Name the structure that matches the spectroscopy data

Question

A molecule with the mol. formula of C6H7N has a UV absorption of 190nm. The IR spectrum has absorbances at 3400 (broad), 3340 (sharp), 2920, 2100, and 650 cm-1. Name the structure that best fits this data.

I am struggling with spectroscopy questions and understanding how to do these types of questions - if anyone could please explain, I would appreciate it.

Answer 1

This is almost certainly too late for the original poster, but this question intrigued me.... Hopefully someone finds this long answer helpful.

The first thing I do with a problem like this is to determine the "degree of unsaturation", or the number of "elements of unsaturation" (same thing). Basically, this is the number of rings and/or pi bonds in the molecule. A molecule with N carbon atoms can maximally hold 2N+2 hydrogen atoms, in the absence of any rings or pi bonds in the molecule. Methane (one carbon) has 4 hydrogens -- 2x1+2. Adding more C atoms allows 2 additional H's per carbon -- ethane is C2H6, propane C3H8 etc. Molecules with a C=C pi bond have 2 fewer H's -- ethene is C2H4 with one pi bond. And C3H6 can either be propene (pi bond) or cyclopropane (ring). Adding one nitrogen only allows ONE additional hydrogen -- methylamine is CH5N (CH3-NH2), ethylamine is C2H7N (CH3-CH2-NH2).  These patterns hold for molecules of any size.

As such, a molecule with 6 C's and 1 N can maximally accommodate 15 H atoms -- 2x6+2 for 6 carbons, with 1 extra for the N. This is only true if there are NO rings or pi bonds -- forming either a ring or pi bond requires removing 2 H atoms to add the extra C-C bond. Since there are only 7 H atoms in the molecule, there must be four "elements of unsaturation" (rings or pi bonds) in the molecule -- 8 H atoms have been "removed", forming one ring or pi bond for each pair removed. This degree of unsaturation frequently corresponds to an aromatic molecule -- benzene and its derivatives have one ring and 3 pi bonds, for a total of 4 elements of unsaturation. So based solely on the formula, you can begin from an assumption that it is a derivative of benzene. (That said, this IS an ASSUMPTION -- it is not necessarily aromatic.... It could have, say, two rings and two pi bonds, or some other combination. But I would only go down that road after exhausting the possible aromatic structures, personally.)  As the previous answer points out, this is consistent with the UV-Vis lambda-max of 190 nm (which is, otherwise, spectacularly unhelpful in determining the structure). It also simplistically makes sense that it might be aromatic, since there are 6 carbon atoms (suggesting a benzene derivative, since benzene has six carbons).

There are very few options for the structure, if the assumption of an aromatic molecule is correct. Specifically, aniline, or aminobenzene (benzeneamine), where one -H on benzene has been replaced with -NH2 (C6H5-NH2). Another option would be an isomer of methylpyridine. In this case, one carbon in the six-membered benzene ring is replaced with a nitrogen, and a -CH3 group is placed somewhere on the ring (there are 3 possible isomers). Google "pyridine" and imagine it with a -CH3 group added.

Unfortunately, neither of these options fits very well with all of the IR data. (As mentioned above, the UV-Vis data is pretty much useless, I think.) The IR data is telling you some things which are present in the molecule, and by omission, implying that certain other things are ABSENT from the molecule. IR is great at indicating BOTH presence AND absence of certain functional groups or "bond types". It behooves one to pay attention to both presence and absence!  Unfortunately, paying close attention to all the IR is telling us leads to a very awkward conclusion....

There are four parts of an IR spectrum which are pertinent to this problem (this is not exhaustive -- I'm only listing IR regions which are pertinent to this problem):

1. 3200-3600 cm-1: O-H or N-H bonds (though as we'll see later, certain C-H bonds can sometimes come in this region)
2. ~3000 cm-1 (+/- about 150): C-H bonds
3. 2100-2200 cm-1: triple bonds
4. Anything less than 1400 cm-1: "fingerprint region"

The absence of peaks at other frequencies means other FGs or features (carbonyl, alkene, etc.) are absent. Working backwards from the last region -- with some exceptions, most absorbances lower than 1400 cm-1 are not particularly helpful in determining the structure of a molecule from its IR. The peak at 650 cm-1 is caused by SOMEthing, but it's probably not helpful trying to figure out what, so I will ignore it.

The third area (2100-2200 for triple bonds) is troubling! Very little else absorbs in this region of the IR, so any time there is a peak mentioned around there, you would be wise to start to think about a triple bond (either a CC triple bond -- alkyne -- or a CN triple -- nitrile). But this immediately causes problems.... A triple bond contains TWO pi bonds, leaving only two other elements of unsaturation in the molecule. That means any structure with this formula that contains a triple bond CANNOT be aromatic, as the aromatic ring requires 4 elements of unsaturation, and the triple bond cannot be incorporated as part of the aromatic ring. That's not the end of the world -- perhaps this suggests our assumption about aromaticity was wrong. That's possible. If so, there are two other elements of unsaturation, aside from the triple bond (since the triple bond contains 2 pi bonds, it counts for 2 elements of unsaturation).

If there really IS a triple bond, it might be a nitrile (CN triple). But those are usually stronger on the IR, and closer to the 2200 cm-1 end (and are sometimes even a little higher than 2200). Peaks closer to 2100 are more often alkynes. Interestingly, terminal alkynes (those which have a -H atom bonded to one of the triply-bonded carbon atoms) will also show a narrow peak (contrast with broad, hydrogen bonded O-H or N-H peaks in the same area) for the sp-hybridized C-H bond on the end of the alkyne. Given the sharp peak at 3340 cm-1 in this spectrum, that lends some credence to the possibility that there is a terminal alkyne in the molecule, even though that contradicts any aromatic structures. (An alkyne would also be consistent with a UV-Vis lambda-max at 190 nm.)

The second region is for C-H bonds, around 3000 cm-1. Almost all organic molecules have lots of C-H bonds, and the vast majority of them involve either sp2 hybridized C or sp3 hybridized C. sp2 C-H bonds are when H atoms are bonded to doubly-bonded carbons (C=C-H), including carbons of aromatic/benzene rings, and sp3 C-H bonds are on saturated, 4-singly-bonded carbon atoms (which could NOT be part of an aromatic ring). The sp2 C-H bonds always come at HIGHER frequency than 3000 -- usually in the range of 3000-3150 cm-1. In contrast, the sp3 (saturated, non-aromatic) C-H bonds come at LOWER frequency -- usually 3000-2850 cm-1. In this list they ONLY indicate 2920 cm-1, which would be in the sp3-hybrid, non-aromatic C-H region. So not only does this indicate the PRESENCE of sp3 C-H bonds, it indicates the ABSENCE of sp2 C-H bonds, since no peaks between 3000-3100 cm-1 are listed.

If that interpretation is true, it pretty much eliminates ANY aromatic structures from consideration (aniline, methylpyridine) as they would have several sp2 C-H bonds. It also indicates the presence of sp3 hybridized carbon atoms with H's attached.

Finally, the O-H/N-H region (3200-3600 cm-1, although some O-H bonds do occur at lower frequencies, and can obscure the C-H region at ~3000 cm-1). O-H and N-H bonds typically participate in hydrogen bonding in IR samples, which causes the peaks in the IR to be broad. Notice that ONE peak in that region in the spectrum is broad (~3400 cm-1), but the other is sharp (3340 cm-1). While most peaks in this frequency range are either O-H or N-H bonds, some C-H bonds can show up here -- specifically, sp-hybridized, terminal alkyne C-H bonds, which occur close to 3300 cm-1. When a terminal alkyne C-H shows up, it will NOT be broad (H-bonded), but will be narrow, or "sharp".

The interpretation of this region is a little ambiguous. At first, given two peaks around 3300-3400, and the presence of a N atom in the molecule, I assumed it was a -NH2 group, since often the two hydrogen atoms on a single N show their own peak. While N-H peaks are typically broadened by H-bonding, the individual peak tips can be sharp, so without visually seeing the spectrum, it's a little hard to know. But looking farther down and finding the 2100 peak, indicating an alkyne (which could have a terminal sp-hybridized C-H), it makes me think the only N-H is the broad 3400 cm-1, while the sharp 3340 cm-1 is a terminal alkyne. The terminal alkyne would also mean the molecule is NOT aromatic -- but that is consistent with the absence of sp2 C-H peaks 3000-3150.

(continued in comment)

Answer 2

While I cannot give you the full answer, I can give you some hints...

1. Look at the UV and IR absorptions and determine what they tell you about the structure.
2. Here you have UV at 190 nm, which tells you there us a conjugated structure, such as a basic benzene ring
3. Then you have IR at various frequencies, which in this case tell you of N-H and/or O-H bonds, triple bonds and C-H bonds (from the benzene ring)
4. Look up the molecular formula on-line, compare against the features above and pick the one that matches all (ore most) of the features.

HTH