[yolo!]

Hello everyone! I am not sure if I am thinking about this problem in the correct manner. For the first part of the question I believe the answer is C. My reasons is as follows: on both CH₃'s the hydrogen's are all equivalent. Thus, i count 1 H to the left and 3 H to the right of CH₂. So n+1=5 so my answer is pentet. For the last part of the question, i believe the Hydrogen on CH would be farthest because it has the least hydrogen's, but i am not sure if there should be more to my answer! Thank you chemical forums

[discodermolide]

A pentet is ok. The chemical shift depends on the magnetic environment around the hydrogen, that is, which atoms it is connected to and how they are connected.

[phth]

There is a sub forum for spectroscopy.  It is not because it has the least hydrogens; it is because it is the least electron dense.  Downfield->less electron density, upfield->more electron density

[yolo!]

Is my rational for a pentet okay for this given problem or is there an issue with it as well? And okay i understand. the hydrogen on CH is furthest downfield because it is the least electron dense, but when taking into account the electron density are you simply looking at the electron around the Hydrogen? i guess i am slightly confused by that still.....if you could explain a little more i would really appreciate it

[discodermolide]

The rational for the pentet is ok.
It is not the electron density around the hydrogen, remember it is attached to a carbon atom. The chemical shift depends on the magnetic environment around the carbon-hydrogen system.

[mjc123]

Pentet??? What happened to quintets? (Do we have to listen to Schubert's C-major String Pentet now?)

[discodermolide]

Was quoting the OP's question, to avoid confusion.

[yolo!]

I've been reading that their is a 3 bond distance rule when looking at NMR is this true? If so the hydrogens on CH2 would be split by the 3 hydrogens on its left and the one to its right on CH. this would obey the 3 bond distance rule, but give me the same answer as a pentet or quintet mjc123 haha!:)

[phth]

For H1 NMR yes, but the experiement can be set up to measure longer coupling

[OrgXemProf]

At the risk of making a mountain out of a molehill....

This thread has been posted for some months now, but it appears that the contributors to this post have not taken into account the fact that 2-butanol possesses a chiral center, which renders the methylene protons (CH3-CHaHb-CHOH-CH3) diastereotopic.

Thus, protons Ha and Hb are expected to display (often only slightly) different chemical shifts. Protons Ha and Hb are mutually coupled and are expected to display Jgem coupling in addition to the Jvic coupling that each proton displays due to spin-spin interactions with adjacent CH3 and CH(OH) protons.

However, it is possible that the difference between chemical shifts of Ha and Hb becomes vanishingly small under the conditions (solvent, concentration, temperature) at which the proton NMR spectrum of 2-butanol has been obtained. This situation represents an example of "accidental degeneracy", and the spectrum would simplify accordingly, thereby permitting the type of first order analysis that the question appears to presuppose.

Additionally, in order to arrive at Choice C ("pentet") as the answer to the question, one also must assume that the magnitude of vicinal coupling between degenerate CH2 protons and the adjacent CH3 group in 2-butanol turns out accidentally to be the same as that between CH2 and the adjacent CH(OH) proton.

A complete proton NMR spectrum of 2-butanol can be viewed by visiting the SDBS online website; reference: 1H NMR Spectrum No. SBDS-507 c/o http://sdbs.db.aist.go.jp/