Advanced Higher Chemistry 2012-2013
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Original post by Zahra Tasleen
20: Neon is a gas and mercury is a liquid (at room temperature), so you can discount those straight away as the other two are solids at room temperature. Phosphorous melts at 44oC and so, at room temperature, it's atoms are vibrating more than sulphur's are, relative to transforming into the liquid state. This is why sulphur has the lowest entropy and the answer is C.
22: If the reaction is exothermic then deltaH must be negative. It can't be A as a negative deltaG needs to occur only if the reaction is to proceed - this doesn't specify if the reaction is exothermic. You can rule out B because entropy is not the same as enthalpy, and you cannot tell the enthalpy from a lone entropy value. You can also rule out D because a reaction cannot thermodynamically proceed if deltaG is positive. This leaves you with C. C is true because deltaG is negative (thus the reaction will spontaneously proceed) and, although deltaS is negative and this will create a positive TdeltaS term, the negative deltaH will still allow a negative value overall for deltaG.
Hope this may have helped.
Original post by Zahra Tasleen
This is pretty simple, look at D again it is a trans isomer, now look at C, have you ever seen that before, does it fit into the category of trans or cis???
Original post by Chemistry Help
Is that of 2002??
Yes, that's the multiple choice. I can try to put up the written part if you give me 10 mins?
The answer for 22 is d
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Posted from TSR Mobile
Original post by Zahra Tasleen
who r u talking to?
Original post by deedee123
for 20 look up the melting points and choose the one with the highest.
Oh? I have seen somewhere on google or idk but that like the one with the lower entropy will be the one that will have the least disorder so it would have less number of electrons?
Neon: 2,8
Mercury: 2,8,8
Sulphur: 2,6
Phosphorus: 2,8,5
And that's how I got C?
So for the lowest standard entropy means the higher the BP. So if it asks the highest standard entropy means the opposite no?
Original post by GermanScientist
Yes, that's the multiple choice. I can try to put up the written part if you give me 10 mins?
Please do
Thank you so muchhh. 
DOriginal post by Chemistry Help
Oh? I have seen somewhere on google or idk but that like the one with the lower entropy will be the one that will have the least disorder so it would have less number of electrons?
Neon: 2,8
Mercury: 2,8,8
Sulphur: 2,6
Phosphorus: 2,8,5
And that's how I got C?
So for the lowest standard entropy means the higher the BP. So if it asks the highest standard entropy means the opposite no?
Neon: 2,8
Mercury: 2,8,8
Sulphur: 2,6
Phosphorus: 2,8,5
And that's how I got C?
So for the lowest standard entropy means the higher the BP. So if it asks the highest standard entropy means the opposite no?
Im not sure it that method will work for every question, I usually think of it as solid having the lowest entropy and then liquid having more and gas having even more. So when referring to that question, you'd be finding the state and the only option which is solid is C.
Hope that helped.
Original post by Zahra Tasleen
Sorry the answers for 22 is c
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This may help you...it might just be beneficial to memorise it. As you now know, [C2H5CO]+ is 57.
Also, the tallest peak is the one that is most favourable to fragment. Really once you have that in this question it falls into place
I suppose if you know it's an ester with 4 carbons, it can only be one of two combinations of alcohol/acid? Combine that with the fact that you have a C2H5 at 29, you can rule out ethyl ethanoate. That leaves either propyl methanoate or methyl propanoate.
Also, the tallest peak is the one that is most favourable to fragment. Really once you have that in this question it falls into place
I suppose if you know it's an ester with 4 carbons, it can only be one of two combinations of alcohol/acid? Combine that with the fact that you have a C2H5 at 29, you can rule out ethyl ethanoate. That leaves either propyl methanoate or methyl propanoate.
Original post by Zahra Tasleen
Chemistry is getting to someone's head....
Original post by Bonzo10
This may help you...it might just be beneficial to memorise it. As you now know, [C2H5CO]+ is 57.
Also, the tallest peak is the one that is most favourable to fragment. Really once you have that in this question it falls into place
I suppose if you know it's an ester with 4 carbons, it can only be one of two combinations of alcohol/acid? Combine that with the fact that you have a C2H5 at 29, you can rule out ethyl ethanoate. That leaves either propyl methanoate or methyl propanoate.
Also, the tallest peak is the one that is most favourable to fragment. Really once you have that in this question it falls into place
I suppose if you know it's an ester with 4 carbons, it can only be one of two combinations of alcohol/acid? Combine that with the fact that you have a C2H5 at 29, you can rule out ethyl ethanoate. That leaves either propyl methanoate or methyl propanoate.
thank youuuu
Original post by SanahAhmed
Im not sure it that method will work for every question, I usually think of it as solid having the lowest entropy and then liquid having more and gas having even more. So when referring to that question, you'd be finding the state and the only option which is solid is C.
Hope that helped.
Hope that helped.
Yes thanks! But how will I know the states if they haven't told us? Lol
Original post by Chemistry Help I understood that there's three differerent environments and it's an ester. I just dont know how to draw it from getting information from the graph 
I understood that there's three differerent environments and it's an ester. I just dont know how to draw it from getting information from the graph Ah right, okay. I will try to help (not promising anything though
). I'll base this off 2009 q11____
You can see on the proton NMR there are three peaks. This means there must be three different hydrogen environments in the molecule, i.e. three surroundings that are different.
You can see that two of the peaks have the same relative intensity, and one is slightly lower. You know that the molecule has 8 hydrogen atoms. Since the peak with a shift of 2.3ppm is about a third smaller than the other two, you can conclude that the peaks represent as follows:
Shift of 1.3ppm - 3 hydrogens in that part of the molecule.
Shift of 2.3ppm - 2 hydrogens in that part of the molecule.
Shift of 3.6ppm - 3 hydrogens in that part of the molecule.
Now, if you head to your good friend - the data booklet, p15 - you can start to place these groups.
A shift of 1.3ppm falls into the 0.9-1.5ppm range and so it is an alkyl group - i.e. carbon with 3 hydrogens that is bound to an R group - another carbon. Hopefully you can see that this is the carbon with the three hydrogens at the very end of the ester away from any oxygens,
A shift of 2.3ppm falls into the 2.0-2.7ppm range and so it is where a carbonyl group is present. As you know it's an ester, you can conclude that the hydrogens in this part are bound to a carbon that has a bond with a carbonyl group. Hopefully you can see that this is the environment where the carbon is bound next to the carbonyl group of the ester.
A shift of 3.6ppm falls into the 3.5-3.9ppm range and so is a CH3 that has a single bond to an oxygen. You can (hopefully) see that this is the methyl part of the ester.
Has this just confused you even more? If so, sorry!
Original post by Chemistry Help
Yes thanks! But how will I know the states if they haven't told us? Lol
Unless it tells you a temperature, always assume room temperature = 298K
Original post by GermanScientist
Ah right, okay. I will try to help (not promising anything though ). I'll base this off 2009 q11
____
You can see on the proton NMR there are three peaks. This means there must be three different hydrogen environments in the molecule, i.e. three surroundings that are different.
You can see that two of the peaks have the same relative intensity, and one is slightly lower. You know that the molecule has 8 hydrogen atoms. Since the peak with a shift of 2.3ppm is about a third smaller than the other two, you can conclude that the peaks represent as follows:
Shift of 1.3ppm - 3 hydrogens in that part of the molecule.
Shift of 2.3ppm - 2 hydrogens in that part of the molecule.
Shift of 3.6ppm - 3 hydrogens in that part of the molecule.
Now, if you head to your good friend - the data booklet, p15 - you can start to place these groups.
A shift of 1.3ppm falls into the 0.9-1.5ppm range and so it is an alkyl group - i.e. carbon with 3 hydrogens that is bound to an R group - another carbon. Hopefully you can see that this is the carbon with the three hydrogens at the very end of the ester away from any oxygens,
A shift of 2.3ppm falls into the 2.0-2.7ppm range and so it is where a carbonyl group is present. As you know it's an ester, you can conclude that the hydrogens in this part are bound to a carbon that has a bond with a carbonyl group. Hopefully you can see that this is the environment where the carbon is bound next to the carbonyl group of the ester.
A shift of 3.6ppm falls into the 3.5-3.9ppm range and so is a CH3 that has a single bond to an oxygen. You can (hopefully) see that this is the methyl part of the ester.
Has this just confused you even more? If so, sorry!
). I'll base this off 2009 q11____
You can see on the proton NMR there are three peaks. This means there must be three different hydrogen environments in the molecule, i.e. three surroundings that are different.
You can see that two of the peaks have the same relative intensity, and one is slightly lower. You know that the molecule has 8 hydrogen atoms. Since the peak with a shift of 2.3ppm is about a third smaller than the other two, you can conclude that the peaks represent as follows:
Shift of 1.3ppm - 3 hydrogens in that part of the molecule.
Shift of 2.3ppm - 2 hydrogens in that part of the molecule.
Shift of 3.6ppm - 3 hydrogens in that part of the molecule.
Now, if you head to your good friend - the data booklet, p15 - you can start to place these groups.
A shift of 1.3ppm falls into the 0.9-1.5ppm range and so it is an alkyl group - i.e. carbon with 3 hydrogens that is bound to an R group - another carbon. Hopefully you can see that this is the carbon with the three hydrogens at the very end of the ester away from any oxygens,
A shift of 2.3ppm falls into the 2.0-2.7ppm range and so it is where a carbonyl group is present. As you know it's an ester, you can conclude that the hydrogens in this part are bound to a carbon that has a bond with a carbonyl group. Hopefully you can see that this is the environment where the carbon is bound next to the carbonyl group of the ester.
A shift of 3.6ppm falls into the 3.5-3.9ppm range and so is a CH3 that has a single bond to an oxygen. You can (hopefully) see that this is the methyl part of the ester.
Has this just confused you even more? If so, sorry!
No it didn't, THANK YOU so much. Like I can't even tell you how much you helped us. THANKS
Original post by GermanScientist
Unless it tells you a temperature, always assume room temperature = 298K
What's at 298K? Solid liquid or gas?
Original post by Chemistry Help
Yes thanks! But how will I know the states if they haven't told us? Lol
You look up them in the data booklet lol
Original post by SanahAhmed
You look up them in the data booklet lol
Which page?
Original post by Chemistry Help
Please do Thank you so muchhh. D
Thank you so muchhh. DHere are the answers to the rest of 2002. I'm sorry - I don't know how to draw out structures on this, so I'll describe as best I can.
___
Section A, Part 2:
31. (a) C, D
(b) E
(c) F
32. (a) E
(b) C
(c) D, F
33. A, C
Section B:
1.
(a) Hydrogen chloride/hydrochloric acid/HCl/HCl(g)
(b) (polar) covalent/covalent molecular.
(c) [Ar] 3d1
(d) 239.5 kJ mol-1
2.
(a)
(i) tetrahedral shape
(ii) bent shape (like the shape of water)
(b) S has lone pairs. Lone pair-lone pair or lone pair-bonding pair repulsion is greater than bonding pair-bonding pair repulsion.
3.
(a) 157.6 kJ mol-1
(b) 245.1 J K-1 mol-1
(c) 643 K
4.
(a) double bond between N and C. F bound to the N. H bound to C. F bound to C in trans position from the F bound to N.
(b) By sideways join/overlap/merging of two parallel atomic orbitals.
(c) 19
(d) 96%
5.
(a)
(i) zero
(ii) rate = k[HNO3]
(iii) 2
(b) H2SO4 and HSO4-
6.
(a) electrophilic
(b) Draw toluene. The Cl can be at any other carbon on the aromatic ring.
(c) (i) Cl2 -> 2Cl.
(ii) Cl. + toluene -> HCl + radical of toluene (not a CH3 but a CH2.)
OR
toluene radical + Cl2 -> Cl. + toluene, but replace an H on the CH3 with a Cl.
(d) benzoic acid
7.
(a) Mn2+ + 4H2O -> MnO4- + 8H+ + 5e-
(b) MnO4- is a coloured ion or coloured solution.
(c) 0.28%
8.
(a) -0.81V
(b) -250.9
9.
(a) 11.3 kJ mol-1
(b) 13 kJ mol -1
(c)
•
not infinite dilution
•
not under standard conditions
•
lack of insulation
•
not carried out under the same conditions
•
heat taken in from the surroundings
•
impurities in the sample
10.
(a) CH3CH2CH2CN or a drawing of it.
(b) hydrolysis
(c) 3.4
(d) H+ ions react with butanoate ions/conjugate base from the salt
11.
(a) proof
(b) C3H6O3
(c)
(i)
(ii) contains (equal amounts of) both optical isomers/racemic mixture/racemates/contains two isomers which cancel out each other's optical activity.
12.
(a)
(b) working, to get answer as myrcene
(c) No double bonds. A bromine atom on top carbon of first double bond, a bromine added to the carbon at the top carbon of the double bond at the bottom and a bromine added to the carbon at the bottom of the double bond at the right hand side. The rest have hydrogens (hope you can understand that
)(d)
(e) Nothing changes in citral, bar the aldehyde becomes a primary alcohol.
13.
(a) shape/structure of sulphanilamide complements that of the active site
or
shape/structure of sulphanilamide similar to that of 4-aminobenzoic acid
(b) antagonist
because it blocks/binds to active site and so prevents 4-aminobenzoic acid from binding.
(c)
(i) The same as the third one down on the table, but remove the H at the furthest right hand side position but keep the bond there.
(ii) It would be ineffective since it does not contain the pharmacophore/CH3 on left hand side of the molecule.
I do not envy the markers that write the marking instructions
Note: The structures are in the pictures.
(edited 10 years ago)
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