What are the key steps to analyzing an NMR spectrum?

[Foghorn]

Homework Statement

Determine the chemical shift, integrated intensity, proton type, multiplicity, J (Hz), and # of protons on the adjacent atoms for this spectra given a 400 MHz NMR.

http://img411.imageshack.us/img411/7629/chem262lnmr1.png

Homework Equations

(n+1) rule?
(n+1) x (m+1) rule?

J = shift difference x NMR MHz

The Attempt at a Solution

Sup,

I'm having trouble analyzing an NMR spectrum of the aforementioned compound. Namely, I'm provided a spectrum with only 5 sets of peaks, but I'm sure it has 7 groups of heterotopic hydrogens.

First, I calculated the ratio of integrals to each other, which I hoped would give me the number of hydrogens represented by each region. 2:1:1:2:3. I have no idea if that's even a correct technique, but it helped me confirm several things. The first peak, upfield at around 2.277 ppm is the methyl group; I've determined that. From there, I'm slightly lost. The singlet at 4.679 ppm is probably the AR-O-CH2, since its integrated intensity corresponds to a ratio of 2 hydrogens.

After that, I'm confused as to why there are so many split peaks. The doublet could correspond to carbon #3's hydrogen adjacent to the methyl group, since the (n+1) rule is followed. Following that, no clue. For example, how could a quintet be produced here? Is it actually a quartet with the singlet from the hydroxyl group of the acid?

Finally, how could I measure the coupling constant (J) for these peaks? A doublet is just the difference between the two peaks (ppm) multiplied by the MHz of the machine (400 MHz in this case), the triplet is the difference of the inner peak and an outer peak multiplied by MHZ (do I provide the range? or just pick one of the differences?); does the same principle apply for the quintet, i.e. find the difference between the inner peak and an outer peak? And singlets do not have a coupling constant, correct?

As you can tell, I am thoroughly confused.

It seems to me that all of the hydrogens are unique. There are no equivalent hydrogens, except those bonded to the same carbon.

 

[TeethWhitener]

Mentors: the image of the NMR spectrum is not attached to this post. Therefore, it's essentially impossible to solve.