1.(i) Ka = [H3O+][F-]/[HF]
6.6e-4 = x^2/0.1
x = [F-] = 8.124e-3 M
Qsp = [Pb+2][F-]^2
Qsp = 0.1*(8.124e-3)^2
Qsp = 6.6e-6 > Ksp (of 2.7e-8, so PbF2 precipitates)
(ii) Well, I'm assuming here that the conditions are identical to (i) except that we're increasing the pH, for example, by adding a strong acid. First we need to find the minimum [F-] required for precipitation of PbF2:
Ksp = [Pb+2][F-]^2
2.7e-8 = 0.1*x^2
x = [F-] = 5.196e-4 M
So from (i), [F-] decreases from 8.124e-3 M to 5.196e-4 M, a change of:
8.124e-3 - 5.196e-4 = 7.604e-3 M
7.604e-3 M will also be the corresponding increase of [HF] (via Le Chatelier's Principle; the surplus [F-] has to go somewhere...)
Ka = [H3O+][F-]/[HF]
6.6e-4 = [H3O+](5.196e-4)/(0.1 + 7.604e-3)
[H3O+] = 0.137 M -> pH = 0.86
Leaving the change in [HF] out of this calculation will give an answer of 0.90M, but, as I've explained, it doesn't quite accurately represent what reactions need to happen.
2. The 2 equations can be balanced like so:
AgCN(s) <-> Ag+ + CN-
H3O+ + CN- <-> HCN
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AgCN(s) + H3O+ <-> Ag+ + HCN (K = Ksp/Ka)
The molar solubility of AgCN would be the equilibrium [Ag+].
From the buffer solution:
Ka = [H3O+][CN-]/[HCN]
6.2e-10 = (1e-3)[CN-]/[HCN]
[CN-]/[HCN] = 6.2e-7
Since most of the CN- from the AgCN would get converted to HCN, the equilibrium [HCN] will also be close to the molar solubility of AgCN:
Ksp/Ka = [Ag+][HCN]/[H3O+]
1.2e-16/6.2e-10 = x^2/1e-3
x = 1.39e-5 M
BedokFunland JC's A Level H2 Chemistry Qns (Part 2)
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KickMe asked :
Phosgene dissociates to form CO and Cl2
COCl2 (g) <----> CO (g) + Cl2 (g)
The degree of dissociation, a, of phosgene is different at different temperature. 9.9g of phosgene is allowed to dissociate in a 10dm3 vessel and the equilibrium mixture attains a pressure of 6.55x10^4 Pa at 633K.
a) Calculate the degree of dissociation, alpha (a), of phosgene under the above conditions.
PV=nRT
P=nRT/V=0.1(8.31)(633)/0.01=5.26x10^4
COCl2 <------> CO + Cl2
Initial 5.26x10^4 0 0
Change -a(5.26x10^4) +a(5.26x10^4) +a(5.26x10^4)
Eqm. (1-a)(5.26x10^4) a(5.26x10^4) a(5.26x10^4)
(1-a)(5.26x10^4)+a(5.26x10^4)+a(5.26x10^4)=6.55x10^4
(1+a)(5.26x10^4)=6.55x10^4
a=0.245
The next part then asks calculate the equilibrium concentration of COCl2, CO and Cl2.
I used PV=nRT
n=PV/(RT)
since concentration= n/v = P/(RT)
is it appropriate to do so?Yes, your working is correct.
As for concentration, unless otherwise specified, it refers to molarity, which is moles / dm3. Hence, your V in your PV=nRT should be in dm3, if you intend n/V to be molarity.
When V is in m3 and P is in Pa (ie. 1 atm = 1.01325 x 10^5 Pa), R = 8.314.
When V is in dm3 and P is in atm (ie. 1 atm = 1.01325 x 10^5 Pa), R = 0.08206. -
kickme posted :
Now that, I know the yellow white solid is aluminium chloride, AlCl3. The next part of the question asks:
When a few drops of water are added to the solid, streamy white fumes are evolved and a white solid remains, which is insoluble in water.
Can I have some clue as to what reaction occurs when water is added to solid aluminium chloride?
First of all, take note that the reaction and equations commonly tested at 'A' levels involve adding excess water to generate an acidic solution of AlCl3(aq), which is really [Al(H2O)5(OH)]2+(aq) + H3O+(aq) + 3Cl-(aq).
However, this question is asking what happens when limiting amounts of water are added to excess AlCl3(s).
Even though the initial reactant AlCl3 may be regarded as covalent, but because the final product is more ionic than covalent, and has the empirical formula Al2O3, it'll be more easier for you to follow mathematically if I illustrate the reaction mechanism using two units of AlCl3 rather than one.
Six (per two units of AlCl3) water nucleophiles / ligands / Lewis bases attack (ie. donate dative bonds to) the (high charge density and thus a strong) Lewis acid Al3+ ion of AlCl3 (empirical formula), thereby eliminating six (per two units of AlCl3) Cl- ions which instantaneously functions as a Lewis / Bonsted base to abstract the six (per two units of AlCl3) protons from the positive formal charged (O atoms of the) six water nucleophiles which attacked earlier, consequently generating the steamy fumes of six HCl(g) molecules and the intermediate Al2(OH)6 which readily dehydrates (due to the high charge density or 'electrophilicity' of the Al3+ ion making it a strong Lewis acid, half of the OH- ligands / nucleophiles / Lewis bases present readily donate a second dative bond and thus loses another proton; furthermore the reaction is exothermic and the high temperature provides additional activation energy *and* favours the positive increase in entropy as predicted by Gibbs free energy during the dehydration process) to generate the insoluble white solid Al2O3 (empirical formula) and three water molecules are simultaneously eliminated (accounting : these three H2O molecules came from the elimination of the 3 OH- ligands/nucleophiles combining with the 3 protons eliminated, when the other 3 OH- ligands/nucleophiles donated a 2nd dative bond to the Al3+ ion).
But because the ions involved are O2- and Al3+, and since the electronegativity difference between oxygen and aluminium is sufficiently large to qualify as predominantly ionic, the so-called 'dative covalent bonds' donated to the Al3+ during the mechanism are now to be regarded as ionic bonds, in the final ionic product Al2O3.
Hydolysis (ie. nucleophilic / Lewis base attack of H2O) :
2 AlCl3 + 6 H2O → 6 HCl + 2Al(OH)3 or Al2(OH)6Dehydration of Al(OH)3 :
2 Al(OH)3 or Al2(OH)6 → Al2O3 + 3 H2OOverall :
2 AlCl3(s) + 6 H2O(l) → Al2O3(s) + 6 HCl(g) + 3 H2O(l)Cancelling the three H2O molecules on both sides of the equation :
2 AlCl3(s) + 3 H2O(l) → Al2O3(s) + 6 HCl(g) -
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Kickme asked :
I have another doubt pertaining to the 'A' level question asked ealier: 2011 'A' level exam qn :
Is PCl5 a simple molecular compound? If so, why is it solid at rtp? Draw it's structure in the solid state and describe the geometries involved.
In your answer, you stated: The P-Cl bond heterolytic dissociation enthalpy is small, compared to the large amount of heat energy that can be evolved from forming the stronger ionic bonds. Hence formation of solid PCl5 is favoured under standard conditions.
I understand that the P-Cl bond dissociation enthalpy is referring to the formation of simple molecular PCl5. But I dont understand how the formation of simple molecular PCl5 requires the bond dissociation of P-Cl.
Unless, simple molecular PCl5 also exists as PCl4+ and PCl6- ions initially, and PCl6- loses a Cl- to donate to PCl4+, thus forming 2 PCl5 molecules.If we begin from the molecular PCl5, then the P-Cl heterolytic (not homolytic) bond dissociation generates the ionic species PCl4+ and PCl6-, as temperature decreases and PCl5 solidifies.
As temperature increases, because molecular PCl5 is able to possess more kinetic energy as liquid and gaseous simple molecules, compared to solid ionic PCl5 (whose atoms can only vibrate about fixed positions), hence the P-Cl heterolytic (not homolytic) bond dissociation now generates the molecular PCl5 from the ionic species.
In other words, whether you're generating the ionic from the molecular, or the molecular from the ionic, you'll need to heterolytically dissociate a P-Cl bond (and of course, form one as well). The molecular and ionic species exist in equilibrium with each other, it's only a matter of who predominates in which temperatures, pressures and solvents.
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Memo wrote to Kickme :
I don't know if you have figured this out. but if this ever come out in exam i would write something like this...
As the total number of electrons per molecule increases from (HCl->HBr->HI), more electrons can be polarised. Thus the strength of the Van der Waals' forces due to induced dipole - induced dipole increases.
However, the polarity of the H-X bond decreases from (H-Cl->H-Br-H-I). This results in weaker Van der Waals' forces due to permanent dipole - permanent dipole attraction.
The increase in strength of the Van der Waals' forces due to induced dipole - induced dipole attractions outweighs the decrease in strength of permanent dipole - permanent dipole attractions.
Hence, the boiling point of (HCl > HBr > HI).Kickme wrote :
Shouldn't it be instead: the induced dipole-induced dipole van der Waals between HI molecules are stronger than the permanent dipole-permanent dipole van der Waals between HCl molecules.
Since the greater electronegativity of Cl over I already shows that HCl has got stronger permanent dipole-permanent dipole interactions over HI.I replied them :
Ok, here's the thing :
Both of you guys (Memo and Kickme) are stuck in the thinking that the strength of the permanent dipole - permanent dipole van der Waals are determined solely by the electronegativity difference, while the strength of instantaneous dipole - induced dipole van der Waals are determined solely by the polarizability of electron charge clouds.
This is what is commonly taught in most JCs, and is perfectly acceptable by Cambridge.
But while acceptable (if you feel comfortable with this concept, go ahead and write Memo's answer, it's perfectly acceptable). it is not completely accurate. If you reread my earlier post, there's another way of looking at it (equally acceptable by Cambridge), which is :
The strength of any of the three types of van der Waals (Keesom, Debye, London dispersion forces), is first and foremost determined by polarizability of electron charge clouds, and next by polarity of electronegativity differences.
If you follow this concept, you'll still end up with the same correct conclusion. Use my concept, or use Memo's. Either way, you get the same conclusion, and full marks from Cambridge. -
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Originally posted by hoay:
The products obtained by cracking an alkane, X, are methane, ethene and propene.
The mole fraction of ethene in the products is 0.5.What is the identity of X?
A C6H14 B C8H18 C C9H20 D C11H24
Start by drawing these 3 smaller hydrocarbon products and piece them together like a jigsaw puzzle. Based on the fact that the longest hydrocarbon obtained is only 3 carbons, obviously the original large alkane must be branched. Based on the fact that 50% of the products are ethene, that means two of the four branches must be ethyl groups. The third branch must be the propyl group. The intersection of the branches, itself generates methane.
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Kai asked :
1. a 0.10M solution of Pb(NO3)2 is added enough HF(g) to make [HF]= 0.10M. (Ksp of PbF2= 2.7e-8 , Ka of HF= 6.6e-4)
(i) Does PbF2 precipitate from this solution?
(ii)What is the minimum pH at which PbF2 precipitates?
2. Calculate the molar solubility of AgCN in a buffer solution with a pH of 3.00. (Ksp of AgCN=1.2e-16 Ka of HCN= 6.2e-10 )
The ans. to these two problems are: 1: (i) yes (ii) 0.90 2: 1.4e-5 mole/LSnowYak replied him :
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Atomos asked :
Originally posted by atomos:Hi all,
I have a question on the electronic configuration of particles, especially when it uses the idea of the stability of 3d^5 4s^1
Some scenarios might be more hypothetical than feasible.
Case 1
What is the final configuration when a particle with initial configuration [Ar] 3d^3 4s^2 gains an electron?
[Ar] 3d^4 4s^2 or [Ar] 3d^5 4s^1?
Case 2
What is the final configuration when a particle with initial configuration [Ar] 3d^6 4s^2 loses two electrons?
[Ar] 3d^6 or [Ar] 3d^5 4s^1?
Too many conflicting approaches on the net, hope to clear things up here.
Thanks in advance!
Case 1 :
You're right, this is more hypothetical than feasible, because vanadium is a metal that is only interested in losing electrons to be oxidized to cationic form, not gaining electrons to be reduced to anionic form.
The short answer is : both configurations are possible, depending on other factors (that you needn't concern yourself with at 'A' levels).
However, the more important lesson here is, the exam-smart candidate will :
If this was a P1 qn, then consider both possible, but (if both options are present, then) assume the more direct answer, ie. 3d4 4s2.
If this was a P2 or P3 qn, then intelligently (ie. with explanation) write both configurations in your answer :
"When a particle with initial configuration [Ar] 3d^3 4s^2 gains an electron, it first attains an electron configuration of [Ar] 3d^4 4s^2, which may (or may not) subsequently rearrange itself (as observed in the example of chromium) to have an electron configuration of [Ar] 3d^5 4s^1, by transferring an electron from its 4s orbital, to one of its 3d orbitals, to achieve a more stable half-filled 3d subshell."
Case 2 :
This is pretty straightforward. No rearrangement takes place (as far as the electron configuration of the Fe2+ ion is concerned, when asked at 'A' levels) because Fe2+ is ionic and readily stabilizes itself in other ways (ie. as an alternative to rearranging its electron configuration) such as by forming ionic bonds, and/or coordination complexes (the electrons-in-boxes diagram will include both its own electrons, as well as those of the coordinate dative bonds donated by the ligands).
Bottomline :
At 'A' levels, when removing electrons from d-block elements, remember to just kick away electrons from the 4s orbital (which is further away from the nucleus, thus easier to remove), before removing any electrons from its 3d orbital (which is nearer to the nucleus, thus more difficult to remove).
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Explain why despite nitrogen and phosphorus both being in Group V, nitrogen forms N2 molecules while phosphorus forms P4 molecules.
Ans : see my website ( google "BedokFunland JC" )
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kickme asked :
A and B are isomers and they have the molecular formula C3H5Cl. B is able to decolourise orange aqueous bromine to form C, while A does not decolourise bromine. When ethanolic KCN is heated with A and B, then subsequently further heated with dilute nitric acid, only A forms a product D. D is able to react with sodium carbonate to give of a colourless gas and an ion, E. When A is heated with limited amount alcoholic ammonia, a white solid F is formed.
Draw all the full displayed formula of A, B, C, D, E and F. Write the balaced equation for all the reaction that has taken place.
A: chlorocyclopropane
B: 3-chloropropene
C: 1-bromo,3-chloropropa-2-ol
D: cyclopropanoic acid
E: I am not sure, but is the ion from the earlier part when A reacts with dilute HNO3 to form NH4+ ion?
F: Is the reaction through polyalkylation? But i do not know how to write the balanced equation for a cycloalkane!
The balanced equation part should be fine except for F. Please help to check if the compound A to F identified is correct. Thank you!
Good attempt, with a couple of errors. Here are the correct answers.A : chlorocyclopropane
B : 1-chloroprop-1-ene or 2-chloroprop-1-ene (the halogen atom must be bonded to a sp2 hybridized C atom, allowing resonance delocalization of a lone pair on the halogen atom to form a pi bond with the C atom, thus giving rise to C-X having partial double bond character, as evidenced by the observed resistance to nucleophilic substitution by the CN- nucleophile)
C : (based on either species B above)
either
1-bromo-1-chloropropan-2-ol (major product) or 2-bromo-1-chloro-propan-1-ol (minor product)
or
1-bromo-2-chloropropan-2-ol (major product) or 2-bromo-2-chloro-propan-1-ol (minor product)D : cyclopropylmethanoic acid
E : sodium cyclopropylmethanoate
F : tetra(cyclopropyl) ammonium chloride (with byproduct of HCl(alc) <---> HCl(g) )
About writing the balanced equation to generate F, you can either write the 4 individual equations (better yet, draw the repeated nucleophilic subsitution mechanism), or a combined overall equation (to generate 3 HCl and 1 RRRRN+Cl-)
Cambridge will accept either, since they didn't specify which was required.
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KickMe then replied :
thank you very much.
just some questions, since the question state that E is an ion, as Nish said, the answer should be cyclopropylmethanoate ion, right?
But since sodium is present, wouldn't sodium react with the ion to form the salt sodium cyclopropylmethanoate?
Another question is about the naming, is cyclopropylmethanoic acid=cyclopropanoic acid?
Cambridge will usually ask you to identify unknown compounds such as E, as a full ionic compound (ie. a metal carboxylate, in this case, sodium cyclopropylmethanoate), rather just as one of the ions present. But yes, if the prelim exam question specifies E as an ion, they're indeed asking for "cyclopropylmethanoate".In aqueous or alcoholic solvent, sodium compounds are soluble, meaning that the Na+ cation and the cyclopropylmethanoate anion, are both ion - permanent dipoled (or for the anion, hydrogen bonded) to the water or alcohol solvent, rather than ionically 'stuck' to each other, which would be the case if the solvent was aprotic non-polar (eg. CCl4, methylbenzene, etc).
No, as Nish pointed out already, the name "cyclopropanoic" is incorrect because it implies only 3 Cs are present. The correct name "cyclopropylmethanoic acid" implies there are 3 Cs + 1 C = 4 Cs present in total, which is indeed the case here.
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To add on another question is: I didn't quite understand why the halogen must be bonded to the sp2 carbon for B.
You said: "B : 1-chloroprop-1-ene or 2-chloroprop-1-ene (the halogen atom must be bonded to a sp2 hybridized C atom, allowing resonance delocalization of a lone pair on the halogen atom to form a pi bond with the C atom, thus giving rise to C-X having partial double bond character, as evidenced by the observed resistance to nucleophilic substitution by the CN- nucleophile)"
It is the same concept or explanation with "why doesn't chlorobenzene undergo hydrolysis? in contrast to chloroalkanes".Most JCs give the orbital overlap explanation, while I personally prefer the resonance delocalization explanation. Cambridge will accept both, as long as you specify that the end-result is that the C-X bond has partial double bond character and is difficult to cleave, and therefore the halobenzene or haloalkene, is resistant to nuclophilic substitution.
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BlackToast asked :
Hi ultima!
regarding the very first post in this thread, Q1, i din't get your explanation so i went to clarify with my teacher. he said i'm supposed to use enthalpy c of soln = enthalpy c of hyd - lattice energy. BUT he din't give me the full explanation, asked me to go home and think. -_-"By Hess Law,
Solution enthalpy = (endothermic) Lattice Dissociation enthalpy + (exothermic) Hydration enthalpy.Your school teacher is using the exothermic version, ie. Lattice Formation enthalpy, in his formula, which will still give you the same correct mathematical answer, but makes less sense, in the context of Hess Law, and therefore I would advise you to use my formula, instead of your school teacher's.
For a full explanation on why Grp II hydroxides become more soluble down the group, but why Grp II carbonates become less soluble from Be to Sr, but mysteriously becomes more soluble again from Sr to Ba, read Jim Clarke's website (Jim Clarke's and Rod Beavon's websites, are the top two most popular 'A' level Chem websites on the internet, you should make the most use of them) :
http://www.chemguide.co.uk/inorganic/group2/problems.html
Even though solubility trends for Group II hydroxides versus carbonates & sulfates, are not required by the H2 syllabus to be memorized, Cambridge can still ask you deduce-and-explain questions on this topic. Students aiming for a distinction, should definitely study the material (Solubility of Group II hydroxides versus Carbonates) on Jim Clarke's website.-----------------------------------------------------------
In explaining why C-X bond is cleaved to produce X radicals and R radicals, instead of cleaving the C-H bond to produce H radicals; besides saying the H radical is much less stable, can i also say because the C-H bond is very strong?
Yes, you can quote relevant data booklet values and say the C-H bond is stronger than the C-X bond, and therefore homolytic cleavage of the C-X bond is energetically favoured over the homolytic cleavage of the C-H bond.-----------------------------------------------------------
So u mean if they say the temperature is cold, and there is an alchohol attached to the side chain of a benzene ring, then the alchohol will get oxidised only?
Yes, this is correct. Usually Cambridge and Prelim papers will specify "heat under reflux with KMnO4", and student is expected to give "benzoic acid" as the product. But a potential trick question, is to "heat under reflux with K2Cr2O7", in which case you only get oxidation of the alcohol (and not the entire side chain), but there will be a % of students who don't read carefully, and wrongly give "benzoic acid" as the answer. -
blacktoast94 asked :
also, is there some kind of framework to follow for explanation questions?
like for example they ask u to derive some structures from the information given, and explain fully.everytime got this kind of question i'll get the strucutres correct but the explanation 0 marks.
besides naming the reaction, what else? issit explain the mechanism?
thanks!
You need to write out (eg. in table format, or in normal writing format, this is optional and up to you), a list of structural deductions, based on the data given in the question.For example :
"Since a positive iodoform reaction occurs when compound A is warmed with alkaline aqueous iodine to generate CHI3, either the -CH(OH)CH3 group or the -C=OCH3 group is present in compound A. However, since the 2,4-DNPH test for carbonyl compounds did not generate an orange precipitate, it can be deduced that the -CH(OH)CH3 group instead of the -C=OCH3 group, is present in compound A."Mechanisms for such "Structural Deductive Elucidation" type are not required, unless specifically asked for in the question. Besides, how many JC students can draw the full iodoform or 2,4-DNPH mechanism? (I do teach these mechanisms to my students for their own deeper understanding, but Cambridge won't require you to memorize or draw them out.)
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'A' level students take note :
Cambridge requires the use of comparative adjectives in your answers to such comparative-type questions.
Originally posted by EiN930:Hi I just want to clarify whether the explanation is correct.
Explain why graphite has a lower melting point than diamond. (1 mark).
The answer is "Graphite has fewer covalent bonds than Diamond."
What if the student wrote this:
"Graphite has weak van der wals forces between the layers. Lesser energy is required to overcome such forces. As a result, it has a lower melting point than diamond".
Is this answer acceptable?
From my own understanding, diamond has 4 covalent bonds between the carbon atoms while graphite has 3 covalent bonds between the carbon atoms and weak van der wals forces. More energy is required to break the extra covalent bond in diamond than the weak van der wals forces between the layers of atoms in graphite.
If the question were to ask, "why is diamond harder than graphite?"To order to mechanically break apart a piece of diamond, a lot more energy is required to overcome the much stronger covalent bonds in diamond. In contrast, in order to mechanically slice away layers of graphite (which we do when we write using pencils), less energy is required to overcome the weaker van der Waals attraction between the graphene layers of graphite.
A comparison (note the use of comparative adjectives more / less / higher / lower / stronger / weaker in my answer above) must be made to contrast the requirements for mechanically breaking apart pieces of both allotropes, rather than just describing the bonding in only either allotrope.
Turns out that to completely melt graphite, you need to overcome both the intramolecular covalent bonds within the graphene layers, as well as the van der Waals forces between the graphene layers.
Still, this question is rather unfair, because this is still not the full picture. Each C atom in graphite is sp2 hybridized and the covalent bonds within the graphene layers is actually stronger (due to partial double bond character) than the covalent bonds within diamond.
So you have greater number of covalent bonds within diamond, versus stronger covalent bonds within graphite. So who wins? Turns out the melting points of diamond and graphite are actually pretty close (despite graphite being soft and diamond behing ultra-hard).
So, this is still not a fair question (so Cambridge is also arguably at fault), because students (especially at 'O' levels) can't be expected to be aware of, and compare the magnitude of energy required, to overcome the greater number of covalent bonds in diamond, versus the stronger covalent bonds in graphite.
Originally posted by EiN930:Okay. Thanks for the update.
This is indeed tricky. So if such qn will to appear, which answer would be better? Or both answers are acceptable? But at this level (I mean O level), should we assume the the covalent bonds are similar in both diamond and graphite?
Or I should explain the ambuigity of the qn to the student?
1. If you feel your student is up to it, then show him the bigger picture, afterwhich you proceed to...2. Instruct him to remember and regurgitate the simplified answer (but still incomplete picture) that "diamond has more covalent bonds than graphite" in the exams.
3. If your student is not up to it, then skip straight to step 2.
Originally posted by EiN930:Thank you :)
Anyway I re-checked the Mark Scheme and the answer given by the mark scheme is:
"Graphite has fewer strong bonds to break."
And according to the examiner reports, it states:
"To compare the two, mention of both diamond and graphite was required. That graphite ‘has weak bonds’ is not true, as the diagram shows. The Examiners required the answer that graphite has some weak bonds but diamond has only strong bonds."
Yes, the examiner report confirms what I said in my previous posts, about usage of comparative adjectives, difference in number of covalent bonds, and strength of covalent bonds (stronger in graphite, weaker in diamond).The so-called 'weak bonds' (in the examiner's report) do not refer to covalent bonds, but van der Waals forces. This should be specified more clearly in the examiner's report, which are to be read by teachers, not students, so Cambridge shouldn't use oversimplified (and technically erroneous) terms like "weak bonds" when referring to van der Waals forces, lest it may confuse some people reading it.
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blacktoast asked :
usually for SN1 mechanism, the nucleophile can attack the carbocation from both sides, hence forming 2 isomers cuz of the chiral centre. but what if there is a tertiary halogenoalkane, and all the groups bonded to it are -CH3? then can the nucleohiple still attack from both sides? how to draw the mechanism and end product?
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Two enantiomers are obtained only if the product contains a chiral carbon. If the carbocation intermediate involved is the trimethylmethyl carbocation, the product would not contain a chiral carbon, and hence does not have optial isomers. Therefore, there would only be one possible product, regardless of which side the nucleophile attacked the trigonal planar carbocation.
blacktoast asked again :so should i still draw the intermediate step whereby the nucleophile attacks from both sides?
No need to draw two mechanism pathways (ie. one from either side of the trigonal planar carbocation attacked), since there is only one product. Just draw the trigonal planar carbocation intermediate being attacked by the nucleophile from either direction.
Students take note :For SN2 mechanism, you need to draw the reactant, transition state, product.
For SN1 mechanism, although there is reactant, transition state, intermediate, transition state, product; but you only need to draw only the reactant, intermediate, product (ie. transition states not required for SN1).
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KickMe asked :
Is excess acidified KCr2O7 implying strong or mild oxidation? No heat is mentioned.
If temperature is not mentioned, then the question is ambiguous, and then the onus would be on you, as the exam-smart student, to specify both possible answers, but with qualification or explanation, "if heated under reflux, the product is..." and "if temperature is kept cold, then the product is..."
Note that only primary alcohols can be oxidized by K2Cr2O7 to give different products. Secondary alcohols will only produce ketones.
Alkyl benzenes and alkenes can only be oxidized by KMnO4, and not K2Cr2O7. -
Kickme asked :
Give the reagents and condition to prepare butan-2-ol from butan-1-ol. What I have in mind is dehydration of butan-1-ol followed by hydrating the alkene formed. But this will mean that butan-1-ol will still be present. Are there other way to synthesize butan-2-ol?
No problem, since the major product of hydration of but-1-ene, will be butan-2-ol.
To understand why (Cambridge may ask you to explain this), draw the full mechanism, and discuss the relative stabilities of the two possible carbocation intermediates.
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kickme asked :
My notes only show the oxidation of primary, secondary and tertiary alcohols but does not mention methanol. If methanol can be oxidised, how can i distinguish methanol from propanol?
Using K2Cr2O7 won't make a difference. Using hot KMnO4 however, will oxidize methanol to methanal to methanoic acid to carbonic(IV) acid, which exists in equilibrium with, and hence can decompose into, CO2 and H2O. Efferversence of CO2 can be thus observed for methanol, but not for propanol. -
Originally posted by Kahynickel:
reaction ralative rate
RCH3 to RCH2Cl 1
R2CH2 to R2CHCl 7
R3CH to R3CCl 21
Using this information, and considering the number of hydrogen atoms of each type
(primary, secondary or tertiary) within the molecule, predict the relative ratio of the
two possible products J and K from the chlorination of 2-methylpropane. Explain your answer.CH3 CH3 CH3
CH3 CH CH3 yielding CH3 C Cl CH3 CH
CH3 CH3 CH2Cl
2-methylpropane J K
SOLUTION:
ratio J / K = 21 / 1
EXPALNATION: According to the information given tertiary hydrogen atom will be repalced in 2-methylpropane to produce J in higher quantity where as there is only ONE primary replaceable hydrogen in the minor product K. Hence the ratio would be
J:K = 21 : 1
Is this the right expalnation or am I missing something UltimaOnline?
This question requires the student to consider both factors in his/her calculations : number of H atoms substitutable and stability of alkyl radical intermediates.
To get J :
(number of H atoms substitutable) x (stability of alkyl radical intermediates)
= 1 x 21 = 21
To get K :
(number of H atoms substitutable) x (stability of alkyl radical intermediates)
= 9 x 1 = 9
Hence ratio of products J to K = 21 / 9 or 7J : 3K
If Cambridge asks "what factors affect product distribution in free radical substitution?" the student should qualitatively explain both factors.
But if Cambridge asks "what is the relative percentages of the expected product distribution?", then :
If the Cambridge exam question does not provide data for the relative stabilities of alkyl radical intermediates (which can be phrased as "relative rates of reaction to generate primary, secondary and tertiary alkyl halides" as seen in the original question above), then the student need only give a quantitative answer based on calculations that only consider "number of H atoms substitutable".
If the Cambridge exam question does provide data for the relative stabilities of alkyl radical intermediates (such as in the original question above), then the student needs to consider both factors "number of H atoms substitutable" and "stability of alkyl radical intermediates" by multiplying the values associated with both factors, to obtain the true product distribution.
Most schools only teach the simpler factor of "number of H atoms substitutable", but this is misleading because in practice, both factors (which you would notice tend to contradict each other : tertiary alkyl halides have the least no. of H atoms to substitute away, but yet tertiary alkyl radicals are the most stabilized by hyperconjugation from the alkyl groups) are equally important and teaches the student that in Chemistry, just as in Real Life, it is myopic and does oneself injustice, to simply consider one side of the story or one particular viewpoint only; one must consider all sides of the story or all relevant viewpoints to understand and appreciate the truer, bigger, more beautiful picture of the Universe and of Life.
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Compound A has the molecular formula C9H12O. Heating A under reflux with acidified KMnO4 produced benzoic acid. Treatment of A with acidified potassium dichromate (VI) produced a ketone B which has a positive result with iodoform test.
Compound A reacts with I2 and aqueous sodium hydroxide to give a yellow solid D and another product which on acidification gave a compound C.
When A was dehydrated, 2 structural isomers E and F were produced, both of which decolourised bromine water. Oxidation of E (but not F) gave compound C as one of the products.
Make deductions to elucidate the structures of compounds A to F.
Answers :A is C6H5CH2CH(OH)CH3
B is C6H5CH2COCH3
C is C6H5CH2COOH
D is CHI3
E is C6H5CH2CHCH2
F is C6H5CHCHCH3 -
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