Chemistry 3/4 2013 Thread

[Jaswinder]

thanks alot ECheong. also attached image is related to

http://www.vcaa.vic.edu.au/Documents/exams/chemistry/2010chem1-w.pdf

From Section B:

Question 2e - if I drew the structure as shown in the attachment, would that still be right?

Thanks 

[ECheong]

I'm afraid not, from the 1H NMR we can see that there are 3 hydrogen environments, the ester you drew only contained 2.

and nw! haha always glad to help

EDIT: Just got word back on question A12

the OH- is a strong nucleophile and so can rip the Cl atom off the choloralkane. On the flip side, HCl is a weak nucleophile, in a reaction betwen HCl and a primary haloalkane, say 1-chloropropane, it would just create dichloropropane.

[lzxnl]

http://www.vcaa.vic.edu.au/Documents/exams/chemistry/2010chem1-w.pdf

From Section A:

Question 12 - Why can't it be A? and could you also explain why D is the correct answer. I tried reading the assessment report and I kinda get it but not too sure.

All from Section B:

Question 2e - if I drew the structure as shown in the attachment, would that still be right?

Question 3a - I wrote NO2 -> NO2+ + e- Why is the answer 2e-?

Question 3b- I said, Spectrum A. Nitrogen dioxide cannot form a peak at 31 m/z. Would this yeild full marks?

Question 7bii - Why is the answer C22H32O2? I get C22H34O2, which I know is incorrect but I don't know why.

Also what exactly is fermentation? Which chemicals undergo fermentation? What are the products of this process?
Thanks 

12. Alkyl chlorides won't react with HCl. That's why A is wrong.
Why won't they react? Let's look at the reaction between chlorine gas and methane.
What happens is that UV light homolytically cleaves the Cl-Cl bond to form chlorine radicals; seven valence electrons each chlorine atom. This is quite reactive; the reaction can be written as Cl.+CH4=>CH3. + HCl
Yay methyl radical. If this thing approaches another Cl. radical, CH3Cl forms.
If this thing approaches Cl2, it forms CH3Cl and Cl.
The radicals vanish when chlorine radicals and methyl radicals all react with each other.
I'm presuming that if you tried to react hydrogen chloride in this way, you would have to try and cleave a H-Cl bond to form a C-Cl bond, and carbon-chlorine bonds aren't that strong, and neither are H-H bonds, which is what you would get if you could substitute the chlorine onto the atom. HCl also interacts with itself better than H2 does, so you have a rather large energy cost which would presumably make it unfavourable.
I also think that if you have hydrogen radicals, they'd react with chlorine radicals anyway.

@ECheong, I don't think HCl can participate in chlorination of alkyl chlorides...HCl is more stable in that form than H2, which would be the product of your substitution reaction.
And it's also that Cl- is a better leaving group than OH-; in the five-membered transition state, Cl- is more likely to fall off.

From a thermodynamics perspective, it would appear to me that HCl and CH4 would be more stable than H2 and CH3Cl
I found some numbers and did the calculations. HCl and CH4 is a lot more stable than H2 and CH3Cl (the standard change in Gibbs energy is +124.86 kJ/mol which is comparable to the standard Gibbs energy of formation of HCl which is -148 kJ/mol or that of CH4, which is -130 kJ/mol; long story cut short, it's very unfavourable like I predicted). There. Numbers.

2e: I presume this is sorted?
Normally I don't use chemical shifts, but if you see one at 9ppm, you should be thinking H-COR, an aldehyde, or maybe R-COOH.

3a. The answer is NO2 + e- => NO2+ + e-. An electron is needed on the left hand side as that electron causes the fragmentation. Although you could argue that by that logic, catalysts should be included in reactions when they're not.

3b. I would say that's fine, but to be safe, you could say that NO2 can't form a peak at 42 either

@ECheong's point about rare isotopes, they wouldn't be present in such large quantities; you wouldn't be able to see such a peak.

7bii We have 6*10^-3 moles of iodine adding to 1*10^3 moles of cervonic acid=>6 double bonds.
22 carbon atoms, saturated acid is C22H44O2. Minus twelve hydrogen atoms from the 6 double bonds=>C22H32O2
Saturated fatty acids are of the from C2nH4nO2 (yes, need the 2n as we only consider even numbers of carbon atoms to form fatty acids; I don't know what C17H34O2 would be )

Fermentation is C6H12O6=>2CH3CH2OH + 2CO2. Anaerobic respiration in plants, or a process to produce bioethanol.

Most of my responses have already been covered...so yeah.

[ECheong]

@ECheong, I don't think HCl can participate in chlorination of alkyl chlorides...HCl is more stable in that form than H2, which would be the product of your substitution reaction.
And it's also that Cl- is a better leaving group than OH-; in the five-membered transition state, Cl- is more likely to fall off.

From a thermodynamics perspective, it would appear to me that HCl and CH4 would be more stable than H2 and CH3Cl
I found some numbers and did the calculations. HCl and CH4 is a lot more stable than H2 and CH3Cl (the standard change in Gibbs energy is +124.86 kJ/mol which is comparable to the standard Gibbs energy of formation of HCl which is -148 kJ/mol or that of CH4, which is -130 kJ/mol; long story cut short, it's very unfavourable like I predicted). There. Numbers.

Ahh that makes sense, I stand corrected!

[Jaswinder]

http://www.vcaa.vic.edu.au/Documents/exams/chemistry/2011chem1-w.pdf

Section 1

Question 8 - Still not sure as to why they chose D over C

Question 9 - I don't quite get the VCAA answers

Question 11 - I understand that the equivalence point is influenced by n(H+) available from each acid and since c x v gives the same answer for both, the equivalence would be the same. However, I was just thinking that a strong acid would mean that it would donate its proton easily? and so would require less NaOH? but I might be completely wrong

Question 12 - Not sure how?

Question 18 - How does mass relate to the energy (frequency) required for vibration?

Question 19 - Why not UV- Vis?

Section B

Question 1 c and d - I think I lack the theoretical knowledge required for these question. Could someone explain how they would answer it? What their thought process is?

Thanks Guys 

[lzxnl]

http://www.vcaa.vic.edu.au/Documents/exams/chemistry/2011chem1-w.pdf

Section 1

Question 8 - Still not sure as to why they chose D over C

Question 9 - I don't quite get the VCAA answers

Question 11 - I understand that the equivalence point is influenced by n(H+) available from each acid and since c x v gives the same answer for both, the equivalence would be the same. However, I was just thinking that a strong acid would mean that it would donate its proton easily? and so would require less NaOH? but I might be completely wrong

Question 12 - Not sure how?

Question 18 - How does mass relate to the energy (frequency) required for vibration?

Question 19 - Why not UV- Vis?

Section B

Question 1 c and d - I think I lack the theoretical knowledge required for these question. Could someone explain how they would answer it? What their thought process is?

Thanks Guys 

8 is admittedly really dodgy.
But you have to take the hint from the question. It's more soluble in water => charged. The amide wouldn't be nearly as soluble as the ionic salt.

For 9, in future, please post the VCAA answers; makes it easier for the poor guy who has to go look it up
OK...this question is funny. The reaction is canola oil plus ethanol gives ethyl stereate and glyercol, with a potassium hydroxide catalyst. The top layer is ethyl stereate, so the bottom layer is everything else; ethanol, glyercol and potassium hydroxide. Hence D is correct. However, depending on the stoichiometry, ethanol may not be present in the bottom layer, so A was accepted. Be mindful though that no reactions ever go to completion, so A is unrealistic.

11 Let's think of it this way
Let A be the acid, and B- be the base.
We have three equilibria here.
HA<=>H+ + A-
B- + H2O <=> OH- + HB
H+ + OH- <=> H2O

Now, B is hydroxide, so we don't need to consider the second equilibrium.
Hydroxide ions will react with effectively ANY H+ ions present (due to the large disparity between the concentrations; [OH-] is around 0.10 M), so the third equilibrium will be driven to the right. Thus, so will the first equilibrium. The net result is that the OH- reacts with HA almost completely. HA's acid strength doesn't matter, as long as HA is a stronger acid than water.

It's true that a strong acid will dissociate more readily. OH- is such a strong base, though, that it can react with essentially any acid.
Think of it this way. For the equilibrium HA<=> H+ + A- and the equilibrium H+ + OH- <=> H2O
Let the equilibrium constants be Ka and 1/Kw
The combined reaction, HA + OH- <=> A- + H2O, has an equilibrium constant of Ka*1/Kw=Ka/Kw
Now...Kw is tiny. 1/Kw is 10^14. Multiplying that by any Ka in the data table and we'll still have a K constant that is quite large.
Thus, same amount of OH- reacts with the acid.

12
Thymol blue is yellow => pH>2.8 (base form)
methyl red is yellow => pH>6.3 (base form)
Phenolphthalein is colourless => pH<8.3 (acid form)
Which pH value satisfies these constraints?

18
This is a very bad question IMO. Larger mass, even with same bond strength => smaller vibration. Think about it. If you have a spring, if the load is heavier, it'll oscillate less.
Although technically, C is correct too because a larger atomic radius means a larger bond length, which means a weaker bond.

19
UV Vis does NOT identify. For starters, the different hydrocarbons may all have similar absorption wavelengths for all we know. Does that help with UV Vis? Not really.
Also, all UV Vis does is determine the absorbance of a particular frequency. It is only useful when we know what we're testing and we have other standards of the known compound to compare absorbances. UV Vis cannot tell us what a compound is. We need to separate it first, with gas chromatography perhaps, and then use something like NMR or mass spec to work out what we have. Hence D.


1. c
Hydrolysed saturated triglyceride => glycerol and three saturated fatty acids.
Glycerol is E. The saturated fatty acid is D.

d
Biodiesel mainly consists of methyl esters. Just remember that.

[Lakkattack]

Sorry this may seem really basic, but what would be a comprehensive good answer when defining renewable energy sources and their advantages and disadvantages?

[Scooby]

Is the potassium hydroxide electrolyte in an alkaline fuel cell (or phosphoric acid in an acid fuel cell) molten or aqueous?

Why do the electrodes in a fuel cell need to be porous? I mean what's wrong with just having hydrogen or oxygen react at the surface of the electrode? (or do they need to be in contact with the electrolyte?)

VCAA 2012 Exam 2 - Q6 c):

So the question says that methanol is oxidised to carbon dioxide and water, but then the equation below shows it being oxidised to carbon dioxide and hydrogen ions... What's going on there?

Also, how are we meant to deduce that oxygen is reduced at the cathode? Why would it have to be oxygen?

If we've got one of the half equations producing or consuming H+, can we assume an acidic electrolyte? (same with hydroxide and a basic electrolyte?)

Thanks

[Scooby]

it doesnt matter. water is only an active reagent in electrolysis (for the most part anyways), so it doesnt matter if its molten or aqueous. Aqueous is probably preferenced due to its relative cheapness.

that

Hmmm... isn't a fuel cell a type of galvanic cell though?

[ECheong]

Exactly, in a galvanic cell (fuel cell) for water to spontaneously react you'd need a pretty strong reductant since water itself is a fairly weak oxidant (low down on the left of the electrochemical series)

[OutstandingInDivination]

Hmmm... isn't a fuel cell a type of galvanic cell though?

precisely Sorry if I'm not really answering your question/confusing you haha! The electrolyte is typically aqueous, according to both my teacher and tutor, but it doesnt effect the effectiveness of the fuel cell. (water is not an active player in galvanic cells right? exactly. the same occurs in fuel cells)

[Scooby]

Ohhh right, I get you! So if we (really hypothetically here) connected a F₂/F^- half cell with a Au/Au⁺ half cell, water would be oxidised in the Au/Au⁺ half cell in favour of gold?


And another thing - why do we never seem to consider that equation where water is oxidised to hydrogen peroxide (or where oxygen is reduced to hydrogen peroxide)? Or maybe this is just me, but I remember at a lecture the lecturer said not to worry about those equations 

Thanks guys

[OutstandingInDivination]

Ohhh right, I get you! So if we (really hypothetically here) connected a F₂/F^- half cell with a Au/Au⁺ half cell, water would be oxidised in the Au/Au⁺ half cell in favour of gold?

EDIT: I stand corrected For a more accurate response, please refer to
nliu1995's post - (im getting rid of mine to avoid confusion!)

not sure about your second question. I would also like to know the logic. I just blindly accepted the rule

[lzxnl]

Is the potassium hydroxide electrolyte in an alkaline fuel cell (or phosphoric acid in an acid fuel cell) molten or aqueous?

Why do the electrodes in a fuel cell need to be porous? I mean what's wrong with just having hydrogen or oxygen react at the surface of the electrode? (or do they need to be in contact with the electrolyte?)

VCAA 2012 Exam 2 - Q6 c):

So the question says that methanol is oxidised to carbon dioxide and water, but then the equation below shows it being oxidised to carbon dioxide and hydrogen ions... What's going on there?

Also, how are we meant to deduce that oxygen is reduced at the cathode? Why would it have to be oxygen?

If we've got one of the half equations producing or consuming H+, can we assume an acidic electrolyte? (same with hydroxide and a basic electrolyte?)

Thanks

Porous...to allow ion flow to the electrodes to complete the circuit.

As for the hydrogen ions, think about what happens when you reduce oxygen gas in acidic conditions. The production of H+ maintains the acidity of the cell.

Why oxygen? Generally fuel cells consume oxygen as the oxidant. It's cheap, "renewable" and readily available. That's been the case for pretty much every fuel cell I've seen in the VCE course.

And yes, for your H+ question. If you had alkaline conditions, you wouldn't produce H+. Instead, you'd consume OH- and produce water as any H+ formed would immediately react with OH- present. Vice versa for OH- in alkaline conditions.

Ohhh right, I get you! So if we (really hypothetically here) connected a F₂/F^- half cell with a Au/Au⁺ half cell, water would be oxidised in the Au/Au⁺ half cell in favour of gold?


And another thing - why do we never seem to consider that equation where water is oxidised to hydrogen peroxide (or where oxygen is reduced to hydrogen peroxide)? Or maybe this is just me, but I remember at a lecture the lecturer said not to worry about those equations 

Thanks guys

You don't worry about hydrogen peroxide because...if you have O2 and H+, oxygen has a higher reduction potential when forming water, so why would it form hydrogen peroxide?
As for oxidising water to hydrogen peroxide...don't think that's happening

If you connected a gold cell to a fluorine cell, yes water would be oxidised instead.
The reactions would be quite interesting...3F2+3 H2O=>O3 + 6HF
Why? Fluorine gas will oxidise water to form oxygen gas. However, it is also strong enough to oxidise oxygen gas in the presence of water to form ozone (that half reaction isn't listed; O3 + 2H+ + 2e- => O2 + 2H2O has a reduction potential of 2.075 according to Wiki). Either reaction could occur, depending on how much oxygen gas is produced.

Au_(s) -> Au²⁺_(aq) + 2e^-
F_2(g) + 2e^- -> 2F^-_(aq)
water wouldn't be involved its almost a given in almost all galvanic cells, that water will NOT take part

EDIT: although, flourine is a funny lil chemical haha itll probably strip electrons off everything haha

not sure about your second question. I would also like to know the logic. I just blindly accepted the rule

Water will be involved as water is a stronger reductant than gold metal. Be slightly careful. In ALMOST all galvanic cells.

[Scooby]

Thanks nliu1995

And just checking, it'd still be possible to oxidise water to hydrogen peroxide if we bubbled fluorine gas into it, yeah?

So we'd essentially have two reactions occurring (one with water being oxidised to hydrogen peroxide and the other with water being oxidised to oxygen and H+), but oxidation to oxygen and H+ would occur to a much greater extent?