Mass Spectrometry and Infrared Spectroscopy Technical Notes

Estimating Carbon Count via M+1 Peak Ratios

• The abundance of the isotope $^{13}C$ is approximately $1.1\%$.
• If a compound contains one carbon atom, the ratio of the intensities of the $M+1$ peak to the $M^+$ peak is expected to be approximately $0.011$ or $1.1\%$.
• As the number of carbon atoms in a molecule increases, the probability of finding a $^{13}C$ isotope increases proportionally:
  • A molecule with two carbons will have an estimated $M+1$ to $M^+$ ratio of $2.2\%$.
  • A molecule with three carbons will have an estimated $M+1$ to $M^+$ ratio of $3.3\%$.
• Formula for approximating carbon count:
  • By knowing the ratio of the intensities of the $M^+$ and $M+1$ peaks, you can approximate the number of carbons by dividing the ratio by $1.1$.
• Case Study: Number 4 in the packet:
  • In this example, the molecular ion ($M^+$) has an intensity of $100$.
  • The intensity of the $M+1$ peak is $12.2$.
  • Dividing $12.2$ by $1.1$ allows for the estimation of the number of carbons in the hydrocarbon or oxygenated system.
• Note on terminology:
  • The terms "molecular ion," "M^+$," and "parent ion" are synonymous and used interchangeably.\n - In some cases, the parent ion is also the "base peak" (the peak with 100 intensity), though this is not always common.\n\n# Halogen Isotope Patterns in Mass Spectrometry\n\n- Mass spectrometry (MS) can identify the presence of specific halogens based on unique isotope abundance patterns.\n- **Chlorine (Cl):**\n - Chlorine has two primary isotopes: ^{35}Cl$and$^{37}Cl.\n - The relative abundance ratio of ^{35}Cl$to$^{37}Cl$is approximately$3:1.\n - While 1.1\%$of carbon is$^{13}C$, significantly more chlorine (roughly$32\%$) is the$^{37}Cl isotope.\n - Presence of chlorine results in an M+2$peak (two mass units away from the parent) rather than just an$M+1 peak.\n - In the mass spectrum (e.g., epichlorohydrin), the M+2$peak will be approximately one-third the height of the parent peak ($M).\n- **Bromine (Br):**\n - Bromine has two primary isotopes: ^{79}Br$and$^{81}Br.\n - These isotopes compete and are two mass units apart, resulting in an M+2 peak pattern.\n - The naturally occurring relative abundances of these two isotopes are nearly equal (1:1 ratio).\n - A brominated compound (e.g., as seen in an example with peaks at m/z = 164$and$m/z = 166) will show two peaks at the far right of the spectrum that are approximately equal in intensity.\n\n# Carbonyl Fragmentation and the McLafferty Rearrangement\n\n- Carbonyl compounds (compounds containing a carbon-oxygen double bond) undergo specific fragmentation patterns in mass spectrometry.\n- **Alpha Cleavage:**\n - This involves breaking a single bond at the alpha position relative to the carbonyl group. A piece of the molecule breaks off right at the carbonyl.\n- **McLafferty Rearrangement:**\n - This is a unique fragmenting pattern commonly observed with ketones that have at least four carbons.\n - It involves a cyclical transition state that results in specific bond-breaking and bond-forming events.\n - **Bonds Broken:** A carbon-carbon single bond is broken between the C2$and$C3 positions (alpha and beta carbons).\n - **Bonds Formed:**\n - A new double (pi) bond is formed between C1$and$C2.\n - The carbon-oxygen double bond becomes a single bond as it picks up a hydrogen atom (proton transfer from the gamma carbon).\n - **Detection:** In the mass spectrometer, this rearrangement results in the detection of a charged "vanillic alcohol" (an enol where the OH is directly attached to a hydrocarbon double bond).\n - **Keto-Enol Tautomerism:** Vanillic alcohols/enols exist in equilibrium with their keto forms. This chemistry is explored further in Chapter 22 but is relevant to identifying fragmented structures.\n\n# Fundamentals of Infrared (IR) Spectroscopy\n\n- IR spectroscopy measures how molecules respond to the infrared portion of the electromagnetic spectrum.\n- Unlike mass spectrometry, which looks at mass-to-charge ratios (m/z), IR spectroscopy looks at frequencies—specifically resonance frequencies.\n- Molecules absorb IR radiation at specific frequencies associated with their specific chemical bonds. This energy causes the bonds to vibrate in various ways:\n - Bending\n - Stretching\n - Rotating\n - These movements are colloquially described as the molecule "dancing."\n- These frequencies allow scientists to associate specific IR signals with particular functional groups and bond behaviors.\n- **Electromagnetic Spectrum Context:**\n - **IR Spectroscopy:** Infrared radiation.\n - **NMR Spectroscopy (Chapter 13):** Radio waves (nuclei responding to radiation).\n - **UV-Visible Spectroscopy (Chapter 14):** Ultraviolet and visible light.\n\n# IR Spectrum Interpretation for Hydrocarbons\n\n- **Alkanes:**\n - These spectra are relatively simple.\n - The most important feature is a group of peaks located just below 3000\,cm^{-1}.\n - Peaks below 1500\,cm^{-1} are in the "fingerprint region," which is typically excluded from basic analysis because it is too complex.\n- **Alkenes:**\n - **C=C$Stretch:** Shows up as a medium-intensity peak around$1600-1700\,cm^{-1}.\n - **sp^2\,C-H$Stretch:** If the alkene has a terminal hydrogen or hydrogens on the double bond, a small peak (described as a "finger") appears just above$3000\,cm^{-1}.\n- **Alkynes:**\n - **C\equiv C$Stretch:** This is a weak stretch found around$2200\,cm^{-1}. This region is unique because very few other functional groups absorb there, making alkynes easy to spot.\n - **sp\,C-H$Stretch:** High frequency, strong, and thin/skinny peak found significantly further to the left (higher frequency) than alkane or alkene$C-H stretches.\n\n# Physical Principles of IR: Hooke's Law and Atomic Mass\n\n- The behavior of a chemical bond in IR can be modeled using Hooke's Law, treating the bond as a harmonic oscillator (two weights attached by a spring).\n- **Bond Strength and Frequency:**\n - As bond order increases (single to double to triple), the bonds get shorter and stronger.\n - Stronger springs (bonds) require more energy to vibrate and thus have higher resonance frequencies.\n - **Trend:**\n - Single bonds: Fingerprint region (< 1500\,cm^{-1}).\n - Double bonds: Around 1600\,cm^{-1}.\n - Triple bonds: Around 2200\,cm^{-1}.\n- **Atomic Mass and Frequency:**\n - Lighter weights attached to a spring result in higher vibration frequencies. \n - Hydrogen is the lightest atom, so bonds involving hydrogen (e.g., C-H$,$O-H$,$N-H) are found in the highest frequency region of the IR spectrum.\n- **Summary of Regions (decreasing frequency):**\n - Bonds to Hydrogen (H): Highest frequency region.\n - Triple Bonds: Second highest frequency region.\n - Double Bonds: Third region.\n - Single Bonds/Bigger Atoms: Lowest frequency (fingerprint region).\n\n# Questions & Discussion\n\n- **Student Question:** Regarding the 100\%$intensity of the parent ion, is it always$100?\n- **Professor Response:** No, it is almost never 100$. Usually, the molecular ion/parent peak is not the base peak. Its intensity is often much less—perhaps under$10$on the y-axis, while the$M+1$would be under$1$. In homework and exams,$100 is often used as the intensity for the parent ion simply to make the calculations easier and clearer for the student.\n- **Student Question:** Can you always see an M+1 peak?\n- **Professor Response:** No. If the molecular ion intensity is extremely small, the M+1 peak may not be visible at all. \n- **Discussion on Problem Solving:** A combined spectroscopy problem (e.g., Question 5) might provide a parent ion mass (e.g., 58$$) and an IR spectrum to determine a compound's formula and structure consisting of carbon, hydrogen, and oxygen.