Homework questions thread

[thushan]

Aldehydes are primary alcohols that have been partially oxidised. If you further oxidise them they will be converted to a carboxylic acid, so A is correct.

AFAIK this isn't in the VCAA Chemistry course.

I don't think there's any reason it shouldn't be on the course. If you look at the oxidation number of C in alcohol, aldehyde and carboxylic acids, it is reasonably clear that those are different steps in oxidation numbers.

That's true, but wouldn't it be a bit unfair to expect students to recognise aldehydes? I'm pretty sure the intention of that question was to convince them that none of A, B or C were primary alcohols or alcohols at all, pushing them to choose D.

I kind of agree with derp over here, I don't think students are expected to distinguish oxidation numbers of two atoms of the same element in the same compound (like CH3-COOH, the two C's have two different oxidation numbers). To work out the distinction between alkanol, aldehyde and alkanoic acid in terms of ox no, they'd need to be introduced to that concept.

Having said that, the study design says 'oxidation of primary alkanols' so it is possible, however unlikely, for them to ask about the aldehyde intermediate.

[ggxoxo]

How many peaks would CH3CH2CH2CH2CH3 produce in a high-res proton NMR?

Would it be 5?

[Tonychet2]

How many peaks would CH3CH2CH2CH2CH3 produce in a high-res proton NMR?

Would it be 5?

Should be three based on symmetry

isnt it one peak due to symmetry?

[soccerboi]

It's 5 peaks on the high res NMR. The middle CH2 group is surrounded by a CH2 to its right and a CH2 to its left, each CH2 has 2 H atoms, so 4 H atoms in total on the neighboring carbon atoms of the middle CH2. Following the n+1 rule, we get 4+1= 5 peaks.

[Tonychet2]

It's 5 peaks. The middle CH2 group is surrounded by a CH2 to its right and a CH2 to its left, each CH2 has 2 H atoms, so 4 H atoms in total on the neighboring carbon atoms of the middle CH2. Following the n+1 rule, we get 4+1= 5 peaks.
[/quo

[soccerboi]

It's 5 peaks. The middle CH2 group is surrounded by a CH2 to its right and a CH2 to its left, each CH2 has 2 H atoms, so 4 H atoms in total on the neighboring carbon atoms of the middle CH2. Following the n+1 rule, we get 4+1= 5 peaks.

thats only if its not in symmetry though

When considering high res peak splitting like in this case, do we consider symmetry? and even with symmetry i see 3 peaks, because the CH2 on one side has 2 H atoms, so wouldn't it be 2+1=3 peaks?

[Tonychet2]

its either 3 or 1 im not sure but i recall reading somewhere when something is in symmetry the peak produced = 1 ? let me ask someone lol

edit: ignore this its wrong

[thushan]

its either 3 or 1 im not sure but i recall reading somewhere when something is in symmetry the peak produced = 1 ? let me ask someone lol

The peak for the protons on the 3rd carbon in pentane will be split into a pentet (5), because the two neighbouring carbons are in the same environment as each other but different to carbon 3, and between them they have 4 protons. So by n+1, a pentet will be formed.

The TOTAL number of peaks in pentane will be 3 - a pentet, a triplet of a quartet (dw about why that is, its to do with multiplicity), and a triplet.

[Tonychet2]

its either 3 or 1 im not sure but i recall reading somewhere when something is in symmetry the peak produced = 1 ? let me ask someone lol

The peak for the protons on the 3rd carbon in pentane will be split into a pentet (5), because the two neighbouring carbons are in the same environment as each other but different to carbon 3, and between them they have 4 protons. So by n+1, a pentet will be formed.

The TOTAL number of peaks in pentane will be 3 - a pentet, a triplet of a quartet (dw about why that is, its to do with multiplicity), and a triplet.

ahhhhhh thx for clearing that up

[stephanieteddy]

Hello,

Was just wondering how to tell if the mobile phase would be polar or nonpolar in HPLC OR TLC.

For example when analysing a sample of coconut oil.

[soccerboi]

Hello,

Was just wondering how to tell if the mobile phase would be polar or nonpolar in HPLC OR TLC.

For example when analysing a sample of coconut oil.

They usually specify in the question, but if not then assume that it's normal phase chromatography (polar stationary phase and non polar mobile phase)

[stephanieteddy]

Hehe, I answered yours and you answered mine! Thank you!

[|ll|lll|]

its either 3 or 1 im not sure but i recall reading somewhere when something is in symmetry the peak produced = 1 ? let me ask someone lol

The peak for the protons on the 3rd carbon in pentane will be split into a pentet (5), because the two neighbouring carbons are in the same environment as each other but different to carbon 3, and between them they have 4 protons. So by n+1, a pentet will be formed.

The TOTAL number of peaks in pentane will be 3 - a pentet, a triplet of a quartet (dw about why that is, its to do with multiplicity), and a triplet.

is it a pentet or a quintet?
The names following it would be sextet, septet, octet? (Just wanted to clarify )

Also, note that there will be three PEAKS, but in the hydrogen atom, it will have five SPLITS.

[thushan]

its either 3 or 1 im not sure but i recall reading somewhere when something is in symmetry the peak produced = 1 ? let me ask someone lol

The peak for the protons on the 3rd carbon in pentane will be split into a pentet (5), because the two neighbouring carbons are in the same environment as each other but different to carbon 3, and between them they have 4 protons. So by n+1, a pentet will be formed.

The TOTAL number of peaks in pentane will be 3 - a pentet, a triplet of a quartet (dw about why that is, its to do with multiplicity), and a triplet.

is it a pentet or a quintet?
The names following it would be sextet, septet, octet? (Just wanted to clarify )

Also, note that there will be three PEAKS, but in the hydrogen atom, it will have five SPLITS.

pentet, sextet, septet and octet

[nina_rox]

Question: 1. For UV spectroscopy does electrons emit when moving from excited to ground?
2. And do we need to know how to do the spitting of for example the CHCl of molecule CHCl2CHClCH2? Apparently the splitting is different (not n=1 rule)