[theanonymous]

For the first structure, what would its IR Spectrum be? I don't get how you can just identify it based on looking at these graphs. *delete me*

Options:

[theanonymous]

The first structure (with the 2 CH3 groups) has to be B or C, according to this website: http://chemistry.tutorvista.com/organic-chemistry/ir-spectroscopy.html

Am I right?

[Dan]

It is probably easiest to use your shift table to assign the carbonyl compounds first, as the carbonyl region is easier to deal with.

[orgopete]

By themselves, I would find this difficult. However, since this is a matching problem, then it becomes easier. I would start with any of the functional groups based upon their stretching frequency and patterns, OH, NH, COOH, C≡CH, C=O. You can eliminate spectra from the possibles and limit the choices. Several of the spectra are distinctive and match their functional group characteristics. For example, I think I looks like a carboxylic acid. What does A look like or what compounds can be eliminated from A? We may need some discussion for F, G, and H.

[theanonymous]

By themselves, I would find this difficult. However, since this is a matching problem, then it becomes easier. I would start with any of the functional groups based upon their stretching frequency and patterns, OH, NH, COOH, C≡CH, C=O. You can eliminate spectra from the possibles and limit the choices. Several of the spectra are distinctive and match their functional group characteristics. For example, I think I looks like a carboxylic acid. What does A look like or what compounds can be eliminated from A? We may need some discussion for F, G, and H.

So Structur 8 would be I, right?

For Diagram A,

This is what I did:

1) Locate the longest stretches/bands

3200-3400 (far left)
2800-3000 (left)

1480-1500 (right)
1000-1100 (far right)

2) Look at Typical Infrared Absorption Frequencies Chart

3200-3400 (far left) - Alcohols/Strong/O-H stretch/Single bonded
2800-3000 (left) - Aldehyde & ketons/C-H stretch/Single bonded

1480-1500 (right) - ?
1000-1100 (far right) - ?

Is Diagram A the 4th compound in the first column??

[theanonymous]

I also used this diagram as a guideline:

[theanonymous]

It is probably easiest to use your shift table to assign the carbonyl compounds first, as the carbonyl region is easier to deal with.

Diagrams F, G, H, and I have carbonyls (C=O)

[theanonymous]

Just to make things easier...

[theanonymous]

Diagram H  has to be Compound 6, 7, 8, or 9 knowing that there is a Carbonyl around 1700.



Can't be Compound 8 since the spectra doesn't have an alcohol stretch. So it has to be Compound 6, 7, or 9

Ok so I'm kind of stuck now.

[theanonymous]

Diagram I has an alcohol (O-H) and a carbonyl (C=O).
The only compound it can be is compound 8

[theanonymous]

Diagram H - strong, sharp stretch at 1750 means it is an ester. So  diagram H is compound 9

[theanonymous]

This is what I did so far...I just don't know about 1, 2, and 6.

This is how I figured out the others:

[Lafleche]

Look at diagram F and G.  They both have a carbonyl group.  You've already assigned the ester and the carboxylic acid to diagram H and I, so you can rule those out.  Now you have an aldehyde and a ketone to assign.  They both look very similar in their IR but the aldehyde will have 2 extra peak around 2750 and 2850 from the H-C=O stretching.  Usually you'll see the band at 2750 better because its not in the C-H sp3 stretching region.  You can see that band in spectra F so that would be your aldehyde.  Spectra G has to be your ketone.

This mean that Diagram G can't be the alkyne you initially assigned it too.  Alkyne's will show C-H stretching at very high wavenumber because they have an Sp1 hydrogen.  I would assign it to E.

You now only have to assign spectrum C and D.  You should be able to figure it out (hint: an alkene show C=C stretching around 1600 and C=C bending in the region of 900 (I'm sorry I forget the exact number here, but since its mono-sub you'll see two bands) and the alkane will have a very simple spectra)

[theanonymous]

Look at diagram F and G.  They both have a carbonyl group.  You've already assigned the ester and the carboxylic acid to diagram H and I, so you can rule those out.  Now you have an aldehyde and a ketone to assign.  They both look very similar in their IR but the aldehyde will have 2 extra peak around 2750 and 2850 from the H-C=O stretching.  Usually you'll see the band at 2750 better because its not in the C-H sp3 stretching region.  You can see that band in spectra F so that would be your aldehyde.  Spectra G has to be your ketone.

This mean that Diagram G can't be the alkyne you initially assigned it too.  Alkyne's will show C-H stretching at very high wavenumber because they have an Sp1 hydrogen.  I would assign it to E.

Wait I'm confused...

What number would you assign Diagram F, E, and G to?

[theanonymous]

So....
You're saying that F and G look very similar because they both have a carbonyl group (C=O) around 1700.
The aldehyde will be the one with 2 extra peaks around 2750 and 2850
Therefore,
F is the aldehyde, and G is the keytone.

So Compound 6 is Diagram F.