A2 Analysis- Combined Techniques

Unless I am mistaken, 74 severely limits possible molecular formulas.

What are possible formulas for one oxygen atom?

What are possible formulas for two oxygen atoms?

Not many of them.

Can someone help me solve this analysis problem?

I've attached what I've done so far (the photocopy isn't great, so I'm happy to clarify anything you can't read).

Thanks,

Dan

Posted from TSR Mobile

You still struggling Dan?

OK - the IR doesn't tell you much. As you note there is an OH and a CO (assuming the sample is pure)

MS - molecular ion at 74 gives you the RMM. Possibly C3H6O2 or C4H10O

The fragment at m/z = 43 could be due to several fragments (as you point out) and so is not that useful. Loss of 31 units suggests CH2OH

13C NMR shows only three carbon environments

1H NMR - I am assuming that the integrals are the integers above each set of peaks. 3H doublet (i.e. adjacent to a single H) - this can only be due to a CH3 group
1H singlet ( i.e. no adjacent H) - could be an OH group, which does not show splitting.
2H doublet (i.e. adjacent to a single H)

The puzzling thing here is the presence of a CH3 group split by a neighbouring proton BUT no multiplet (quartet) due to this proton. The only answer lies in the messy signal at about 2.7. I believe this to be a multiplet due to the CH group that has split the CH3.

However, this now gives more than 6 protons, suggesting C4H10O

This is then adjacent to a CH2 group producing the 2H doublet.

The only explanation I can see is if your integrals are incorrect (perhaps the numbers are not integrals)

(CH₃)₂CHCH₂OH

... fits all of the other information...

You still struggling Dan?

OK - the IR doesn't tell you much. As you note there is an OH and a CO (assuming the sample is pure)

MS - molecular ion at 74 gives you the RMM. Possibly C3H6O2 or C4H10O

The fragment at m/z = 43 could be due to several fragments (as you point out) and so is not that useful. Loss of 31 units suggests CH2OH

13C NMR shows only three carbon environments

1H NMR - I am assuming that the integrals are the integers above each set of peaks. 3H doublet (i.e. adjacent to a single H) - this can only be due to a CH3 group
1H singlet ( i.e. no adjacent H) - could be an OH group, which does not show splitting.
2H doublet (i.e. adjacent to a single H)

The puzzling thing here is the presence of a CH3 group split by a neighbouring proton BUT no multiplet (quartet) due to this proton. The only answer lies in the messy signal at about 2.7. I believe this to be a multiplet due to the CH group that has split the CH3.

However, this now gives more than 6 protons, suggesting C4H10O

This is then adjacent to a CH2 group producing the 2H doublet.

The only explanation I can see is if your integrals are incorrect (perhaps the numbers are not integrals)

(CH₃)₂CHCH₂OH

... fits all of the other information...

Sorry can you just help me identify another IR spec again?



The numbers are impossible to read on the photocopy but I have made the best guess I can however I'm not sure if the peak on the far right is an O-H group.

Thanks

Posted from TSR Mobile

IR spec are often worse than useless. They give you an 'idea' about what bonds are present in the molecule, but nothing definitive.

The only worthwhile peaks are the carbonyl and hydrogen bonded OH ...

IR spec are often worse than useless.

So true! You can see exactly what you want to see if you look hard enough with IR spectra.

It's not unheard of for some people to try and identify every peak in a fingerprint region to a functional group

IR spec are often worse than useless. They give you an 'idea' about what bonds are present in the molecule, but nothing definitive.

The only worthwhile peaks are the carbonyl and hydrogen bonded OH ...

Sorry for all the questions. I'm doing another combined techniques question and have got stuck again.

IR suggests O-H at ~3000 and a C=O at ~1750, therefore it could be a carboxylic acid.

The reading at ~165 on the C13 backs this up

However there is nothing between 11.0-12.0 on the H-nmr which means it cannot be a carboxylic acid, so the O-H must be a phenol

The MR is 151 and the m/z is 105

Where do I go from here?

Thanks,

Dan

That's not really broad enough for O-H, it looks more like C-H stretching i.e. not very helpful. Yeah, the C=O stretch is significant.

See what I mean that with IR? You can pretty much see whatever you want to see NMR and MS are much more useful, I'd strongly suggest you use these two first then use IR to confirm you're interpretation.



It suggests some kind of C=O environment



It's not COOH...



No, 151 is the [M+1]+ peak, 150 is the M+ ion you're interested in. The m/z is a scale, not the tallest peak But it is significant because 105 = 77(benzene) + 28 (16O + 12C) suggesting that you have a C6H5CO+ fragment which goes along with what we've deduced so far - a C=O group



You've miscounted the number of carbon environments - every peak is a different environment, unlike 1H-NMR. A small peak is distinctive of a carbon with no hydrogens directly attached.

In your 1H-NMR you have two peaks at around 7-8 ppm (you should have counted from 0 on the left. so your ppms are 1 off ) these are your aromatic peaks from the benzene ring. You've correctly identified the CH2CH3 pattern form the triplet-quartet pair

Thank you so much for all your help.

So with the carbon environments from the¹³C, do I need to use the n+1 rule to deduce from the 7 peaks that there are 8 separate carbon environments?

So there are the two fragments (-CH2CH3 and C6H5CO+) which leaves me with an MR of 17 for whatever is left.

How do I go about finishing the molecule?

No problem



No simpler than that, n+1 is only for 1H-NMR 7 peaks = 7 different carbon environments



C6H5CO+ = 105 peak in your MS, CH2CH3 = 29 from 1H-NMR. Leaves you with 16.



You are pretty close now I think you labelled one of your carbon NMR peaks as C-O, that's correct too. Hint hint

Is this it?

Perfect

Brilliant, again thanks for your help

Btw, is the secret to these questions just practice?