Note the use of different curved arrows to show single electron shifts compared with electron pair shifts.ģ-Pentanol shows three significant fragment ions. The first two fragmentation paths lead to even-electron ions, and the elimination (path #3) gives an odd-electron ion. These are summarized in the following diagram, where the green shaded box at the top displays examples of such "localized" molecular ions. By localizing the reactive moiety, certain fragmentation processes will be favored. After all, it is easier to remove (ionize) a non-bonding electron than one that is part of a covalent bond. This influence is thought to occur because of a "localization" of the radical cation component of the molecular ion on the heteroatom. The presence of a functional group, particularly one having a heteroatom Y with non-bonding valence electrons (Y = N, O, S, X etc.), can dramatically alter the fragmentation pattern of a compound. Alcohols, ethers and highly branched alkanes generally show the greatest tendency toward fragmentation.
Among simple organic compounds, the most stable molecular ions are those from aromatic rings, other conjugated pi-electron systems and cycloalkanes. Fortunately, most organic compounds give mass spectra that include a molecular ion, and those that do not often respond successfully to the use of milder ionization conditions. Without a molecular ion peak as a reference, the difficulty of interpreting a mass spectrum increases markedly. The nature of the fragments often provides a clue to the molecular structure, but if the molecular ion has a lifetime of less than a few microseconds it will not survive long enough to be observed. The fragmentation of molecular ions into an assortment of fragment ions is a mixed blessing. These get lost into the normal equilibrium which exists wherever you have water molecules - heavy or otherwise. You might wonder what happens to the positive ion in the first equation and the OD - in the second one. The negative ion formed is most likely to bump into a simple deuterium oxide molecule to regenerate the alcohol - except that now the -OH group has turned into an -OD group.ĭeuterium atoms don't produce peaks in the same region of an NMR spectrum as ordinary hydrogen atoms, and so the peak disappears. The fact that here we've got "heavy water" makes no difference to that. The hydrogen on the -OH group transfers to one of the lone pairs on the oxygen of the water molecule. All alcohols, such as ethanol, are very, very slightly acidic. The reason for the loss of the peak lies in the interaction between the deuterium oxide and the alcohol. If you measure an NMR spectrum for an alcohol like ethanol, and then add a few drops of deuterium oxide, D 2O, to the solution, allow it to settle and then re-measure the spectrum, the -OH peak disappears! By comparing the two spectra, you can tell immediately which peak was due to the -OH group. Since C-H bonds don't hydrogen bond very well, you don't see that phenomenon in an ether, and an O-H peak is very easy to distinguish in the IR spectrum. Instead of seeing one sharp peak, you see a whole lot of them all smeared out into one broad blob. Because protons are shared to varying extent with neighboring oxygens, the covalent O-H bonds in a sample of alcohol all vibrate at slightly different frequencies and show up at slightly different positions in the IR spectrum. The rounded shape of most O-H stretching modes occurs because of hydrogen bonding between different hydroxy groups. Peak shapes are sometimes very useful in recognizing what kind of bond is present. Source: SDBSWeb : (National Institute of Advanced Industrial Science and Technology of Japan, 14 July 2008) O-H peaks are usually very broad like this one.įigure IR8. there is a very large peak around 3400 cm -1.there is a strong C-O stretching mode near 1000 cm -1.there are sp 3 C-H stretching and CH 2 bending modes at 29 cm -1.If you look at an IR spectrum of 1-butanol, you will see: