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

last SG · 2010-07-26T23:40:35+00:00

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2010 'A' Level H2 Chemistry Paper 3.

Alcohol J, CxHyOH, is a volatile fungal metabolite whose presence when detected in air can indicate hidden fungal attack on the timbers of a house.

When 0.10 cm^3 of liquid J was dissolved in an inert solvent and an excess of sodium metal added, 10.9cm^3 of gas(measured at 298K) was produced.

When 0.10cm^3 of liquid J was combusted in an excess of oxygen in an enclosed vessel, the volume of gas (measured at 298K) was reduced by 54.4cm^3. The addition of an excess of NaOH(aq) caused a further reaction in gas volume of 109cm^3 (measured at 298K).

Use these data to calculate values for x and y in the molecular formula CxHyOH for J.

TPJC Chem teachers' solution :

In the first experiment, the gas evolved in the reaction with Na metal is hydrogen.
Moles of H₂(g) = 10.9/24000 = 0.000454 mol

Þ moles of liquid J present = 0.000454 x 2 = 0.000908 mol

In the 2^nd experiment, moles of CO₂ evolved = 109/24 000 = 0.00454 mol

Þ moles CO₂ : moles J (C_xH_yOH)

5 : 1 Þ x = 5

Moles of O₂ used in the reaction = (109 + 54.4)/24000 = 0.00681 mol
Note that since CO₂ was formed, it ‘masked’ the O₂ that was used up
Þ Reduction in volume of gas = (vol O₂ used up) – (vol CO₂ formed)
Þ vol O₂ used up = (reduction in vol) + (vol CO₂ formed)

moles O₂ : moles J (C_xH_yOH)
7.5 : 1

Þ C₅H_yOH + 15/2 O₂ ® 5CO₂ + ((y+1)/2)H₂O

16 = 10 + (y+1)/2 Þ y = 11

ii) J is an alcohol that reacts with K₂Cr₂O₇ Þ J is either primary or secondary alcohol.

J can be dehydrated to alkene Þ J contains the -CH2-CH(OH)- structure

Since alkene K is oxidized into ethanoic acid and propanone

Þ K is CH3CH=C(CH3)2

Þ J is CH3CH(OH)CH(CH3)2

SAJC Chem teacher's solution :

When 0.10 cm3 of J is added to sodium metal, hydrogen gas is evolved.

CxHyOH + Na <-----> CxHyO-Na+ + (1/2) H2

No. of moles of hydrogen gas = 10.9 / 24000 = 4.542 x 10-4 mol

No. of moles of alcohol J = 2 x no. of moles of hydrogen = 9.084 x 10-4 mol

When NaOH is added to J, the carbon dioxide gas produced from the combustion is being absorbed.

No. of moles of CO2 = 10.9 / 24000 = 4.542 x 10-3 mol

Since all the C in CO2 is produced from combustion of J, we can find the value of x when finding out the ratio of the no. of moles of CO2 produced to that of alcohol J present initially.

No. of moles of CO2 / No. of moles of J = 4.542 x 10-3 / 9.084 x 10-4 = 5

Therefore, x = 5

When J is combusted in an excess of oxygen, the reduction in the volume of gas is due to the consumption of oxygen to produce water. This reduction in volume of gas is the difference between the volume of oxygen gas consumed in combustion and the volume of carbon dioxide gas produced.

Volume of gas being consumed= 54.4 cm3

No. of moles of gas being consumed = 54.4 / 24000 = 2.267 x 10-3 mol

Taking the ratio between the no. of moles of gas consumed and the no. of moles of alcohol J,

Ratio = 2.267 x 10-3 / 9.084 x 10-4 = 2.5 mol

Since we have determined x = 5 above, we can simplify the equation to,

C5HyOH (l) + ((19+y)/4) O2 (g) 5CO2 (g) + ((y+1)/2) H2O (l)

Hence, difference in no. of moles of gases on combustion = ((19+y)/4) – 5 = 2.5

y = 11

Alcohol J : C5H11OH

Observation : J reacts with acidified K2Cr2O7, J has molecular formula C5H11OH
Deduction : J is oxidised [1/2]. J is a primary or secondary alcohol.

Observation : J dehydrated to form K, K reacts with hot KMnO4 to give equimolar mixture of ethanoic acid and propanone.
Observation : Oxidative cleavage
K has the structure (CH3)2C= and J has the structure CH(CH3)=
K is 2-methylbut-2-ene.
J is 3-methylbutan-2-ol.

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In a reply to a post made by "forgive me" :

Regarding F- vs I-. your conclusions are wrong, and your reasons are also wrong. Further cryptic hints : why does a base behave as a base? what makes a nucleophile a nucleophile? What determines basic strength, and what determines nucleophilic strength?

As for your "amino acids : pH, pKa and pI" qn, to be honest, it'll be more effective to ask your JC teacher or tuition teacher (if you have one) about it. Students need at least one entire tuition lesson to properly understand and correctly apply "amino acids : pH, pKa and pI". It's unfeasible to expect a single forum post to thoroughly enlighten you on this topic.

Nonetheless, I won't leave you at the deep end of the pool. I'll give one (but just one) attempt at explaining this to you here. But it'll be less confusing if you have a tuition teacher to explain this to you in person ( JC teachers are so overworked and tired-out that they will mostly just brush you off with a "no need to know, don't ask so much", which is why there is a real need for tuition teachers).

So is there a need to know all about "amino acids : pH, pKa and pI"? Tis a grey area, in which Cambridge can argue that JC students should be able to deduce the answers to questions such as the one in the 2010 exam, even though JCs don't usually teach much about this topic. It always helps to be prepared to go a little (or if you've the aptitude for it, a lot) beyond the basic syllabus requirement, especially if you're committed to securing a distinction.

Peruse these links (open 3 browsing windows simultaneously).
http://themedicalbiochemistrypage.org/amino-acids.html
https://www.bio.cmu.edu/courses/03231/LecF04/Lec05/aa_titration.pdf
http://wwwchem.csustan.edu/chem4400/aapi.htm

The 3rd link above, notice that the isoelectric point must be such that the alpha ammonium group remains protonated, while the R group amine group remains deprotonated, hence the pH of the solution must be more acidic than 9.6 (the alpha ammonium pKa), but more alkaline than 6.0 (the R group ammonium pKa), hence the isoelectric point will be midpoint of the two pKa values, or (6+9.6)/2 = 7.8

UltimaOnline SG

Hi sorry to trouble again. can you explain the difference in TM and the s-block metals in their reactivity in water to me?thanks.

S-block metals are generally more reactive with water (with the reactivity increasing down the group) and react in a redox reaction with water (the metal being oxidized and water being reduced).

D-block metals are generally unreactive with water, as the effective nuclear charge for the valence electrons are stronger (in turn because additional electrons enter into the d orbitals, which are more diffused than s orbitals and hence provide poorer shielding*), and therefore the oxidation potentials for oxidizing d-block metal atoms into their cationic states, are generally not as positive (ie. not as reactive) as that of s-block metals.

* Sometimes you need to say "d-orbitals provide poor shielding", while at other times you need to say "d-orbitals provide effective shielding", much like The Accountant Joke**, depending on the question. For instance, the former statement would be used to explain why the d-block metals generally have a smaller radius than s-block metals (d orbitals provide poorer shielding compared to s and p orbitals of the same electron shell). The latter statement would be used to explain why the atomic and ionization energies of the d-block metals remain similar (d orbitals provide more effective shielding compared to the s and p orbitals of the next electron shell). Why do we care about the next electron shell? Because unlike s-block metals of (say) period 4, for the d-block metals in period 4, together with the additional proton (as you go from left to right of the Periodic Table), the additional electron goes into the 3-d orbital rather than the 4-d orbital (for a period 4 s-block metal, the additional electron goes into the 4s orbital).

** The Accountant Joke :

A businessman was interviewing job applications for the position of manager of a large division. He quickly devised a test for choosing the most suitable candidate. He simply asked each applicant this question, "What is two plus two?"
The first interviewee was a journalist. His answer was, "Twenty-two".
The second was a social worker. She said, "I don't know the answer but I'm very glad that we had the opportunity to discuss it."
The third applicant was an engineer. He pulled out a slide rule and came up with an answer "somewhere between 3.999 and 4.001."
Next came an attorney. He stated that "in the case of Jenkins vs. the Department of the Treasury, two plus two was proven to be four."
Finally, the businessman interviewed an accountant. When he asked him what two plus two was, the accountant got up from his chair, went over to the door, closed it, came back and sat down. Leaning across the desk, he said in a low voice, "How much do you want it to be, boss?" The accountant got the job.

Source :
http://www.accountingcoursesonline.com/accountantjokes.html

UltimaOnline SG

Originally Posted by forgiveme :
Rate of decomposition of h202 can be determine by withdrawing fixed Vol of small sample of soln at known time into conical flask obtain cold dil h2so4 and titrating it with kmno4. Deduce rate of rxn of h2o2 is first order. I dun understand how come can use graph of kmno4 to find deduce. Ans write condition of h2o2 = mol of kmno4. Bt conc of h2o2 = to tat meh ?

Does HCOOH under give orange ppt On 2,4 dnph since the other side is a carbonyl?

Kinetics :
You're right that the mole ratio isn't 1 : 1. But the mole ratio is directly proportional. So plotting a graph of [KMnO4] is akin to plotting a graph of [H2O2], and a constant half-life is required to prove 1st order kinetics.

Organic :
Once the carbonyl group is bonded to a OH, it's considered a carboxylic acid group and not a carbonyl compound (ie. ketones and aldehydes), even though the C=O carbonyl group is present, as carboxylic acid groups have a different reactivity and hence takes priority. FYI, it is the resonance stabilization of the carbonyl group in carboxylic acids, esters and amides, that reduce the electrophilicity of the carbonyl C atom, to the extent that the weak nucleophile 2,4-DNPH (since it has two electron withdrawing by induction and by resonance nitro groups) is unable to react with carboxylic acids, esters and amides.

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2007 H2 Chemistry Paper 3 Qn 3 (b)(ii)

My question is what exactly is the information given in Data 1. Enthalphy change of formation of Ca2+ (aq)= -543 and f- (aq)= -333 ??

This data is given together with enthalphy change of atomisation of Ca(s), enthalphy change affinity of flourine atom and enthalphy change of formation of CaF2(s).

Do I have to use information from data booklet as well, as it is not stated in the question ?

2006 H2 Chemistry Paper 3Qn 5

Write a balance equation when KI(aq) is added to a solution of Pb2+(aq). A bright yellow precipitate is formed. Further addition of KI(aq) causes the ppt to redissolve.

My Answer:

2KI(s) + Pb2+(aq) ----PbI2(s)+ 2K+(aq)

Yellow ppt of PbI2 is formed.

PbI2 (s) + I- (aq)----[PbI3]- (aq)

Further addition of KI causes the formation complex ion which is soluble.

However, the formula of the complex ion given in the TYS was [PbI4]-. Does that mean my answer is wrong cause I wasn't taught about this in my school. Is there a way of coming up with the right complex ions.

Thanks In Advance :)

Qn 1. The data given in the question suffices (for drawing a Hess Law enthalpy cycle), you do not need any further info from the Data Booklet. But if any exam qn appears to give inadequate data, feel free to quote values from Data Booklet to help you answer the question. It's just that in this instance, further info is not required.

Qn2. Different JCs teach different content. If you intend to get a distinction, you cannot rely merely on notes provided by any JC (even so-called top JCs), but you must be prepared to peruse multiple sources. Your fictitious tri-iodo complex ion will not score the mark. You must give the formula for the tetra-iodo plumbate(II) ion.

No single source (eg. any single JC's notes) will give an exhaustive list of complex ions, their formulae, their geometries and their colours. Cambridge deliberately does not release any list, to reserve the right to ask on obscure complex ions to test the candidate's ability to extrapolate and deduce probable formulae, geometries and colours.

You can find a more complete list of complex ions, their formulae, their geometries and their colours on Jim Clarke's website and Rod Beavon's website.

UltimaOnline SG

Thanks for reply UltimaOnline.

If you don’t mind me asking, what is the equation of the enthalpy change of formation of
the Ca2+ ion or f- ion stated in the qn. Cause I tot enthalpy change of formation is only for substance and not ions. Maybe i am mistaken it is not even enthalpy change of formation, then what is it ?

By Hess Law, formation enthalpy for Ca2+(aq) ion = atomization enthalpy + 1st ionization enthalpy* + 2nd ionization enthalpy + hydration enthalpy.

* There is a technical difference between enthalpy and energy, but for 'A' level purposes these may be used interchangeably, eg. bond enthalpy = bond energy. Enthalpy change is heat change under constant pressure, energy change is heat change under constant volume. If a reaction generates gaseous products, some of the energy is used to push away the atmosphere, and enthalpy will be slightly less exothermic than energy change.

UltimaOnline SG

Amino acids :

The most acidic group (ie. largest Ka, or smallest pKa) is the alpha carboxylic acid group, followed by the 'R' group carboxylic acid group, followed by the alpha ammonium group, followed by the 'R' group ammonium group (weakest acid).

The most basic group (ie. largest Kb, or smallest pKb) is the 'R' group amine group, followed by the alpha amine group, followed by the 'R' group carboxylate group, followed by the alpha carboxylate group (weakest base).

You should be able to explain why, based on "electron-donating by induction 'R' group" and "electron-withdrawing by induction alpha carbon".

UltimaOnline SG

NH3 is a stronger field ligand (not quite the same meaning as stronger ligand, there is some correlation although there are also exceptions, eg. H2O is a stronger field ligand than OH-, although OH- is a stronger ligand than H2O) compared to H2O, which means replacing H2O ligands with NH3 ligands, results in a larger magnitude of d-d* splitting of the metal ion's orbitals. Exactly what effect this has on color, is a case-by-case matter.

Case 1

[Cr(H2O)6]3+ absorbs longer wavelength green light to give a violet coloured complex (although in practice, the counter ion present often substitutes away one or more of the water ligands, and you get green colour).

[Cr(NH3)6]3+ absorbs shorter wavelength violet light to give a yellow coloured complex.

Case 2

[Co(H2O)6]3+ absorbs longer wavelength yellow-violet light to give a blue-green coloured complex.

[Co(NH3)6]3+ absorbs shorter wavelength blue-ultraviolet light to give a yellow-orange coloured complex.

Case 3

[Cu(H2O)6]2+ absorbs longer wavelength orange light to give a blue coloured complex.

[Cu(NH3)4(H2O)2]2+ absorbs shorter wavelength yellow light, to give a deeper blue coloured complex.

UltimaOnline SG

hello
may i know why is pcl5 solid while pcl3 is liquid? thanks

Because PCl5 at room temperature and pressure exists as an giant ionic lattice structure of [PCl4]+ cations and [PCl6]- anions.

Test Yourself :

Draw the structure of both cation and anion, stating their electron and ionic geometries. Work out the Oxidation States of phosphorus in both ions.

You need to use the BedokFunland JC formula for Oxidation State = Formal Charge + Electronegativity consideration, in order to correctly answer the Singapore-Cambridge A level 2011 P3 Qn, "What is the OS of Br in RCONHBr?"

UltimaOnline SG

Thanks for the reply! Not sure if anyone had asked this before. For electrochem cell, polarity of anode is negative while cathode is positive. How come the polarity is reversed for the case of electrolytic cell?
Anode is oxidation, so electrons are released into the electrolyte hence it is negative electrode and vice versa for cathode.

Red Cat riding/killing/punching/hugging/kissing/loving An Ox.

Reduction occurs at the Cathode.
At the Anode, Oxidation occurs.

In an electrolytic cell, electrons flow from the power source to the electrode, where some species (usually a cation) accepts electrons to be reduced. Hence, this electrode is called the cathode. And as far as you (the species accepting the electron to be reduced) are concerned, the cathode is your source of negatively charged electrons, hence the cathode is the negative electrode, and therefore the anode is the positive electrode.

In a Galvanic aka Voltaic cell, electrons flow from the anode where oxidation occurs. Think of any oxidation half-equation, eg. Al ---> Al3+ + 3e-. See the electrons being released? It flows from the anode to the light bulb. As far as you (the light bulb accepting the electrons to light up) are concerned, the anode is your source of negatively charged electrons, hence the anode is the negative electrode, and therefore the cathode is the positive electrode.

UltimaOnline SG

oh.. Forgot that water is a polar solvent as well. For that qn, is it correct to give the disubstituted pdt instead of the trisubstituted one?
Hmm. The ester undergoes basic hydrolysis to produce Rcoo- and Roh?
One more qn.. For 08 paper/2 qn 3c, is it okay to use nabh4 instead of lialh4 in dry ether?

No. You have to give the tri-halogenated product.

Don't forget that any alcohol generated from the hydrolysis will be further oxidized as well. This is the tricky bit.

NaBH4 can only be used to reduce aldehydes and ketones, absolutely nothing else. NaBH4 is a weaker reducing agent compared to LiAlH4 because of the difference in size of their 'balls'. Translation : because Al has one entire electron shell compared to B, thus the valence electron charge clouds and hydride ions of AlH4- are more polarizable and available for nucleophilic attack, compared to that of BH4-.

Both LiAlH4 and NaBH4 cannot reduce alkenes to alkanes because LiAlH4 and NaBH4 are not gay. Guys only like girls, guys hate other guys. Because all species here (alkenes, and hydride ions from LiAlH4 and NaBH4) are all healthy, full-blooded horny males (ie. nucleophiles) pumped full of testosterone, and will only seek out attractive fertile chio-bu (ie. electrophiles) to attack. Electron-rich nucleophiles (eg. alkenes) repel other electron-rich nucleophiles (eg. H- ions from LiAlH4 and NaBH4).

UltimaOnline SG

okay. That qn was reduction of ketone to alcohol so i guess nabh4 can be used. Smile
See if my explanation is right:
alh4- is easier to be polarised due to the greater size of electron cloud of al as compared to b. The hydride ions are nucleophiles as they are negatively charged. Are the electrophiles the compound that are going to be reduced? Is the mechanism similar to tat of nuc sub of RX? Btw, why is reduction of alkenes not electrophilic addition?

EDIT: my qn should be is the mechanism of reduction of carbonyls similar to that of nuc addition? It shouldnt be nuc sub since there is no double bond in the pdt..

Yes, that's right. Whether it's CN- (from NaCN) or H- (from LiAlH4), or R- (from Grignard reagent), the mechanism is exactly the same : guys like girls, so nucleophiles like electrophiles, the nucleophiles shoot out their 'balls' (ie. lone pair) for attack (ie. donate a dative bond) to the electrophile (eg. ketone or aldehyde or carboxylic acid or ester) being reduced.

Alkenes to alkanes, is indeed electrophilic addition, but because the term "hydrogenation" is more precise and informative, you must give "hydrogenation" rather than "electrophilic addition" for converting an alkene to an alkane.

UltimaOnline SG

thanks. For the last part, i went to wiki and it said it is not electro add. Not sure if my reasoning is sound:
In hydrogenation, pt/ni as act a heterogenous catalyst, thus h2 and alkene molecules will be adsorb at the surface of the cat. In the case of electro addition, there is an unequal sharing of e, but hydrogen has only 1 electron, thus partial +ve/-ve charges cannot be formed. Also, there is no carbocation intermediate, thus hydrogenation is not electrophilic addition

You're confusing mechanism with reaction type. "Electrophilic addition" is not a mechanism.

The mechanism for hydrogenation of alkenes require a transition metal catalyst, therefore the mechanism is described as "catalytic hydrogenation".

But as long as alkene is the nucleophile (guy) here, and hydrogen (when interacting with the transition metal catalyst) behaves as the electrophile (girl) here, the reaction type is still considered "electrophilic addition", since you're adding the smaller electrophile (girl) to the larger nucleophile (guy).

As for your statement "since hydrogen has only 1 electron, thus partial +ve/-ve charges cannot be formed" shows you're confusing partial charges with formal charges and ionic charges.

Edit : You made one more glaring error.

Wikipedia says hydrogenation is an electrophilic addition. Look under "Typical electrophilic additions" :

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

UltimaOnline SG

2011 H2 'A' Level Exams Paper 2 Qn.

Qn : Why is CCl4 inert to hydrolysis while SiCl4 is supsceptible to hydrolysis?

Ans : 2 reasons.

Reason #1 : Because unlike C, Si has vacant energetically accessible 3d orbitals to use to expand their octet, and as such, is able to accept dative bonds from water nucleophiles without needing to first cleave existing Si-Cl covalent bonds, resulting in a lower activation energy required for the hydrolysis.

Reason #2 : The Si-Cl bonds are longer (but not weaker! See Data Booklet bond energies) compared to the C-Cl bonds. Accordingly, the partially (or delta) negatively charged Cl atoms in SiCl4 provide poorer shielding against the incoming water nucleophiles, compared to the Cl atoms in CCl4.

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