[xshadow]

Hi

I've a doubt about the signal I get from this molecule with ¹H NMR

CH₃-CH₂-CH₂-OH

I mean the  signal  of -CH₂ , where C is the C-2 carbon,the one in the middle of the carbon skeleton


Which signal will it give?

- a sextet?  (3+ 2 hydrong...but all those H need to  be equivalent )
- a triplet of doublet
- a doublet of triplet 


Because near that -CH₂ I have:
- CH₃₃
-₂CH₂OH

but these two groups of hydrogen I think they aren't equivalent...is it correct?
Because I've seen a H-NMR sprectrum and they consider that signal as a sestet (so all hydrong equivalent?!?!) ...but are they truly equivalent? 

[Babcock_Hall]

You are correct that the two groups you indicated are not equivalent.  However, they may couple with the -CH₂- group in question with a coupling constant of the same magnitude.  That would produce a simpler-looking pattern.  However, I don't believe that you are correctly applying the n+1 rule.  If the methyl group were the only group that coupled to the group in question, you would obtain a quartet, for example.

[xshadow]

You are correct that the two groups you indicated are not equivalent.  However, they may couple with the -CH₂- group in question with a coupling constant of the same magnitude.  That would produce a simpler-looking pattern.  However, I don't believe that you are correctly applying the n+1 rule.  If the methyl group were the only group that coupled to the group in question, you would obtain a quartet, for example.

For the last phrase IF I suppose that  the hydrogens of -CH₃and the ones of -CH₂ bonded to OH are equivalent (ie same j- copupling)....the central  carbon would  see 5 equivalent carbon ---> so a sextet?

Why is it wrong?
thanks


....and  what  type of  signal  I have in this NMR sprectum? 
I see  six signal around it ..but a sextet has a different intensity ratio  (the two central lines have the should have the same lenght)

[Babcock_Hall]

Suppose that a for a given -CH₂-, there is coupling to two nonequivalent H nuclei with two different coupling constants, we would observe a doublet of doublets.  The same reasoning can be applied to more nuclei.  Let us now assume that for carbon-2 of 1-propanol that there are two distinct coupling constants, to the hydrogen nuclei on C-1 and C-3.  One should ideally see a quartet of triplets or a triplet of quartets (not a triplet of doublets or a doublet of triplets).  However, it seems plausible that many of the lines will overlap, but in a way that depends on the relative sizes of the two coupling constants.  If the coupling constants were identical, then one should see a sextet.