CHE 231 Module 4 and 5

higher energy light

(UV-visible) causes electronic excitation

lower energy radiation

(infrared) causes vibrational excitation

common fragmentation behavior of alkyl halids, ethers, alcohols, and ketones (3)

1/ a bond between carbon and a more electronegative atom breaks heterolytically 2/ a bond between carbon and an atom of similar electronegativity breaks homolytically 3/ the bonds most likely to break are the weakest bonds and those that lead to formation of the most stable cation.

mass spectrometry (2)

1/ an analytical technique that allows us to measure the masses of atoms and molecules. 2/ an electron is ejected from the compound, thereby forming a molecular ion

nitrogen rule (4)

1/ an odd molecular ion generally indicates that a compound contains nitrogen 2/ effect called nitrogen rule: a compound with an odd molecular ion contains an odd number of N atoms 3/ a compound that contains an even number of N atoms gives an even molecular ion 4/ two "street" drugs that mimic the effect of heroin illustrate this

remainder for rounding and calculating the molecular formula (3)

1/ if remainder is 0.4 or less, then round down 2/ if remainder is 0.7 or more, then round up 3/ if remainder is between 0.4-0.7, then consider two different carbon counts for formula candidates and use other data to narrow down the correct formula

characteristics of an IR spectrum (4)

1/ in an ir spectrometer, light passes through a sample 2/ frequencies that match the vibrational frequencies are absorbed and the remaining light is transmitted to a detector 3/ an ir spectrum is a plot of the amount of transmitted light versus its wavenumber 4/ most bonds in organic molecules absorb in the region of 4000 cm to 400 cm

gas chromatography-mass spectrometry (3)

1/ mixtures of compounds can be analyzed using gas chromatography and mass spectrometry (GC-MS) at the same time 2/ the sample is injected into a gas chromatograph, and the various components of the mixture travel through the column at different rates, based on their boiling points 3/ the mass spectrometer records the mass spectrum for each component of the mixture

determining molecular formula with heteroatoms present

1/ subtract M+ number by amu 2/ consider CnHn+m, solve for n with M+/13 = n 3/ solve for m, M+ -(n*13) 4/ solve the base formula, if CnHn+m 5/ add heteroatom

bending vibrations (4)

1/ symmetric in-plane bend (Scissor) 2/ asymmetric in-plane bend (rock) 3/ symmetric out-of-plane bend (twist) 4/ asymmetric out-of-plane bend (wag)

types of stretching vibrations (2)

1/ symmetric stretch 2/ asymmetric stretch

absorption of electromagnetic radiation

1/ when electromagnetic radiation strikes a molecule, some wavelengths, but not all, are absorbed 2/ for absorption to occur, the energy of the photon must match the difference between two energy states in the molecule (ground state to excited state) 3/ the larger the energy difference between two states, the higher the energy of radiation needed for absorption

functional group region

4000-1400 cm-1

which m/z is more abundant for isopentane?

57

how mass spectrometry works (step 2/5)

an electron gun ionises molecules in the sample by knocking out electrons, producing positive ions. some molecules break into smaller ions and fragments

how mass spectrometry works (step 4/5)

as the positive ions pass through the magnetic field, they are deflected. lighter ions are deflected more than heavier ions, as are those with higher charges.

heterolytically

bonds between atoms of different electronegativities break heterolytically

homolytically

bonds between atoms of similar electronegativities break homolytically

carbon-bromine bond

breaks heterolytically

molecular ion peak

can be used to determine the mass of the molecule, but fragment ions can also provide information on chemical structure

fingerprint region

contains a complex set of absorptions, which are unique to each compound. though these are hard to interpret visually, by comparison with references they allow identification of specific compounds. 1400-600 cm-1

index of hydrogen deficiency

determines if you have rings or pi-bonds, in other words the degree of unsaturation IHD = [2C+2-H-X+N]/2 (X = halogens)

stretching and bending vibration

each stretching and bending vibration occurs at a characteristic wavenumber

molecular ion (M)

gives the molecular mass of the compound

carbon oxygen bond breaks

heterolytically

the carbon-chloride bonds breaks

heterolytically

high energy (frequency and wavelength)

high frequency, short wavelength

the carbon-carbon bond breaks

homolytically

e =

hv = hc/λ

correcting for bromine (determining MF)

if you see an M+2 peak, with a 1:1 ratio M:M+2 1 / use rule of thirteen 2/ shift rule of thirteen formula by shifting carbon mass to hydrogens so that it is greater than 7 H 3/ subtract C6H7 and add Br to get mass

correcting for chlorine (determining MF)

if you see an M+2 peak, with a 3:1 ratio M:M+2 (recall Cl = 35 amu = c2H11) 1/ use rule of thirteen 2/ shift rule of thirteen formula by shifting carbon mass to hydrogens so that it is greater than 11 H 3/ subtract C2H11 and add Cl to get mass

correcting for nitrogen (determining MF)

if your mass is odd 1 / use rule of thirteen 2/ shift rule of thirteen formula by shifting carbon mass to hydrogens so that it is greater than 2H 3/ subtract CH2 and add N to get mass

low energy (frequency and wavelength)

low frequency, long wavelength

infrared frequencies

make up a portion of the electromagnetic spectrum. if a range of infrared frequencies are shone through an organic compound, some of the frequencies are absorbed by the chemical bonds within the compound. different chemical bonds absorb different frequencies of infrared radiation. there are a number of characteristic absorptions which allow functional groups (the parts of a compound which give it its particular reactivity) to be identified. this graphic shows a number of these absorptions.

most important peak in a mass spectrum

molecular ion peak

isopentane

more likely to lose a methyl group because it forms a secondary carbocation

stability and abundance

more stable fragments means more abundance

stretching vibration

occurs along the line of the bond

a mass spectrometer

only positively charged species reach the recorder

electromagnetic radiation

radiant energy having dual properties of both waves and particles

infrared spectroscopy

technique allows chemists to identify characteristic groups of atoms (functional groups) present in molecules.

wavelength

the distance from one point on a wave to the same point on an adjacent wave

base peak

the fragment in greatest abundance

ethers

the fragmentation pattern of an ether is similar to that of an alkyl halide

m/z

the mass-to-charge ratio of the fragment because z=1

loss of a water from an alcohol

the molecular ion of an alcohol is often not seen because if it has a hydrogen on a y carbon, it readily loses water. in this fragmentation, two bonds are broken.

polarity and absorption

the more polar the bond, the more intense the absorption

fragmentation of the molecular ion

the more stable the fragments, the more abundant they will be. C-2--C-3 fragmentation forms more stable fragments.

wave number

the number of waves in 1 cm

frequency (v)

the number of waves passing per unit time. it is reported in cycles per second (S-1), which is also called hertz (hx)

pentane vs isopentane stability

the peak at m/z = 57 is more abundant for isopentane than for pentane because a secondary carbocation is more stable than a primary carbocation

how mass spectrometry works (step 3/5)

the positive ions generated are passed through an electric field which accelerates them into a magnetic field generated by an electromagnet.

how mass spectrometry works (step 5/5)

the positive ions hit a charged plate and accept electrons, creating a signal. the more ions that hit, the greater the signal. the output is a complex stick diagram.

how mass spectrometry works (step 1/5)

the sample is introduced to the mass spectrometer. only very small samples are required. a heater is often present to vaporize the sample.

determining molecular formula from MS

use the rule of thirteen; this is particularly effective for hydrocarbons. if heteroatoms are present you must take them into account; recall you can spot Cl and Br in the MS based on the M+2 rule, and N based on the nitrogen rule. so how do you incorporate them into formula?

the rule of thirteen

when a molecular mass, M+ is known, a base formula can be generated. since the mass of CnHn+m (where n=1, m=0) equals 13 amu, dividing M+ by 13 gives the max number of possible carbons (M+/13) = n