Module 6 - chapter 29 P2

Question 1

How does mass spectrometry separate atoms?

Answer 1

Atoms are separated according to mass. Fragments of the molecular ion are formed in the mass spectrometer .

Question 2

What is the molecular ion in mass spectrometry?

Answer 2

the positive ion when the molecule has lost one electron (M+), this gives the molecular mass of the compound

Question 3

How is the molecular ion shown on the mass spectrum?

Answer 3

the clearest peak with the highest m/z value (mass-to-charge ratio) to the right on the spectrum

Question 4

What causes a small peak to the right of the molecular ion on a mass spectrum?

Answer 4

the presence of carbon-13 in a small percentage of the compound, causing the molecular mass to be greater, this peak is the M+1 peak

Question 5

what is fragmentation in mass spectrometry?

Answer 5

when a molecular ion breaks down into fragment ions and radicals, only the positive ions are detected by the mass spectrometer (as the radicals are uncharged so not deflected by the magnetic field) and so fragments can be pieced together to determine the structure of the molecular ion (parent molecule)

Question 6

what is the calculation for the number of carbon atoms in the molecule using M+1?

Answer 6

height of M+1 peak / height of M peak x100

Question 7

how does infrared spectroscopy work>

Answer 7

radiation within the 200-4000cm-1 range is passed through a sample and the energy is absorbed by the bonds at a specific frequency of radiation - this is the bond’s natural frequency. the absorption of IR causes the bonds to vibrate

Question 8

how are bond types identified from IR spectra?

Answer 8

the frequency of radiation absorbed by the bond is the frequency of the bond so the radiation absorbed on the spectra can be compared to the radiation frequencies absorbed by different types of bonds to identify them

Question 9

what is the IR frequency for C=O bond?

Answer 9

sharp peak in given range, close to 1700cm-1

Question 10

what indicates an O-H bond on IR spectra?

Answer 10

a broad peak in the fingerprint region, !!be aware that all molecules have a peak in the 2850-3100cm-1 region due to C-H so don’t confuse this

Question 11

what are uses for IR spectroscopy?

Answer 11

breathalysers, monitoring gases causing air pollution

Question 12

how can you identify a carboxylic acid group in IR spectroscopy?

Answer 12

peaks in C=O, O-H, and C-O regions

Question 13

what is elemental analysis?

Answer 13

the use of percentage composition data to determine the empirical formula of compounds

Question 14

how do you calculate Rf value?

Answer 14

Rf = distance moved by component/ distance moved by solvent front, ALL Rf VALUES SHOULD BE LESS THAN ONE

Question 15

how do you interpret gas chromatograms?

Answer 15

• compare retention times for known components to identify components present in the sample
• area under the peak (peak integrations) to determine the concentrations of components in the sample

Question 16

what is retention time in gas chromatography?

Answer 16

the time taken for each component to travel through the capillary column, shorter retention time = less soluble in stationary phase as more soluble components move slower through the column

Question 17

how do you determine the concentrations of components in a sample using peak integration?

Answer 17

external calibration

1) prepare standard solutions of known concentrations of the compound being tested
2) obtain gas chromatograms for each standard solution
3) plot a calibration curve of peak area against concentration
4) obtain a gas chromatogram of the compound being investigated under the same conditions
5) use the calibration curve to measure the concentration of the compound

Question 18

test for alkenes?

Answer 18

add bromine water, bromine water decolourised as added across double bond - orange to colourless

Question 19

test for haloalkanes?

Answer 19

add silver nitrate and ethanol and heat to 50C in water bath, precipitate of silver halide formed
chloro = white precipitate,
bromo =cream ppt,
iodo = yellow ppt

Question 20

how do the rates of hydrolysis change in haloalkanes?

Answer 20

chloro silver halide precipitate forms more slowly, bromo forms quicker than chloro but slower than iodo, iodo forms the quickest.

because Cl-Cl bond stronger so more energy required to break so takes longer

Question 21

test for primary/secondary alcohols and aldehydes?

Answer 21

acidified (with dilute sulfuric acid) potassium dichromate (VI) K2Cr2O7 and warm in a water bath, orange to green

dichromate (VI) ions reduced to chromium (III) ions as dichromate is oxidising agent

primary alcohol = oxidised to aldehyde then carboxylic acid

secondary alcohol = oxidised to aldehyde then ketone

tertiary alcohol can’t be oxidised (no hydrogens present on C with OH, 2H must be removed to form C=O) so no colour change

Question 22

test for carboxylic acids/

Answer 22

aqueous sodium carbonate (CO3^2-), effervescence,

neutralising with a carbonate produces CO2

Question 23

how does carbon-13 NMR work?

Answer 23

radiation is absorbed by the nuclei of the atoms and measured using a spectrometer. the absorptions of energy are detected and displayed on the spectra in reference to TMS

Question 24

what details does C-13 NMR give about molecules?

Answer 24

• the number of carbon-13 environments from the number of peaks
• the types of carbon environments from the carbon shifts

Question 25

what does the peak at 0ppm on a C-13 spectra show?

Answer 25

reference peak, peak for TMS

Question 26

how can carbons in the same environment have one with a slightly higher chemical shift?

Answer 26

carbons nearer to an electronegative atom/molecule (e.g. O) will have a slightly higher chemical shift due to deshielding

Question 27

what details does high resolution proton NMR spectroscopy give about compounds?

Answer 27

• number of proton environments (remember if no protons on C then no proton environment) - number of peaks - not including TMS peak! same enivronment = same peak (only one)
• number of protons on adjacent atoms - splitting pattern due to spin-spin coupling- n+1 rule so e.g. triplet has 2 on adjacent atom - only occurs if don’t have same environment
• number of protons in environment = area under peak (relative peak areas)
• types of proton environments present - chemical shift

Question 28

Use of TMS as the reference compound for chemical shifts

Answer 28

ppm = number of Hz more than 100mil Hz resonance frequency for TMS,

• lowest chemical shift value bc protons as shielded as possible (tetrahedral) and shielding decreases shift values
• soluble in most organic solvents
• chemically inert
• protons in identical environments

Question 29

how does shielding/deshielding effect chemical shift values in proton NMR?

Answer 29

shielding- orbiting electrons shield, exposes protons to a smaller magnetic field (by setting up their own magnetic field in opposition) so a lower frequency is required for resonance, so chemical shift lower

deshielding - electronegative atoms attract electrons away so proton experiences stronger magnetic field, frequency required for resonance is higher, higher chemical shift

Question 30

why are deuterated solvents used in proton NMR?

Answer 30

so the solvent doesn’t interfere with the spectra produced,
deuterium = isotope of hydrogen with +1 neutron so doesn’t spin/create magnetic field so doesn’t change spectra

deuterium replaces H in solvent

Question 31

example of deuterated solvent

Answer 31

CDCL3

Question 32

why are O-H and N-H groups difficult to identify on proton NMR spectra?

Answer 32

appear as singlets which could be interpreted as a single proton attached to a carbon with no other protons because the protons are labile (not permanently attached to other atom) so not bonded long enough for spin-spin coupling with adjacent protons to form a splitting pattern

Question 33

how can O-H and N-H groups be identified in proton NMR?

Answer 33

deuterate the sample with heavy water (D2O), this replaces all protons with D (proton exchange) so a peak isn’t formed for the OH/NH environment

Question 34

what is the base peak on mass spec?

Answer 34

peak with highest relative intensity, greatest abundance of one fragment