1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

IR spectroscopy

  1. May 3, 2016 #1
    1. The problem statement, all variables and given/known data
    upload_2016-5-3_11-49-6.png
    I want to put these in order of increasing wavenumber for the absorptions of the carbonyl group in an nmr spectrum

    2. Relevant equations


    3. The attempt at a solution
    I know that the third one is the highest, the NH ketoamine is the lowest and last one is the second lowest but I'm unsure about the first. Which one has higher absorption?
     
  2. jcsd
  3. May 8, 2016 #2
    Thanks for the post! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?
     
  4. May 16, 2016 #3
    Recall that IR works by exciting bonding electrons (for lack of a better term) and this causes an increase in rotational and vibrational energy, as well as bond stretching and bending. Also recall that electronegative atoms tend to shift things downfield, the more electronegative something is, the farther downfield it is shifted. So looking at what you've posted we have (from left to right) cyclohexanone, 3-cyclohexen-1-one, δ-valerolacetone, δ-valerolactam, and 2-cyclohexen-1-one (I'm not 100% on this, but I think we can call 3-cyclohexen-1-one just 3-cyclohexenone since the carbonyl carbon is always the first carbon in these molecules, maybe someone with a stronger knowledge of IUPAC nomenclature can verify this). The links are links to the NIST website with the relevant IR Spectra.

    Right away you should see that the cyclic ether and cyclic amide (lactones and lactams respectively) will be shifted further than the usubstituted cyclohexanone. Typically carbonyl C=O stretches occur in the range of 1670 to 1820 or there about. Cyclohexanone has a peak at about 1710. Now valerolacetone has a broad peak about 1730 to about 1750. This is due to the electronegative oxygen atoms. Valerolactam has a forked peak at about 1710 and 1720. As you can see the more electronegative oxygen shifts this peak further than nitrogen does.

    These are the kinds of things you should be thinking about when you look at IR stretches, what kinds of substitutions are going on and how they will move that peak away from your unsubstituted molecule (cyclohexanone in this case). Also you should consider the types of motions available to your molecule. If the bond can rotate, vibrate, and stretch/bend then it will be able to dissipate that energy better than a bond that cannot rotate or bend and that will affect how far something is shifted. For instance cyclohexanone shows a peak about 1710, 2-hexanone has a peak around 1720, and hexanal has a peak around 1750.
     
    Last edited: May 16, 2016
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: IR spectroscopy
  1. Spectroscopy questions (Replies: 2)

  2. Help With Spectroscopy (Replies: 1)

Loading...