[H2 Chemistry] - Misc TYS qns
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TYS2014/P3/5
Why is H2+Ni @150degrees able to reduce the benzene ring but not the amine and carboxylic acid?
Is there any decsrboxylation reaction in the syllabus we need to know? Like some sodalime reaction to decarboylate a carboxylic acid
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Originally posted by BCME:
TYS2014/P3/5
Why is H2+Ni @150degrees able to reduce the benzene ring but not the amine and carboxylic acid?
Is there any decarboxylation reaction in the syllabus we need to know? Like some sodalime reaction to decarboylate a carboxylic acid
H2+Ni @150degrees is able to reduce the benzene ring *as well as* the nitrile group, but not the carboxylic acid.
First of all, although hydrogenation/reduction of the benzene ring is not in the H2 syllabus, but Cambridge *did* give you the Mr of both Z and Omega, which will enable you to deduce the hydrogenation/reduction of the benzene ring.
Next, in the H2 syllabus you ought to have memorized that heating with H2 gas over Ni catalyst at 140 deg C can reduce nitrile to amine (CS Toh pg 244). And also in the H2 syllabus you ought to have memorized that under no circumstances can H2 gas reduce carboxylic acid to aldehyde or primary alcohol. LiAlH4 (in dry ether) is required for that.
Finally, both hydrogenation of the benzene ring to cyclohexane and hydrogenation of nitrile to amine is easier to achieve using H2 gas, compared to reduction of carboxylic acid using H2 gas (which cannot be done), because the former 2 are simpler addition reactions (ie. adding H2(g) to reactant to get the product), while the latter involves not just addition of H, but also requires elimination of an O atom, which requires chemical reagents (to carry away the O) in a solvated environment (ie. not just gaseous H2).
Decarboxylations aren't in the H2 syllabus, but Cambridge can (and have, in recent years) give you sufficient info in the question to deduce that decarboxylation has occurred.
Two notable decarboxylation reactions that are useful for A level students to be familiar with, are the use of soda lime, and heating beta-keto-carboxylic acids (Singapore-Cambridge H2 Chemistry A Level 2012 Paper 3 Qn 5d).
As a challenge for H2 Chem students, try drawing out the decarboxylation mechanisms for both reactions described above.
Additional decarboxylation reactions that Cambridge can use in data-based questions, are the Hunsdiecker reaction and the Kochi reaction. -
TYS 2010/P3/3(d)
Why do we use 24dm3 as the avogadro constant to calculate the number of moles of gases evolved?Is there any instances where the number of moles of gaseous molecules are not calculated using avgadro's constant? like us the Mr of the gases?
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TYS 2014/P2/4(b)
Between Ge and As, and Br and Kr, why is the difference in 1st IE due to "relative constant shielding effect"? And what is that about? -
TYS 2014/P2/5(b)
The phenylamine will be neutralised in alkaline solution to give Ph-NH2, since the N is quite electronegative and hence would be able to form permanent dipole permanent dipole, why is it considered to have low solubility? -
TYS 2014/P2/1(c)
Double the amount of H2O, double the amount of heat, mass of product dobles, then why change in temperature and maximum temperature don't change?
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TYS 2014/P3/4(d)(i)
Is it possible for the answer to be acid base reaction of the phenol in tyrosine and the compound in (i)? And also if there is acid base reaction, will an ionic bond be formed since there will be an anion and a cation?
TYS 2014/P3/4(b)(i)
Can use LiAlH4 to reduce followed by protonation?
TYS2010/P2/5
Is imine group considered and amino acid? also what is the name of the bond formed? Like can is be considered an amide or peptide bond?
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TYS 2013/P1/33
why is option D considered an amino acid when there is no proper amino can carbonyl group? (if I’m not wrong it is an imine group? reference to TYS2010/P2/5)
Also, since the question ask for hydrogen bond to be formed, then ALL the options would be able to form hydrogen bond because that is a characteristic of the secondary structure of protein, then is there something wrong with this question?
And to cause a “bend” in the alpha helix would mean that the R group is bulky am i right? then from option A B and C, only C would cause the bend or some form of distortion in the alpha helix right?
What is the difference between the Alpha helix and the Beta pleated sheet? like why would one occur instead of the other?
TYS 2013/P1/28
why is the formation of disulfide bridge considered an oxidation reaction? Like why is condensation wrong? (am i right to say that H2 is evolved in the formation of the disulfide bridge)
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Delta G (Gibbs free energy) = 0
when there is equilibrium or state change like from H2O(l) <=> H2O(g),
but from H20(s) to liquid the entropy will change so is the Gibbs free energy of that reaction not 0?
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Originally posted by BCML:
TYS 2010/P3/3(d)
Why do we use 24dm3 as the avogadro constant to calculate the number of moles of gases evolved?Is there any instances where the number of moles of gaseous molecules are not calculated using avgadro's constant? like us the Mr of the gases?
In questions like these, it doesn't matter whether you use 24dm3 (which is actually for room temperature at 20 deg C) or using PV=nRT to calculate the more accurate 24.465 (dm)3/mol at 25 °C and 1 atm, or 24.789 (dm)3/mol at 25 °C and 1 bar (Cambridge accepts either 1 atm or 1 bar for standard conditions, as long as clearly specified).
Because in questions like these, you're dealing with ratios, so your final answer for the value of x and y, and hence the formula of the alcohol, is unaffected by which molar volume you use.
Also, often in questions that do not involve PV=nRT, Cambridge will specify "you may assume molar volume of gas under these conditions to be 24dm3", in which case since Cambridge gives you permission to use this shortcut (ie. no need to calculate using PV=nRT), you should use it to save time. -
Originally posted by BCML:
TYS 2014/P2/4(b)
Between Ge and As, and Br and Kr, why is the difference in 1st IE due to "relative constant shielding effect"? And what is that about?
Shielding effect (ie. electrons repel electrons) is most pronounced from inner shell electrons (repelling outer shell electrons). Hence for elements in the same period, the valence shell electrons to be removed experience almost the same shielding effect (since additional shielding effect from inter-electron repulsion within the valence shell itself is relatively negligible). -
Originally posted by BCML:
TYS 2014/P2/5(b)
The phenylamine will be neutralised in alkaline solution to give Ph-NH2, since the N is quite electronegative and hence would be able to form permanent dipole permanent dipole, why is it considered to have low solubility?
Because of the hydrophobic, non-polar benzene ring.
In protonated conjugate acid ionic form, strong ion - permanent dipole interactions makes the phenylammonium ion too soluble in water, hence inseparable from water. Since you want to isolate phenylamine (ie. separate it from water), you need to add alkali to deprotonate phenylammonium to obtain phenylamine, which is significantly less soluble in water, and has a different boiling point from water (184 °C vs 100 °C), and hence can be separated out via fractional / steam / vacuum distillation. -
Originally posted by BCML:
TYS 2014/P2/1(c)
Double the amount of H2O, double the amount of heat, mass of product dobles, then why change in temperature and maximum temperature don't change?
It's true that double the moles of H2O is generated, and hence double the heat is given off. But since the volumes of the water solvent (which absorbs the heat) in acid + alkali, is doubled, this also means the total (doubled) heat given off is also spread out and absorbed over the total (doubled) volume of water solvent, and thus the maximum change in temperature will still be the same. -
Originally posted by BCML:
TYS 2014/P3/4(d)(i)
Is it possible for the answer to be acid base reaction of the phenol in tyrosine and the compound in (i)? And also if there is acid base reaction, will an ionic bond be formed since there will be an anion and a cation?
TYS 2014/P3/4(b)(i)
Can use LiAlH4 to reduce followed by protonation?
TYS2010/P2/5
Is imine group considered and amino acid? also what is the name of the bond formed? Like can is be considered an amide or peptide bond?
TYS 2014/P3/4(d)(i)
No, "acid-base" reaction is not accepted, because the question is asking for an "interaction" not "reaction" and furthermore the question defines the "interactions" as that responsible for the tertiary structure of the protein.
Yes, the required answer is ionic interaction with the aspartate ion (deprotonated conjugate base form at pH 7).
TYS 2014/P3/4(b)(i)
Cannot, coz -CN would be reduced to -NH2.
TYS2010/P2/5
No, an imino carboxylic acid is an imino carboxylic acid, not an amino carboxylic acid. And no, imino carboxylic acids do not form peptide bonds or groups, as an imine group is significantly less nucleophilic compared to an amine group, due to the lone pair residing in a sp2 hybridized orbital, which has greater % s-orbital character, compared to the sp3 hybridized orbital of an amine group (assuming lone pair isn't delocalized by resonance).
Note however, that for this question, Cambridge is loosely (and obsoletely and arguably erroneously) including a secondary amine group in the definition of an "imine group". A secondary amine group, is after all, still an amine group (ie. lone pair residing in an sp3 hybridized orbital, assuming lone pair isn't delocalized by resonance). As such, it is sufficiently nucleophilic to generate the peptide group, and so, yes in this context (as per Cambridge's loose, obsolete and erroneous definition of an imino acid), the so-called imino acid does form the peptide group and peptide bond, in the protein.
In biochemistry contexts, real or actual alpha-imino acids (ie. properly defined as carboxylic acids with the C=N imine group on the alpha C atom, rather than Cambridge's secondary amine group trying to pass of as an imine group) involved in protein synthesis, are usually metabolic intermediates converted from one particular proteinogenic amino acid, catalyzed by a specific enzyme, and subsequently converted into another particular proteinogenic amino acid, catalyzed by another specific enzyme. -
Originally posted by BCML:
TYS 2013/P1/30
why is option D considered an amino acid when there is no proper amino can carbonyl group? (if I’m not wrong it is an imine group? reference to TYS2010/P2/5)
Also, since the question ask for hydrogen bond to be formed, then ALL the options would be able to form hydrogen bond because that is a characteristic of the secondary structure of protein, then is there something wrong with this question?
And to cause a “bend� in the alpha helix would mean that the R group is bulky am i right? then from option A B and C, only C would cause the bend or some form of distortion in the alpha helix right?
What is the difference between the Alpha helix and the Beta pleated sheet? like why would one occur instead of the other?
TYS 2013/P1/28
why is the formation of disulfide bridge considered an oxidation reaction? Like why is condensation wrong? (am i right to say that H2 is evolved in the formation of the disulfide bridge)
TYS 2013/P1/30
As explained previously, proline is more correctly considered an alpha-amino acid with a secondary amine group, rather than an alpha-imino acid (since it doesn't have a true C=N imine group).
You must be able to interpret the Cambridge phrasing correctly. The question is asking for you to verify the capacity for both ends (ie. the amine end and the carboxylic acid end) of each amino acid (in the 4 options) to each form a peptide group that are both capable of both accepting H bonds (ie. by the -C=O peptide group) and donating H bonds (ie. from the -N-H peptide group).
Thus, the answer is proline (option D), because upon nucleophilic acyl substitution and condensation to generate the peptide group, the secondary amine has lost it's only H atom (eliminated as H2O, enzyme catalyzed in biosynthesis), and as such the peptide group generated at this end, does not contain a H atom, and thus is incapable of donating a H bond.
The standard usual hydrogen bond for secondary structure of proteins, involve the -C=O group acting as hydrogen bond acceptor, and -N-H peptide group acting as hydrogen bond donor.
Consequently, the proline amino acid residue's peptide group's inability to donate a H bond, together with the steric hindrance effect caused by the exceptional conformational rigidity of proline (in turn due to the distinctive cyclic structure of proline's R group side chain), results in proline acting as a structural disruptor in the middle of secondary structures such as alpha helices and beta sheets.
As to your loaded question on why some sections of a polypeptide chain, preferentially form alpha helices while others form beta sheets, this goes beyond both H2 Chemistry and H2 Biology. Of course it has to do with the primary structure (of which the secondary, tertiary and quaternary structures build upon), or the sequence of amino acids in the section of the polypeptide chain, and the configuration of different R groups of the different amino acid resides exerting a compound effect on whether the alpha helix or beta sheet would be more thermodynamically favorable.
Certain amino acid residues favor forming the alpha helix (ie. high alpha-helical propensities), while other amino acid residues favor forming the beta sheet (ie. low alpha-helical propensities), depending on the amino acid's R group. Check out the following table on amino acid alpha-helical propensities :
https://en.wikipedia.org/wiki/Alpha_helix#Table_of_standard_amino_acid_alpha-helical_propensities -
Originally posted by BCML:
TYS 2013/P1/28
why is the formation of disulfide bridge considered an oxidation reaction? Like why is condensation wrong? (am i right to say that H2 is evolved in the formation of the disulfide bridge)
There are 2 reasons why this reaction is more properly labeled as a redox reaction, rather than a condensation reaction.
1stly, because sulfur is more electronegative than H, the OS of the S atoms become more negative when adding H atoms to break the disulfide bond or bridge into 2 separate thiol groups of cysteine amino acid, hence this is a reduction reaction. And when the disulfide bond or bridge is generated from the removal of H atoms from 2 separate thiol groups of cysteine amino acid, the OS of the S atoms become more positive, hence this is an oxidation reaction.
(Note : the actual numerical OS values of the S atoms before and after the disulfide bond forming / breaking redox reaction, depends on which electronegativity system you ascribe to : some electronegativity systems regard S to be more electronegative than C, while other electronegativity systems regard C to be more electronegative than S. The different electronegativity systems, named after the different chemists using different experimental criteria and data, are Pauling's, Mulliken's, Allred–Rochow's, Sanderson's, and Allen's.)
2ndly, a condensation reaction (in organic chem, not physical chem) refers to the actual elimination of 1 small molecule to join 2 molecules together. You do *not* eliminate an actual H2 molecule when forming the disulfide bond or bridge. Rather, the mechanism (whether enzyme-catalyzed in biosynthesis, or laboratory / industrial synthesis) involves eliminating 2 H+ ions and 2 electrons (instead of a H2 molecule), and hence this does not strictly qualify as a condensation reaction. -
Originally posted by BCML:
Delta G (Gibbs free energy) = 0
when there is equilibrium or state change like from H2O(l) <=> H2O(g),
but from H2O(s) to liquid the entropy will change so is the Gibbs free energy of that reaction not 0?
At 273K (ie. the melting / freezing point of water), the kinetic rate and thermodynamic feasibility of both the freezing and melting processes are the same, hence Gibbs free energy change for both the forward (eg. melting) and backward (eg. freezing) reactions are both zero.
It's true that the exact values for entropy change and (even) enthalpy change of a particular reaction depends on its temperature; but for A level purposes, in order to simplify calculations, you have no choice but to make the assumption (and Cambridge has indeed asked before in past A level papers, "State the assumptions in carrying out your calculations") that the variation in entropy and/or enthalpy across the temperature range (involved in the question) is negligible, when plugging in the delta S and delta H values into your delta G formula. -
General Conceptual Questions:
(1) Is there resonance in CH3CO3- anion? (I'm guessing no, but I don't know why.)
(2) Why is it a "reflux" for KCN(alc) reactions?
A Level Questions:
2013P3Q2(b)[ii]
The answer key says "NH3 acts as a reducing agent as it reduced HNO2 to N2, itself being oxidised to N2." It's a bit confusing to me. Is the thing being reduced N as well? Is this a disproportionation reaction?
2013P3Q2(d)[ii]
"Benzylamine does not decolourise reddish-brown bromine water and does not form white ppt."
Why does it not decolourise reddish-brown bromine water? Is it not a ring activated by a NH2 group that can have Br groups attached to it?
2013P3Q3(c)[ii]
The answer said that "KMnO4 reacts with SO2 to form colourless products" then proceeded to say "a permanent pale pink colur is seen at the end-point". Is it colourless or pale pink? (Saw both colourless and pale pink in a lot of other Qs too, very confused on this front.)
2014P3Q5(c)[i]
Why is H2O not oxidised instead? The answer placed E value of Ni2+/Ni instead, but E value of H2O (-0.83) is more negative than Ni2+/Ni (-0.25)?
2014P3Q5(d)
"silver will drop to the base as anode sludge."
What is anode sludge and how does it work? Why does it not stay at the anode? (This is very confusing for me because even though it's a alloy with many components oxidising away gradually, I don't see how it will just fall off? *confused face* *brings out hot tea to calm self* Sorry if the question is kinda dumb, haha.)
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Thanks :)
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Originally posted by Shanflopstoground:
General Conceptual Questions:
(1) Is there resonance in CH3CO3- anion? (I'm guessing no, but I don't know why.)
(2) Why is it a "reflux" for KCN(alc) reactions?
A Level Questions:
2013P3Q2(b)[ii]
The answer key says "NH3 acts as a reducing agent as it reduced HNO2 to N2, itself being oxidised to N2." It's a bit confusing to me. Is the thing being reduced N as well? Is this a disproportionation reaction?
2013P3Q2(d)[ii]
"Benzylamine does not decolourise reddish-brown bromine water and does not form white ppt."
Why does it not decolourise reddish-brown bromine water? Is it not a ring activated by a NH2 group that can have Br groups attached to it?
2013P3Q3(c)[ii]
The answer said that "KMnO4 reacts with SO2 to form colourless products" then proceeded to say "a permanent pale pink colur is seen at the end-point". Is it colourless or pale pink? (Saw both colourless and pale pink in a lot of other Qs too, very confused on this front.)
2014P3Q5(c)[i]
Why is H2O not oxidised instead? The answer placed E value of Ni2+/Ni instead, but E value of H2O (-0.83) is more negative than Ni2+/Ni (-0.25)?
2014P3Q5(d)
"silver will drop to the base as anode sludge."
What is anode sludge and how does it work? Why does it not stay at the anode? (This is very confusing for me because even though it's a alloy with many components oxidising away gradually, I don't see how it will just fall off? *confused face* *brings out hot tea to calm self* Sorry if the question is kinda dumb, haha.)
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Thanks :)
General Conceptual Questions:
Q1. No, and there are 2 ways of looking at why not. 1stly, in a peroxycarboxylate ion (conjugate base), the central O atom is sp3 hybridized (2 lone pairs + 2 bond pairs), hence no unhybridized p orbital available to overlap sideways, required for resonance delocalization of p or pi electrons to occur.
2ndly, if you try to delocalize the -ve formal charge on the terminal O atom, by using lone pair form bond pair with the central O atom, the central O atom will have it's octet configuration 'kena violated' (recall period 2 elements do not have vacant, energetically accessible 3d orbitals to accommodate an expanded octet).
Q2. Yes, you need to know how to draw, label and describe the use of the reflux apparatus (when it came out in the Singapore A levels a few years ago, most Singapore JC students couldn't do this question). There are 2 main types of reflux setups that can be tested at A levels : simple reflux, and reflux with distillation.
Heating is required for most organic chemistry reactions (other than those whose reactants or products are vulnerable to thermal decomposition or undesired side reactions, then room temperature or warming is required instead of heating), as most organic chemistry reactions involve breaking of covalent bonds and hence involve high Ea.
However, heating alone without reflux (ie. the use of a condenser) causes problems : organic reactants and solvents, being simple covalent molecular with weak intermolecular van der Waals are volatitle (ie. low boiling point). As such, either the solvent, or one or more of reactant(s), may vaporize and escape, instead of remaining in the reaction mixture, which sabotages the reaction by preventing it from continuing and/or reaching equilibrium yield for the product.
Hence heating under reflux (ie. with the use of a condenser) has 2 main purposes :
1) Ensuring the solvent does not vaporize away and cause the reaction vessel to boil dry, stopping the reaction. In addition, as a particular solvent has its own fixed boiling point, hence by careful choice of solvent, you can control the temperature within a very narrow range, to allow the reaction will proceed at the temperature ideal for that particular reaction. The constant boiling action also serves to continuously mix the solution, ensuring the reactants can continue to effectively mix and collide and thus reach equilibrium yield for the product.
2) Ensuring that the organic, volatile reactants (1 or more of which may have boiling points below the solvent and products) do not vaporize away and cause the reaction to slow down and stop, which prevents reaching the maximum equilibrium yield possible. -
Originally posted by Shanflopstoground:
General Conceptual Questions:
(1) Is there resonance in CH3CO3- anion? (I'm guessing no, but I don't know why.)
(2) Why is it a "reflux" for KCN(alc) reactions?
A Level Questions:
2013P3Q2(b)[ii]
The answer key says "NH3 acts as a reducing agent as it reduced HNO2 to N2, itself being oxidised to N2." It's a bit confusing to me. Is the thing being reduced N as well? Is this a disproportionation reaction?
2013P3Q2(d)[ii]
"Benzylamine does not decolourise reddish-brown bromine water and does not form white ppt."
Why does it not decolourise reddish-brown bromine water? Is it not a ring activated by a NH2 group that can have Br groups attached to it?
2013P3Q3(c)[ii]
The answer said that "KMnO4 reacts with SO2 to form colourless products" then proceeded to say "a permanent pale pink colur is seen at the end-point". Is it colourless or pale pink? (Saw both colourless and pale pink in a lot of other Qs too, very confused on this front.)
2014P3Q5(c)[i]
Why is H2O not oxidised instead? The answer placed E value of Ni2+/Ni instead, but E value of H2O (-0.83) is more negative than Ni2+/Ni (-0.25)?
2014P3Q5(d)
"silver will drop to the base as anode sludge."
What is anode sludge and how does it work? Why does it not stay at the anode? (This is very confusing for me because even though it's a alloy with many components oxidising away gradually, I don't see how it will just fall off? *confused face* *brings out hot tea to calm self* Sorry if the question is kinda dumb, haha.)
_________________________________________________
Thanks :)
A Level Questions :
2013P3Q2(b)[ii] : This is not a disproportionation reaction, but a symproportionation or comproportionation reaction. It does not change the fact that NH3 is being oxidized and hence is the reducing agent, and that HNO2 is being reduced and hence is the oxidizing agent.
2013P3Q2(d)[ii] : This molecule isn't phenylamine, it is benzylamine. The NH2 group cannot donate electrons by resonance into the benzene ring, as it is inhibited or blocked by the sp3 C atom, and hence resonance delocalizaton of the N atom's lone pair to the benzene ring sp2 C atom (remember that resonance requires sideways overlap of unhybridized p orbitals) cannot occur without 'violating' the octet of the sp3 C atom.
2013P3Q3(c)[ii] : Purple MnO4- is reduced to Mn2+, which is a very pale pink colour, which in diluted solutions appear colorless. The end-point is indicated by the first sighting of a permanent pink colour (even after swirling the conical flask). This "permanent pink colour" is *not* due to the very pale pink colour of Mn2+, but instead is due to the last drop (from the burette) of the purple KMnO4 which indicates the end-point has been reached, and that the last drop is an 'extra' drop of KMnO4 in excess which doesn't react away (since the analyte has already been fully oxidized), hence the "permanent pink colour".
And if you're wondering "but won't that mean the volume reading won't be accurate, since there's an 'extra' drop in excess?", then 2 things : 1stly, that single extra drop is relatively inconsequential and negligible, compared to the other inevitable experimental limitations : limitations of measuring instruments, uncertainty values, human errors, systematic errors, and random errors. 2ndly, remember that the end-point only approximates equivalence point. We can only claim the value obtained to be the end-point, which we can only hope (with minimal experimental error) to be sufficiently close to the true equivalence point. Hence to ensure reliability and reproducibility, you have to repeat the titration several times until you obtain concordant titres.
2014P3Q5(c)[i] : Eh walau you. -_-". -0.83V is the standard reduction potential of H2O to H2 gas. Since the standard reduction potential of Cu2+ to Cu is +0.34V (ie. reduction potential more positive than -0.83V the standard reduction potential of H2O to H2), obviously Cu2+ wins and is hence reduced to Cu at the cathode (instead of H2O being reduced to H2 gas). If you wanna compare the standard oxidation potential of Ni to Ni2+ which is +0.25V, to the standard oxidation potential of H2O to O2 which is -1.23V, then obviously Ni wins (ie. oxidation potential more positive) and is hence oxidized to Ni2+ at the anode (instead of H2O being oxidized to O2 gas). And you can use +0.83V only if you want to oxidize H2 gas (in alkaline solution) to H2O, but wherefore cometh the H2 gas (and the required OH-(aq) ions) in this question?
2014P3Q5(d) : "Anode sludge" is the term given for the gooey disguisting stuff (ie. sludge) that is deposited below the anode. Why does it fall off and drop down? Because metals are insoluble in water *and* denser than water. Imagine you have a cake with many small stones embedded in it. As the cake is being eaten away by ants, won't the stones fall? As the Cu atoms in the block of impure Cu ore rock (which is the anode) is being oxidized to *aqueous* Cu2+ (ie. cations are soluble (unlike solid metals in atomic state) thanks to ion - permanent dipole interactions can dissolve and become aqueous, making the block of impure Cu ore rock which is the anode appear to dissolve away), the less reactive (ie. less electropositive and hence harder to oxidize) metals such as silver, remaining as the unoxidized solid metal, and solid metals being insoluble in water and denser than water, naturally drop down from the impure copper block of ore rock (ie. the anode), which together with any unreactive (organic or inorganic) material left over from the ore rock (eg. sand, soil, etc), becomes the gooey disgusting mess below the anode, which we call "anode sludge". -
AJC_P2_2013
1(a)(iii) Is there any other solvent that is suitable besides the one given in the solution?
Also we required to know the functions of ether?
HCJC_P2_2013
1(b) How do we know that the precipitation of barium sulfate is exothermic and sodium sulfate is not?
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Originally posted by BCME:
AJC_P2_2013
1(a)(iii) Is there any other solvent that is suitable besides the one given in the solution?
Also we required to know the functions of ether?
HCJC_P2_2013
1(b) How do we know that the precipitation of barium sulfate is exothermic and sodium sulfate is not?
ACJC lah, not AJC.
Grignard reagents, like LiAlH4, must be anhydrous and kept in dry ether, because they are sources of C- and H- strong Lewis bases, which are strong nucleophiles and strong Bronsted-Lowry bases, and hence will readily abstract protons H+ from moist air (which will result in your expensive Grignard and LiAlH4 reagents being wasted, and worse, pose a safety flammable explosive hazard due to the H2 generated for LiAlH4) if not kept safely solvated in a dry solvent.
You need to know that ether is an excellent solvent to use because it is inert (it is aprotic due to a lack of a partial positive H atom and thus non-acidic; and it is non-nucleophilic and non-basic as it's already bonded to 2 R groups which are non-viable leaving groups) and it's a good balance of polar (the partial negative charged O atom) and non-polar (the hydrophobic alkyl groups) and although it can't donate H bonds, it can receive H bonds, and consequently it is an excellent inert solvent which can dissolve almost all organic molecules (both polar and non-polar, both protic and aprotic), including phenol and phenylamine.
HCJC_P2_2013 - Wah biangz eh, all sodium salts are soluble lah (everything covered in O levels is examinable at A levels!). So obviously no sodium ppt forms. BaSO4 is famously insoluble (O levels!), and (as part of the H2 syllabus, you should know this mah) obviously ionic precipitations are exothermic due to the formation of (large no. of strong ionic bonds holding together the) giant ionic lattice structure that you see (with your naked eye) as a precipitate. -
When calculations on TM are done, when do we include the hydrated H2O or the anhydrous salt? Eg 2013 P2 Qn 2(c)(iii) we use the Mr of anhydrous salt
2013 Paper 2
Q3(a)(i) can reagent A be Na metal instead of NaOH?
Q4(a)(i) When we compare between main group and TM can we use NC increase across the group and SE remains relatively constant to compare ENC and from there explain the atomic radii? Since all the elements are in the same period?2013 Paper 3
Q1(c)(i) When we draw the mechanism do we use equilibrium arrow? Sincs that is given in the question?2012 Paper 2
Q2(a)(ii) Can we say that nitrogen containing hydrocarbon is combusted to produce NO gas?
Q2(c)(i) How do we know that delta G is supposed to be measured in joules?
Q5(a)(ii) Do we have to describe the mechanism by steps like nucleophilic attack and regeneration of catalyst in the mechanism? -
Originally posted by BCML:
When calculations on TM are done, when do we include the hydrated H2O or the anhydrous salt? Eg 2013 P2 Qn 2(c)(iii) we use the Mr of anhydrous salt
2013 Paper 2
Q3(a)(i) can reagent A be Na metal instead of NaOH?
Q4(a)(i) When we compare between main group and TM can we use NC increase across the group and SE remains relatively constant to compare ENC and from there explain the atomic radii? Since all the elements are in the same period?2013 Paper 3
Q1(c)(i) When we draw the mechanism do we use equilibrium arrow? Sincs that is given in the question?2012 Paper 2
Q2(a)(ii) Can we say that nitrogen containing hydrocarbon is combusted to produce NO gas?
Q2(c)(i) How do we know that delta G is supposed to be measured in joules?
Q5(a)(ii) Do we have to describe the mechanism by steps like nucleophilic attack and regeneration of catalyst in the mechanism?
2013 P2 Qn 2(c)(iii) - If the question doesn't specify "hydrated", you've to use the molar mass of the anhydrous form.
2013 Paper 2 Q3(a)(i) - You've only a small chance that Cambridge might show you mercy and give you that mark, for 3 reasons. 1stly, Cambridge is testing to see if the student knows that phenol, being more acidic than alcohols, can be deprotonated just by the use of NaOH(aq) alone. 2ndly, NaOH is a lot cheaper and less hazardous than Na. You should always only use the cheaper and safer reagent, if both reagents have the same effect. 3rdly, beyond the H2 syllabus, Na being such a strong reducing agent, can actually (under certain circumstances or with certain co-reagents) reduce the benzene ring (eg. Birch reduction).
Q4(a)(i) - Acronyms are not acceptable by Cambridge in the A levels. Don't over-use them even in your own self-revision. No, your answer is relevant only for period 2 and 3 elements excluding d block metals. For this question, you need to specify that 3d orbitals (being more diffused) provide relatively poor shielding for the 4s orbitals electrons, which are consequently more strongly electrostatically attracted to the positively charged nucleus, of which Fe and Cu have 6 and 9 more protons compared to Ca.
2013 Paper 3 Q1(c)(i) - Cambridge won't give a damn either way.
2012 Paper 2 Q2(a)(ii) - You won't get any marks for your answer. Because in the 1st place by definition of hydrocarbon, there is no nitrogen present (immediately zero marks for this part of the question liao). Next, even if some organic fuels contained nitrogen (eg. amine group, nitrile group, amide group, imine group, etc), that isn't the main mechanism (which Cambridge is testing you for) by which oxides of nitrogen are formed. It's the atmospheric N2 which reacts with atmospheric O2 at high temperatures that is being tested here. Be sure you mention the strong NN triple bond, the high activation energy involved, and the high temperatures present in combustion engines.
Q2(c)(i) - If before today you didn't know, after today you know liao. Yes you're supposed to have memorized this in the H2 syllabus lah!
Q5(a)(ii) - As long as you draw out the full mechanism correctly (yes including regeneration of the catalyst), you don't need to describe anything in words. -
TJC_2013_P2
Q1 (iii) How do we know what chemical reagents A and B are?
For B can we use LiALH4? Or must it be something that can react with h20?
Thanks!
