What mechanism and reaction will vinyl benzene + Br2(ethanolic) undergo? Im guessing electrophilic addition of Br2 and the vinyl group. Is ethanolic purely a solvent like say CCl4? or is there a special reason for using an alcoholic medium which we usually encounter in elimination rxns for alkyl halide.
The common reason for using ethanol instead of aqueous or CCl4, is because many reactions involve one ionic (eg. KCN) and one non-polar (eg. aklyl halide) reactant, which will not be miscible if you use aqueous or CCl4 solvents. Instead, a middle-man which is part polar and part non-polar, such as alcohol, is used. And ethanol is the least toxic of the alcohols, and also the most balanced in terms of polarity. Thus ethanol is usually the solvent of choice.
The other reason for using ethanolic solvent, say ethanolic OH- as opposed to aqueous OH-, is to control the behaviour (ie. reactivity) of the OH- ion. It can function as either a nucleophile (guy) or base (guy). As long as you shoot out your balls to grab someone else, you're a horny guy. But the girl can be either a petite H+ ion (really just a proton) or a voluptuous alkyl electrophile.
With aqueous solvent, the OH- is better stabilized by more extensive H-bonding, and is thus less desperate to protonate itself (which it does so to stabilize itself), and will behave more as a nucleophile (rather than a base). With ethanolic solvent, it is less stabilized by less extensive H-bonding, and will be more desperate to stabilize itself, and thus behave more as a base (rather than nucleophile). I could go into more explicit analogies here (in terms of variant sexual behaviours), but I'm sure that won't be necessary.
With vinyl benzene and Br2 (ethanolic), the major product will be the addition product of Br+ and Br- across the double bond, and the minor product will be the addition product of Br+ and CH3CH2OH across the double bond, afterwhich the CH3CH2OH+ will lose a H+ and become the CH3CH2O alkoxy side chain on the alkyl side chain of the benzene ring.
kickme :
One thing I do not understand is what reaction aryl halide C6H5Cl, is going through when reacting with hot ethanolic KOH. You mentioned hydrolysis(addition of steam), but it is surprising to see from my notes that the only reaction mentioned under aryl halides is nitration.
Hydrolysis is not addition of steam. That's called hydration (for organic chem context; in physical & inorganic chem, hydration = solvation where water is the solvent). Hydrolysis in (physical + organic + inorganic) chemistry means "reaction with water", and in organic chem, can refer to OH- nucleophile in lieu of water nucleophile.
Aryl halides undergo electrophilic aromatic substitution (EAS), of which nitration is but one such EAS reaction.
kickme :
Another thing I wish to clarify about suggesting of difference in behaviour is that, am I right to say that the question simply wants us to see the difference in reactivity between Aryl Halides and alkylbenzene?
Behaviour = Reactivity.
kickme :
Okay... So aryl halides undergo electrophilic substitution reactions. And hydrolysis(reaction with water) is one of them. Electrophilic substitution involves the addition of an electrophile. I suppose the water molecule will break into H+ and OH-, so is the electrophile H+?
No you've misunderstood, nitration is an example of EAS, hydrolysis is not.
Hydrolysis of alkyl halides as well as aryl halides involves SN1 or SN2 substitution (for simplified 'A' level purposes anyway; at higher levels the mechanism for hydrolysis of aryl halides is elimination-addition and involves a benzyne intermediate), but in the 1st place for 'A' level purposes, you must state that aryl halides do not undergo hydrolysis under normal conditions, due to the C-X partial double bond character.
vluos lamented :
Woah whats the last paragraph about haha! clueless >.< btw on a side note, what happens when a compound with 2 double bonds such as 1,3-butadiene is oxidised strongly? how to determine the products?
When heated under reflux with hot acidified (or alkaline, followed by acidification) KMnO4, the diene 2HC=CH-CH=CH2 will undergo oxidative cleavage to generate carbonic(IV) acid + ethandial + carbonic(IV) acid, which exists in equilibrium with, and hence can decompose into CO2 + ethandial + CO2, which will be further oxidized to CO2 + ethandioic acid + CO2, which will be further oxidized to CO2 + carbonic(IV) acid + carbonic(IV) acid + CO2, which exists in equilibrium with, and hence can decompose into CO2 + CO2 + CO2 + CO2.
Hence the final product is just CO2.
Terminal =CH2 groups are oxidized to carbonic(IV) acid H2CO3, which exists in equilibrium with, and hence decomposes into CO2 and H2O. Most JCs just teach you to memorize that =CH2 groups are oxidized directly to CO2, which is also acceptable by Cambridge.
=CH-CH= groups are oxidized by hot KMnO4 (if cold KMnO4, you get diol) to ethandial HCOCOH, which is further oxidized by KMnO4 to ethandioic acid HOOCCOOH, which is further oxidized by KMnO4 to carbonic(IV) acid H2CO3, which exists in equilibrium with, and hence decomposes into CO2 and H2O.
Notice that CO2 is a gas at rtp (and at higher temperatures used in reflux) and thus readily leaves the reaction mixture, pulling the position of equilibrium (as predicted by Le Chatelier's principle) over to the RHS to generate more CO2.
Nish asked :
If Cambridge sets a question whereby phenylpropene/or any other more complexed compounds for that matter (with C=C further away from benzene) undergoes oxidation, does that alkyl chain necessarily always become -COOH? or will the major product have just the C=C oxidised, without necessarily forming a -COOH?
The part of the molecule beyond the C=C bond will be cleaved oxidatively like any normal alkene (to generate ketone or carboxylic acid), while the part of the molecule bonded to the benzene ring, will become benzoic acid and one or more moles of CO2 (balance accordingly).
kickme posted :
Okay, I get it now, so Aryl Halides undergo nucleophilic substitution reactions in addition to electrophilic substitution. I wonder why nucleophilic substitution was not included in my notes for aryl halides. Thanks for clearing that up!
Now, to apply the knowledge to the question, C6H5Cl doesn't react with hot ethanolic KOH due to the C-X partial double bond character( But i notice that this is a HOT ethanolic KOH solution... So should I say nucleophilic substitution reaction can occur? whereby the nucleophile is OH- ion)
So, in conclusion, the rate of reaction of C6H5Cl is lower than that of C6H5CH2CH2Cl
EDIT: I note that you mentioned conditions as 350 deg C and 200 atm required for nucleophilic substituion of Aryl Halides to occur. So I assume that no reaction take place between hot ethanolic KOH and C6H5Cl.
Yes, nucleophilic substitution does not readily occur with aryl halides (and again I repeat : for 'A' level purposes, you must state that aryl halides do [B][U]not[/U][/B] undergo hydrolysis or any other nucleophilic substitution under normal conditions, due to the C-X partial double bond character.
This is why your JC lecture notes did not mention nucleophilic substitution for aryl halides, but only electrophilic aromatic substitution (it's annoying how JCs often leave out the word "aromatic", personally I would penalize all those JC teachers who leave out that word, but Cambridge accepts with or without the word "aromatic"... still, it's better to write the full name "electrophilic aromatic substitution" instead of "electrophilic substitution" for reactions in which the benzene ring is the nucleophile).
Another annoying and lousy teaching by JCs is :
Remember how in 2011 Paper 3, there was a question which asked what is the type of reaction involved in generating an amide?
Many JCs teach "acylation" which is ambiguous and can refer to either Friedal-Crafts acylation (which is actually electrophilic aromatic substitution) or nucleophilic acyl substitution.
The correct answer (to describe reactions which generate amides) is... and the exam-smart candidate will give all three answers (because they're all correctly describing different aspects of the same reaction) :
Nucleophilic acyl substutition; addition-elimination; condensation.
Some JCs only teach one or two out of the three keywords above. You should give all 3 answers, if you're exam-smart and want to secure your marks.
You can not say "esterification" because here an amide, not ester, is generated.
Yes, the rate of reaction of C6H5Cl is obviously and certainly much lower than that of C6H5CH2CH2Cl, and Cambridge expects you to be able to explain why (I already explained it many times on this forum).
kickme posted :
Okay, now I know what nucleophilic substitution rxn was not included.
Now on to the structure of the answer, Firstly I would talk about the possible reactions that the reagents can undergo, and explain why C6H5Cl do not undergo nucleophilic substitution due to the partial double bond character of C-X and thus conclude the rate of reactions of the reagents based on the number of possible reactions the reagent can undergo.
Right.
FYI, nitration is a type of electrophilic aromatic substitution (EAS) in which the benzene ring is the guy (nucleophile) and the nitronium cation is the girl (electrophile). You add on one NO2+ electrophile and you lose one H+, so it's substituting away a H+ for a NO2+ electrophile. And since the aromatic benzene is the nucleophile here, therefore we call this electrophilic aromatic substutition.
Blacktoast94 asked :
some questions to ask, pls help thank u!
1. "down the group II, the elements dissolve in water with increasing ease." why is this so? is it because of lattice energy?
2. when they say "higher lattice energy," i always can't tell if they mean higher as in the absolute value or higher in terms of the negative value. does it mean easier or harder to dissolve? ._.
3. how to explain fully if they ask why a ring structure restricts the formation of cis-trans isomers?
4. explain fully why the hydrogen radical cannot be formed?
5. the enthalpy change of combustion for alkanes gets more and more negative as the carbon chain length increases. what is the explanation for that?
6. just to clarify, is it true that ethanolic KOH/NAOH can not only do elimination, but also can do hydrolysis?
7. explain why in SN2, the optical isomerism is reversed.
8. also, is K or Kr larger in atomic radius? across the period the radius is supposed to get smaller, but my lecture notes says because van der waal's radii is measured for the group 0 elements, therefore it has the largest atomic radius. the answer from the book however, is K has the larger atomic radius.
Q1.
For solubility of group II metals itself, look at oxidation potentials in Data Booklet. Bear in mind that by Hess Law, oxidation enthalpy = ionization enthalpy + hydration enthalpy.
For solubility of group II hydroxides versus carbonates & sulfates, read :
http://www.chemguide.co.uk/inorganic/group2/problems.html
Q2.
By default, Cambridge will always refer to the magnitude of the lattice energy. If this was Paper 3 or Paper 2, then you should be exam-smart and use the full terms "endothermic lattice dissociation energy" versus "exothermic lattice formation energy" (similarly for "endothermic bond dissociation energy" versus "exothermic bond formation energy" when describing covalent bond energies).
Q3.
For small rings, only the so-called cis-isomer exists, because the trans-isomer will break the ring. For larger rings, the trans-isomer will exist, but may cause some degree of angle strain.
Q4.
The hydrogen radical can indeed be formed, just that it's much less stable than most other radicals, and is therefore not encountered in the H2 syllabus, as other radicals (eg. alkyl, halogen, etc) would be preferentially formed instead during initiation and propagation steps.
Q5.
Because we're not talking about more exothermic per C atom, but more exothermic per hydrocarbon molecule. The more C atoms present in a single molecule, the more bonds are broken (in reactants) but also formed (in products), and since combustion has proven to beexothermic, the magnitude of exothermic energy transfer always exceeds the magnitude of endothermic energy transfer, with this difference in magnitude increasing with an increasing number of C atoms per molecule.
Q6.
Both are possible. In practice, all 4 reactions, SN1 vs SN2 vs E1 vs E2 are always in competition with each other. It's only in the simplified 'A' level syllabus that you simply state the major product when ethanol solvent is used, is the elimination product rather than substitution product. But in a tricky Cambridge question, you may still have to consider other (minor) possible products.
Q7.
(If this was a Paper 3 or Paper 2 question, then)
Draw the diagram of the SN2 mechanism to clearly illustrate the inversion of configuration which occus during SN2 reaction, to secure full marks for such a question.
Q8.
In a sense, it's true that you cannot make a fair comparison because Group 0 aka Group VIII noble gases only have van der Waals radius, while Group I metals only have atomic and ionic radius. But for simplified 'A' level purposes, just state that you expect the Kr atom to have a smaller atomic radius than K, because across the period, there is increasing effective nuclear charge.
Q11. Ans is C. (ambiguous and debatable, cockanaden Cambridge, but this is the most likely the answer Cambridge wants, based on how Cambridge thinks. Notice Cambridge emphasized that the pH "increases very slightly" which is absolutely correct when applied to a buffer).
Q13. Ans is A. (ambiguous and debatable, cockanaden Cambridge, but this is the most likely the answer Cambridge wants, based on how Cambridge thinks. Actually, all 4 options are wrong. The y-axis should be (Vn-Vt) as the raw data, which should then be explained as being directly proportional the the molarity of the diazonium ion [C6H5N2+]. The 1/t is only relevant, but even then not at all necessary, if the volume of the gaseous product rather than molarity of aqueous reactant is used as the y-axis).
Q23. Ans is B. (ambiguous and debatable, cockanaden Cambridge, but this is the most likely the answer Cambridge wants, based on how Cambridge thinks. Ideally, exact temperature and pressure should be given. But in the absence of such data, Cambridge's intention is to see if the student recognizes the reactivity of alkenes, which does indeed undergo hydration with H+ / H2O / heat, just more slowly if the temperature and pressure isn't high enough)
(My answers above are based on Choobie's typed out qns and options, I do not have the Question Paper)
Increasingly so in recent years (see 2010 paper), there will be a few such problematic questions in which even JC teachers will disagree with each other as to which is the best answer. Thus, these are arguably not fair questions to ask students. But Cambridge-SEAB would say, "it's a bell-curve, so if it's not fair to everyone, it's still fair! *evil laughter*"
Anyone wants to check any more answers, eg. Q21, 27, 31, 33, 38, please type out the full question here for me to read and comment on.
Choobie's typed out qns are :
11. Soft drinks often have sodium citrate added to them to act as a buffer.
Which statement about the buffer solutions is correct?
A.The pH of a buffer solution changes slightly when very large amounts of acid or base are added.
B. The pH of a buffer solution increases very slightly when small amounts of acid are added.
C.The pH of a buffer solution increases very slightly when small amounts of base are added.
D. The pH of a buffer solution remains unchanged when small amounts of acid or base are added13. The rate of decomposition of the diazonium cation, C6H5N2+ + h2o -> C6h5oh + h+ + n2 can be followed by measuring the time taken for the same volume of nitrogen to be produced from a range of diazonium cation concentrations.
To find the order of reaction with respect to the diazonium cation, which would be the most suitable graph to plot using the data?
A. [C6H5N2+] against time
B. [C6H5N2+] against 1/time
C. volume N2 against time
D. volume N2 against 1/time23. The diagram shows a reaction (Benzene - Ch2Co2 - cyclyhexene) -> (Conditions are Steam + H+) to form Z.
What final products could be Z.
A. Benzene-Ch2CooH + oh-cyclohexene
B. Benzene-ch2cooh + oh-cyclohexane-oh
Q38. Ans is 1 only.
Nish posted :
38: A catalytic converter is part of the exhaust system of many modern cars.
Which reactiond occur in a catalytic converter?1. 2CO + 2NO => 2CO2 + N2
2. CO2 + NO => CO + NO2
3. 2SO2 + 2NO => 2SO3 + N2
Q1.
Since this form of TiO2 is given (by the qn) to be ionic, hence the cation here is Ti4+. Since Ti is [Ar] 3d2 4s2, hence Ti 4+ is [Ar] 3d0 4s0, which is simply [Ar] which has 18 electrons. Hence answer is A.
Q31.
Ans is 1 only.
Q32.
Ans is 1 and 2 only.
RelicDemon posted :
The Sun light-induced photolysis of water is being investigated as a useful source of the pollution-free fuel hydrogen.
2H20 => 2H2 + O2
It has been found that anatase, one of the three crystalline forms of the ionic compound TiO2, is a good catalyst for this reaction.
How many electrons are associated with each titanium ion in the analase lattice?
A 18 B 19 C 20 D 22
31 Which statements about relative molecular mass are correct?1 It is the sum of the relative atomic masses of all the atoms within the molecule.
2 It is the ratio of the average mass of a molecule to the mass of a carbon-12 atom
3 It is the ratio of the mass of 1 mol of molecules to the mass of 1 mol of hydrogen-1 atoms
32 Which conditions are necessary when an electrode potential is measured using a standard hydrogen electrode as the reference electrode?1 the use of hydrogen gas at 101kPa (1 atm)
2 measurement of the emf when the current delivered by the cell is effectively zero
3 a pH of 1.0 for the solution at the hydrogen electrode
Qn (number?)
Answer is 1, 2 and 3.
Mn : +7 to +4
Cu : +2 to +1
Ag : +1 to 0
helloying posted :
Baeyer's reagent is used to detect carbon-carbon double bonds: a dilute alkaline soution of KMnO4 loses its pink colour in the presence of such a bond.
IN which tests does a positive result involve a reduction in the oxidation number of the metal present in the reagent?
1. Baeyer's test
2. fehling
3. tollens
Q21. As I said many times before (on these forums), the lone pair on the Cl atom delocalizes by resonance to generate a pi bond with the sp2 C atom of the alkene, giving the C-X bond in the resonance hybrid some partial double bond character, which makes it difficult for the Cl (leaving group) atom to be eliminated when a nucleophile (guy) tries to attack the electrophilic (girl) partial positive atom (partially positively charged, due to the electron-withdrawing effect of the more electronegative halogen atom).
A secondary reason (which if Cambridge included it in the options, would then make this a ambiguous question with more than one arguable answer... but Cambridge didn't include this reason in the options) is that "guys only like girls, guys don't like guys" the pi-bond of the electron-rich alkene nucleophile repels the electron-rich OH- (or any other attacking) nucleophile.
So the answer is obviously : C.
Option A doesn't explain why, it just states a fact.
Option B is contradictory. If both are electron withdrawing (one by resonance and one by induction), then the C atom is destabilized, not stabilized.
Option D ambiguously states a fact (the restricted rotation is due to the partial double bond character, which is due to option C anyway) that doesn't answer the question directly (ie. doesn't explains the non-reactivity).
Darko123 posted :
Q21. Why chloroethene is unreactive to nucleophiles ?(what explains lack of reactivity of chloroethene)
A. Substituted alkenes undergo only electro add
B. The carbon double bond and cl atom are both e- withdtawing which stabilises it
C. The electrons on the cl attom delocalise into the pi bond
D. Presence of pi bond prevent free rotation of c-cl bond thus decreasing reacytivity?
Q27 is rather unfair. Though, evidently Cambridge knows this (that it's rather unfair), and thus gave an added hint in the form of solubility. So this question is testing the candidate for his/her understanding of solubilities, as much as (or even moreso than) testing for the candidates knowledge of reactivities of amines and amides.
What happens is that, after reacting PCl5 (electrophile) followed by NH3 (nucleophile), the intermediate is a benzene ring with two ortho substitutents : amide and amine. With these two ortho substitutents in close proximity, with sufficient heat and over sufficient time, the nucleophilic amine group is able to nucleophilically attack the electrophilic amide group (type of reaction : nucleophilic acyl substitution; mechanism : addition-elmination; alternative description : condensation), occuring simultaneously with a proton transfer reaction to generate product A and NH3(g).
helloying asked :
What is the ans for qn 2011 P1 29?
Q29. Ans is B.
Alpha carboxylic acids is always the most acidic (hence pKa 2.1), followed by 'R' group carboxylic acid (hence pKa 4.1), followed by alpha ammonium group (hence pKa 9.5), followed by 'R' group ammonium group (which is not present in this particular amino acid).
Since pH 7 is more acidic than the pKa value, hence the alpha ammonium group remains protonated.
Since pH 7 is more basic than both pKa values, hence both the alpha carboxylic acid group and the 'R' group carboxylic acid groups, are deprotonated.
Q18. Ans is A, since melting point corresponds to fusion enthalpy.
Q36. Ans is 1, 2 and 3.
Options 1 and 2 are direct byproducts, generated from the redox reaction between HBr and H2SO4 (Group VII chemistry). But option 3 is arguable, since it would be a byproduct of a byproduct. It's possible for the primary alcohol to be dehydrated to an alkene (in presence of concentrated H2SO4), and subsequently re-hydrated to generate a secondary alcohol. Not a fair question.
Q40. Ans is 1 and 2 only.
The 'R' groups of the 3 amino acids are not capable of ionic interactions, only H-bonding. Van der Waals interactions are always present across all molecules.
v09876 asked :
How come for question 36, the question was "a mixture of propan-1-ol and sodium bromide...", but the equation below isn't NaBr, it was HBr.
Br- from NaBr will be protonated by H2SO4 to generate HBr, which in turn reacts with the alcohol via a protonation and nucleophilic substitution reaction, to generate the alkyl halide.
hellothere1234 requested :
Can you pls post the ans here? Just type the letters or sth that would really help thanks!!
2011 'A' Level H2 Exam Paper 1
Q1. Ans is A
Q2. Ans is D
Q3. Ans is C
Q5. Ans is C
Q7. Ans is D
Q8. Ans is C
Q9. Ans is A
Q11. Ans is C
Q13. Ans is A
Q17. Ans is C
Q18. Ans is A
Q20. Ans is B
Q21. Ans is C
Q22. Ans is B
Q23. Ans is B
Q25. Ans is C
Q26. Ans is A
Q27. Ans is A
Q29. Ans is B
Q31. Ans is D (1 only)
Q32. Ans is B (1 and 2 only)
Q33. Ans is A (1, 2 and 3)
Q34. Ans is B (1 and 2 only)
Q35. Ans is C (2 and 3 only)
Q36. Ans is A (1, 2 and 3)
Q37. Ans is D (1 only)
Q38. Ans is D (1 only)
Q39. Ans is A (1, 2 and 3)
Q40. Ans is B (1 and 2 only)
Kickme asked :
I understand that C6H5CH2Cl cannot undergo an elimination reaction due to no adjacent carbon present. But how do I answer the question: Explain why C6H5CH2Cl cannot undergo an elimination reaction.
You just gave the answer yourself. If you're looking for a more professional way to phrase the answer, here goes :
C6H5CH2Cl cannot undergo an E1 or E2 elimination of the hydrogen halide, because there is no beta-proton available to be abstracted by the base. The only reactions possible for C6H5CH2Cl are electrophilic aromatic substitution (in which the benzene ring, although slightly deactivated by the electron-withdrawing CH2Cl substituent, functions as the nucleophile), nucleophilic aliphatic substitution (ie. SN2 or SN1; SN1 is possible despite being a primary alkyl halide because the benzene ring can donate electrons by resonance to stabilize the carbocation), and oxidation (either by hot KMnO4, or by combustion).
Kickme asked :
RCl is a saturated alkyl chloride.
RCl reacts with NaOH and a concentration vs time graph is shown.
From the graph, a conclusion that the order of reaction w.r.t. NaOH is zero order and order of reaction w.r.t. RCl is first order.
Is RCl most likely to be a primary, secondary or tertiary alkyl halide?
I would say it is most likely to be tertiary because alkyl groups are electron donating groups which helps to stabalise the positive charge on the carbocation if the reaction takes place via the Sn1 route.
Correct. Also mention the steric hinderance experienced by the incoming nucleophile from the 3 bulky alkyl groups, results in greater feasibility of the SN1 mechanism.
And even if the question doesn't explicitly ask for it, if you've time to do so, draw the SN1 mechanism anyway, with explanatory annotations (of the aforementioned sterics and electronics arguments). That's the exam-smart thing to do.
Posted by kickme
oh.. so steric hindrance answers the part whereby the reaction occurs via the Sn2 route. How do we determine exactly which route the reaction take place via?
DIdn't you just finish your 'A' levels? If you continue to study Chemistry at higher levels, you'll learn to analyze in greater detail the factors the determine the relative % of products for a reaction, eg. one in which SN1 vs SN2 vs E1 vs E2 mechanisms are in competition with each other.
But to answer your question more directly, if it's not one single route, you've to open your mind to the possibility that it's several routes occurring simultaneously, to generate several products simultaneously. It's just a matter of (especially for 'A' levels) identifying the single most major route, mechanism and product.
For SN1 vs SN2, use electronics (electron-donating alkyl groups to stabilize carbocation intermediate) and sterics (steric hinderance by bulky alkyl groups on the incoming nucleophile).
Both electronics and sterics collective predict SN2 for primary alkyl halides, SN1 for tertiary alkyl halides, and a mixture of both SN2 and SN1 for secondary alkyl halides.
E1 and E2 mechanisms are not required for the H2 syllabus. If asked to draw elimination (eg. elimination of HX from alkyl halide using ethanolic NaOH), either E1 or E2 mechanisms will be accepted by Cambridge.
KickMe asked :
I was asked to calculate the enthalpy changes of 2 steps, A and B.
Step A had an enthalpy change of -21kJmol^-1 whereas Step B had an enthalpy change of +70kJmol^-1.
Following which, I was asked to determine which step, A or B, would be more likely to occur considering the answer to the earlier part.
My guess is that the step with a negative enthalpy change would be more likely
Yes, simply because endothermic reactions generally have a significantly higher activation energy compared to exothermic reactions.
Also state that exothermic reactions are more energetically and thermodynamically favourable (just as positive entropy changes are more thermodynamically favourable compared to negative entropy changes).
Kickme asked :
I was asked to calculate the temperature for which a rxn is spontaneous.
What I did was gibbs free energy<0 and so did the calculation and got the temperature range.
The next part of the question asks: State any assumptions made in the calculations.
For 'A' level purposes, the assumption is that the enthalpy and entropy change values remain the same at different temperatures. In reality, these values do change with temperature (in addition to the temperature variable T, itself changing as well).
'A' Level (or challenging 'O' Level) Chemistry Calculations Qn.
Originally posted by hoay:0.144 g of an aluminium compound X react with an excess of water, to produce a gas. This gas burns completely in O2 to form H2O and 72 cm3 of CO2 only. The volume of CO2 was measured at room temperature and pressure.
What could be the formula of X?
A Al2C3 B Al3C4 C Al4C3 D Al5C3
UltimaOnline's solution :
Experimentally, moles of C present = 3x10^-3
Molar mass of AlxCy = 27x + 12y
Moles of AlxCy present = sample mass / molar mass = 0.144 / (27x + 12y)
By stoichiometry, moles of C present = moles of (AlxCy) x (y) = y (0.144 / (27x + 12y))
Which can be equated to 3x10^-3 found earlier.
Simplifying the mathematical expression, we obtain x/y = 4/3.
Hence the answer is C, aluminium carbide.
why iodine is a solid at rtp but hydrogen iodide is a gas?
Because iodine (although non-polar) has very polarizable electron charge clouds (due to it's large no. of electrons and large atomic radius) and hence relatively stronger instantaneous dipole - induced dipole van der Waals (ie. London dispersion forces), in comparison to hydrogen iodide, which may be polar, but still has weaker permanent dipole - permanent dipole van der Waals (ie. Keesom forces) because it has a lot less electrons and H is much smaller, and accordingly it's molecular electron charge clouds are hence much less polarizable.
Polarity is therefore outweighed by no. of electrons, size of atom/molecule and hence polarizability of electron charge clouds. Permanent dipole - permanent dipole van der Waals (ie. Keesom forces) is only stronger than instantaneous dipole - induced dipole van der Waals (ie. London dispersion forces) in cases where the no. of electrons, size of atom/molecule, and hence hence polarizability of electron charge clouds, are comparable or similar.
draw an isomer of octane which contains chiral carbons but is optically inactive. is that possible?
Yes, such molecules are called meso compounds (assuming we're not talking about a racemic mixture of both enantiomers), which contain one or more chiral carbons, but are optically inactive due to either an internal plane of symmetry, and/or an internal center of inversion.
KickMe asked :
When dry chlorine is passed over heated aluminium foil in a hard glass tube, a vapour is produced which condenses to a yellow-white solid on the cooler parts of the tube. At low temperatures, the vapour has the empirical formula. AlCl3, and an Mr of 267. I understand that a dimer was formed. However, I do not know what the yellow-white solid is. Can you tell me?
The yellow-white solid is simply solid aluminium chloride, which has a (sheet-like layered cubic) covalent polymeric structure or ionic lattice structure (both description adjectives are acceptable and correctly describe the same structure, because there is both ionic and covalent character present in the bonds) in the solid state, as opposed to a simple covalent molecular structure in the liquid state (molecular dimer) or in the gaseous states (dimers at lower temperatures, monomers at higher temperatures).
The H2 syllabus only requires you to be able to draw the structures in the liquid and gaseous states (ie. the molecular dimer and monomer), so it's unlikely you'll be asked to draw the structure of AlCl3 in the solid state, but you should at least be aware that in the solid state it has a ionic or polymeric lattice structure, and not simple molecular.
As your chemistry basics would have told you, simple molecular compounds are usually liquid or gaseous, not solid.
Check out an image of AlCl3 in the solid state here on Rod Beavon's webpage :
http://www.rod.beavon.clara.net/aluminiu.htm
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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.
Answer :
PCl5 exists as a simple molecular compound in the liquid state, but exists as ionic PCl4+ PCl6- in the solid state. The ionic geometries are tetrahedral and octahedral respectively for cation and anion.
The P-Cl bond 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.
Simple molecular compounds (eg. water, methane, etc) are almost always liquid or gaseous at rtp, with the notable exception of iodine, for which the large no. of electrons and the large atomic radius, results in highly polarizable electron charge clouds, resulting in extensive and relatively strong instantaneous dipole - induced dipole van der Waals (ie. London dispersion forces), and hence iodine is a simple covalent molecular solid at room temperature.
kickme asked :
I was asked to indicate clear the type of hybridisation on each carbon atom of a hydrocarbon. I know the hybridisation state of the carbon atoms. However, I am rather unsure of how I should present the hybridisation state.
Is it alright for me to draw the entire structure of the hydrocarbon out and just note the hybridisation state at the side of the carbon atom by writing sp, sp2 or sp3? Or is there a proper chemistry way of presenting such an answer?
That's fine. You can also circle the individual C atoms and use long arrows to link it to your written answers "sp3" etc at the side or below the structure if you like.
Bottomline, as long as it's clear and unambiguous communication, it's good Chemistry, whether at 'A' levels or at any level (including professional career Chemists).