ochem week 6 part 2

Introduction to Mass Spectrometry

  • Mass spectrometry (MS) is a key analytical tool used to determine the composition of substances, including identifying unknown chemicals.

  • The primary goal of mass spectrometry is to help identify the molecular weight and formula of compounds based on their fragmentation patterns.

Fundamentals of Mass Spectrometry

  • The process begins with ionizing the molecule, causing it to break into fragments.

  • Fragments hit a detector, providing a spectrum displaying a distribution of these pieces.

  • Molecular ion: The intact molecule that reaches the detector without fragmentation is often the heaviest ion detected.

  • Base peak: The most abundant ion present, representing 100% relative abundance against which all other peaks are measured.

Key Concepts in Interpretation

  • Fragmentation Patterns: These are unique to different molecules and help infer the structure.

    • For example, a methyl group (CH3) has a typical mass contribution of 15 amu.

  • m/z Ratio: This refers to the mass-to-charge ratio (mass of ion divided by its charge), crucial for identifying fragments.

Analyzing Mass Spectra

  • The largest peak on the x-axis of a spectrum corresponds to the molecular weight. For instance, an m/z value of 44 indicates the molecular weight of the entire molecule under analysis.

  • When analyzing a spectrum, one often starts from the highest m/z value (molecular ion) and works down to identify potential fragments.

Breaking Down Molecules

  • Molecules can fragment in predictable ways, often breaking off smaller alkyl groups (e.g., CH3, CH2) leading to characteristic mass differences in the spectrum.

  • Homologues: Identifying peak patterns can reveal types of substituents within the molecule, such as alkyl chains.

Understanding Peaks and Molecular Structure

  • The mass spectrum does not show all possible masses; only those fragments formed during ionization.

    • Some numbers may not appear due to the structure of the molecule itself.

  • Important numbers include:

    • Base peak: Indicates the most frequently seen fragment in relative abundance.

    • Molecular Ion Peak: The highest m/z value, representing the preserved whole molecule.

Patterns of Substitution

  • The presence of certain groups can dictate fragmentation behavior:

    • A methyl group contributing 15 amu affects how other fragments are interpreted.

    • Larger groups or additional carbons (like when adding to a base peak) indicate the complexity of the molecule.

The Rule of 13 for Molecular Formula Calculation

  • To ascertain the molecular formula, divide the molecular weight by 13. This determines the number of core carbon atoms, helping to structure the formula:

    • From example: If molecular ion mass = 138, then:

      • 138 / 13 = 10.61 -> approximately 10 carbons and a remainder indicating additional hydrogens.

  • Molecules containing heteroatoms (O, N, etc.) require modifications via substitutions when calculating the final molecular formula.

Handling Heteroatoms and Degrees of Unsaturation

  • Heteroatoms shift the regular patterns:

    • Oxygen: Does not subtract directly from the hydrogen count; it maintains balance in formulas.

    • Nitrogens: Contribute extra hydrogen equivalents, thus impacting the degree of saturation calculations.

  • Degrees of Unsaturation refers to the missing hydrogens from a theoretical fully saturated compound (C_nH_(2n+2)). Each pi bond or ring introduces a degree of saturation.

Summary of Breakdown in Analysis Methods

  • Utilize the molecular ion and base peak to gauge overall structure and presence of functional groups.

  • Employing the Rule of 13 and calculating degrees of unsaturation enhances understanding of molecular configuration.

  • The interplay of carbon, nitrogen, and other atom contributions results in unique spectral appearances that require careful analysis.

Conclusion

ochem week 6 part 2

make the up the direction where most of the h's are located as down or dashes

draw out the molecule to have it make more sense in the diretion it's heading

if down is on right, make directionof drawn idea go right

whichever you lay down with an equal amount of hydrogens on either side doesn't necessarily matter, since the isgnificance is turning the flat 2d image into a 3d olane

back to chapter 5

mass spectrometry- provides info about main info for molecule besides stereoisomers, fome times functional groups and some info for constitutional isomers

base peak is most abundant thing that makes it to end of the detector. always reaches (or attempts to reach) 100%

spectra can give hints as to a molecule we may have

with radical ions, it may split apart with one with a positive charge and the other with the free radical

reading spectra tells of molecular formula, degree of unsaturation, base peak and mi peak, presence of cl and br, and gragmentation pattern (skeletal structure and presence of carbonyls, alcohols, and amines)

TOFIND: molecular formula from specta

step1: MW (m/z)/13= how many carbons are present

step 2: the remainder + the number of carbons= how many hydrogens are present

ex- mw is 128

128/13= 9 with a remainder of 11

formula is c9H9+11 so C9H20

(you can also use the decimal form, then times the decimal by 13 to get the molecular formula)

only works if it's hydrocarbons

TO FIND: molecular formula with another atom (alocohol)

1-2: same steps as hydrocarbon

step 3: replace 1 carbon and X hydrogens correlating to its atomic mass

(C9H20-> O weighs 16 amu, as does CH4, so replace 1 carbon and 4 hydrogens with 1 O)

(IR may help determine the presence of another molecule

OR subtract its molecularweight of other molecule before beginning steps 1 and 2

with multiple of a different compound (COOH or carboxylic acid) you can just substract twice to count for 2 o's

degree of unsaturation

nitrogen often brings another hydrogen, changing the molecular formula AND the nitrogen adds another degree of unsaturation by 1 more hydroen

formula changes to CnH2n+2+#N

pi bond= 1 degree of unsaturation, ring is 1 degree of unsaturation

JUST COUNT PI AND RINGS FOR DEGREES OF UNSATURATION