hobbitle's CHEM10004 Questions Thread

[lzxnl]

Hey guys, this refers to the attached question 10.3.

I don't understand what part b) is asking.

I know that we use Ksp = [Cu][OH]^2 and can solve for [Cu] from there, because that's what our tutor said to do, but I don't understand why. I think I don't really get the concept of a solubility product (I guess it's a measure of how soluble something is... but soluble in what? Is it always the inverse of Kc, or is that just if the ionic equation goes from ion --> solid instead of solid--> ion?)

And this is for Question 10.5:

I am getting confused with concentrations. 
So we are starting off with 0.1M of the oxidant, cool.
So when 1% of the oxidant is used, does that mean 1% of 0.1 M has been used? So the concentration is 0.099 M, when 1% has been used up? Feels like an idiotic question but I'm just confused.

The Ksp is the equilibrium constant for the dissociation of copper (II) hydroxide, which is Cu(OH)₂(s) => Cu²⁺(aq) + 2OH^-(aq)
Now, work out an equilibrium constant expression from this reaction, disregarding the solid in the equilibrium (as we never consider solids). What do you get?

And I think your interpretation is right for the second question, although I'm confused too. Wording isn't awfully clear

[hobbitle]

Thanks heaps mate I have another one, see attached.

I don't have the answers but I feel like mine is wrong. Explanation:

Reduction: 3Pb2+ + 6e- --> 3Pb (s) / -0.13V
Oxidation: 2Al (s) --> 2Al3+ + 6e- / -(-1.66V)

E(standard conditions) = 1.53V
Q = [Al3+]^2 / [Pb2+]^3

Because there is the 3:2 molar ratio between the generation of Al3+ and the consumption of Pb2+, I figure that [Al3+] goes to 1.6M and [Pb2+] will go to 0.1 M.

I plug this into Nernst eqn at standard conditions:

Ecell = E - 0.0592/6*log (1.6^2 / 0.1^3)
= 1.496 V

This answer to me seems way too high? Shouldn't it be nearing zero? What have I done wrong, I feel like the issue is with my interpretation of the concentration changes?

[lzxnl]

Thanks heaps mate I have another one, see attached.

I don't have the answers but I feel like mine is wrong. Explanation:

Reduction: 3Pb2+ + 6e- --> 3Pb (s) / -0.13V
Oxidation: 2Al (s) --> 2Al3+ + 6e- / -(-1.66V)

E(standard conditions) = 1.53V
Q = [Al3+]^2 / [Pb2+]^3

Because there is the 3:2 molar ratio between the generation of Al3+ and the consumption of Pb2+, I figure that [Al3+] goes to 1.6M and [Pb2+] will go to 0.1 M.

I plug this into Nernst eqn at standard conditions:

Ecell = E - 0.0592/6*log (1.6^2 / 0.1^3)
= 1.496 V

This answer to me seems way too high? Shouldn't it be nearing zero? What have I done wrong, I feel like the issue is with my interpretation of the concentration changes?

It's certainly higher than the original voltage. Did you expect the voltage to change a lot? It doesn't. There is a log there that means any deviations are going to be slow.

[hobbitle]

It's certainly higher than the original voltage. Did you expect the voltage to change a lot? It doesn't. There is a log there that means any deviations are going to be slow.

Sorry, I don't understand. My result was lower than the initial voltage, just not by much.
I'm not sure, I thought the Ecell = 0 was when the system reached equilibrium. And because there is almost no reactant left at a concentration of 0.1 M I figure it should be closer to equilibrium by this stage....

[lzxnl]

Sorry, I don't understand. My result was lower than the initial voltage, just not by much.
I'm not sure, I thought the Ecell = 0 was when the system reached equilibrium. And because there is almost no reactant left at a concentration of 0.1 M I figure it should be closer to equilibrium by this stage....

Yeah. My bad. I just realised I typed something really stupid. It IS lower than the original voltage; I just can't read

Work out the equilibrium constant for the reaction first. That should give you an idea for how ridiculous this equilibrium is.

[hobbitle]

Yeah. My bad. I just realised I typed something really stupid. It IS lower than the original voltage; I just can't read

Work out the equilibrium constant for the reaction first. That should give you an idea for how ridiculous this equilibrium is.

Thanks man. They released the answers last night and turns out I was correct.

[hobbitle]

Hey guys, I have a bunch of questions.  See attached!

1) The attached organic chem reaction mechanism. 
QUESTION: Which one of these mechanisms IS NOT AN ACCEPTED MECHANISM for a reaction involving a hydroxide ion?
I felt like they were all OK? The only one that was questionable to me was perhaps D because of the tertiary carbon, but OH is small and strong so I figured it's still OK?

2) QUESTION: For the reaction 2HI --> H2 + I2, what is the rate of reaction with respects to the reactants and products?
Would it be -d[HI]/dt = (1/2)d[H2]/dt = (1/2)d[I2]/dt?
I'm mostly asking from a POV of ratios and negative signs.

3) QUESTION: How many isomers (geometric and optical) are possible for the octahedral complex ion [Mn(OH2)3Cl3]?
I said just two, fac/mer. Is this correct?

4) See other organic chem attachment.  QUESTION: Draw the organic product from this reaction.
Do you get the two benzenes breaking away from the cycle ring and you end up with 2 x benzene molecules each with 2 x COOH substituents?

5) Question B6 attached.
I’m confused, sometimes when there are two intermediates, we ignore one and use SSA or pre-equilibrium. 
But here there are two intermediates but we are accounting for both.
When do we use SSA or pre-equilibrium or both intermediates to solve these problems?
I seem to have 3 ways of solving them (SSA, pre-equilibrium and using both intermediates) but never know which method to use.

[mahler004]

Hey guys, I have a bunch of questions.  See attached!

1) The attached organic chem reaction mechanism. 
QUESTION: Which one of these mechanisms IS NOT AN ACCEPTED MECHANISM for a reaction involving a hydroxide ion?
I felt like they were all OK? The only one that was questionable to me was perhaps D because of the tertiary carbon, but OH is small and strong so I figured it's still OK?

2) QUESTION: For the reaction 2HI --> H2 + I2, what is the rate of reaction with respects to the reactants and products?
Would it be -d[HI]/dt = (1/2)d[H2]/dt = (1/2)d[I2]/dt?
I'm mostly asking from a POV of ratios and negative signs.

3) QUESTION: How many isomers (geometric and optical) are possible for the octahedral complex ion [Mn(OH2)3Cl3]?
I said just two, fac/mer. Is this correct?

4) See other organic chem attachment.  QUESTION: Draw the organic product from this reaction.
Do you get the two benzenes breaking away from the cycle ring and you end up with 2 x benzene molecules each with 2 x COOH substituents?

5) Question B6 attached.
I’m confused, sometimes when there are two intermediates, we ignore one and use SSA or pre-equilibrium. 
But here there are two intermediates but we are accounting for both.
When do we use SSA or pre-equilibrium or both intermediates to solve these problems?
I seem to have 3 ways of solving them (SSA, pre-equilibrium and using both intermediates) but never know which method to use.

1) Yep, you're right, it's D, it shows it as a SN2 mechanism while it would go SN1 (for the reason you said.)

3) Just fac/mer irrc

4) Agreed - you'd get the cleavage of the ring and two subsisted benzenes, as you said. I could swear that they don't cover this in first year? (I had to check my third year notes.) (look here.)

[lzxnl]

1) Yep, you're right, it's D, it shows it as a SN2 mechanism while it would go SN1 (for the reason you said.)

3) Just fac/mer irrc

4) Agreed - you'd get the cleavage of the ring and two subsisted benzenes, as you said. I could swear that they don't cover this in first year? (I had to check my third year notes.) (look here.)

4 is implicitly mentioned in first year; if there are benzylic hydrogens, KMnO4 will oxidise the alkyl side group to a carboxylic acid.

Hey guys, I have a bunch of questions.  See attached!

1) The attached organic chem reaction mechanism. 
QUESTION: Which one of these mechanisms IS NOT AN ACCEPTED MECHANISM for a reaction involving a hydroxide ion?
I felt like they were all OK? The only one that was questionable to me was perhaps D because of the tertiary carbon, but OH is small and strong so I figured it's still OK?

2) QUESTION: For the reaction 2HI --> H2 + I2, what is the rate of reaction with respects to the reactants and products?
Would it be -d[HI]/dt = (1/2)d[H2]/dt = (1/2)d[I2]/dt?
I'm mostly asking from a POV of ratios and negative signs.

3) QUESTION: How many isomers (geometric and optical) are possible for the octahedral complex ion [Mn(OH2)3Cl3]?
I said just two, fac/mer. Is this correct?

4) See other organic chem attachment.  QUESTION: Draw the organic product from this reaction.
Do you get the two benzenes breaking away from the cycle ring and you end up with 2 x benzene molecules each with 2 x COOH substituents?

5) Question B6 attached.
I’m confused, sometimes when there are two intermediates, we ignore one and use SSA or pre-equilibrium. 
But here there are two intermediates but we are accounting for both.
When do we use SSA or pre-equilibrium or both intermediates to solve these problems?
I seem to have 3 ways of solving them (SSA, pre-equilibrium and using both intermediates) but never know which method to use.

For 2, you have it the wrong way around. It should be -d[HI]/dt = 2d[H2]/dt = 2d[I2]/dt because the HI is being used twice as fast as either of the other two is being formed.

As for the kinetics problem, let's look at the equilibrium arrows and the relative rates of each step. Step 1 is fast and reversible, so presumably it gets to equilibrium quickly. Hence use a pre-equilibrium assumption that the forward and back rates are the same.
The second reaction is irreversible, but slow. Note how the third reaction, however, uses up any hydrogen atoms that are formed in step 2 quickly? Use the steady state approximation on the hydrogen atoms.

[hobbitle]

1) Yep, you're right, it's D, it shows it as a SN2 mechanism while it would go SN1 (for the reason you said.)

Cool yeah so my tutor said that it wouldn't go Sn2 because there are too many hydrogens in the way, so it's going to want to go Elimination predominantly.  But presumably there will also be a proportion of Sn1, but no Sn2.

4) Agreed - you'd get the cleavage of the ring and two subsisted benzenes, as you said. I could swear that they don't cover this in first year? (I had to check my third year notes.) (look here.)

Oh great link! Thanks.  Pretty much we just have been told that an alkyl substitute on a benzene will be cleaved if oxidised and replaced with a COOH.  No idea of the mechanism or whatever.

For 2, you have it the wrong way around. It should be -d[HI]/dt = 2d[H2]/dt = 2d[I2]/dt because the HI is being used twice as fast as either of the other two is being formed.

Thankyouuuuu.  This makes sense now.

As for the kinetics problem, let's look at the equilibrium arrows and the relative rates of each step. Step 1 is fast and reversible, so presumably it gets to equilibrium quickly. Hence use a pre-equilibrium assumption that the forward and back rates are the same.
The second reaction is irreversible, but slow. Note how the third reaction, however, uses up any hydrogen atoms that are formed in step 2 quickly? Use the steady state approximation on the hydrogen atoms.

Riiiight okay so you can use a combination of pre-equilibrium assumption and SSA for intermediates to solve these - I kept thinking it was one or the other. In this particular question, apparently, we aren't supposed to assume pre-equilibrium (ridiculous I know but she says if we are to assume it she will write it explicitly... not sure how much more explicit than FAST and equilibrium arrows you can get but anyway).... talked to tutor about this today also and she clarified some things too.  Cheers, you guys are the best.

[hobbitle]

Hi guys, a question.

Attached Q A5, what is the type of reaction happening at step 3 and why is D the answer?

Other attachment (iii): What is making the double bond on the end turn into a single bond? I understand that the carbonyl is undergoing reduction (ketone to a secondary alcohol) but I don't understand why the double bond changes in this reaction?

[lzxnl]

SOCl2 + ROH => RCl + HCl + SO2
Unfortunately you need to know that. What happens is that the alcohol oxygen acts as a nucleophile to attack the sulfur in SOCl2. The OH proton disappears and the SOCl2, which is itself almost like an acid chloride, ejects one of the chlorines (you know how acid chlorides react easily with nucleophiles because of the good chloride leaving group? Same idea here).

So you go from SOCl2 + ROH to RO-SOCl + H+ + Cl-
Now the chloride formed attacks the carbon that is bonded to the hydroxyl oxygen. The chloride donates an electron pair to that carbon, which donates an electron pair to the oxygen (breaking the C-O bond), the oxygen sends that electron pair to the sulfur (making a O=S pi bond) and the sulfur in turn gives this pair of electrons to the chlorine it's bonded to, thus breaking the S-Cl bond and giving you another Cl- ion. Hence the above equation.


As for your second question, there is something called conjugate addition in which the beta carbon in an alpha-beta unsaturated ketone is electrophilic. The hydride has probably attacked at that carbon instead. A bit dodgy of them to ask you that though.

[hobbitle]

Thanks man. Yeah, they didn't really cover either of those things in the content this year, I think - I just wanted to be extra sure. I hope they don't give us that conjugate addition thing. THANKYOU!