[Violagirl]

I have attached IR and NMR spectrums of isopentyl acetate for an experiment that I had to do this past week in synthesizing it for a Fischer Esterification reaction we had to do in my Organic Chem lab. I've been having a hard time with analyzing them. For the NMR, I have to determine the J figures, something I never had to do in lecture and determine where the different H's correspond to the different peaks. For IR, I just have to determine which bands belong to the different parts of isopentyl acetate. I know that the OH group is going to be above 3000 cm. I know that there is also an ester and a carbonyl group, the ester should be around 1300-1000 cm and the ester carbonyl should be around 1765-1735 cm. It looked like the OH group appeared at 3464 and the carbonyl group appeared at 1743. However, for the ester, it looks like there are bands/peaks between 1388-1057, could everything between this area indicate the prescence of the ester? Any help is greatly appreciated!! Also on the NMR, a peak appears at 7.2, could it be water?

[discodermolide]

There is no OH group in your compound. The NMR signal at 7.2 is chloroform. Where is the rest of the NMR you only show a part of it

[Violagirl]

It is all there though, I clicked the link below the picture of it and was able to see all of the peaks from when it was scanned.

[Violagirl]

Also thanks for letting me know about there being no OH, I just looked up the structure of it and found that to be true. 

[discodermolide]

It is all there though, I clicked the link below the picture of it and was able to see all of the peaks from when it was scanned.

Sorry It was my screen resolution it somehow changed:((

[Violagirl]

Oh no that's ok, thanks for looking at it!  The main uncertainity I have is calculating the J values from the NMR spectrum. 

[discodermolide]

Oh no that's ok, thanks for looking at it!  The main uncertainity I have is calculating the J values from the NMR spectrum. 

The scale corresponds to 200Hz, each division  then = 20Hz, i.e. 1-2 =20Hz, 2-3=20Hz, etc. Measure this distance in centimeters, lets say 10cm (to make it easy), so 10cm = 20Hz. Take the triplet, measure the distance between the middle of the two small peaks and do the maths. Lets say 5cm (to make it easy) then you have 10 cm = 20Hz, 5cm = 10Hz. That is the J value. Just work out your own values.
Sorry I can't draw it but I can't seem to upload figures to this site.

[arit]

The peaks in the spectrum have already been picked, so to figure out
the J-couplings you just need to do simple subtractions.

Let's say you want to know the J-coupling value for the triplet at around 4 ppm:
The peak locations are 4.121, 4.087 and 4.053. The coupling to one of the
protons causing the splitting is (4.121-4.087)*200 = 6.8 Hz. For the other
one (4.087-4.053)*200 = 6.8 Hz, also. Turns out that the triplet is due to
coupling to two equivalent protons (CH2 most likely).

[Violagirl]

Thank you both so much!! Appreciate the *delete me*