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

last SG · 2010-07-26T23:40:35+00:00

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2011 H2 'A' Level Paper 2 Qn.

Qn : Draw the appartus required for heating under reflux.

Ans : http://www.rod.beavon.clara.net/reflux.htm

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Vluos posted :

Yep I'm really confused over these definitions and its differences, could someone explain them to me? Thanks!
1. Induction/Inductive effect (isnt this just polarity??)
2. Conjugation
3. Hyperconjugation
4. Resonance
5. Delocalisation

There. is. a. reason. why. Wikipedia. exists.

http://en.wikipedia.org/wiki/Inductive_effect
http://en.wikipedia.org/wiki/Resonance_(chemistry)http://en.wikipedia.org/wiki/Hyperconjugation

Not that any of these concepts will make a difference to your Paper 1.

Inductive effects can be due to two reasons :

1) Formal charges, if any. Eg. In NH4+, each H atom is extra-strongly delta +ve, due to the +ve formal charge on the N atom, causing the N atom to be extra-strongly electron-withdrawing by induction.

2) Electronegativity differences, in the absence of formal charges. Eg. in NH3, each H atom is strongly delta +ve, due to the greater electronegativity of the N atom compared to the H atom, causing the N atom to be strongly electron-withdrawing by induction.

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1. 3 compouds A,B,C have the same molecular formula C6H12. Compounds A and B react with H2 in the presence of Ni to yield pdts with molecular formula C6H14. Vigorous oxidation of A yields only one organic product with structural formula CH3COCH3. Under similar conditions, B yields (CH3)2CHCOCH3 as the only organic product. The compound C does not absorb H2. It also does not decolourise acidified KMnO4. Determine structures of A, B and C

2. An organic compound A with a molecular formula C9H12 does not decolourise bromine, however upon addition of iron filing, compound B is formed. When A is strongly heated with acidified potassium permanganate, a colourless gas C is evoled and diprotic acid D is obtained.
A) deduce structures of A B C D, explain how you obtained the structures
B) Suggest possible isomers of A which will provide the same set of observations

Solutions :

Q1.
A is (CH3)2C=C(CH3)2.
B is (CH3)2CHC(CH3)=CH2
C is cyclohexane.

Q2.
A is a benzene with two substituents (one methyl and one ethyl) on either the ortho, para or meta positions to each other. Hence there are 3 possible answers for A.
B will be A (whichever isomer you gave as the answer) having undergone an electrophilic aromatic substitution with the Br+ electrophile substituting away a proton on the benzene ring. The Fe is oxidized to Fe3+ (with Br2 reduced to Br-) and consequently forming the Lewis acid catalyst FeBr3 to generate the positive formal charged and thus reactive Br+ electrophile that can be readily attacked by the (relatively) weak benzene ring nucleophile, despite not having any strong activating (ie. electron-donating by resonance) substitutents present..
C is carbon dioxide.
D is ortho or para or meta benzene-dicarboxylic acid (depending on your answer for A).

Ethylbenzene will simply be oxidized (only by hot KMnO4, but not K2Cr2O7) to benzoic acid and 1 CO2.
Propylbenzene will simply be oxidized (only by hot KMnO4, but not K2Cr2O7) to benzoic acid and 2 CO2, and so on.

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1. Pyrogallol is a white crystalline powder and a powerful reducing agent that is used to absorb oxygen. When 30 cm3 of a gaseous hydrocarbon is exploded with 200 cm3 of oxygen, there is a contraction of 90 cm3. On further treatment with alkaline pyrogallol, there was a reduction of 20 cm3. What is the molecular formula of the hydrocarbon? All volumes are measured at r.t.p.

2. When a large current was passed through acidified aqueous copper (II) sulfate, there was simultaneous liberation, at the cathode, of x mol of copper and y dm3 of hydrogen measured at s.t.p. How many moles of electrons, in terms of x and y, passed through the solution?

3. To determine the mass of arsenic present in a sample of pesticide, all the arsenic was first converted to AsO4^(3-). 1.25 x 10^(-3) moles of AgNO3 was then added to precipitate AsO4^(3-) as Ag3AsO4. The excess Ag+ ions needed 3.64 cm3 of 0.054 mol dm^(-3) KSCN to form AgSCN. Calculate the mass of arsenic (Ar=74.9) present in the sample of pesticide, correct to 3 decimal places.

4. Which of the following is/are chemically unstable when left to stand in air?
1 A solution of potassium hexacyanoferrate(III)
2 A solution of iron (II) chloride
3 A mixture of aqueous sodium hydroxide and iron (II) sulfate

A. 1, 2 and 3 are correct.
B. 1 and 2 are correct
C. 2 and 3 are correct
D. only 1 is correct.

Hints / Partial Solutions :

Q1.
Such questions are often ambiguous. But assuming that the contraction refers to overall contraction, but at rtp, then 230 - 90 = 140cm3 refers to excess O2 + CO2 generated. Since pyrogallol removes O2, the further reduction of 20cm3 tells you that 20cm3 is excess O2, hence CO2 generated is 140-20 = 120cm3. From this data you can determine the formula of the hydrocarbon, using stoichiometric ratios of C to H in the products.

Q2.
Each mole of Cu2+ requires 2 moles of electrons to reduce. Hence x moles of Cu deposited = 2x moles of electrons.
Each ydm3 of hydrogen gas = y/22.4 moles of hydrogen gas, which requires 2(y/22.4) moles of electrons to generate (since 2H+ + 2e- ---> H2)
Hence answer = add up total number of electrons in terms of x and y.

Q3.
Total Ag+ - excess Ag+ = Ag+ which precipitated out as Ag3AsO4.
Hence moles of Ag+ used / 3 = moles of AsO4 3- = moles of As. Sample mass = moles x molar mass.

Q4.
1 A solution of potassium hexacyanoferrate(III)
2 A solution of iron (II) chloride
3 A mixture of aqueous sodium hydroxide and iron (II) sulfate

Check Data Booklet to see if the standard cell potential for any of these 3 species (including subspecies such as Fe2+ and Cl-, within the FeCl2 species for example) being oxidized by atmospheric oxygen is feasible under standard conditions.
Two out of three of the options above, should react with atmospheric oxygen.

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Easy 'A' Level Qn / Challenging 'O' Level Qn

FA 1 contains Fe3+ ions. In an experiment it was that 25cm3 of FA 1 required 18cm3 of .2 mol/dm3 acidified KMnO4(aq) for a complete reaction. what is the volume of 0.150 mol/dm3 acidfied K2Cr2O7(aq) needed to react with 25 cm3 of FA 1?

Solution :

Find moles of KMnO4 used.

Write balanced reduction half-equation for MnO4- to Mn2+.

Apply stoichiometry to find moles of Fe2+ present, by considering moles of electrons transferred.

Write balanced reduction half-equation for Cr2O7 2- to Cr3+.

Based on moles of Fe2+ present (found earlier; this question is kind enough to give you the same volume of FA1 used in both experiments, I would have changed the volume to make the question more fun for my students),

Apply stoichiometry to find moles of Cr2O7 2- required, by considering moles of electrons transferred.

Using Molarity = Moles / Volume (dm3), solve for volume of K2Cr2O7 required.

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fafahahalala asked :

Hey can someone help me out with a few tys mcq questions? Thank you very much.

Q1) when two liquids are mixed, heat may be evolved if intermolecular bonds formed are stronger than those broken, even if there is no chemical reaction. Which pair of liquids, when mixed, will give out heat? (Chem bonding Tys nov 2006 q4)

A) ch2cl2 and (ch3)2co
B) chcl3 and c6h14
C) ccl4 and (ch3)2co
D) ccl4 and ch3ch2oh

I've 2 copies of Tys with 2 different answers A , C. So I don't really know which is the correct ans. Mind explaining too? Thanks alot.

Q2) after doing MCQs for organic chem. I realized quite a few tys questions required us to assume that phenol can straight away react with acyl chloride without the need to generate the phenoxide ion by reacting with naoh(Aq). Case in point are questions (n2010/I/q39) , (n2008/I/q27). So do we just assume that phenol can react with acyl chloride? Btw all the above q can be found in the Ty's section carboxylic acids and derivatives.

Q3) nov2009/I/29 , nitrogen compounds. It's the q on insulin. I couldn't place the question in this post as this structure is very complicated. The ans is b. my question is why the amino group of the side chain does not break up even with 6 mol of hcl? Don't tell me they broke up initially and form back via condensation..

Thanks alot. :)

Q1.
By elimination, only A has a chance of being exothermic. For comparable number of electrons, permanent dipole - permanent dipole van der Waals (ie. Keesom forces) may be considered stronger than permanent dipole - induced dipole van der Waals (ie. Debye forces) may be considered stronger than instantaneous dipole - induced dipole van der Waals (ie. London Dispersion forces).

Q2.
Acyl halides are indeed so feminine & chio (ie. electrophilic) that even guys (ie. nucleophiles) with weak balls (ie. not so electron-rich) are able to sexually (ie. chemically) react with. Deprotonating phenol to generate the stronger nucleophile phenoxide ion (ie. O atom with a -ve formal charge instead of merely a -ve partial charge) may in theory lower Ea and increase rate of reaction, but you've to be careful not to use excess NaOH (and it must be anhydrous!), else you'll be wasting your precious acyl halide chiobu that will be brutally ravaged away by your hired OH- ion mercenaries / soldiers / gangsters, as well as the horny water molecules if you had used aqueous NaOH.

Q3.
It does indeed react; the correct answer is C.

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kickme begged for help :

Covert CH3CH2Cl to CH3CH2CH2NH2.

I understand that the NH2 can be obtained by nucleophilic substitution by reacting it with ammonia. But I dont know how I could obtain another CH2

Another conversion I am having trouble with is : Covert CH3CH2CH2Br to CH3COOH (I need a hint)

Q1. Attack with CN- nucleophile, then reduce to amine.

Q2. Generate alkene by eliminating HBr, then oxidative cleavage with hot KMnO4.

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Enthuboy_93 posted :

Q3.
Why in these two compounds the carbon atom in the reactant change from being sp3 hybridised to sp2 hybridised in the intermediate?

I. ethylbenzene + CL2 in UV light
II. C6H5C(CH3)2i + NaOH(aq), heat

Reaction I is free radical substitution, during which the sp3 C atom enters into a hybridization geometry directly in between sp2 and sp3. Whoever said (eg. prelim paper answer, JC lecture notes, JC teacher, etc) the sp3 C becomes fully sp2 during the intermediate, got it wrong.

Reaction II is SN1 nucleophilic aliphatic substitution, sp3 C atom temporarily becomes a sp2 C carbocation intermediate, after the halogen leaving group is eliminated.

Enthuboy_93 posted :

Q4.
Chlorine, in the presence of aluminium chloride, undergoes a substitution reaction with benzene forming chlorobenzene. When (1.1- dichloroethyl)benzene reacts with iodine monochloride, ICl, in the presence of aluminium chloride, a similar reaction occurs and compound U is formed.

Which product will be present in the greatest yield when U is heated under reflux with aqueous sodium hydroxide? All options contain benzene ring.

A. CL-C6H4-COCH3
B. HO-C6H4-COCH3
C.i-C6H4-COCH3
D.HO-C6H4-CCL2CH3

Answer is C. Due to resonance delocalization of the lone pair on the I atom (directly bonded to the benzene ring) forming a pi bond with the benzene ring, the C-I bond (of the resonance hybrid) has partial double bond character and does not cleave readily in the presence of OH- nucleophiles (extremely high temperature and pressure is required to hydrolyze aryl halides). A secondary reason for the resistance of aryl halides against hydrolysis, is due to [Guys only like Girls, Guys don't like Guys] the repulsion between the two electron-rich nucleophilic species : the benzene ring and OH- ion.

The OH- nucleophile substitutes away both chlorine atoms of the geminal dihalide group in the alkyl side chain on the benzene ring, generating a geminal diol, which due to the close proximity of the two OH groups, readily undergoes dehydration to generate the aryl ketone product.

Enthuboy_93 posted :

5. CH3CH2OH + NaBr, heat can give CH3Ch2Br?

No, you'll need HBr or PBr3 to do the job. Mechanism for the latter involves the alcohol guy (nucleophile) attacking the PBr3 girl (electrophile). Mechanism for the former involves upgrading the poor leaving group OH into an excellent leaving group H2O+ by simple protonation (that's the role of the H+ in HBr), so that the Br- guy (nucleophile) can attack the partial +vely charged C atom girl (electrophile).

Did you know?
The +ve formal charged O atom is even more electron-withdrawing than usual, and enhances the magnitude of the partial +ve charge on the C atom, making it more electrophilic in this way, in addition to itself (the H2O+ group) being a far more viable leaving group for elimination, compared to the OH (since OH- has high charge density and is thus unstable).

Enthuboy_93 posted :

6. C6H5C(CH3)2 + KMnO4, H+ and heat can give C6H5COOH?

First of all, your reactant is faulty. You probably meant C6H5C(CH3)3 or C6H5CH(CH3)2. If the former, then no, KMnO4 is helpless and cannot oxidize it, as there is no H atom available to be removed. If the latter, then yes, heated under reflux with KMnO4 will generate benzoic acid.

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fafahahalala is back with :

Sorry. I've another question.

It's a q from section b of mcq.
A-1,2,3 correct. B-1,2 correct. C-2,3 correct. D-1 correct only.

Nov2008/I/q35: (Chem periodicity)
The following represents the electronic configuration of both a group ii and a group vii anion.
1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6
The radius of the anion is approximately twice that of the cation.

Which reasons explain the difference in size?

1) the cation has a greater nuclear charge than the anion.
2) there is more electron shielding in the anion than in the cation.
3) on forming the anion from its atom, the extra electron repulsion makes the ion much larger.

Lol sadly I've 2 Differing answers to this q again. A or D.
Which is the correct one?

(Correct) Answer :

Only 1 is true.
The shielding effect (point 2) which is the electron repulsion (point 3) is exactly the same for both, since the electron configuration is the same for both. Point 3 tries to confuse the student, because anions are of course larger than the atom, but in this question we're comparing the sizes of two different ions, and not either ion against it's original atomic form.

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kickme was taken by surprise :

Hi, it's you again! A very big thank you. So, another C atom can be obtained by reaction with CN, interesting!

You're totally welcome. Another way to add on more C atoms, is to use a Grignard reagent, which is essentially a C (carbon or alkyl) nucleophile.

http://en.wikipedia.org/wiki/Grignard_reagent

Cambridge had asked on Grignard reagents (as a data-based question) in last year's 2010 'A' levels. It's certainly worth reading up on.

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Hey. Some qns regarding tys mcq:
1) Is it right to assume that phenol does react with an acyl chloride to form an ester? I know that phenol is a weak nucleophile and it must be converted to a phenoxide ion. So the reaction btw phenol and acyl chloride proceeds at a slower rate than phenoxide ion and acyl chloride?
2)07/p1 qn 3 An excess of cold water was added to 0.3mol of a chloride in third period. 0.6molof hcl was formed. What was the chloride? Ans is pcl5.i thought pcl5 reacts in limited water to form pocl3 and 2hcl? Why cant the ans be mgcl2?

Q1.
Acyl halides are indeed so feminine & chio (ie. electrophilic) that even guys (ie. nucleophiles) with weak balls (ie. not so electron-rich) are able to sexually (ie. chemically) react with. Deprotonating phenol to generate the stronger nucleophile phenoxide ion (ie. O atom with a -ve formal charge instead of merely a -ve partial charge) may in theory lower Ea and increase rate of reaction, but you've to be careful not to use excess NaOH (and it must be anhydrous!), else you'll be wasting your precious acyl halide chiobu that will be brutally ravaged away by your hired OH- ion mercenaries / soldiers / gangsters, as well as the horny water molecules if you had used aqueous NaOH.

Q2.
The charge density of Mg2+ is not sufficiently high to generate 2x the moles of HCl. It only undergoes partial hydrolysis to generate a small fraction (relative to the moles of MgCl2) of HCl.

Temperature is actually more important than limited/excess water. (Just as temperature is the more important factor when determining oxidation of alkenes to diols versus oxidative cleavage, over other factors such as pH and concentration).

Hence in cold water, 1 mole of PCl5 reacts with 1 mole of H2O to generate 1 mole of POCl3 and 2 moles of HCl.

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Forgiveme asked :

help me pls TYVVVVM :qmarkshy:

QNS 30 right , how come nitro benzene the carbon is sp3 ?

QNS 12 ermm , how do i start this ?

QNS 2 ? is it find no. of mole den times 6.02x10^23 ? still cant find .

qns 7; how come half life is not 0.144 ..i got tat lei ..

qns 13 ; isnt the higher the Ksp the least soluble ?

Q30. There is a sp3 hybridized C atom only in the intermediate, which is not shown in the question (only the reactants and products are shown)..

Q12. Do an ICE table. Solving for x, you get Equilibrium pressures of 2, 14, 14. Using the kinetics formula (your JC may not have taught you this formula, so learn this now) of [Desired Molarity] / [Initial Molarity] = (1/2)^n where n = number of half-lives for an overall 1st order reaction. You get n = 3, hence time taken = 3 x 20min = 60min.

Q2. You must first find the molarity of pure water. Molarity = moles / volume. In 1dm3, moles = sample mass / molar mass = 1kg / 18g = 55.56 moles. Hence [H2O] in pure water = 55.56 mol/dm3. These steps (described above) are also required for the challenging exam question : what is the pKa of H2O at 25 deg C? Ans : 15.74.

Q7. There is no such thing as "half-life of an experiment". Option 1 is sheer nonsense.

Q13. You can compare Ksp of two species to determine their relative solubilities, only if their stoichiometry is the same, eg. AgCl vs AgBr, or PbCl2 vs PbBr2. When the stoichiometry is different, eg. AgCl vs PbBr2, you cannot just look at their Ksp values. You must work out their molar solubilities (calculated from Ksp) and/or check which species will have its ionic product exceeding Ksp, at any given concentration of the common ion.

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Nish asked :

O can i ask, does k2cr2o72- oxidise alkyl benzene to form benzoic acid? thx!

No. K2Cr2O7 does not oxidize alkenes, alkyl benzenes, methanoic acid and ethandioic acid. Only KMnO4 can oxidise these classes of compounds.

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Rainbowlovex asked :

Use of data booklet is relevant to this question.

In the aromas red sands aquifer, the drinking water source for part of California, there are high levels of soluble, toxic chromium(VI) compounds.
Which compound in the aquifer's sands is most likely to be responsible for the formation of the chromium(VI) compounds from the sparingly soluble chromium(III) bearing rocks?

A) Al2O3
B) CuO
C) Fe2O3
D) ZnO

Answer is (C)

For such questions, when there are more than one possible answer, choose the best answer. In this context, to oxidize chromium from an OS of +3 to +6, choose the strongest oxidizing agent among the options, ie. the species with the most +ve reduction potential.

The reduction potential of Fe3+ to Fe2+, is more positive, than the reduction potentials of Al3+ to Al, Cu2+ to Cu, and Zn2+ to Zn.

-----------------------------------------------------

Bonus Question

Rainbowlovex said :

Okay! Thank you! But can I ask you an extra qn I'm curious about? " I have checked the data booklet and found that the e values are lower than that of chronium. Is it due to solubility or acidity?" Are Fe2O3 and CuO soluble?

All the oxides listed here are insoluble (only ZnO is very slightly soluble). Although the Data Booklet redox potentials are for aqueous ions, but if all are insoluble, then it's still a fair comparison using the aqueous redox potentials.

Although redox potentials, solubilities and acidities are all related (because these are all properties determined by effective nuclear charge, electronegativities, ionization energies, charge densities, lattice dissociation enthalpies, hydration enthalpies, etc), but there is no simple or direct correlation between them at 'A' levels.

Perhaps this related question will interest you :

Generally, the more reactive a metal, the more positive it's oxidation potential. So why is it that K is more reactive than Li (explain why this is so), but yet the oxidation potential of Li to Li+ is more positive than the oxidation potential of K to K+ (explain why this is so)?

Bonus Answer :

K is more reactive than Li because the 1st ionization energy of K is less endothermic than the 1st IE of Li. This is due to the greater atomic radius and shielding effect of the K atom, allowing it to lose a valence electron more readily.

However, oxidation potentials as given in the Data Booklet, is defined as for the aqueous state. And by Hess Law, oxidation enthalpy = ionization enthalpy + hydration enthalpy.

The 1st IE of K to K+ may be more favourable (ie. less endothermic) than that for Li, but the significantly more favourable (ie. more exothermic) formation of stronger ion - permanent dipole bonds between the higher charge density Li+ ion and polar water molecules during the hydration process, more than compensates for the more endothermic 1st IE of Li, compared to for K.

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ACJC 08
1. An equimolar mixture of A and B at an initial total pressure of 2 atm was allowed to reach equilibrium. The equation for the reaction is
A(g) + 2B(g) <---> 2C(g)
At equilibrium the total pressure was reduced to 1.8 atm. What is the value of Kp?

Solution :

A + 2B <---> 2C

1 | 1 | 0

-x | -2x | +2x

1-x | 1-2x | 2x

Since total = (1-x) + (1-2x) + (2x) = 2-x = 1.8 atm, hence x = 0.2

Hence Equilibrium partial pressures = 0.8 | 0.6 | 0.4

Hence Kp = (Pc)^2 / [(Pa)^1 x (Pb)^2] = 5.556 x 10^-1

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NJC 09
2. Which sequence represents the solutions in increasing order of their pH?
given Ka of ethanoic acid is 1.75x10^(-5)

Solution W: 10^(-8) mol dm^-3 HCl
Solution X: 10^(-3) mol dm^-3 CH3COONa
Solution Y: 10^(-6) mol dm^-3 CH3COOH

Answer :

At such low molarity of HCl, you have to add the [H+] generated from the auto-ionization (aka auto-dissociation) of water, to the [H+] generated from the dissociation of HCl.

Doing so will get you the correct answer of (most acidic) Y, W, X (least acidic).

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kickme exuberantly declared :

Thanks!!!!!!!! we can say that any alkyl groups attached to benzene will be oxidised to -COOH under strong oxidation.

Except tertiary alkyl groups eg. C6H5C(CH3)3, since these have no H atom available to be removed, and thus cannot be oxidized by KMnO4.

The only way to oxidize such compounds (and in fact, *any* organic compound, eg. ketones, that cannot be further oxidized by chemical oxidizing agents), is to burn it alive. Students often forget that combustion is also considered oxidation.

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