition, there is an additional (albeit weaker) buffer region (unique to weak acid - weak base titrations) that spans from the 1st buffer region (ie. near half-equivalence point) all the way to the 2nd buffer region (ie. near double-equivalence point), including the titration equivalence point, since both cation and anion of the salt at equivalence point, are themselves a weak acid and a weak base.

If Cambridge were to ask Singapore students to calculate the pH at equivalence point of a weak acid - weak base titration, 99.99% of Singapore H2 Chem students (and even many JC teachers and private tutors) would die. But if you're interested to challenge yourself, try solving such a question. Hint : you'll need to solve simultaneous equations which takes into account the simultaneous hydrolysis of both cation and anion of the salt.

UltimaOnline SG

okay i see. for cyclo alkanes, only substitutents exhibit cis trans only. for cycloalkenes, the whole molecule cis or trans about the double bond.

Of course, it's possible for cycloalkenes to have geometric isomerism for both the double bond as well as substituents. In which case (for A levels), u just need so specify whether the cis-trans isomerism refers to the substituents or the double bond, eg. "This molecule is cis in regard to the double bond, but trans in regard to the 2 Cl atoms on the ring". Cambridge won't require you to name such complicated molecules with more than 1 set of cis-trans isomerism present simultaneously.

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(Planning qns) In starch-iodine titration, why must starch indicator only be added near the titration end point and not that the start of titration?

If starch solution is added at the start of the titration, the insoluble dark blue colored triiodide - starch clathrate precipitate formed will cause a time delay in releasing the triiodide ion (to be reduced by the thiosulfate ion), which may cause your titer volume and hence your titration results to be inaccurate. In other words, you may already have added enough thiosulfate solution from the burette to reach end point (which approximates equivalence point), but the dark blue color may remain for several seconds longer, fooling you into adding another couple of drops of thiosulfate solution from the burette, ruining the accuracy of your titration results.

This is much less of a problem if the starch solution is added only near the end of the titration, as there will be a much smaller amount of triiodide ions left to be released from the decomposition of the much smaller amount of dark blue colored triiodide - starch clathrate precipitate, which will take much less time (almost instantaneous) and hence poses little risk of error to your titer volume and titration results.

UltimaOnline SG

there's no methanal generated anywhere in the mechanism is it?

e.g. george chong organic book page 83

is it that the alkene is oxidised to ethanedioic acid then h2co3 then further to CO2 and H2O?

The CH2 side of a terminal alkene is oxidized to methanal which is further oxidized to methanoic acid which is further oxidized to carbonic acid which decomposes to CO2 and H2O, provided you heat under reflux with MNO4-/H+.

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2011 TYS Paper 1

Qns 8: How do we know that the conditions are cold such that Cl2 can disproportonate in NaOH? Is it because as long as the reaction takes place in rtp, we assume that the conditions are cold?

2011 TYS Paper 2

Qns 4c(iii): Do we use the bond energies found in 4c(ii) to answer? Or the bond energies are not relevant in this qns?

Qns 6a(ii): Based on reaction 1 alone, shouldn't the functional group be an acyl chloride since it can react with silver nitrate immediately to produce a white ppt? So why is Halogenoalkane the answer when they have to undergo additional steps after adding silver nitrate?

Q8. Yes.

Q4ciii. The stability of CCl4 over SiCl4 is *DESPITE* the stronger Si-Cl bonds (as a challenging A grade question, Cambridge can ask you to suggest at least 2 different possible explanations for this anomaly of C-Cl bond being weaker than Si-Cl bond despite C and Si being in period 2 and 3 respectively ; I expect my BedokFunland JC students to be able to rise up to Cambridge's challenge!).

So the bond enthalpies are not irrelevant, they are meant for Cambridge to (using A level exam questions) educate the student who understands what's going on, "Oh yah hor, how come Si-Cl bond stronger, yet SiCl4 kena hydrolyzed more readily leh? Chemistry damn interesting sia! I love Chem!"

The reasons for the stability of CCl4 over SiCl4 is covered in the basic H2 syllabus (although Singapore JCs usually teach only 1 simplified reason, while I teach my BedokFunland JC students (at least those who ask me) 3 different reasons for this).

Q6aii). Acyl chloride will give a white ppt at rtp. The qn specified warming is necessary, thus alkyl chloride.

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pg 112 , same book as above

lone pair of O resides in a sp3 hybridised orbital, can undergo delocalisation? i thought it's only for lone pairs residing in unhybridised p orbitals

That's exactly right. Since the lone pair on the O atom in phenol *is* indeed delocalized by resonance into the benzene ring, obviously the lone pair doesn't occupy an sp3 hybridized orbital, geddit?

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page 102 organic made easy

top of page

what does he mean by conjugated ring exhibits a stabilisation stronger than conjugation alone?

does he meant that/ he is comparing conjugation between 2 double bond , or 1 double-single-double bond, compared to rings with multiple alternating single and double bonds?

He means that as compared to conjugation in straight chain, complete conjugation throughout a ring (with same no. of double bonds in the resonance contributors) confers greater stability (this is known as aromaticity). Why? Count the no. of partial double bonds in the resonance hybrid for straight chain vs ring.

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yet another factor favouring hydrolysis of SiCl4 greater than CCl4 ?

delta +ve on Si greater than C. bond str Si-Cl > C-Cl. but SiCl4 much higher propensity to be hydrolysed

Minor contributive point. Doesn't make it to the top 3 reasons.

And instead of overthinking and obsessing over conceptual stuff like these (which the rest of the Singapore JC cohort just memorize and move on, efficiently albeit blindly ; your desire to focus more deeply on understanding is commendable and serves you well up to a certain easy-to-accidentally-pass point, beyond which being anally obsessive is like missing the forest for the trees, allowing you to win the battle but lose the war), and you J2 students really ought to be practicing as many Prelim papers as you can right now. You'll be far better off (as far as your A level results and hence the quality of the rest of your life are concerned) focusing on doing and asking about Pelim paper qns here, instead of OCDing about conceptual ideas too much.

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2008 p2 qn 1

noradrenaline qn

qn : draw the pdt when noradrenaline reacts with bromine dissolved in a suitable solvent

super open to interpretation?

if Br2 dissolve in water, will have multi sub

if Br2 dissolve in CCl4, will have mono sub

pls take a look at pg 233 of organic made easy book, therein writes Br2 in CCl4 will only have mono substitution. Thanks ultima.

If you're a BedokFunland JC student, you'll be exam-smart enough to write out both answers, with qualifications, ie. "If this solvent is used, this product is obtained ; if that solvent is used, that product is obtained". Cambridge examiner will exclaim, "Excellent student! Must be BedokFunland JC!"

UltimaOnline SG

Hi UltimaOnline,

I've a few questions here to consult you.

Q1: IJC/09/P1/39

In which sequence are the compounds listed in order of increasing ease of hydrolysis?

Choice 3 (chlorobenzene, chloromethane, chloroethane) is not a correct choice.

Remarks: In ascertaining the ease of hydrolysis, I'll first have to see the C-X bond strength. Since it's the same in this choice, I'll have to consider either (I) Resonance within the reactant and (II) stability of the intermediate products.

In this case, shouldn't 3 also be correct? What's the difference in the electron donating effect between the methyl and ethyl substituent?

Q2: In relation to VJC/09/P1/29:

Remarks: How do I establish the relative acidity of an alcohol (CH3CH2OH) to that of an acidic salt (CH3CH2NH3+Cl-)?

Q3: VJC/09/P1/23

Remarks: It must have been my blind spot, but why is A not an acceptable answer as well?

Thank you very much Ultima!

Hi Gohby, no prob.

Q1. The diff btwn Me and Et groups, is the stability (and hence Ea required) of the intermediate (if SN1) or transition state (if SN2). Et group is more strongly electron-donating by induction, which can better stabilize the intermediate or transition state.

Q2. Acidic salts (notice the adjective?) are always more acidic than alcohol (which is considered neutral, and even less acidic than neutral water).

Q3. A is the *only* acceptable answer.

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i thought need to have NaOH (aq) warm, to react with chloroalkane before there will be free Cl- to react with Ag+ in AgCl? is that a printing error, or how can the little amount of OH- present in AgNO3(aq) react with the chloroalkane to produce AgCl ah?

OH- being a stronger nucleophile than H2O increases the rate of reaction (hydrolysis then ionic precipitation). But the reaction will still proceed (albeit a lot more slowly) at rtp and at neutral pH. The exothermic ionic precipitation shifts the position of equilibrium over to the RHS and makes the reaction thermodynamically feasible. But the kinetics is influenced by temperature and strength of nucleophile.

UltimaOnline SG

Hi ultimaonline.

I recently did a question on why pka of peroxyethanoic acid(as shown in picture) is more than pka of ethanoic acid. my school answer is :The negative charge on oxygen in CH3CO2-
is very effectively dispersed through delocalisation within the ‒COO- π system. However, the anion CH3CO3- is not able to delocalise its electrons
over the C=O group. Hence, CH3CO2‒ is more stable than CH3CO3-.

May I ask if it is correct to say that since the extra o atom in the peroxyethanoic acid has p orbital, resonance of pi electron cloud can also form in the ‒COOO- group in the conjugate base of peroxyethanoic acid , the lone pair of the extra o atom increases electron density of pi electron cloud, hence intensifies the negative charge of conjugate base , making it more unstable, extent of dissociation of peroxyethanoic acid is lesser, ka is smaller, pka is bigger. correct me if im wrong in my reasoning. thank you.

Hi Rapidestlime, to fully understand the answer to this question, why the negative formal charge on the terminal O atom in COOO- cannot be delocalized by resonance, you have to go beyond just simply the idea of sideways overlapping of p orbitals making resonance delocalization of pi electrons and hence delocalization of a formal charge possible.

You actually have to draw out the resonance contributors in each case, ie. COO- vs COOO-, to explain why resonance delocalization of the negative formal charge is not possible in COOO-. Because unless it involves complete delocalization around a ring, ie. aromaticity, otherwise, lone pairs (in unhybridized p orbitals) cannot delocalize into being lone pairs on other atoms by resonance, lone pairs can only at most delocalize into pi bond pairs.

In the case of COOO-, the central O atom's lone pair occupying a unhybridized p orbital can indeed delocalize by resonance as a pi bond pair with the acyl sp2 C atom, which can indeed accept the pi bond pair (which happens in the resonance stabilized COO- conjugate base) without having to expand its octet (which it can't do so, being in period 2), because the C=O pi bond can simultaneously delocalize into a lone pair on the O atom.

However, the lone pair on the terminal O atom bearing the negative formal charge in the COOO- conjugate base, *cannot* at the same time delocalize by resonance to form a pi bond with the central O atom in any resonance contributor (ie. regardless of whether the central O atom has delocalized it's lone pair unto the acyl C atom or not), because the central O atom is unable to accommodate an expanded octet configuration (forcing the central O atom to accept a pi bond from the terminal O- atom will result in the central O atom having either 2 lone pairs + 3 bond pairs in a resonance contributor, or 1 lone pair + 4 bond pairs in another resonance contributor, which cannot be done because period 2 elements only have 1 x 2s orbital and 3 x 2p orbitals, and do not have vacant, energetically accessible 3d orbitals to accommodate an expanded octet).

If (like all other H2 students) you've no experience with drawing resonance contributors and hybrids, the above will be tough for you to understand. It's best if you get your school teacher or private tutor to draw them out for you.

Of course, Cambridge will be delighted if you are able to do so (ie. draw out the resonance contributors and hybrids) as part of your explanation, but at the very least, giving your school's (ie. Singapore JCs') simpler answer will be sufficient to obtain the 1 mark for this question. No worries, when the question goes beyond the H2 syllabus and it's not fair to everyone*, it's still fair : it's a bell-curve afterall.

*Of course, H3 Chem, Olympiad Chem, and BedokFunland JC students will know how to draw out resonance contributors and hybrids. It's still fair because they've put in additional effort to understand Chem deeper. Why else join H3 Chem, Olympiad Chem, and BedokFunland JC?

UltimaOnline SG

Hi ultimaonline.

Very good, Rapidestlime. That is correct.

UltimaOnline SG

RI 2015 Prelim

Paper 1

Q12. How do I infer option B?

Q36. Why is option 2 wrong? Cl is reduced frm 0 in cl2 to -1 in cl- and oxidised to +5 in cl03-, so technically it is reduced right?

Paper 2

Q4bi. Why is the rate determining step not the step involving the formation of nitronium ion? Generation of nitronium ion involves 2 neutral molecules while the next step involves a positively charged ion so isn't the reaction faster in second step?

2015 RJC P1 Q12. Vol of vessel = 2dm3, hence 0.1mol = 0.05M. Concordantly, at 4 mins, molarity increases from 0.05 to 0.10.

2015 RJC P1 Q36. Yes, but "reduced" implies "reduced only" while here Cl is "disproportionated", ie. "both reduced and oxidized". Blame RJC teacher if you disagree, but even Cambridge often 'chut' ambiguous stunts like these, and Singapore JC teachers often disagree on the best answer.

2015 RJC P2 Q4bi. Because even temporarily losing aromaticity is a hugely destabilizing, reluctant step.

2015 HCI P1 Q25. The MCQ asks for "best describes". NO2 isn't itself a greenhouse gas (that would be N2O, not NO2), it only contributes to the formation of tropospheric ozone which is a greenhouse gas (in contrast, the beneficial ozone layer refers to stratospheric ozone). Instead, NO2 pollutant is far more damaging to humans and the environment as a major contributor to acid rain.

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2012 TYS Paper 2

Qns 5b(i): The answer that my school gave was that "All molecules present in the same chirality, that us, only one enantiomer is present." But can I just say that the lactic acid contains chiral carbons that's why it causes the rotation of th eplane polarised light?

Your school teacher's answer and your own answer are both incomplete. You also need to define what a chiral carbon is, and/or what an optically active molecule entails, as part of your answer.

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MJC 2015 Prelims Paper 2

1e) For the safety hazard, can I write chemical reagents like ethanal are corrosive when it is in contact with the skin and eyes?

3a) Where did the ans get 2/4 from? I am also quite weak at approaching this type of qns where do you start?

3d) Why doesnt the ans include bond length? How do you know when the ans requires you the write bond length or bond energy?

Will I be penalised if I write PCl5 at rtp instead of anhydrous PCl5 at rtp?

Q1e) Ethanal is toxic and carcinogenic, not corrosive.
https://en.wikipedia.org/wiki/Acetaldehyde#Toxicity

Q3a) Instead of blindly memorizing and applying a mathematical formula, use common sense. For a uninegatively charged species : when mass is 37g, deflection is merely -1 degrees. Hence when mass is 1g (ie. much lighter), deflection should concordantly be -37 degrees (ie. larger magnitude deflected). Since mass of He is 4g, hence deflection should be -9.25 degrees (if uninegatively charged). Since He nuclei is dipositively charged, hence deflection should be +18.5 degrees.

Q3d) Bond length here is only indirectly relevant (since bond length is inversely proportional to bond strength, ceteris paribus). It is bond strength which is directly relevant. Hence, there is no need to (ie. no marks awarded for) mentioning bond length.

Q4ai) It's ok if you don't specify "anhydrous", as long as you do *NOT* specify aqueous (aq) (which would be marked wrong).

Same for all reagents which need to be anhydrous, eg. LiAlH4 reducing agent, acyl halides for nucleophilic acyl substitution, Grignard nucleophilic & reducing reagents, AlCl3 or FeBr3 Lewis acid catalysts for halogenation via electrophilic aromatic substitution and for Friedel-Crafts alkylation & acylation, etc.

UltimaOnline SG

pg 57 and 58 made easy inorganic book

hydrolysis of both Mg2+ and Al3+ produces 1 mol of HCl.

However, pH of soln containing Al3+ is lower than Mg2+. means concentration of H+ is higher for the hydrolysis of Al3+ compared to Mg2+.

so what does 1 mol entails? it doesn't really mean equal amounts of a particular substance, even if two seperate solns have 1 mol of that particular substance? always must look at position of equilibrium, which one lies further to the right and, which equilibrium has its forward reaction of a greater extent to the right?

one way is to use ICE table?

You cockanaden! If hydrolysis of both Mg2+ and Al3+ produces 1 mol of HCl, wouldn't the pH be the same?!? Obviously the extent of hydrolysis is significantly greater for AlCl3, and lesser for MgCl2, hence the different pH values. Cambridge will require you to explain why, and the balanced equations will (of course) be required.

If a single source or book confuses you, cross-reference with other sources and books, to come to a better understanding of the subject matter. If you did so in this case, and you *still* came to the cockanaden conclusion that "hydrolysis of both Mg2+ and Al3+ produces 1 mol of HCl", then you might as well voluntarily withdraw from the 2016 A level exam, save yourself the pain of waiting for your U grade results in March 2017.

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