(Standard) 'A' Level Qn.
(Challenging) 'O' Level Qn.
For gases, there exists a formula that describes the relationship of the variables of Pressure, Volume, no. of moles, Temperature :
P V = n R T (where R is a constant mathematical value).
If these units (pressure in Pa; volume in m3) are used, the value of the gas constant R is 8.314. If these units (pressure in atm; volume in dm3) are used, the value of the gas constant R is 0.08206. Temperature is always measured in Kelvins (where 298K = 25°C; 273K = 0°C).
(Note : in the actual 'A' level exam, use the Data Booklet's printed 3-significant-figures value given for gas constant R = 8.31; pressure in Pa, volume in m3)
a) A bubble of gas (initial diameter 1.585 cm; mass 2.0973 x 10-2 g) that contains twice as much oxygen as it does carbon dioxide, emerges from a photosynthesizing aquatic plant (there’s a plesiosaur reptile aka “Loch Ness monster” right next to the plant, btw) at the bottom of a lake, where the temperature is really cold and the pressure is 6.4 times greater than at the lake’s surface; and the bubble rises until it reaches the surface of the lake (which happens to be at sea level, and at that moment is equivalent to room temperature), and bursts “pop!”.
(i) Calculate the diameter (in cm) of the bubble just before it bursts.
(ii) Calculate the body temperature (in °C) of the plesiosaur at the time the bubble emerges.
b) At sea level, standard atmospheric pressure (ie. 1.01325 x 105 Pa) causes mercury in a dish to rise 760 mm up a glass column. A mixture of two alkanes (with molar masses 16.0 g and 30.0 g respectively) is stored in a container at 294 mmHg. The gases undergo complete combustion to produce CO2 that has a pressure of 356 mmHg when measured at the same temperature and volume as the original mixture. Calculate the percentage composition of the mixture.
aii) 6 deg C
b) 21.1% and 78.9%Originally posted by Bigcable22:
as for my question first hor i usedp = 101000pa
V=(4.6 X 10^ -5) m^3
my answer i get like 106.7 but the answer is 106.4
Your working is ok. It's a silly issue about significant figures.
Student answers are slightly different when you use
1) 3 sig figs throughout
2) 4 sig figs for intermediate working and 3 sig figs for final answer
3) stored-in-memory calculator's 10 sig fig values for all calculations throughout.
Furthermore, in this context of gas laws,
R = 8.314472 (pressure in Pa, volume in m3)
R = 0.0820574587 (pressure in atm, volume in dm3)
But when you use 3 sig fig values for (arguably "intemediate") calculations, of R = 8.31 or 0.0821, you end up with a slightly different 3 sig fig answer.
In the actual 'A' level exams, Cambridge should not penalize students for this issue in the mark scheme (as this is not an issue of Chemistry, but mathematical 'nomenclature'), as long as the student doesn't try to annoy the marker by spamming 10 sig fig values on the answer booklet.
Nonetheless, with regards to gas constant R, since it's included in the Data Booklet, use the value of R = 8.31 as stated in the Data Booklet.
Similarly, although 1 atm = 1.01325 x 10^5 Pa, but if you use the 3 sig fig value of 1.01 x 10^5, you will end up with a slightly different answer.
(You should be as consistent as possible; Bigcable22 you used 4 sig fig for gas constant R but used 3 sig fig for converting atm to Pa; either use 4 sig fig each, or 3 sig fig each.)Edited by UltimaOnline 23 Mar `09, 9:09AM
'A' Level Qn. (For conceptual understanding purposes only; not directly tested in 'A' level exams).
For most reactions involving gases (in which there are unequal moles of gas on LHS vs RHS, and the forward or backward reaction is either endothermic or exothermic), changing pressure (by changing volume), as well as changing temperature, will will result in a "shift of position of equilibrium". Yet, in terms of Qc (reaction quotion) and Kc (equilibrium constant), there is a fundamental difference, in using pressure versus temperature, to "shift the position of equilibrium". Explain.
In the case of changing pressure (by changing volume), Kc has not changed. Qc has changed, and is now lesser (or greater) than Kc, and hence we say "the (unchanged) position of equilibrium (Kc) has shifted to the right (or left)", relative to our (new) Qc value.
In the case of changing temperature, Qc has not changed. Kc has changed, and if the forward reaction is endothermic (or exothermic), Kc is now greater (or lesser) than Qc, and we say "the (new) position of equilibrium (Kc) has shifted to the right (or left)", relative to our (unchanged) Qc value.
'A' Level Qn.
Given that the pyridine molecule is the benzene ring, but with an N atom replace a C atom, ie. C5N1H6.
1) Draw the displayed structural formula for dipyridineiodine(I) nitrate(V).
2) Explain the effect of adding KI to dipyridineiodine(I) nitrate(V).
Regarding the structure of dipyridineiodine(I) cation; first notice that this is a cation, so an electron has been lost from the species, and from the aromatic nature of pyridine, you would be able to correctly deduce that the electron has been lost from iodine.
Because nitrogen does not have empty d orbitals to expand its octet, (and) in order to preserve aromaticity, pyridine would act as a nucleophile attacking the I+ ion (because the lone pair on N is not delocalized into the benzene ring, it's a strong nucleophile and base). Hence the dipyridineiodine(I) species would see the iodine atom having two bond pairs, one with each pyridine.
So each N atom would have a postive formal charge (having donated its lone pair for dative bonding), and the iodine atom, having 3 lone pairs (recall that it has lost an electron) and 2 bond pairs, giving the iodine atom a negative formal charge.
Overall ionic charge = sum of formal charges = (+1) + (+1) + (-1) = +1 cationic charge.
Oxidation State (O.S.) of iodine = formal charge + electronegativity consideration = (-1) + (+1) + (+1) (because the iodine has 2 covalent bond pairs with more electronegative nitrogen atoms) = +1
Because the O.S. of iodine in dipyridineiodine(I) cation is +1, hence the stock name dipyridineiodine(I).
Check your displayed structural formula for the dipyridineiodine(I) cation :
Each of the N atoms has
(i) a formal charge of +1 (4 valence electrons (close to nucleus) on a Grp V element; donated to form a dative bond in a nucleophilic attack)
(ii) an oxidation state of -3 (ie. (+1) + 4(-1); N is more electronegative compared to I and C)
(iii) a stable octet (4 bond pairs = 8 valence electrons (in total)).
The I atom has
(i) a formal charge of -1 (8 valence electrons (close to nucleus) on a Grp VII element; initially lost an electron but subsequently accepted 2 electrons in the form of 2 dative bonds)
(ii) an oxidation state of +1 (ie. (-1) + 2(+1) ; N is more electronegative compared to I)
(iii) expanded its octet (2 bond pairs + 3 lone pairs = 10 valence electrons (in total). Iodine is in period 5 and can expand its octet by accepting electrons into its vacant 5d orbitals / subshell).
To draw the nitrate(V) ion, NO3 -
(do not confuse with the nitrate(III) ion, aka nitrite ion)
Central N atom bonded to 3 O atoms.
Since uninegative ionic charge, hence at least 1 uninegative formal charge, obviously on O.
So we have 1 singly bonded -ve formal charged O atom (ie. 1 bond, 3 lone pairs)
Assuming no other formal charges, the other two O atoms are doubly bonded (ie. 2 bond, 2 lone pairs).
However, this would violate N's octet (since N doesn't have energetically accessible vacant 3d orbitals to expand its octet).
Hence, shift one of the pi bonds to become a lone pair on one of the neutral O atoms.
So now we have 2 singly bonded O atoms, each with a uninegative formal charge.
And the N atom now has its stable octet and a unipositive formal charge (ie. 4 bonds, 0 lone pairs).
One of bonds from the N to O- atoms is a dative covalent bond, which explains the presence of two atoms with opposite formal charges (positive formal charge on the donor, negative formal charge on the recipient).
The negative formal charge on the other singly bonded O atom is the result of the loss of a proton (from its conjugate acid, nitric(V) acid).
>>> 1) Explain the effect of adding KI to dipyridineiodine(I) nitrate. (Hint. redox) <<<
The O.S. of I in KI is -1, while the O.S. of I in dipyridineiodine(I) nitrate is +1. So the probable resulting redox reaction is quite easily deduceable - molecular iodine aka diiodine (O.S. of zero) is produced, along with pyridine (in its protonated conjugate acid form if an acidic protic solvent is used) and potassium nitrate(V) (as spectator ions).Edited by UltimaOnline 09 Feb `10, 4:48PM
'A' Level Qn.
Are the following compounds covalent or ionic?
(*Not required at 'A' levels; for students' interest only)
Solution & Comments :
Aluminium chloride is a well known example of a compound that is predominantly covalent (how much ionic character its structure has, depends on whether it is in the solid, liquid or gas states/phases), due to the relatively small electronegativity difference between aluminium and chloride. (In contrast, aluminium oxide is predominantly ionic with covalent character, which explains its amphoteric properties; 'A' level students should attempt drawing out the hydrolysis mechanisms of covalent vs ionic oxides).
( Did you know? Aluminium chloride exists in 3 different structural forms, depending on temperature and pressure - as an ionic crystal lattice structure in the solid state (see Rod Beavon's webpage on AlCl3), as a dimer (Al2Cl6 molecules) with dative covalent bonds (to allow Al to achieve a stable octet) in the liquid and moderate temperature gas phases, and as a monomer (AlCl3 molecules) in higher temperature gas phase. Visit Wikipedia's http://en.wikipedia.org/wiki/Aluminium_chloride )
Oxides of titanium, are a much more interesting matter. Titanium has an electronegativity rating relatively similar to aluminium, to be precise - Al (1.5) is slightly more electronegative than Ti (1.3). After electronegativity, there is secondary factor* that determines the extent of covalent vs ionic character in such compounds - its actual structure, in turn due to the valencies or stiochiometries of the compound.
(*'A' level H2/H3/H1 students will only be examined on the primary factor of electronegativity difference. Anything beyond this is for your own further reading only.)
Oxides of titanium illustrate this principle rather nicely.
>>> The nature of the chemical bond in three titanium oxides of different crystal structure and different formal oxidation state has been studied by means of the ab initio cluster-model approach. The covalent and ionic contributions to the bond have been measured from different theoretical techniques. All the analysis is consistent with an increasing of covalence in the TiO, Ti2O3, and TiO2 series as expected from chemical intuition. Moreover, the use of the ab initio cluster-model approach combined with different theoretical techniques has permitted us to quantify the degree of ionic character, showing that while TiO can approximately be described as an ionic compound, TiO2 is better viewed as a rather covalent oxide. <<<
PubChem's (structural diagrams) on :Edited by UltimaOnline 23 Jun `09, 11:04AM
'A' Level Qn.
Why do alkenes undergo electrophilic addition while aldehydes/ketones undergo nucleophilic addition?
In every single case of nucleophile-electrophile reaction in Organic Chemistry, it is always the nucleophile attacking the electrophile (ie. the electron rich-nucleophile sends out its available lone pairs to become (dative) bond pairs with the electron-deficient electrophile target).
In the case of alkenes (whose electron-rich pi bond functions as the nucleophile), the electrophile (eg. HCl --> H+ + Cl-; H+ is the electrophile initally added; Cl- (nucleophile) attacks the carbocation (electrophile) only later) is added on to the alkene (nucleophile) - hence "electrophilic addition".
In the case of aldehydes/ketones (whose electron-poor, partially positively charged carbon (thanks to the electronegative oxygen) of the carbonyl group, functions as the electrophile), the nucleophile (eg. HCN --> H+ + CN-; CN- is the nucleophile initially added; H+ is the proton accepted by the CO- base only later) is added on to the aldehyde/ketone (electrophile) - hence "nucleophilic addition".Edited by UltimaOnline 04 Apr `09, 6:12PM
'A' & 'O' Levels Qn.
A perfectly healthy gentleman was taking an evening stroll in the forest. Suddenly, the sky rapidly darkened with heavy cloud cover, lightning flashed in the distance accompanied by booming thunderclaps. The man spotted a deserted coal mine and began to walk briskly towards it, thinking it might shelter him from the storm. As the man neared the entrance of the coal mine, he suddenly collapsed and died. Assume you are young Kindaichi, and solve the murderous mystery.
During the atmospheric buildup of a thunderstorm, the air pressure at ground level drops rapidly. The atmospheric pressure differential (inside and outside the mine) causes toxic gases present in the mine to rush out.
Specifically, the toxic carbon monoxide (commonly present in varying amounts in coal mines; why? because within the mines there is very limited air (and hence oxygen) present for any combustion that occurs, accidentally or deliberately, within the mines) is a powerful ligand that has a much higher affinity with the haemoglobin protein present in erythrocytes (red blood cells), as compared to oxygen.
Hence ligand exchange occurs, and carboxyhaemoblogin (not carmabinohaemoglobin) is generated instead of oxyhaemoglobin. Because carbon monoxide is a much more powerful ligand (due to its much greater instability) compard to oxygen, the formation of carboxyhaemoglobin is permanent, and consequently red blood cells are unable to absorb and transport oxygen from the alveoli of the lungs to the rest of the body cells & tissues (the sensitive brain is the first to suffer; hence initial symptoms of dizziness, headaches and unconsciousness before death), and the victim's brain and body tissues suffer from oxygen deprivation/starvation and eventually death.
Our victim fell unconsious as a result of oxygen deprivation/starvaton, and because he collapsed right at the entrance of the lots-of-carbon-monoxide-rushing-out-coal-mine (and no one else was in the vicinity to drag his unconscious body away in time), hence more and more of his oxyhaemoglobin became replaced with carboxyhaemoglobin, until he died.
For more information on "carbon monoxide poisoning" and "carboxyhaemoglobin", visit Wikipedia's :Edited by UltimaOnline 31 Jul `09, 3:15PM
'A' & 'O' Level Qn.Originally posted by d3sT1nY:
When we talk about acid, according to Arrhenius definition, acid is a substance that produces H+ ions when in aqueous state. So this definition limits acid to substances that can dissolve in water.
According to Bronsted-Lowry definition, acid is a proton donor. Acid is said to be dissociated after donating the proton (H+). So it encompasses acid that doesn't dissolve in water. But doesn't it also means that this definition limits acids to substances that contain the proton? Then what about those acidic oxides (non-metal oxides)?
Cl2O5(s) + 2NaOH(aq) ----> 2NaClO3(aq) + H2O(l)
I suppose the Cl2O5 is in solid state to start with. Correct me if I'm wrong.
An acidic oxide is not necessarily an acid by itself (depending on which definition of acid you use). To be precise, an acidic oxide undergoes hydrolysis to form an (Arrhenius and Bronsted-Lowry) acid.
In your case, the hydrolysis of dichlorine pentoxide Cl2O5 yields chloric(V) acid HClO3, which then reacts with NaOH in a typical proton transfer acid-base reaction.
In terms of Arrhenius,
HClO3 (or Cl2O5) is the Arrhenius acid, while NaOH (or Na or Na2O) is the Arrhenius base.
In terms of Bronsted-Lowry,
HClO3 (from Cl2O5 + H2O) is the Bronsted-Lowry acid, while OH- (from NaOH) is the Bronsted-Lowry base.
In terms of Lewis,
H+ (from HClO3) is the Lewis acid, while OH- (from NaOH) is the Lewis base.
For a list of acidic oxides and the aqueous acids they generate upon hydrolysis, visit Wikipedia's
'A' Level Qn.Originally posted by d3sT1nY:
Methylamine is more basic than trimethylamine, less basic than dimethylamine. Electron-donating group on the N makes the electrons more available for protonation, so with an increase in the number of electron-donating group, it makes the amine more basic. Instead, trimethylamine is less basic than methylamine though it contains more methyl (electron-donating) group.
I remembered it has something to do with trimethylamine not being able to form hydrogen bonds with water thus the lower basicity. Or is it due to steric hindrance?
The stability of a conjugate acid (ie. when a base is protonated), depends on stabilization by induction, resonance, and solvation (hydration).
Recall that bond forming is exothermic (ie. products less energy hence more stable than reactants; energy lost to surroundings), and therefore favourable (enthalpic) intermolecular interactions such as ion-dipole interactions, hydrogen bonds, dipole-dipole interactions, all stabilize the species.
In the case of ammonium and alkyl ammonium cations (eg. protonated conjugate acid forms of methylamine, dimethylamine and trimethylamine), there is no resonance stabilization to speak of.
Purely in terms of induction effects (ie. the electron donating alkyl groups (in turn because C is more electronegative than H) stabilizing the positive formal charge on the N atom (a Grp V element with 0 lone pairs and 4 bond pairs) after protonation of the ammonia or alkylamine base), the greatest stabilization conferred by induction would be experienced in the tertiary alkyl ammonium cation, followed by secondary alkyl ammonium cation, followed by primary alkyl ammonium cation, and lastly the ammonium cation.
Purely in terms of solvation (hydration) effects, specifically hydrogen bonding with water molecules, the greatest stabilization conferred by solvation would be experienced in the ammonium cation (4 H-bonds), followed by primary alkyl ammonium cation (3 H-bonds), followed by secondary alkyl ammonium cation (2 H-bonds), and lastly the tertiary ammonium cation (1 H-bond).
(Note that hydrogen bonds rather than ion-dipole interactions, exist for the protonated cation in aqueous solution; the positive formal charged N atom is surrounded by hydrogens and/or alkyl groups tetrahedrally, preventing effective ion-dipole interactions with the polar water solvent molecules.)
Acknowledgements & Recommendation : Graham Patrick's "Organic Chemistry (Bios Instant Notes)" available from NUS co-op, Clementi bookstore, Amazon.com, etc. - Google's Book Search snapshot of page 92 ("Solvation effects").
Just as in real-life, it is often the (fascinating) case in Chemistry, that we have to consider opposing patterns/trends (for instance, solubility of Grp II compounds, boiling points of hydrogen halides, etc), and apply our theoretical conceptual understanding to explain observed experimental data.
In other words, based on experimental evidence, we know that such-and-such principle outweighs such-and-such principle at such-and-such a point. (A good case in point would be enthalpy vs entropy in solubility of Grp II carbonates - theoretically, enthalpy dictates carbonates become less soluble down Grp II; theoretically, entropy dictates that carbonates become more soluble down Grp II; based on experimental evidence, we know that enthalpy effect outweighs entropy effect from Be to Sr, and entropy effect outweighs enthalpy effect from Sr to Ba. Visit Jim Clark's ChemGuide.co.uk's "Solubility of Grp II compounds").
Chemistry (and all Sciences) teaches us that is an overall complex balance of (under varying circumstances, often) opposing principles/patterns/trends, and for the wise with the eyes to see, the mind to comprehend and the heart to feel; this may well serve as a ethical, moral, philosophical and spiritual lesson for humanity.
That in truth (or as Einstein would say, "relative leading edge truth"), there is no "right" and "wrong" (ie. dogma), but there is Ethics ("do unto others as you would like others to do unto you") and there is CosmoEthics ("that which results in the greatest and highest good for all beings involved and ultimately for the entire universe").Edited by UltimaOnline 07 Apr `09, 4:13AM
'A' Level Qn.
Mass of ammonium chloride added = 23.41g
Change in pH (upon adding acid) from 8.9 to 8.868 is minimal due to buffer action.
Significant pH values of graph : 11.27 (0cm3), 9.24 (25cm3), 5.12(50cm3), approaches 0.699 (large excess of HCl).Edited by UltimaOnline 19 Jun `09, 9:41PM
'A' Level Qn.
'O' Level Qn (Challenging).
3 identical coins, made up of 2 metals, alloy.
1st one dissolved in concentrated nitric acid to form a blue solution.
2nd one added to excess diluted sulphuric acid, reddish brown precipitate and a colourless solution is formed.
Then it was filtered and ammonia solution added to the filtrate. White precipitate formed which dissolved in excess ammonia solution.
3rd one added to silver nitrate solution, after several hours, the solution turns pale blue and a grey solid is formed on the coin.
Identify all substances described above, including all relevant equations.
1st experiment - copper(II) nitrate (aq) + zinc nitrate (aq)
2nd experiment - zinc sulfate (aq) + copper metal (s); zinc hydroxide (s); tetraaminezinc(II) complex ions (aq)
3rd experiment - copper(II) nitrate (aq) + zinc nitrate (aq) + silver metal (s)
'A' Level Qn.Originally posted by PyroPunk:
this is my first time i googled for chem qns and stumble on this hmwk forum! lol
anw, just to clarify a doubt;
If aqueous bromine (aka bromine water) is added to 4-aminophenol, what will be the organic product(s)?
If according to lecture notes, since -NH2 and -OH are 2,4-directing saturated EDGs, thus the remaining 4 positions (2,3,5,6) will be subbed with Br...?
Just want to know if it's possible for all 6 position to be filled up, like... very crowded in the benzene ring... ok... maybe follow the directing of the stronger grp instead??? dunno what im talking sry
Thx in advance~~~!
>>> If aqueous bromine (aka bromine water) is added to 4-aminophenol, what will be the organic product(s)? <<<
The phenolic hydroxy group and the amino group are located para to each other. Both the phenolic hydroxy group and amino group, withdraw electrons by induction but donate electrons by resonance (and resonance effects outweigh induction effects here), and are hence (because they donate electrons by resonance) ortho-para directing.
NH2 is a (only slightly) stronger activating substituent compared to OH, because while they both donate a lone pair by resonance, O is more electronegative than N, and hence OH withdraws by induction (only slightly) more than NH2.
So based on electronics, you might expect the incoming electrophile to be directed ortho to NH2 rather than OH (para positions are out, for obvious reasons).
However, NH2 presents a (only slightly) greater steric hinderance compared to OH, due to the additional hydrogen on NH2.
So based on sterics, you might expect the incoming electrophile to be directed ortho to OH rather than NH2.
In practice, and without any further info given by the question on the yield ratios, you may state that a mixture of isomeric products are obtained. Directing effects largely cancel out in this particular case (for reasons discussed above).
At lower temperatures and/or with limiting bromine, you would obtain a mixture of mono or di bromination, rather than tri or tetra bromination, due to steric hinderance and/or limiting bromine.
At much higher temperatures and with excess bromine, in theory you would be able to achieve tetrabromination. That is to say, in theory you can obtain an end-product of 4-amino-2,3,5,6-tetrabromophenol a.k.a. 4-hydroxy-2,3,5,6-tetrabromoaniline.
There is certainly steric hinderance, but both phenolic and amino groups strongly activate the benzene ring towards electrophilic aromatic substitution (even though the bromine substituents added on are slightly deactiving in nature as they withdraw electrons by induction a little stronger than they donate electrons by resonance); and bulky molecules (such as hexabromobenzene, methyltetrabromophenol, etc) are indeed known to exist, and are regularly synthesized by chemical industries.
Having said all this, remember that ultimately, theory is supposed to HELP you, not limit you. Be open to all possibilities, and use your theoretical knowledge to help you interpret understand experimental findings and observations of the universe around you.
As an 'A' level examination candidate, if given such a question (which is unlikely in the actual 'A' level exam; more likely the directing effects of substituents will be clear cut, and/or experimental isomeric provided by the question), then your duty is to predict and explain (just about everything discussed above, ideally) based on your own understanding of organic chemistry principles, of electronics, directing effects, sterics and temperature (ie. activation energies), what the possible products are, and why.Edited by UltimaOnline 15 Apr `09, 12:06PM
'O' & 'A' Level Qn
(Simplified version that can be done at 'O' levels; adapted from H2 Specimen Paper III Q5)
The monoprotic (monobasic) hydroxycarboxylic acid Z, derived from an alarm pheromone used by the African ant, has the molecular formula CH3RC(OH)COOH (where R refers to an alkyl group).
It is known that Z contains 2 chiral carbons (where a “chiral carbon” refers to a carbon atom bonded to 4 different groups).
A solution formed by dissolving 1.0g of Z in 250cm3 of water, had a pH of 2.73; a 25.0cm3 aliquot of this solution required 6.85cm3 of 0.10 mol/dm3 NaOH for neutralization.
Draw the structure of compound Z.
2-hydroxy-2,3-dimethyl-pentanoic acidEdited by UltimaOnline 11 Jul `09, 11:50PM
'A' Level Qn.
The aromatic compound ortho-hydroxyphenol a.k.a. 2-hydroxyphenol is mixed with 1,2-diiodoethane, and refluxed with ethanolic (alcoholic) NaOH. The final product has an empirical formula of C4H4O.
Draw the mechanism of the reaction pathway leading to the final product.
Electrophilic aromatic substitution occurs, and the 1,2-diiodoethane substitutes away a H para to (either one of the two) phenolic OH groups. (Para, not ortho, because of steric hinderance).
The iodine leaves to avoid violating the octet on the electrophilic carbon (of the 1,2-diiodoethane). Note that iodide ion is very stable (due to large ionic radius to stabilize the -ve charge), hence iodine is an excellent leaving group.
When refluxed with alcoholic OH-, the OH- ion acts as a base under alcoholic solvent conditions (as opposed to acting as a nucleophile under aqueous conditions; why? remember that the more unstable the stronger the base, the more polarizable the stronger the nucleophile; NH2- is a stronger base than OH-, while SH- is a stronger nucleophile than OH-), and hence an elimination reaction occurs, and a pi bond (ie. alkene) is formed.
The final product is therefore 2-hydroxy-4-ethene-phenol.Edited by UltimaOnline 21 Apr `09, 12:33PM
'A' Level Qn.
Draw the mechanism to explain the reaction between sodium hydride and water.
Lone pair on H- attacks partially +ve H of H-O-H. Sigma bond between H and O cleaves to become a lone pair on O, to avoid violating H's duplet. End products : hydrogen gas, sodium hydroxide.Edited by UltimaOnline 03 May `09, 12:34AM
'A' Level Qn.
Distinguish phenylmethanal from phenylethanal.
Acidified KMnO4. Although a redox reaction occurs for both compounds, one will additionally result in the production of a covalent oxide gas (that gives a ppt with limewater). Guess which?Edited by UltimaOnline 03 May `09, 12:37AM
'A' Level Qn (but beyond H2/H3 syllabus requirements).
What we call 'hydrogen' usually refers to its most common isotope - protium. Protium hydrogen has molar mass of 1g, while deuterium hydrogen has molar mass of 2g, and tritium hydrogen has molar mass of 3g.
Q1. If a C atom is bonded to 2 different R groups, a protium hydrogen, and a deuterium hydrogen, ie. RR'CDH, is the carbon considered chiral, and does the compound rotate plane polarized light? In other words, do isotopic variants count towards enantiomeric / optical isomerism?
Q2. Usually, isotopic variation results only in the change of physical properties, and not chemical properties. However, heavy water (D2O or dideuterium monoxide) is biochemically toxic (even fatal) to humans and all living organisms. Explain.
Q1. Yes, the carbon is considered chiral, and the compound rotates plane polarized light, but only at a very small angle.
Q2. See Wikipedia on Heavy Water : http://en.wikipedia.org/wiki/Heavy_water
'A' Level Qn.
Because the question is flawed or inadequate in phrasing (asking only for 'stereoisomers' implies no structural isomers are formed, a false assumption), hence there is no satisfactory answer. It may well be that all the available MCQ choices are wrong in the Prelim paper.
Notice that there are two possible structural isomers (the double bond at the bottom of the ring applies to both structural isomers, and isn't capable of cis-trans or E-Z, so ignore it in your counting), both of which have diastereomeric ("geometric") isomers - one is capable of cis/trans and E/Z, the other capable of E/Z only (many JCs wrongly teach students to artificially force cis/trans isomeric categorization for many alkenes that can actually only have E/Z. Technically, cis-trans isomers exist only when an alkene has two substituents and two hydrogens attached to the sp2 hybridized carbons).
After dehydration by excess conc H2SO4, only one of the two abovementioned resulting structural isomers have a single stereocenter / chiral center.
So if the question was reiterated or rephrased to "how many isomers in total (include both structural and stereoisomers) may be generated?", then the correct answer would be
structural isomer #1 - E - R
structural isomer #1 - E - S
structural isomer #1 - Z - R
structural isomer #1 - Z - S
structural isomer #2 - E
structural isomer #2 - Z
structural isomer #1 - cis - R
structural isomer #1 - cis - S
structural isomer #1 - trans - R
structural isomer #1 - trans - S
structural isomer #2 - E
structural isomer #2 - Z
structural isomer #1 - cis - (+)
structural isomer #1 - cis - (-)
structural isomer #1 - trans - (+)
structural isomer #1 - trans - (-)
structural isomer #2 - E
structural isomer #2 - Z
structural isomer #1 - E - (+)
structural isomer #1 - E - (-)
structural isomer #1 - Z - (+)
structural isomer #1 - Z - (-)
structural isomer #2 - E
structural isomer #2 - ZEdited by UltimaOnline 28 Apr `09, 11:46PM
'A' Lvl Qn.
I. Molecular iodine is non-polar, hence shouldn't be soluble in water (a polar solvent). Accordingly and therefore, what is the major species in "aqueous iodine solution"?
II. Is "two bond pairs, zero lone pairs" the only scenario in which you have a linear molecular geometry? Explain.
I2 + I- --> I3-
It is the triiodide ion which is most soluble (ion-dipole interactions with water solvent), and is hence the majority species in "aqueous iodine" solutions.
The electron geometry for I3- is : 5 electron pairs hence trigonal bipyramidal. The molecular geometry for I3- is : 3 lone pairs 2 bond pairs hence linear.
Google/Wikipedia for more information.Edited by UltimaOnline 03 May `09, 12:31AM
'O' & 'A' Level Qn.Originally posted by donkhead333:
so..i had this question (will try to describe as clearly as possible, cant find diagram)
2 beakers, A and B, contains a bar of zinc and iron each along with indicators phenolphthalein(detects alkaline solution) and potassium hexacyanoferrate (detects presence of Fe2+: will turn blue, otherwise colorless). the metals are linked by a wire and a voltage is observed. a salt bridge connects the 2 beakers together, with the salt bridge being partially submerged in both beakers' indicators.
the question was "would any color change be observed at the iron electrode?EYA"
so, i reasoned out this way
zinc is more reactive than iron.
zinc will reduce iron, so iron will gain electrons.
Fe + 2e- -> Fe2- (wtf?)
the answer is that color change will be detected (meaning there is Fe2+ present), but im quite lost as to how Fe2+ is actually obtained......
pls feel free to clear any misconceptions, i'm not very adept with this topic
UltimaOnline's comments :
The question is obviously incomplete (thus the answer is open to possibilities) because the electrolytes in both compartments (connected by the salt bridge) are not specified.
If instead, we modify the question to :
2 beakers, A and B, contain a bar of zinc and iron each along with indicators phenolphthalein (detects alkaline solution) and potassium hexacyanoferrate (detects presence of Fe2+: will turn blue, otherwise colorless). The metals are linked by a wire and a voltage is observed. A salt bridge connects the 2 beakers together, with the salt bridge being partially submerged in both beakers' indicators. A colour change was observed at the cathodic electrolyte. Suggest and explain two possible scenarios (including identities of electrolytes) that would allow this.
One possible scenario, would have the cathodic electrolyte as an aqueous Fe3+ solution.
Assuming unimolarities (ie. standard conditions), the standard reduction potential of the cathodic half equation Fe3+ + e- --> Fe2+ is +0.77V; the standard oxidation potential of the anodic half equation Zn --> Zn2+ + 2e- is +0.76V; hence a current would certainly flow and the standard cell potential would be +1.53V.
3[Fe(CN)6]3- + 4Fe2+ --> Fe4[Fe(CN)6]3
Prussian Red --> Prussian Blue
In such a scenario, the colour change at the cathodic electrolyte would be due to Prussian Red reacting with Fe2+(aq) to generate Prussian Blue.
Another possible scenario would have the cathodic electrolyte as an aqueous acidic solution.
Assuming unimolarities (ie. standard conditions), the standard reduction potential of the cathodic half equation 2H+ + 2e- --> H2 is +0.00V; the standard oxidation potential of the anodic half equation Zn --> Zn2+ + 2e- is +0.76V; hence a current would certainly flow and the standard cell potential would be +0.76V.
Notice that H+ ions came from the solvent water, generating OH- ions as a consequence.
In such a scenario, the colour change at the cathodic electrolyte would be due to phenolphthalein reacting with the OH- ions; specifically, di-deprotonation and structural change from a non-conjugated, tetrahedral center and absorbing UV frequencies (hence colourless) to a highly conjugated, trigonal planar center and absorbing blue-green frequencies (hence pink/fuchsia).
Further reading on the fascinating chemistry of
PrussianRed / PrussianBlue :
Phenolphthalein :Edited by UltimaOnline 03 May `09, 1:13AM
'O' & 'A' Level Qn.
A sample of copper(II) sulfate pentahydrate, CuSO4.5H2O was dissolved to form copper(II) sulfate solution. 25.0cm^3 of this solution was added to 30cm^3 of excess potassium iodide solution. The iodine liberated was titrated with 22.50cm^3 of 0.108 mol/dm^3 sodium thiosulphate solution. What was the sample mass of the copper(II) sulfate pentahydrate, per dm^3 of the solution?
1) Write the balanced redox between Cu2+ and I-, to generate Cu+ and I2 ---> eqn 1.
2) Write the balanced redox between S2O3 2- and I2, to generate S4O6 2- and I- ---> eqn 2.
3) Based on no. of moles of thiosulfate used, find no. of moles of iodine present (use eqn 2).
4) Based on no. of moles of iodine present, find moles of Cu2+ (use eqn 1).
5) Based on volume of solution, find molarity of CuSO4.
6) State that molarity (ie. no. of moles per dm3) of Cu2+ = no. of moles of copper(II) sulfate pentahydrate added to 1dm3.
7) Based on no. of moles of copper(II) sulfate pentahydrate, find sample mass of copper(II) sulfate pentahydrate added.
'O' & 'A' Level Qn.
What is the difference between the process described by these two equations?
CO2(g) + aq ---> ???
CO2(g) + H2O(l) ---> ???
CO2(g) + aq ---> CO2(aq)
This represents the hydration process (if it was a soluble ionic compound, it would be (s) to (aq) = lattice dissociation + hydration = solution process) of gaseous carbon dioxide, to form aqueous carbon dioxide, CO2(aq). This refers to carbon dioxide molecules that have interactions with water solvent, specifically :
a) permanent dipole - induced dipole interactions (since linear CO2 molecule is overall non-polar)
b) permanent dipole - permanent dipole interactions (since in the linear CO2 molecule, the partial positively charged C (of CO2) is exposed/available for electrostatic attraction with the partial negatively charged O of water, and the partial negatively charged O (of CO2) is exposed/available for electrostatic attraction with the partial positively charged H of water).
CO2(g) + H2O(l) ---> H2CO3(aq)
This represents the hydrolysis reaction (ie. a reaction between a particular species (eg. ion, molecule, etc) and water) of carbon dioxide, to generate carbonic acid.
The mechanism is as follows :
Nucleophilic O of H2O attacks C of CO2.
Pi bond shifts to become a lone pair on O.
Proton transfer from H2O+ (+ve formal charge) to O- (-ve formal charge).
Hence, (HO)CO(OH) or carbonic acid, H2CO3(aq) is generated.Edited by UltimaOnline 11 Jul `09, 11:46PM
'A' Level O.C. (Organic Chem) Qns.
Q1. What is the difference between using lithium aluminium hydride LiAlH4, versus using sodium borohydride NaBH4, as a reducing agent?
Q2. LiAlH4 must be used in dry ether, followed by solvating the product of the reduction in a protic solvent (eg. via hydrolysis with water). Explain why these must be done in two separate steps.
Q3. LiAlH4 can be used to reduce almost all organic compounds, except two notable exceptions. One of them is nitrobenzene, for which the recommended reduction pathway involves Sn/Zn/Fe and concentrated HCl, or alternatively hydrogen gas with nickel catalyst. Using LiAlH4 would result in undesirable side-reactions, which are beyond the interest of the H2 syllabus. What is the other class of organic compounds, for which you cannot use LiAlH4 to reduce? Explain why LiAlH4 fails to reduce this class of compounds.
1. Because Al is one electron shell larger than B (the Li+ and Na+ are counterbalancing spectator ions), hence the electron density of LiAlH4 is more polarizable, and hence a more ready/effective source of hydride anion nucleophiles, and therefore a stronger reducing agent, as compared to NaBH4. NaBH4 can only reduce aldehydes and ketones, while LiAlH4 can be used to reduce almost all organic compounds (other than the two in Q3), including carboxylic acids, acyl chlorides, esters, nitriles and amides.
2. Firstly, understand why a protic solvent is REQUIRED after the reduction, then we look at why AFTER the reduction. As the hydride anion nucleophiles attack the partial positively charged carbonyl carbon, the pi electron density / pi-bond shifts up to become a lone pair on carbonyl oxygen. To stabilize this negative formal charged oxygen (forming an alkoxide anion), we need to give it a source of protons. Usually water is used for this, hence some JCs' notes write it as "LiAlH4 (in dry ether) followed by hydrolysis", since hydrolysis simply means the reaction of a species with water; what happens here is a proton transfer from water to the alkoxide anion, to stabilize it.
Why add water only AFTER the reduction? Obviously, if you attempt to use wet or aqueous LiAlH4 (as opposed to in DRY ether), the protons H+ from water would combine with the hydride anions H- from LiAlH4, and you would lose your precious hydride anions (your ammunition for reduction!) in the form of fast escaping hydrogen gas!
1. LiAlH4 (in dry ether)
Hence, we write it in two steps : -------------------->
3. C=C alkenes cannot be reduced by LiAlH4, since the double C=C bonds are electron rich and nucleophilic, and the H- hydride anions from LiAlH4 are also electron rich and nucleophilic. Hence they repel each other. Even if not for the repulsion (eg. if you have high activation energies to overcome the repulsion), no reaction would occur because in Organic Chemistry, reactions occur because electrons always naturally flow from electron-rich to electron-poor, in forming dative covalent bonds as nucleophiles attack electrophiles (note that so-called 'electrophilic attacks' is nevertheless still about a nucleophile attacking (ie. forming a dative covalent bond with) an electrophile; but some pple use the term 'electrophilic attack' simply because the electrophile is added to the substrate nucleophile; macham the substrate kena 'attacked'; I punch you then I cry "boohoo teacher he punch me" to be precise "his face punch my fist").
Incidentally, unsaturated aromatic compounds such as benzene, cannot be reduced by LiAlH4 for the same reason it cannot reduce C=C unsaturated alkenes. The benzene ring is electron-rich, and will have no reaction with hydride anion nucleophiles. To reduce the benzene ring into cyclohexane, you will need to use hydrogen and nickel catalyst, at 200 deg C and 30 atm. Such extreme conditions are needed to overcome the strong resonance stabilization energy* of benzene.
A typical H2 exam question related to this is found in the H2 Specimen paper, "Explain why even though the benzene ring is unsaturated, it does not usually undergo addition reactions". The answer, of course, is : Benzene is resonance stabilized by the delocalization of pi electrons within the 6-carbon aromatic ring. Electrophilic addition that would destroy the exceptional stability of the aromatic ring is thus resisted. The official SEAB Mark Scheme iterates it as : "The delocalised electrons give increased stability / the resonance energy stabilises benzene".Edited by UltimaOnline 12 May `09, 1:52PM