9
$\begingroup$

Question 3:

enter image description here

It said $m/z = 122,$ and $m/z = 124$ is in a $3:1$ ratio, so I figured that meant that chlorine is present. Then I thought $m/z$ was the actual compound's molecular mass.

So I used the rule of 13, and did:

Chlorine's molar mass = 35

122 - 35 = 87

Using the rule of 13

87/13 = 6C + 9H/13

So the molecular formula is $\ce{C6H9Cl}.$ But the integral values I rounded are: 2,2,2,2 and 3 adding up to 11. Did I use the wrong mass to charge ratio? What am I missing?

$\endgroup$
  • $\begingroup$ I would check for consistency with the proton spectrum: that suggests you have 11 H or a multiple thereof. $\endgroup$ – Buck Thorn Oct 18 at 8:26
  • $\begingroup$ How do I know for sure if there are exactly 11 H? $\endgroup$ – Mohamed Oct 18 at 8:45
  • $\begingroup$ Well, I posted an explanation as an answer... $\endgroup$ – Buck Thorn Oct 18 at 9:27
10
$\begingroup$

Being an NMR fan myself I would inspect that NMR spectrum:

enter image description here

The integrals suggest you have 11 $\ce{^1H}$ or a multiple thereof (the number under each peak is the normalized integral, which is proportional to the number of protons represented by the multiplet). That leaves you with $\pu{87 Da -11 Da}=\pu{76 Da}$ to explain. If you throw in an oxygen you get $\pu{60 Da}$, which is neatly divisible by 12, for a formula $\ce{C5H11OCl}$. Next use the NMR spectrum again to infer the connectivity. The rightmost integrated multiplets (multiplet and triplet) are typical of $\ce{-CH2CH3}$ with $\ce{CH2}$ split by methyl and something else. From the integrals we see clearly there is only one methyl group, and there are no lone H which seems to eliminate the possibility of carbonyl or hydroxyl groups, suggesting an ether.

In the end I just "cheated" (it's 2019 after all) and used the online NMR simulator predictor:

enter image description here

This nice piece of software lets you tinker quickly with different structures and inspect assignments interactively, helping you develop your intuition.

Next you need to verify the MS and IR spectra, but the NMR spectrum is a giveaway that this is the correct hit.

$\endgroup$
  • $\begingroup$ ‘verify the IR spectrum’? I thought IR spectra nowadays were only recorded because some journal editors complain if they are not there; then a couple of wavenumbers are written down and the spectrum filed. $\endgroup$ – Jan Oct 18 at 11:27
  • $\begingroup$ @Jan Yes, well, in that case consider it "fine print". Evidently this is homework and the OP should incorporate all available info into the answer :-) $\endgroup$ – Buck Thorn Oct 18 at 12:05
10
$\begingroup$

I would probably also use the method Buck has suggested, but let’s say the NMR broke down or somebody is measuring a $\ce{^13C}$ of $\pu{2.5mg}$ meaning it will be blocked until tomorrow; in this case, we can still extract more information from the mass spectrum.

In addition to the molecule peak at 122, you have:

  • a chlorine-containing fragment $m/z=93$
  • a chlorine-containing fragment $m/z=63$
  • a chlorine-free fragment $m/z=73$
  • and some more stuff at lower masses that doesn’t analyse itself quite as easily.

Those three peaks should derive from fragmentation processes, so we can get an idea of what groups we have from looking at the mass difference (i.e. the bit that fragmented away).

For the peak at 93: $122-93=29$. The chlorine atom is retained. The most common groups to leave are methyl, ethyl etc., and 29 happens to be exactly the mass of $\ce{CH3CH2}$ (methyl is 15, methylene is 14; you internalise those numbers pretty quickly). Therefore, your molecule should have an ethyl group somewhere.

The chlorine-free peak at 73: $122-73=49$. That’s an uncommon number at first sight but remember we lost a chlorine. Taking that into account, $49-35=14$ and we find another methylene group. So we can say that there is a terminal $\ce{CH2Cl-{}}$ group.

Finally, the peak at 63: $122-63=59$. That again doesn’t say everything at once, but what happens if we remove the 29 we found earlier for the ethyl group? $59-29=30$. 30 could mean two methyl groups but that’s somewhat unlikely given how few fragments we have overall—especially since we would also expect a peak with a mass difference of 15 for a single methyl group. But $30=14+16$, meaning there could be a $\ce{-CH2-O-{}}$ group attached to the ethyl group. Other possibilities for 16 might be $\ce{NH2}$, but the molecule disobeys the odd-nitrogen rule and two amino groups doesn’t look good with the data.

If we take what we have, we have a strong suspicion that there might be a $\ce{CH3-CH2-CH2-O-{}}$ group and a $\ce{CH2Cl-{}}$ group. Adding these together gives an intermediate mass of $59+49=108$ or 14 missing from the complete molecule. It seems reasonable to assume a $\ce{CH2}$ group connecting the fragments.

We should consider this all a working hypothesis until we can then take a look at the NMR spectrum which beautifully confirms that molecular structure.

$\endgroup$
7
$\begingroup$

I like both answers provided before me where one has used exclusive use of internet to suggest structure by NMR spectrum, and the other has used thorough analysis of mass spectrum. Although these two are valuable techniques, I feel OP needs to analyse step by step analysis of spectra given to predict the structure since he/her seemingly in a graduate course, Spectroscopic Analysis of Organic Compound, or one similar to that. Hence OP needs knowledge of how to analyse given spectra to determine relevant structure.

Thus, I'd like to give some insight to provide that knowledge. I usually start with $\ce{^1H}$-NMR since it gives the most information. The given NMR is showing 5 resonances indicating 5 different protons in 2:2:2:2:3 ratio, according to the corresponding integral values given. By that information, you can guess there are at least 11 protons and 5 carbons are in the molecule to be analyzed. Therefore, the molecule possibly contains $\ce{C5H11}$- portion, which counts for $\pu{71D}$. There are three -$\ce{CH2}$- resonances present in the downfield of the NMR spectrum, each having chemical shifts greater than $\pu{3 ppm}$. This clearly suggest that at least one side of each of these -$\ce{CH2}$- groups attached to a electron withdrawing group (such as oxygen or halide) making the corresponding proton deshielded than ordinary alkyl group.

Mass spectra of that compound has shown that it must contain one $\ce{Cl}$ atom, as OP correctly guessed (peaks with 3:1 ration intensities with $2 \: m/z$ difference). It also has shown molecular peak is $122 \: m/z$ and $124 \: m/z$ for $\ce{^{35}Cl}$ and $\ce{^{37}Cl}$ fragments. Therefore, $\ce{C5H11^{35}Cl}$- fragment gives $\pu{(71+35) D}=\pu{106D}$. Thus, the rest is $\pu{(122-106) D}=\pu{16 D}$, which indicates the presence of $\ce{O}$ in the molecule. Therefore, molecular formula of the molecule would be $\ce{C5H11ClO}$, which accounts for $\pu{122 D}$ (for $\ce{C5H11^{35}ClO}$). Based on these values, you can predict the NMR resonance as follows:

enter image description here

Since two electron withdrawing groups in the molecule are $\ce{O}$ and $\ce{Cl}$, we can predict that $\mathrm{EWG^{I}}$ is $\ce{Cl}$ and $\mathrm{EWG^{II}}$ is $\ce{O}$, based on the chemical shift values of peaks at ~$\pu{3.9 ppm}$ and ~$\pu{3.6 ppm}$. Therefore, it can be suggested that $\ce{Cl-CH2-CH2-O}$- group is present in the molecule. The third deshelded peak at ~$\pu{3.3 ppm}$ suggests the presence of $\ce{R-CH2-CH2-O}$- group in the molecule as well. Thus, conventional wisdom would suggest the molecular structure $\ce{Cl-CH2-CH2-O-CH2-CH2-R}$ for the molecule. Based on two shielded resonances at ~$(\pu{1.5 ppm},\: sextet)$ and ~$(\pu{0.9 ppm},\: t)$ suggest that $\ce{R}$- group would be $\ce{CH3}$-, based on chemical shifts of $1.5$ and $\pu{0.9 ppm}$, and integral ($3.07$) and splitting pattern of resonance at $\pu{0.9 ppm}$. Therefore, the tentative structure of the molecule is $\ce{Cl-CH2-CH2-O-CH2-CH2-CH3}$ as given in the diagram above. The given mass spectrum confirms this structure as shown in fragmentation illustrated in following diagram and elsewhere in one of the answers:

Structure by MS

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.