[rtkt01]

 Here's the question.

 Consider the H NMR spectrum of very pure 1-propanol, CH3-CH2-CH2-OH

                                                                            a     b     c     d 

Assuming the maximum multiplicity of signals and non-superposition of peaks, what is the number of peaks to be observed theoretically, in the order: a,b,c,d

 
  I got 3,6,3,1 by 2+1 for A , 3+2+1 for B, and 2+1 for C.
  It's not one of choices while this question's actually a multiple choice question.

  Did I misunderstand anything?

 
 

[Fridushka]

Yeah as i know its right, because as i know the rule applies for vicinal proton, i.e. protons on vicinal carbon, so there is nothin related with OH, and its right for B that u took from both sides..and sure A and C are right
For me everything seems to be oki 

Gd luck 

[sjb]

Whilst you're right that n+1 would make 6 for the protons at b, bear in mind that protons a and c are in different environments, so you'd theoretically see a triplet of quartets (or the other way around), so in fact 12 signals if you have "the maximum multiplicity of signals and non-superposition of peaks", and so on... How many for c now?

[rtkt01]

Whilst you're right that n+1 would make 6 for the protons at b, bear in mind that protons a and c are in different environments, so you'd theoretically see a triplet of quartets (or the other way around), so in fact 12 signals if you have "the maximum multiplicity of signals and non-superposition of peaks", and so on... How many for c now?

  Is it 3, 12 , 3 , 1 ?

[orgopete]

The hydrogens are chemically equivalent so I don't agree with the 12 peaks thing. I might imagine that you are leaving out the OH hydrogen. The signal for c should be 4 peaks and d should be 3 peaks. If the NMR is run in normal CDCl₃, meaning in the presence of acid, then a rapid exchange takes place so the OH loses the coupling with the neighboring peak. It is in that case that you would observe the peaks originally suggested by the original poster. If the CDCl₃ were carefully purified to remove residual HCl, then the additional coupling will be observable. This is generally be found in d₆-DMSO also.

[rtkt01]

The hydrogens are chemically equivalent so I don't agree with the 12 peaks thing. I might imagine that you are leaving out the OH hydrogen. The signal for c should be 4 peaks and d should be 3 peaks. If the NMR is run in normal CDCl₃, meaning in the presence of acid, then a rapid exchange takes place so the OH loses the coupling with the neighboring peak. It is in that case that you would observe the peaks originally suggested by the original poster. If the CDCl₃ were carefully purified to remove residual HCl, then the additional coupling will be observable. This is generally be found in d₆-DMSO also.

The answer must be 3, 6 , 4 , 3 then.  Thanks everyone for the inputs.

[sjb]

The hydrogens are chemically equivalent so I don't agree with the 12 peaks thing.

Are you sure they're equivalent? After all, replacing a hydrogen with deuterium on carbons "c" and "a" results in 1-deuteriopropan-1-ol and 3-deuteriopropan-1-ol respectively, which are not the same compound. Of course, there is also the option of longer range stuff, but I have ignored that as ⁴J coupling is very weak in non conjugated carbon chains, in my experience.