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I want NMR questions

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Kyle_S-C

That's because the spectrometer was old as EvS pointed out. :p:

You should still be able to work it out, think about how many hydrogen environments there are and what that means about the symmetry.

If it helps, here's some additional info/hints:

- There are 7 carbon environments.
- Broad peaks are ones which exchange with the solvent (typically O-H or N-H).
- Look at the integrations...their relative heights indicate how many protons correspond to each peak.

Yeah, the problem is I can see the triplet and quartet at the start. However the blur between 6.5 and 8 is making it difficult to see the splitting pattern of the two peaks there.

Reply 41

Kyle_S-C

Mathematician!

Yeah, the problem is I can see the triplet and quartet at the start. However the blur between 6.5 and 8 is making it difficult to see the splitting pattern of the two peaks there.

The major splitting in both is a doublet. There's some other splitting, but that's not due to neighbouring hydrogens which is why it's a bit messy.

Reply 42

username165864

I've got as far as knowing there is a CH3CH2 group. However, I can't go any further... :sad:

Reply 43

Kyle_S-C

Mathematician!

I've got as far as knowing there is a CH3CH2 group. However, I can't go any further... :sad:

Yeah, it is a bit of a hard one in retrospect...I can't remember how much you guys do on chemical shifts either. :s-smilie:

Reply 44

username165864

Wait, is the structure:

CH_3-(CH_2)_3 - CH (NO_2) - (CH_2)_3-CH_3

Reply 45

EierVonSatan

Mathematician!

Wait, is the structure:

CH_3-(CH_2)_3 - CH (NO_2) - (CH_2)_3-CH_3

Afraid not - peaks about 6-8 ppm generally mean the Hydrogens's are on carbons with double bonds (so alkene/aromatic)

Reply 46

username165864

EierVonSatan

Afraid not - peaks about 6-8 ppm generally mean the Hydrogens's are on carbons with double bonds (so alkene/aromatic)

Yeah, plus there are WAY too many Hs in my structure :tongue:

I realised it was incorrect a few minutes ago, but started watching something on iPlayer lol.

Reply 47

Kyle_S-C

Mathematician!

Wait, is the structure:

CH_3-(CH_2)_3 - CH (NO_2) - (CH_2)_3-CH_3

That's certainly a clever way of making it symmetric...the problem is that it's actually got an aromatic group in there which I'm guessing you guys don't cover (yeah, I can't remember A level).

Another rather important clue I forgot to give you is that it contains an ester...you could (just about) tell that from an IR peak we were given.

I've put the solution in a few spoilers below:

Spoiler

I must confess, we did cheat a bit and guessed at what it was...and yeah, I need to go revise. :p:

Reply 48

EierVonSatan

Yeah the ester bit was quite important Kyle :tongue:

I had: Br-C6H4-C(=N-NH2)-CH2-CH3 (i.e. the hydrazone to make up the 7th carbon environment and the Br to satisfy the symmetry)

Reply 49

Kyle_S-C

EierVonSatan

Yeah the ester bit was quite important Kyle :tongue:

I had: Br-C6H4-C(=N-NH2)-CH2-CH3 (i.e. the hydrazone to make up the 7th carbon environment and the Br to satisfy the symmetry)

Bah, we were given a C13 spectrum with 5 peaks in, the two quaternaries were just not there...at least I didn't give out incorrect information. :p:

Reply 50

EierVonSatan

Kyle_S-C

Bah, we were given a C13 spectrum with 5 peaks in, the two quaternaries were just not there...at least I didn't give out incorrect information. :p:

there are three quaternaries in benzocaine :hand:

Reply 51

username165864

Hehe, I continued a while ago, and got upto there being a benzene ring. I thought the structure was CH3CH2-C6H3(NO2)(CH3)

(With NO2 on first carbon, CH3CH2 on 2nd and CH3 on 5th).