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Chemistry Predictions/Thoughts (3 Viewers)

Goldattacker40000

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Also for the question where they found the nh4 in the fertiliser i had no clue what to do. i had 1.55g or smth. I didnt know what to do at the step when they boiled the ammonia with NaOH
I think I got a mass of ammonium that was in between 0.1 and 1 but i dont remember the exact mass.
 

Luukas.2

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PV = nRT is the way to go for Q27 (CO2 production from fermentation) because the gas volumes on the data sheet only apply to temperatures at 273 K and 298 K. Alternatively, if you used one of those (since the pressure matched) you would then need to adjust the volume using a Charles' Law calculation.

Question 21 is badly written as there is (technically) no correct answer. Butanone and butanal are NOT functional group isomers as they have the same functional group - carbonyl. They are different classes of compounds (a ketone and an aldehyde), but that is not the same thing as classifying them as having different functional groups. In the same way, 1-butanol, 2-butanol, and 2-methyl-2-propanol all have the same functional group (a hydroxyl) but have some differences in chemistry due to the different positioning of the functional group. In this case, though, we call all of them alcohols and consider them part of the same class.

I doubt it will cause a problem - what they wanted was obvious - but it is technically wrong.
 

carrotsss

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PV = nRT is the way to go for Q27 (CO2 production from fermentation) because the gas volumes on the data sheet only apply to temperatures at 273 K and 298 K. Alternatively, if you used one of those (since the pressure matched) you would then need to adjust the volume using a Charles' Law calculation.

Question 21 is badly written as there is (technically) no correct answer. Butanone and butanal are NOT functional group isomers as they have the same functional group - carbonyl. They are different classes of compounds (a ketone and an aldehyde), but that is not the same thing as classifying them as having different functional groups. In the same way, 1-butanol, 2-butanol, and 2-methyl-2-propanol all have the same functional group (a hydroxyl) but have some differences in chemistry due to the different positioning of the functional group. In this case, though, we call all of them alcohols and consider them part of the same class.

I doubt it will cause a problem - what they wanted was obvious - but it is technically wrong.
I’ve noticed ketones and aldehydes being treated as functional group isomers in trial papers asw so im not really sure, it’s definitely odd since they’re both carbonyl but maybe they’re treated as functional group isomers in the syllabus or something?
 

Luukas.2

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I’ve noticed ketones and aldehydes being treated as functional group isomers in trial papers asw so im not really sure, it’s definitely odd since they’re both carbonyl but maybe they’re treated as functional group isomers in the syllabus or something?
Well, the syllabus and the Department of Education / NESA / etc do things their own way... and trials often have questionable content... but functional groups and compound classes are not the same thing.

And the HSC has a track record of being, shall we say, less than 100% on the accuracy front. The cadaverine 1H NMR spectrum from a couple of years ago, for example... the mass spectrum in question 28 (compound R) in this year's paper is a touch misleading in my opinion, too.
 

Luukas.2

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And there was Q31, which refers to a species that I'm doubtful exists.

Copper(II) ions do react with lactate anions to make a compound known as copper(II) lactate, that can be thought of as a salt or as a coordination complex: [Cu(C3H5O3)2] - that is, [Cu(CH3CHOHCO2)2].

This year's HSC exam, however, claims a cationic complex (with the lactic acid in its protonated form) as [Cu(C3H5O3)2]2+, or as [Cu(CH3CHOHCOOH)2]2+.

Carboxylic acids form complexes and salts as anions, not in their protonated forms. The species of the composition given seems highly improbable, made more questionable by the disregard for the acidity of the uncomplexed lactic acid (with its pKa of 3.86).
 

Luukas.2

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For compound R how did you know where to put the two chlorines? And what was the answer
The addition of chlorine to an alkene produces a dichloroalkane.

The fact that the mass spectrum appears to show only one is misleading. The product should be 1,2-dichloropropane.

The mass spectrum of the compound can be seen at this link: https://webbook.nist.gov/cgi/cbook.cgi?ID=C78875&Mask=200#Mass-Spec

You can see that it matches the spectrum provided in the exam, but with the lowest intensity peaks missing.

Taking the 35Cl : 37Cl ratio as 3 : 1 yields the parent (molecular) ion, [C3H6Cl2]+ . at m/z = 112, 114, and 116 in a 9 : 6 : 1 ratio.
 

=)(=

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The addition of chlorine to an alkene produces a dichloroalkane.

The fact that the mass spectrum appears to show only one is misleading. The product should be 1,2-dichloropropane.

The mass spectrum of the compound can be seen at this link: https://webbook.nist.gov/cgi/cbook.cgi?ID=C78875&Mask=200#Mass-Spec

You can see that it matches the spectrum provided in the exam, but with the lowest intensity peaks missing.

Taking the 35Cl : 37Cl ratio as 3 : 1 yields the parent (molecular) ion, [C3H6Cl2]+ . at m/z = 112, 114, and 116 in a 9 : 6 : 1 ratio.
doesnt 1,2-dichloroprop-1-ene fit cause you could assume Q might have a double bond and since doesnt specify excess cl2 it could only have 1 double bond being saturated also 1,2-dichloroprop-1-ene has 32 percent carbon by mass and has molecular ion peak at 114 it also has a fragment at 15 and 62?
 

Jane Mao

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Solutions ?
Surely multiple choice at least ;)
Rather than wait for some genius person who got everything right, how bout we just compile everyone’s answers.
Here’s what I had for everything
1. D
2. C
3. B
4. D
5. D
6. C
7. A
8. B
9. C
10. A
11. C
12. B
13. A
14. D
15. B
16. B
17. A
the full paper isn’t uploaded yet so I can’t check this but I think I got all A’s for 18, 19, and 20 (I rushed multiple choice so this could be wrong)
Now if anything looks wrong just tell me and I’ll edit this
 
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