Bazza's 3/4 chemistry questions

[Aurelian]

I'm feeling pretty lazy at the moment, so I might skip a few questions/answer more briefly than I'd like to lol. I'm tired =(

1) You'll actually learn about this in physics in U4, but I'll try and point you in the right direction for now. I wouldn't worry about knowing everything about this though. For now, I'll try and satisfy you with just "when something 'absorbs' a frequency of light, at least in this context, we typically mean that the light is converted to heat" - does that help? Thus, when white light shines on something red, everything but red is absorbed, the thing gets hotter, and red light is reflected back. This is why black cars get so hot - all the light is absorbed and converted to heat! (And 'none' is re-emitted).

2) Has been answered

3) Lots and yes. Don't worry about this now though. In the past, on exams, questions have typically mainly concerned the interpretation of data, but I would not be surprised at all if they sprung highly conceptual questions about chromatography/spectroscopy this year - especially as the whole 'reading a graph' thing is getting kinda old. If I remember correctly, we had a taster of greater conceptual difficulty in last year's midyear (something about caffeine or something, I can't remember). It wasn't a particularly hard question, but it shows they're open to those types of questions. I certainly remember hoping prior to the U3 exam that there'd be difficult questions of this kind about nmr especially, because I knew very few people actually properly understood what what going on

Just as an aside, I really wouldn't bother trying to properly learn spectroscopy and chromatography yet... I didn't at all. Focus on the stuff easier to self-learn, if you feel like you need to do stuff =)

4) I've forgotten the technical definition of the major peak for NMR. I'd have to get up and check, and ceebs atm (I will do it though! I'm just really comfortable on my bed right now... )

5) I'm not sure why you think it should be 460nm. 425nm has the greatest absorbance?

[WhoTookMyUsername]

2) Propanone is a ketone, which has a C=O bond

4) Just the highest peak in a cluster

5) My worked answers also say 425nm. Could be that Heinemann solutions are wrong, wouldn't be the first time

I'm feeling pretty lazy at the moment, so I might skip a few questions/answer more briefly than I'd like to lol. I'm tired =(

1) You'll actually learn about this in physics in U4, but I'll try and point you in the right direction for now. I wouldn't worry about knowing everything about this though. For now, I'll try and satisfy you with just "when something 'absorbs' a frequency of light, at least in this context, we typically mean that the light is converted to heat" - does that help? Thus, when white light shines on something red, everything but red is absorbed, the thing gets hotter, and red light is reflected back. This is why black cars get so hot - all the light is absorbed and converted to heat! (And 'none' is re-emitted).

4) I've forgotten the technical definition of the major peak for NMR. I'd have to get up and check, and ceebs atm (I will do it though! I'm just really comfortable on my bed right now... )

5) I'm not sure why you think it should be 460nm. 425nm has the greatest absorbance?

Thanks guys
Aurelian - with Q 5 at 460 NM isn't the highest peak the highest absorbance??
Aurelian and nobby - For Q2 does that mean the info in my textbook is saying locatin is in "... (Carboxylic acids, esters)" but not limited to?
With Q4 Nobby, in heinemann chpater 7 Q 18b) the answers seem to indiciate that the chemical shift is the furthest line from TMS... I'm not sure why :S
Aurelian - here is the response i recieved from my teacher, i think he's saying something slightly different :S

Good to see you thinking through the concepts in U3. Your basic premises are correct.
When metal ions in solution are sprayed into a flame, electrons of the metal absorb small packets of energy (ie a quantum of E) equal to the difference in E between the energy levels so that the electrons are able to "jump" to the higher E level. We say the electrons are in an excited state. This can be observed in an absorption spectrum - black lines on a visible light spectrum background corresponding to wavelengths in the visible part of the spectrum equal to the E absorbed by the electrons.
The colored flame observed is the quantum of E released when the electrons drop back to their ground state. This can also be seen in an emission spectrum, where the colored lines on a black background correspond to the E released by the electrons . Your textbook shows these spectra.


Colored solutions are different. The first thing to note is that most colored solutions are of transition metal ions. Most group 1 & 2 metals form colorless solutions, whereas for example copper or cobalt or iron ion solutions are colored. So, a solution of sodium ions is colorless but produces a yellow flame. Copper ion solutions are generally blue but will produce a green flame.
The color of the solutions is as you stated due to the solutions absorbing E corresponding to
wavelength components of visible light. However, the explanation for this is the electron configuration of the transition metals. Due to the influence of the anions bonded to the cations, the electrons in say the 3d subshell move into different orbitals within the subshell, by absorbing the E from the visible light spectrum corresponding to the very small E difference between the orbitals. They are not in an excited state and therefore not dropping back to ground state as with the previous paragraph.  The exact reasons for this is beyond the yr 12 course.

[Aurelian]

@ Q5 - oh sorry, I read the graph lazily - you are correct, I'd say Heinmann is just wrong.

As for your teachers email, what he's saying isn't contradictory to what I'm saying, just to start. I didn't really want to go into what he's saying because some of it is quite unnecessary for summer holidays... The main difference between our explanations is that he's explaining things strictly within the context of Unit 3 chemical analysis techniques. I was just explaining the general idea of light absorption for a macro-scale object.

However, I will add this for AAS, which I think might help. You have this gas cloud, you shine light on it from one direction, it absorbs a specific frequency of that light, and the rest just passes through. Then it re-emits that frequency - but it does so in all directions, not just the direction the original incident light was heading. The result is that a black line shows up in your absorption spectrum, because the light of that frequency which actually makes it there is of greatly reduced intensity. Most people in the state just view it as some light gets completely 'absorbed' and somehow disappears.

[WhoTookMyUsername]

Oooooo, my bad, i read the txt q wrong
Aurelian and nobby - For Q2 does that mean the info in my textbook is saying locatin is in "... (Carboxylic acids, esters)" but not limited to?

[Aurelian]

Oooooo, my bad, i read the txt q wrong
Aurelian and nobby - For Q2 does that mean the info in my textbook is saying locatin is in "... (Carboxylic acids, esters)" but not limited to?

Sorry, for Q2 what exactly is your question? I didn't really read it first time round cos I thought it'd been answered, but going back now I'm not sure what exactly you're asking haha

[WhoTookMyUsername]

2) looking at the IR spectrum of proanone  ( CH3COCH3 there is a peak at around 1800. Looking at the table for infrared absorbance bands it seems to indiciate the CO bond is the stretched one here, but the table also says the location for this stretch is in carboxylic acids and esters, of which propanone is neither. What is going on here?



That's the table in my book, just clearing up does the book mean for 1680 - 1750 that the bond is found E.G. in carboxylic acids and esters ? (given it's in propanone a ketone but "ketone" isn't in the table??)

[Aurelian]

2) looking at the IR spectrum of proanone  ( CH3COCH3 there is a peak at around 1800. Looking at the table for infrared absorbance bands it seems to indiciate the CO bond is the stretched one here, but the table also says the location for this stretch is in carboxylic acids and esters, of which propanone is neither. What is going on here?

That's the table in my book, just clearing up does the book mean for 1680 - 1750 that the bond is found E.G. in carboxylic acids and esters ? (given it's in propanone a ketone but "ketone" isn't in the table??)

Yeah look, I haven't had a look at the spectrum myself, but I'd just take the absorption band at 1800cm-1 to be the C=O bond of the ketone, corresponding to that table's "1680-1750cm-1" specification. Just because the table has "found in carboxylic acids, esters" doesn't necessarily exclude ketones. You'll notice ketones aren't mentioned anywhere in the table - Heinmann may well have purposely left them out of that table on account of the fact that they aren't strictly on the course. Additionally, in VCAA's data booklet, the same range for the C=O has no functional group restriction on it.

At any rate, as Nobby said, it's very likely that the C=O bond of a ketone is going to have an absorption range very similar to the same bond in a carboxylic acid or an ester. If anything, perhaps it being a ketone accounts for the 1800cm-1 (i.e., falling slightly outside the range given by the table)!

[WhoTookMyUsername]

Do we need to show a double bond (=) in the semi struc formula? e.g. CH=CH (or is it HC=CH (:S)

2) Do we need to know about thermal cracking in U3?

3) Do amino acids contain and amine func group or amino func group? (what's the difference and why)

[Hellrocks]

1) No
2) It isn't hard, I remember I learnt it anyway
3) Strictly amino because there is a carboxyl in the molecule so the amine is a secondary functional group and called by its pre fix name.

[WhoTookMyUsername]

When do i use amu and when do i use g/mol as the unit?

[charmanderp]

http://www.chemguide.co.uk/organicprops/alkanes/cracking.html

Just read that, it covers both thermal cracking and catalytic (use of catalysts as opposed to heat) cracking of alkanes.

[WhoTookMyUsername]

if you need to use the answer to a) in b) but you've rounded a off to say 2 SF, but have an answer to 5 SF, should you use the 5 SF in your calculations for b)/?

[Hellrocks]

Yes

[charmanderp]

if you need to use the answer to a) in b) but you've rounded a off to say 2 SF, but have an answer to 5 SF, should you use the 5 SF in your calculations for b)/?

Good question!

Yes; even though practice exam answers and VCAA examiners reports answers suggest otherwise, they will always allow for a slight variation given that you used this method, which is the correct one.

Although it'd be interesting to see what method some of the 50 getters used, and what the norm is at university level.

[Hellrocks]

They accept both, but I always prefer to use the most accurate previous answer to calculate my new one.