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VCE Chemistry > organic compound analysis > Flashcards

organic compound analysis Flashcards

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1
Q

m/z

A

mass-to-charge ratio

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2
Q

mass spectromery uses

A

calculate molecular weights of sample’s contents
identify unknown compounds and their physical structure
distinguish between isomers

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3
Q

stages in mass spectronmeter (instrument)

A

1) sample is vaporised
2) ionisation: bombarded w electrons -> framgents
3) deflection by magnetic field
4) detection

only positively chared particles are measured and compound must be (g)

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4
Q

how are compounds ionised in mass spectrometry (show an equation)

A

bombardment with electrons knock of an electron
eg X+ e- -> [X]+ + 2e-

(need to be able to show ionisation equation)

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5
Q

fragmentation

A

breaking of almost any bond in a molecular ion into a fragment and an alkyl group
eg M+ -> X+ (FRAGMENT) + R.

dot in R. represents unpaired electron. R. is not +vely charged

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6
Q

base peak
molecular ion peak

A

base= most abundent fragment
molecular/parent= highest Mr

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7
Q

IR spectroscopy function and principles

A

principle= use infrared radiation to vibrate covalent bonds making the bonds bend/stretch
function= info about covalent bonds, functional groups and structure/ type of molecule

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8
Q

possible uses of IR spec

A
  • identify org compounds
  • design new drug
  • analyse proteins
  • testing blood alcohol
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9
Q

relationship between bond strength and the absorption of IR radiaion

A

weaker bonds require a lower fq of radiation to vibrate bonds

bond that vibrates a higher fq adsorbs IR radiation with a higher wave#

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10
Q

relationship between mass and IR absorption

A

atoms with higher masses absorb lower frequency radiaion

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11
Q

how to tell IR with most amount of C-H bonds/ longest chain

A

more peaks in fingerprint region

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12
Q

NMR

A

Nuclear Magnetic Resonance spectroscopy => determine entire structure/ provides structural information
magnetic field interacts with nuclear spin of atoms
- short powerful radio pulse absorbed by nuclei -> flip to high nrg spin status
- spectrum produced when nuclei return to low nrg spin states releasing a pulse of nrg

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13
Q

fts of NMR

A

form of absorbance specroscopy
v fast
v expensive and hazardous bc of strong magnetic field
qualitative data

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14
Q

chemical shift

A

distance from standard peak (far right given by standard compound TMS= tetra methyl saline which has lowest known chemical shift=0)

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15
Q

why is the chemical shift different for different hydrogen environments

A

Because diffent hydrogen environments have different atoms surrounding its nucleus, this affects the amount of chemical shielding experienced thefore affecting the energy emitted from the nucleus spinning

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16
Q

area under peak

A

ratio of hydrogens in environment

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17
Q

splitting patern

A

number of hydrogens on neighbouring carbons (only H NMR)

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18
Q

number of peaks=

A

number of neighbouring H atoms (n) + 1
1 neighbouring = doublet ect

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19
Q

chromarography

A

technique to separate substances in a mixture
- stationary phase
- mobile phase

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20
Q

Sample _ _ to the stationary phase
and _ _ to the mobile phase

A

adsorbs and desorbs

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21
Q

which molecules are more attracted to the polar stationary phase

A

polar molecules

22
Q

stationary phase

A

solid onto which compounds of a sample adsorb

23
Q

mobile phase

A

phase that flows through stationary phase carrying components of a sample with it

24
Q

adsoption
desorpion

A

ad= components adhere to the stationary phase
de= component released from the stationary phase and dissolves into mobile phase

25
Q

what effects the rate substances move along the stationary phase

A

solubility in mobile phase (higher solubility -> shorter reention time)
strength of adsorption to stationary phase (stronger -> longer retention time)

26
Q

which moleculesa are more attracted to polar solvents

A

polar molecules

27
Q

how to determine which sample has most strongly adsorbed to the stationary phase

A

the sample which has moved the least amount of distance through the column due to strong attraction to the stationary phase

28
Q

how to determine the component which is least atracted to the stationary phase

q

A

the component that has moved the greatest distance through the column. That component would have adsorbed ino the stationary pahse and desorbed back into the mobile rate at a faster rate than the other components in the system

29
Q

explain separation of two different components in chromatography

A

-> separation occurs as components move with different rates of adsorpion to the stationaruy phase and desorption to the moile phase
-> one component move least distance bc more strongly adsorbs to stationary phase
-> other component moves most distance bc less strongly adsorb to stationary phase and more readily desorbs to mobile phase

30
Q

retention time

A

amount of time a sample stays in the COLUMN
longer rt if sample
- adsorbs more strongly to stationary phase
- are bigger tf carried more slowly by mobile phase

31
Q

position of peaks in HPLC chromotograms

A

represent RT

32
Q

area under chromatogram peak

A

link to concentration of component in sample, greater amount of component, greater AREA under peak

33
Q

fatcors affecting rt

A

length of ccolumn
temperature
surface area of stationary phase
identity/composition of stationary and mobile phase

34
Q

stardard solution

A

solution with accurately known concenration

35
Q

HPLC

high performance liquid chromatography

A

measure large compounds
stationary phase= fine particle waxy solids
mobile phase= pure solvent pumped under high pressure that helps move substance thru densly packed column
measures rt

q

36
Q

interpolation

A

values exist within range of measurements

reliable

37
Q

primary standards

A

used to make standard solution:
- pure form
- known chemical formula
- easy to store without deterioating/reacting w atmosphere
- high Mr
- inexpensive

eg anhydrous Na2CO3

38
Q

equivalence point

A

neither reactant is in excess and reactants have reacted in exact molar ratio

39
Q

why is it at equivalence point, pH is not always 7

A

due to acid-base properties of conjugate products that might be formed

40
Q

what can affect the titre required to achieve the endpoint

(for ans, assuming acid in conicle flask and base in burette)

A

dilution of base in burette (change in conc of base)
change in type of acid ie monoprotic vs diprotic (change in conc of H ions)

NOT DILUTION OF ACID OR USE OF WEAKER/STRONGER ACID

41
Q

finding mol of compound in vol flask from aliquot

A

[mol of aliquot] divided by [ml of aliquot] multiplied by [ml of flask]

42
Q

rf value calc

A

distance travelled from origin / distance travelled by solvent front from origin

43
Q

products of esterification reactions

A

WATER and ester

44
Q

extrapolation

A

values outside range of data tf predictions are unreliable

45
Q

how to explain whether selction of indicator is appropriate (structure)

A
  • state equivalence point (below/above/at 7)
  • “a pH indicator that has a pH range similar below/above/at 7 is appropriate)
  • state indicator range and whether it is inside/outside steepest section of pH curve tf not/appropriate
  • suggest beter indicator if not
46
Q

standard soln

A

solution w an accurately known concentration

47
Q

%w/w

A

g per 100g of mixture

48
Q

%w/v

A

g per 100mL of soln

49
Q

%v/v

A

mL per 100mL of soln

50
Q

%v/v

A

mL per 100mL of soln

51
Q

ppm

A

microgram per gram of mixture

52
Q

ppb

A

g in a billion grams

VCE Chemistry (8 decks)

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