I think there will be two pairs of doublets, since the hydrogens adjacent to the $\ce{OCH3}$ and those adjacent to the $\ce{OH}$ group will be in different environments. The answer, however, is that there'll be a singlet for 4 Hydrogens.

To give some context, I am extremely unused to the proton NMR of benzene derivatives, and any (introductory) links telling me a bit about it would be appreciated. I'm aware of the theory behind NMR spectrosocopy, the 'n+1' rule, and the ways to tell if two Hydrogens are magnetically equivalent.

  • 1
    $\begingroup$ Two peaks won't get separated, because of similar environment, I guess; you might get sth more detailed then singlet with hi-res though. $\endgroup$ – Mithoron Apr 25 '17 at 14:43
  • $\begingroup$ But one has OCH3 and the other has an OH.. $\endgroup$ – John Apr 25 '17 at 14:55
  • 1
    $\begingroup$ And that is very similar. $\endgroup$ – Mithoron Apr 25 '17 at 15:06
  • $\begingroup$ I see.. but if it were something like a OCH2CH2Ch3? What about a Fluorine directly attached? Basically, at what 'point' do I know if it is similar enough for the 2 doublets to become a singlet? $\endgroup$ – John Apr 25 '17 at 15:29
  • $\begingroup$ @Mithoron According to a very rough estimation (chat.stackexchange.com/rooms/57692/p-methoxyphenol-nmr); around 600 MHz (and above) might offer a separation. $\endgroup$ – Buttonwood Apr 25 '17 at 16:22

In theory, this structure is expected to provide four different signals:

  • a broad singlet, due to the potential exchange of the phenolic OH
  • a singlet accounting for three protons of the methyl group
  • and signals for the remaining four aryl protons ... which may be tricky

The chemical shift of the aryl protons is influenced by the two, para standing towards each other, subsitutents; here: OH and methoxy. With their electronic properties, they influence the electron density of the aromatic ring, and eventually alter the location of the signals of the aryl H's. The centre of theses signals may be predicted by empirical increment rules which many textbooks about NMR include, like Silverstein, for example. Depending on the experimental conditions, accidentally they may simplify to a signal which only looks like a singlet:

enter image description here

(source, recorded in $\ce{CDCl3}$ at a Lamor frequency of 89.56 MHz)

Addition: The distance between the aryl H's and the phenol or the methoxy group (at least three bonds away) and their very similar electronic influence on the aromatic ring may render the aryl H's very similar. So it takes a spectrometer with higher measurement resolution (measurement points per ppm) and higher magnetic field to discern them. Rayment et al. published more recently in J. Org. Chem., 2012, 77, 7052–70602012 (doi 10.1021/jo301363h) data recorded again in $\ce{CDCl3}$, on an instrument with stronger magnetic field. To quote:

1H NMR (400 MHz, CDCl3) δH 3.78 (3H, s), 4.77 (1H, br s), 6.76−6.82 (4H, m)

enter image description here

Even here, some discern, yet likely the two doublets overlap a bit, as a prediction by ChemDoodle suggests:

enter image description here

(the other doublet corresponds to the second pair of aryl H's). Note the spread of the scale, too. While I imagine even more advanced NMR spectrometer may eventually resolve the doublets (600 MHz is not so uncommon, and a few of about 1 GHz are around, too), I doubt someone will invest the more precious measurement time on these to delve more into detail and publish the then very high resolution 1H NMR, instead of collecting 1H NMR data of proteins and other natural products.

Addendum and correction: Following the comment by @long, I reread the relevant pages in Silverstein again. Stating the for aryl H's would yield two doublets is wrong.

There are three types of coupling each of the aryl protons in p-methoxy phenol participates which are illustrated below:

enter image description here

(Source for the values of coupling constants: Silverstein, loc. cit., 6th edition, 1998, p. 212)

In blue is the proton which signal is of concern. In the top row, it is the left in ortho position towards the methoxy group. This proton couples with the close ortho standing proton (only three bonds away, hence $^3J$), typically with a coupling constant of $\pu{9 Hz}$, hence yielding according the $(n + 1)$ rule a doublet. Simultaneously the proton marked in blue couples with the meta standing one with a typical coupling constant of $\pu{3 Hz}$; hence the previous doublet would become a doublet of doublet (dd). Even farther distant, the para standing proton, hence coupling with an even lesser coupling constant which not every NMR spectrometer is capable to resolve -- but in principle yielding ddd, a doublet of doublet of doublet -- solely for this proton, up to $2 \times 2 \times 2 = \pu{8 signals}$ may become observable.

Following the theory further, the proton marked in blue of the bottom row is not identical to the one marked in the top row, and consequently, this proton may yield up to 8 observable signals, its own ddd. This already adds up to 16 lines, which however is not (yet) the end of nitpicking theory as -- according to Silverstein's textbook -- the aryl Hs on the left hand are not equivalent to the ones on the right and a total of 32 lines possibly to observe.

With the graphical example provided by the Silverstein book for p-chloro nitrobenzene

enter image description here

which will serve for a follow-up question by mine, to record the aryl H's in a signal in shape like a singlet was by chance.

| improve this answer | |
  • $\begingroup$ But why do two doublets appear as a singlet? The resolution isn't high enough? $\endgroup$ – John Apr 25 '17 at 14:50
  • $\begingroup$ Thank you. It makes some sense now, but could you read my comment on the question? Also, this may be unrelated, but why does the strength of the magnetic field affect the 'separation' of the doublets? $\endgroup$ – John Apr 25 '17 at 15:32
  • $\begingroup$ @Saad better here : chat.stackexchange.com/rooms/57692/p-methoxyphenol-nmr $\endgroup$ – Buttonwood Apr 25 '17 at 15:35
  • $\begingroup$ Suffice to say that the chemical shifts are coincidental within the natural linewidth. Sometimes that just happens. Under different conditions (different Bo, solvent, temperature, concentration etc), there may be more of a discernible difference. Note that this is an AA'BB' spin system, not an AB spin system. Also the splitting diagrams taken from Libretexts do not represent an AB spin system - they are for an AX system. Only when the chemical shift difference is comparable to the coupling constant (Δ δ>~3J) does the system begin to show second order effects. $\endgroup$ – long Apr 25 '17 at 22:03
  • 1
    $\begingroup$ Also, the expected splitting for a p-disubstituted benzene is not a pair of doublets as you suggest - AA'BB' are classically stereotypical and even at lower fields typically display 6 lines. You should not ignore these extra lines as they are important diagnostic features of the spin system that can help characterise the molecule. $\endgroup$ – long Apr 25 '17 at 22:06

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.