more questions

[bec]

that's interesting - but luckily, proline is the only one like that that's in the data booklet....

stupid question but can TLC ever be used for quantitative analysis?

[Collin Li]

Nope.

[chem-nerd]

it can - but with the availability of more accurate, reliable and less time consuming alternatives it rarely is

[bec]

when a question asks you to state how many peaks on an NMR, do you include the peak for TMS?

[Collin Li]

it can - but with the availability of more accurate, reliable and less time consuming alternatives it rarely is

How would you go about quantitative analysis with TLC? How would you determine a concentration of something that just remains adsorbed into the stationary phase?

[Collin Li]

when a question asks you to state how many peaks on an NMR, do you include the peak for TMS?

Depends. I've seen questions where they fail to show the peak for TMS at all, so you probably wouldn't bother.

[doboman]

When they ask how many peaks on a HPLC/GLC chromatogram, do you include the first eulet (i think thats how u spell it) which is the moblie phase?

[chem-nerd]

http://www.chromatography-online.org/quant/Quantitative-Analysis-by-TLC.html


So it's not a particularly accurate way to do quantitative analysis but it can give you an indication as to how much of each component is present.

[bec]

When they ask how many peaks on a HPLC/GLC chromatogram, do you include the first eulet (i think thats how u spell it) which is the moblie phase?

would they ever ask for that though? because doesn't the number of peaks just correspond to the number of components being analysed?

another question- how do diprotic acids work? given that we haven't done equilibriums etc, is it possible for us to work out the pH of a 0.005M solution of H2SO4? in the solutions to the prac exam where i found this question, they just multiplied [H2SO4] by 2 - is this correct, or is it assuming that the ionization of HSO₄^- is 100% complete?

[shinny]

stupid question but can TLC ever be used for quantitative analysis?

the nelson textbook says theres something called preparative TLC where u use a larger stationary phase, allowing u to scrape off the components and gravimetrically quantify them, but yeh, its quite limited in its accuracy and uses and typically ild ignore it since no other sources make any mention of this technique

when a question asks you to state how many peaks on an NMR, do you include the peak for TMS?

from what ive seen, no, since they omit it, and im fairly sure in an exam, theyld make room for an answer including tms, and one without

When they ask how many peaks on a HPLC/GLC chromatogram, do you include the first eulet (i think thats how u spell it) which is the moblie phase?

would they ever ask for that though? because doesn't the number of peaks just correspond to the number of components being analysed?

another question- how do diprotic acids work? given that we haven't done equilibriums etc, is it possible for us to work out the pH of a 0.005M solution of H2SO4? in the solutions to the prac exam where i found this question, they just multiplied [H2SO4] by 2 - is this correct, or is it assuming that the ionization of HSO₄^- is 100% complete?

never actually thought about that, but is it because the ions are dissociated anyway so it doesnt matter? since all pH is measuring is the H+ concentration and regardless of whether its present as HSO4-(aq) or H3O+(aq) doesnt matter i think, but im quite uncertain and someone perhaps a bit more knowledgeable than me could correct me. but either way for our purposes in unit3, yes, its just n(H+)=2n(H2SO4)

[bec]

thanks so much for all this help
i have a list of qs that i don't understand, including some that don't really seem directly related to the course - can anyone tell me if i need to know them?

1. Does an amine functional group necessarily have to be NH2? Or can it just be NH?
Eg. Does thymine have an amine functional group? (the neap book says it does...)

2. The boiling point of hexane is 69oC and octane is 107oC A mixture of hexane and octane is
heated in a laboratory distillation apparatus. The temperature of the mixture will: (answer is D)
A. rise at a linear rate if the apparatus is kept insulated
B. rise quickly until it reaches 69oC and then rise more slowly as the hexane is boiling
C. rise quickly until it reaches 69oC and then rise more quickly as the hexane is boiling
D. rise quickly and then remain at 69oC until all the hexane evaporates

3. [basically, my question is: do we need to know how to figure out the pH of diprotic or triprotic acids?] A 20.0 mL sample of acid is delivered to a laboratory for analysis. The identity of the acid,
and its concentration, is not known. The 20.0 mL sample is made up to the mark with water in
a 250 mL volumetric flask. 25 mL aliquots are then titrated against 0.05 M sodium carbonate, Na2CO3 solution. Methyl red is chosen as the indicator and a sharp transition occurs at an average titre of 15.6 mL.
b. Complete the table, showing possible concentrations of the acid.
If the acid is:  concentration of diluted acid      concentration of original acid
Monoprotic
Diprotic
triprotic

4. Why would hex-1-ene only have 3 peaks? TSSM says “one peak for CH, one for CH2, one for CH3” – but doesn’t it depend on what’s bonded to the adjacent carbon as well?

Q5. In a proton NMR of CH3CH2CH2OH, the peak at chemical shift 1.6 is split into 6 finer peaks, i.e. a sextet. Referring to the semi-structural formula of the compound briefly explain why this occurs.
Answer:
The sextet peak set is the peak for the two hydrogens on the CH2 between CH3 and the other CH2. i.e. CH3CH2CH2OH. These two H atoms have 5 (3 on the adjacent CH3 and 2 on the adjacent CH2) neighbouring H atoms and so their peak is split into 5 + 1 = 6 finer peaks.
do we need to know about peak splitting due to things bonded to carbons on EITHER SIDE of a proton environment, or just the one side?

6. Which compound is a member of the same homologous series as
1 -chloropropane?

A. 1-chloropropene
B. 1-chlorobutane
C. 1-bromopropane
D. 1,1-dichloropropane
(answer is d: why not b)

thanks!

[shinny]

1. not sure
2. its D due to the nature of how boiling works; up until 69 degrees, all the heat is used as thermal energy to heat up the mixture, but at 69 degrees, all the heat will be absorbed by the bonds in hexane until it all turns into a gas - you should have learnt something similar regarding water in unit4 i think. do the words latent heat of vaporisation ring a bell?
3. that question doesnt really have anything to do about pH; its equation writing.
Na2CO3+2HX-->H2O+CO2+2NaX
Na2CO3+H2X-->H2O+CO2+Na2X
3Na2CO3+2H3X-->3H2O+3CO2+Na3X2
use those mole ratios and ull get the answer. to make it faster, just solve for monoprotic, and divide the answer accordingly since concentration and mol are proportional
4. i think theyre wrong
5. yes we need to know about either side, but isnt it quite simple? just add them. there are actually cases where u need to multiply though, but ive never seen it come up in an actual vce question so dont worry - just stick with the n+1 rule
6. its b; theyre wrong again

[Collin Li]

Does an amine functional group necessarily have to be NH2? Or can it just be NH?

An amine group is just a nitrogen that has 3 single bonds and a lone pair of electrons. Usually you will encounter -NH2, but it is not uncommon to see -NH bonded to two carbons. You would call that one a "secondary amine," although that terminology is not on the course. Even a nitrogen connected to 3 carbons is an amine group.

[Collin Li]

Why would hex-1-ene only have 3 peaks? TSSM says “one peak for CH, one for CH2, one for CH3” – but doesn’t it depend on what’s bonded to the adjacent carbon as well?

TSSM is wrong, yes.

Do we need to know about peak splitting due to things bonded to carbons on EITHER SIDE of a proton environment, or just the one side?

Maybe. I'd lean towards yes. Every chemist has to learn how to use NMR nowadays, so they might as well learn it properly. You also have to know that splitting can only occur from different groups. Take butane for example:

CH3CH2CH2CH3

The -CH2 protons will only have a splitting of 3+1 = 4, not (3+2)+1 = 6, because the adjacent -CH2 groups do not split this group.

[shinny]

You would call that one a "secondary amine," although that terminology is not on the course.

well i guess that explains what they mean by N-H (primary amines) in the IR table in the data booklet