2016 H2 Chemistry JC1 & 2 students post your questions here

last SG · 2016-02-03T02:17:14+00:00

Hello, would like to ask a questionAs to why c=c is stronger than c-c bond, is it because of the additional pi bond, or because the sigma bond in c=c is shor...

nicolemantou SG

HCI 2014 paper 2
Q4a) why doesnt the reagent dilute HNO3 need conc H2SO4 for electrophilic sub? Why dilute HNO3 dont react with compound C?

UltimaOnline SG

HCI 2014 paper 2
Q4a) why doesnt the reagent dilute HNO3 need conc H2SO4 for electrophilic sub? Why dilute HNO3 dont react with compound C?

If the reagent were conc HNO3 + conc H2SO4 (ie. strong nitrating agent), then all 4 molecules would be nitrated. The question deliberately specified dilute HNO3 (ie. weak nitrating agent), to see if you can tell which benzene rings are sufficiently activated to react with dilute HNO3 (ie. weak nitrating agent).

Only D is sufficiently activated, because of the 2 electron-donating by resonance groups. If you're wondering why only mononitration occurs, it's because while OH and OR are both activating as they donate electrons by resonance, OH is a stronger activator than OR (why? I've already explained this somewhere among my 100+ questions on my BedokFunland JC website, go ask your school teacher or private tutor if you don't know), *and* NO2 group is strongly deactivating (ie. electron-withdrawing by *both* induction and resonance). Concordantly, while D is the only molecule sufficiently activated to undergo nitration with dilute HNO3 (ie. weak nitrating agent), but only mononitration occurs, ortho to the OH group.

C is not sufficiently activated (actually it's not activated at all), because by induction, the vinyl side-chain's sp2 C atom (ie. not being an sp3 C atom) does not donate electrons by induction. By resonance, the vinyl side-chain is in equal measure electron-donating and electron-withdrawing (vis-à-vis the benzene ring), and is hence neither. Concordantly, C is neither an activated or deactivated benzene ring, and requires conc HNO3 + conc H2SO4 (ie. strong nitrating agent) to undergo nitration.

gohby SG

Hi UltimaOnline, I have 2 questions here:

Q1: In a question which requires “state what you would observe when A is converted to B” or “state the observations”, would a response of “white fumes are evolved” (instead of “white fumes of HCl (g) evolved”) suffice? I am of the opinion that given one cannot possibly observe the identity of the fumes evolved, a response of “white fumes are evolved” is sufficient and correct, and any further details as to the identity of the gas is superfluous (and wrong).

Q2: NYJC/2012/P1/Q18 [Ans: D]

When I mix A & B or A & C together, I will obtain AgCl (s), which will be the white ppt soluble in ammonia as shown in row 1. However, I do not understand the observations shown in row 2. Mixing A & C should produce iron (II) nitrate and silver chloride - so where is there no white ppt produced, as per row 1, and why is there a grey ppt (most likely it is iron) formed?

Thank you, UltimaOnline :)

UltimaOnline SG

Hi UltimaOnline, I have 2 questions here:

Q2: NYJC/2012/P1/Q18 [Ans: D]

Thank you, UltimaOnline :)

Hi Gohby, no prob :)

Q1. Yes, you're correct.

Q2. Ag+ from AgCl(s) is reduced to Ag(s) a grey ppt insoluble in NH3(aq), and Fe2+(aq) is oxidized to Fe3+(aq).

gohby SG

Hi UltimaOnline - I've some further enquiries here:

[YJC 2012 Prelim]

I get that there are 5 dative bonds in this complex, but how does that point to the electronic configuration of the cobalt cation, given that in this case, I think it would have been the 1x4s, 3x4p and 1x4d orbitals which will hybridise to accept the lone pairs from the dative bonding?

NYJC/2012/P1/Q17

My thoughts: When I add iron (II) sulphate to bromine and chlorine, a redox reaction will occur and Fe3+ & chloride/bromide will be formed. When I then add sodium hydroxide to the mixture, iron (III) hydroxide will be formed. However, how can I distinguish between the solution containing chloride and that containing bromide? According to the E⦵ values, neither chloride nor bromide will be able to reduce iron (III) hydroxide to iron (II) hydroxide.

NJC/2012/P1/Q15 [Ans: B]

Upon heating with barium carbonate, the barium oxide formed will react vigorously with water - however, what is the gas evolved during the "effervescence" stated in the question? Unless the question used the term "effervescence" to describe the vigorous reaction - which would be an inaccurate characterisation, no?

Thank you for your help. :)

UltimaOnline SG

Hi UltimaOnline - I've some further enquiries here:

[YJC 2012 Prelim]

NYJC/2012/P1/Q17

NJC/2012/P1/Q15 [Ans: B]

Thank you for your help. :)

NJC Qn : this question has come up time and time again, because it is, as you suspect, a faulty question. While there is a slight difference in the color of solution (because color isn't just dependent on cation, but on ligand substitution equilibria as well), it is not sufficient to reliably distinguish between iron(III) chloride vs bromide. And while the differing magnitudes of thermodynamic feasibility translates into differing extents of equilibrium positions (and thus differing amounts of ppt obtained), this isn't an appropriate pedagogical assessment context for the H2 syllabus. In summary, this is a faulty question, but which by elimination still finds that option to be the most feasible (for MCQs, choose the 'most correct' answer, even if it isn't 100% correct).

YJC Qn : for such questions in the context of the A level H2 syllabus, instead of considering the transition metal orbitals in which the dative bond electrons reside in, treat such questions as simply asking you to deduce the OS of the metal cation, and concordantly, the electron-configuration. In other words, disregard the dative bonds residing in the transition metal orbitals.

NJC Qn : BaCO3 is significantly more stable than CaCO3, and at temperatures which decompose CaCO3, BaCO3 remains undecomposed and thus results in "effervescence" when mixed with HCl(aq).

No prob, Gohby :)

Ng.keebin SG

from qifei: in the reaction between mg and hcl, why is the effective collisions between mg atoms and hydrogen ions but not mg and cl- ions?

UltimaOnline SG

from qifei: in the reaction between mg and hcl, why is the effective collisions between mg atoms and hydrogen ions but not mg and cl- ions?

Because the reaction is *not* an ionic precipitation reaction (since MgCl2 is soluble), but instead it is a redox reaction (H+ from acid reduced to H2 gas, Mg oxidized to Mg2+) between Mg and H+ (whether from HCl(aq) or H2SO4(aq) etc).

Edited to add : and it also can't be a redox reaction between Mg and Cl-, because both Mg and Cl- can only be oxidized, but not reduced further. A redox reaction needs someone to be oxidized, and someone to be reduced. Mg and Cl- are both reducing agents, so no reaction btwn them.

Flying grenade SG

theophilus SG

Yo is there anyway to get ready for private candidate practical paper 5 for chem?? Or is it smokable? And because private candidates do a different practical paper, are they marked similarly? :O Because major handicap as I heard in JCs they tend to get like full marks or sth for SPA

UltimaOnline SG

There isn't much you can do to prepare for Pract Paper. With a solid theoretical foundation (ie. as long as you know your Chem), just follow the Pract Paper instructions carefully, record down all observations carefully, etc.

The only feasible way to prepare for Pract Paper, is by preparing for your Planning Qn, for obvious reasons (in fact, for the new syllabus, the Planning Qn is part of the Pract Paper, no longer Paper 2). For this purpose, buying and studying Chan Kim Seng and Jeanne Tan's H2 Chem Planning Book is the best (and kinda only) way for you to prepare for your Planning Qn + Pract Paper.

And no worries about losing out to school candidates. In practice (pun), all things considered (I won't go into details), they do not have any advantage (or disadvantage) over private candidates.

MightyBiscuits SG

Hi UltimaOnline, in George CHong's organic chem book pg 43, it gives an example of FRS in the methyl group in methylbenzene which makes me confused as I had always thought that FRS only occurs in alkanes.

If free radical substitution occurs can occur in alkene, how does it work?

Also for George Chong's Organic chem book pg 45, I dont get how the ratio is formed. Thanks

UltimaOnline SG

If free radical substitution occurs can occur in alkene, how does it work?

Also for George Chong's Organic chem book pg 45, I dont get how the ratio is formed. Thanks

Alkyl side-chain of benzene is considered an 'alkane'.

To determine product distribution of halogenation via free radical substitution, mathematically combine both factors :

No. of H atoms substitutable x Stability of alkyl radical intermediate (also known as Kinetic Reactivity of Pri vs Sec vs Tert H atoms)

Take the halogenation of butane for instance. In terms of "No. of H atoms substitutable", you'd expect 1-halobutane to be the major product (since 6 H atoms vs 4 H atoms). In terms of "Stability of alkyl radical intermediate", you'd expect 2-halobutane to be the major product (since the secondary alkyl radical intermediate to generate secondary alkyl halide 2-halobutane, is more stable than the primary alkyl radical intermediate to generate primary alkyl halide, 1-halobutane ; the more stable the intermediate, the lower the Ea, hence the faster the rate of reaction, resulting in the major product).

Overall, to determine which factor outweighs which to result in the major product, you'd have to combine both factors mathematically (don't memorize the relative stabilities or reactivities, the exam question will provide it, as this factor depends on the specific halogen, as well as specific temperature).

senga SG

2011 TYS question 6 part c v and d ii.

Which other reaction confirms W is aromatic? Explain your answer.

Why can't I use reaction 5 since reaction 5 proves that there is phenol functional group?

Explain clearly why you have placed each of the functional groups in their particular positions.

Why can't the chlorine atom be attached to the benzene ring? Why didn't they specify the adding of NaOH or KOH for reaction 1? Sorry if this question seemed too simple but I'm really confused over here. I thought NaOH/KOH is needed to substitute the chlorine no matter whether it is halogenoalkane or halogenoarene.

Edit. I think I get the answer. Reaction 3 specify that organic product produced has Mr of 138. Which means chlorine is removed. If reaction 3 is not there, can I assume that chlorine is attached to the ring instead of the alkyl group? The answer sheets say that reaction 1 shows that chlorine is attached to the alkyl group instead of the benzene ring, which doesn't make sense for me.

Do I need to remember how to produce phenol from benzene?

UltimaOnline SG

2011 TYS question 6 part c v and d ii.

Which other reaction confirms W is aromatic? Explain your answer.

Why can't I use reaction 5 since reaction 5 proves that there is phenol functional group?

Explain clearly why you have placed each of the functional groups in their particular positions.

Do I need to remember how to produce phenol from benzene?

Reaction 3 is a better choice, as only alkyl benzenes can be oxidized by KMnO4 to hydroxybenzoic acid. Reaction 5 is a weaker reason, as it could imply an acyl halide side chain (although the degree of unsaturation won't fit, but most H2 students won't know how to check for this anyway).

Without OH-(aq), only alkyl halides can be hydrolyzed upon warming (adding OH- would of course speed up the reaction). If the Cl atom was attached to the benzene ring as you suggest, then due to the C-Cl bond strengthened by having partial double bond character in the resonance hybrid, concordantly and consequently not only would OH-(aq) be required (now you understand why Cambridge didn't use OH-(aq)? now you understand the answer sheet?), but a much higher temperature is required (the question specified 'warm'), and besides, the nucleophilic aromatic substitution hydrolysis (via the elimination-addition mechanism involving a benzyne intermediate) of aryl halides to phenol is beyond the H2 A level syllabus.

At University and professional levels, there are many different ways to generate phenol from benzene, but none of these are taught at A levels. In other words, within the H2 A level syllabus, there is no (within syllabus) method you can generate phenol from benzene. Of course, if you're aiming for a distinction A grade, it's always useful to expose yourself beyond the syllabus.

MADstud SG

Hi all, I know this is quite random (and not related to JC) but can anyone help us fill up the below survey form?

We are a group of students from MAD School trying to get survey answered for our marketing project.

Very short one, just 1-2 minutes can finish and capitaland vouchers and other gifts to be won!

Thanks a million!

https://www.surveymonkey.com/r/SW7G9YY

Tvllpit SG

Nice sharing but i can give you advice if you are looking for any homework help you can take online homework services provider help.

Help me with my homework

gohby SG

Hi UltimaOnline,

I would like to seek your help with regard to the following questions:

Q1: DHS/2015/P1/Q14

What’s wrong with Choice B, given that acid strength increases from HCl to HI? Interestingly when I researched on the pH of these compounds in a fixed concentrations their pH are the same. (Source: http://www.aqion.de/site/191)

Q2: DHS/2015/P1/Q34 [Ans B]

Why is the answer not A? I think choice 3 is correct - vanadium oxide gets reduced before being oxidised. (Source: http://www.chemguide.co.uk/inorganic/transition/vanadium.html) Or am I missing out on something here?

Q3: VJC/2015/P1/Q3 [Ans C]

Which one of the following is a possible configuration of a stable M3+ ion in the ground

state?

Remarks: I do not understand this question. Are they asking for the configuration of M or M3+? If it’s the latter, then why is there a need to add the words “in the ground state”?

Q4: VJC/2015/P1/Q14 [Ans C]

Remarks: I thought the answer should have been B instead. This is because at B, since the concentration of lead (II) containing species is at its lowest, [Pb2+] should be at its lowest at V. I didn’t think C was the correct answer, because at any point between 0cm3 to Vcm3, although the ionic product was initially greater than the solubility product, the ionic product being equal to the solubility product at any point through the formation of the ppt.

Many thanks, UltimaOnline for your help! :)

UltimaOnline SG

Hi UltimaOnline,

I would like to seek your help with regard to the following questions:

Q1: DHS/2015/P1/Q14

Q2: DHS/2015/P1/Q34 [Ans B]

Q3: VJC/2015/P1/Q3 [Ans C]

Which one of the following is a possible configuration of a stable M3+ ion in the ground

state?

Remarks: I do not understand this question. Are they asking for the configuration of M or M3+? If it’s the latter, then why is there a need to add the words “in the ground state”?

Q4: VJC/2015/P1/Q14 [Ans C]

Many thanks, UltimaOnline for your help! :)

Hi Gohby, no prob ;)

Q1: DHS/2015/P1/Q14. While acid strength (indicated by pKa) increases from HCl to HI, but due to the leveling effect, in aqueous solutions they are all strong acids which dissociate completely and hence result in the same pH.

Q2: DHS/2015/P1/Q34. You're right. Perhaps (lame, but) the DHS teacher question-setter was nitpicking that "likely" should be replaced with "definitely", since the maximum OS of V is +5 (based on electron configuration).

Q3: VJC/2015/P1/Q3. The question is asking for the configuration of M3+. "Ground state" must always be specified, otherwise there would be many permutations and combinations possible for excited states.

Q4: VJC/2015/P1/Q11 (it's Q11, not Q14). "Ionic product" is always intended to be "initial before precipitation". This is a somewhat unfair question, as the mathematics involved goes beyond A levels, but conceptually still examinable as a sadistic distinction question.

The VJC teacher question-setter's thinking (and feeling rather proud of him/herself for setting such a tricky question, no doubt) is that in sufficiently large excess of halide ions, not only would all the precipitate dissolve, but even the molarity of aqueous Pb2+ ion, would eventually decrease to become even less than at point V, due to large excess of halide ions pulling the position of equilibrium over to the RHS to form the coordination complex ion, ie. [PbCl4]2- and [PbI4]2-, which are not considered as Pb2+(aq).

gohby SG

Hi UltimaOnline,

I would like to seek your help on some of the HCI questions here:

HCI/2015/P1/28 [Ans A]

Remarks: I think the answer is C instead. PCl5 reacts with the alcohol and COOH (and not the phenol) and it reacts with sulphuric due to the close proximity between the COOH and OH.

HCI 2015/P1/25 [Ans C]

Remarks: I’m not entirely sure why the answer is C, but here are my deductions:

A is wrong because even though NO converts CO and unburnt hydrocarbons in the catalytic converter, the carbon dioxide formed is not harmless because it’s a greenhouse gas.

B is wrong because it should be N2O.

D is wrong because the reaction named is dissolution.

I referred to wikipedia to evaluate choice C [https://en.wikipedia.org/wiki/Acid_rain]:

In the gas phase sulfur dioxide is oxidized by reaction with the hydroxyl radical via an intermolecular reaction:

SO2 + OH· → HOSO2·

which is followed by:

HOSO2· + O2 → HO2· + SO3

In the presence of water, sulfur trioxide (SO3) is converted rapidly to sulfuric acid:

SO3 (g) + H2O (l) → H2SO4 (aq)

Nitrogen dioxide reacts with OH to form nitric acid:

NO2 + OH· → HNO3

However I’ve some disquiet about this choice:

Firstly, such detailed knowledge and pathway of acid rain formation seems to be beyond the reach of the syllabus.

Secondly, even if the student possesses knowledge on how exactly acid rain is formed, I wouldn’t say it catalyses the formation of acid rain from atmospheric sulphuric dioxide. This is because sulphuric acid is formed from sulphur dioxide, and the acid contributing to acid rain by nitrogen dioxide is nitric acid. Hence nitrogen dioxide doesn’t catalyse the formation of acid rain from sulphuric dioxide (unless it somehow forms sulphuric acid).

HCI 2015/P1/18 [Ans C]

I can accept the answer because that forms an integral part of the definition of transition element. However, what’s wrong with B as the answer?

HCI 2015/P1/19 [Ans A]

Is there a manner for students to make an intelligent guess/extrapolate from their knowledge in the syllabus to obtain the answer without understanding and appreciating their individual mechanisms? Don’t think the Grignard reagent in step 1 is part of the syllabus so I’m thinking there should be a more efficient way of solving this question.

Many thanks Ultima :)

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