More information can be found in Chapter 12 in your textbook for CHEM 3750/ 3770 and on pages in your laboratory manual.

Transcription

1 CHEM 3780 rganic Chemistry II Infrared Spectroscopy and Mass Spectrometry Review More information can be found in Chapter 12 in your textbook for CHEM 3750/ 3770 and on pages in your laboratory manual. Identify the functional group(s) in the following molecules: H H Functional Group(s) Functional Group(s) H C N H3C C C CH 3 H Primary Use of IR spectroscopy is functional group identification. rganic compounds absorb energy in the infrared region (IR) of the electromagnetic spectrum. Bonds are not rigid links rather, one can compare them to tiny springs with varying degrees of stiffness that connect the atoms together. These vibrations include: stretching, bending, twisting, etc. Infrared energy can cause the atoms of organic compounds to vibrate faster and with increased amplitude about the covalent bonds that connect them. The most important factors determining where a chemical bond will absorb in the IR region are: o The bond order (single, double, or triple bond) i.e., the stiffness of the spring o The types of atoms joined by the bond (i.e., the mass (m 1, m 2 )

2 Step 1: Unsaturation / Elements of Unsaturation / Degrees of Unsaturation: Calculate the number of unsaturations (rings or multiple bonds) to limit the number of possible structures. The degree of unsaturation indicates whether or not the compound has one or more double bonds or rings, or even a triple bond. There are many different ways to calculate it: *Note the following relationship between carbon and hydrogen: alkanes - CnH2n+2; alkenes and cyclohexane - CnH2n; alkynes - CnH2n-2. Try these formulae by analyzing ethane, ethene, and ethyne. *Note the following rules when heteroatoms (atoms other than carbon and hydrogen) are present: organohalogens, containing C, H, X where X = F, Cl, Br, I; oxygen containing compounds, containing C, H, ; and nitrogen containing compounds, containing C, H, N. f course there are molecules that have all of the above, i.e. C, H,, X, N. Take the formula that you are given and Add the number of halogens to the number of hydrogens Ignore the number of oxygens Subtract the number of nitrogens from the number of hydrogens For example: C5H9N is equivalent to C5H8. Compare the latter with a fully saturated molecule. According to CnH2n+2 an alkane with 5 carbons should have 12 hydrogens. So, half of the hydrogen difference gives you the element(s)/degree(s) of unsaturation: (12-8) / 2 = 2. Thus, C5H9N has two degrees of unsaturation and could contain one of the following: a) a triple bond; b) two double bonds or c) one double bond and a ring. r take the formula and use the equation below Number of Unsaturations = {2(#C) #H - #Hal + #N}/2 For example: C5H9N would yield: {2x }/2 = 2 Step 2: In even rather simple, mono-functional molecules there are so many IR bands that it is not feasible to assign every band in an IR spectrum. Don t try to assign or interpret every peak in the spectrum. Concentrate on learning the major bands and recognizing their presence and absence in any given spectrum. It will be helpful to you to have the following band regions committed to memory if you have not already done so.

3 cm -1 N H stretch primary and secondary amines cm -1 H stretch alcohols, a broad, strong band cm -1 C H stretch alkenes cm -1 C H stretch alkanes cm -1 C= stretch ketones, aldehydes, esters cm -1 C=C stretch alkenes Some things to look for: Are any or all to the right of 3000? Are any or all to the left of 3000? alkyl groups (present in most organic molecules) a C=C bond or aromatic group in the molecule Look for a carbonyl in the region If there is such a band: Is an H band also present? ( cm -1 ) a carboxylic acid group Is a C band also present? ( cm -1 ) an ester Is an aldehydic C H band also present? ( cm -1 ) an aldehyde Is an N H band also present? ( cm -1 ) an amide Are none of the above present? a ketone Look for a broad H band in the region cm -1. If there is such a band: Is an H band present? an alcohol or phenol Look for a single or double sharp N H band in the region cm -1. If there is such a band: Are there two bands? a primary amine Is there only one band? a secondary amine ther structural features to check for: Are there C stretches? ( cm -1 ) Is there a C=C stretching band? ( cm -1 ) Are there aromatic stretches? ( , cm -1 ) Is there a C C band? ( cm -1 ) Are there -N 2 bands? ( , cm -1 ) an ether (or an ester if there is a carbonyl band too) an alkene an aromatic an alkyne a nitro compound MST IMPRTANT THING T REMEMBER... When analyzing an IR spectrum avoid the temptation to try to assign every peak

4 IR Spectroscopy C-C and C-H Bonds sp 3 C C weak, not useful sp 2 C=C cm -1 sp 2 C=C (aryl) cm -1 sp C C cm -1 sp 3 C H cm -1 sp 2 C H cm -1 sp C H 3300 cm -1 C(CH 3 ) cm -1 (two peaks) Alcohols and Amines H or N H cm -1 C or C N cm -1 Ethers C cm -1 strong Ketones (saturated) C= cm -1 Aldehydes C= cm -1 C() H cm -1 (weak) cm -1 (weak) Carboxylic acids C= cm -1 C() H cm -1 Acid Chlorides C= 1800 cm -1 Acid anhydrides C= cm -1 C cm -1 Esters C= cm -1 C() R cm -1 Amides C= cm -1 C()N H cm -1 Nitrile C N cm -1

5 Interpreting Mass Spectra A simple spectrum that of methanol is shown below. CH 3 H +. (the molecular ion) and fragment ions appear in this spectrum. Major peaks are shown in the table next to the spectrum. The x-axis of this bar graph is the increasing m/z ratio. The y-axis is the relative abundance of each ion, which is related to the number of times an ion of that m/z ratio strikes the detector. Assignment of relative abundance begins by assigning the most abundant ion a relative abundance of 100% (CH 2 H + in this spectrum). All other ions are shown as a percentage of that most abundant ion. For example, there is approximately 64% of the ion CH + compared with the ion CH 2 H + in this spectrum. The y-axis may also be shown as abundance (not relative). Relative abundance is a way to directly compare spectra produced at different times or using different instruments. Molecular ion (M.+ ): If the molecular ion appears, it will be the highest mass in the spectrum (except for isotope peaks discussed below). This peak will represent the molecular weight of the compound. Its appearance depends on the stability of the compound (think carbocation stability). Double bonds, cyclic structures and aromatic rings stabilize the molecular ion and increase the probability of its appearance. Reference Spectra: Mass spectral patterns are reproducible. The mass spectra of many compounds have been published and may be used to identify unknowns. Instrument computers generally contain spectral libraries which can be searched for matches. Fragmentation: General rules of fragmentation exist and are helpful to predict or interpret the fragmentation pattern produced by a compound. Functional groups and overall structure determine how some portions of molecules will resist fragmenting, while other portions will fragment easily. A detailed discussion of those rules is beyond the scope of this introduction, and further information may be found in your organic textbook or in mass spectrometry reference books. Isotopes: Isotopes occur in compounds analyzed by mass spectrometry in the same abundances that they occur in nature. A few of the isotopes commonly encountered in the analyses of organic compounds are below along with an example of how they can aid in peak identification. Methyl Bromide: An example of how isotopes can aid in peak identification. The ratio of peaks containing 79 Br and its isotope 81 Br (100/98) confirms the presence of bromine in the compound.

6 Steps to interpret a mass spectrum 1. Look for the molecular ion peak. This peak (if it appears) will be the highest mass peak in the spectrum, except for isotope peaks. Nominal MW (meaning=rounded off) will be an even number for compounds containing only C, H,, S, Si. Nominal MW will be an odd number if the compound also contains an odd number of N (1,3,...) 2. Try to calculate the molecular formula: The isotope peaks can be very useful, and are best explained with an example. Carbon 12 has an isotope, carbon 13. Their abundances are 12 C=100%, 13 C=1.1%. This means that for every 100 (12) C atoms there are 1.1 (13) C atoms. If a compound contains 6 carbons, then each atom has a 1.1% abundance of (13) C. Therefore, if the molecular ion peak is 100%, then the isotope peak (1 mass unit higher) would be 6x1.1%=6.6%. If the molecular ion peak is not 100% then you can calculate the relative abundance of the isotope peak to the ion peak. For example, if the molecular ion peak were 34% and the isotope peak 2.3%: (2.3/34)x100 = 6.8%. 6.8% is the relative abundance of the isotope peak to the ion peak. Next, divide the relative abundance by the isotope abundance: 6.8/1.1=6 carbons. Follow this order when looking for information provided by isotopes: (A simplified table of isotopes is provided in the introduction, more detailed tables can be found in chemistry texts.) Look for A+2 elements:, Si, S, Cl, Br Look for A+1 elements: C, N "A" elements: H, F, P, I Common Fragments and Ions for Mass Spectrometry Fragment lost (or ion present) m/z loss Fragment lost (or ion present) m/z loss Fragment lost (or ion present) m/z loss CH 3 M-15 NH 2 M-16 H M-17 CN M-26 CH 2 =CH 2 M-28 CH=NH, C M-28 CH 3 CH 2, CH M-29 N M-30 CH 3 M-31 CH 2 H M-31 CH 2 -CH=CH 2 M-41 CH 3 C= M-43 CH 3 CH 2 CH 2 M-43 C(=)NH 2 M-44 CH 2 CH 3 M-45 H(C=) M-45 CH 3 CH(H) M-45 N 2 M-46 CH 3 CH 2 C() M-57 CH 2 =C(H)NH 2 M-59 CH 2 =C(H) 2 M-60 C 5 H 5 M-65 Ph M-77 PhCH 2, tropilium ion M-91 Ph(C=) M-105 PhCH 2 CH 2 M-105 I M-127