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Organic Chemistry - MCAT > Class Two > Flashcards

Class Two Flashcards

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1
Q

what does solubility depend on

A

polarity of the solute and the polarity of the solvent

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2
Q

polar molecules are soluble in..

A

polar solvents

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3
Q

how to extract organic amines

A

with acidic water solution

protonates the basic functional group (forms a positively charged ion) → cationic salt

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4
Q

how to extract carboxylic acids

A

extraction with a dilute weak base → anionic salt

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5
Q

how to extract phenols

A

sodium hydroxide → converts phenols into their anionic salts

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6
Q

what does a separatory funnel allow for

A

allows for you to extract a specific substance by separating the other ones

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7
Q

uses of chromatography

A

identification purposes

purification

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8
Q

thin layer chromatography

A

compounds are separated based on differing polarities

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9
Q

how does TLC work

A

a mobile liquid phases ascends a thin layer of absorbent that is coated onto a supporting material (glass plate)

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10
Q

purpose of absorbant in TLC

A

acts as a polar stationary phase for the sample to interact with

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11
Q

why does separation occur in TLC

A

because different compounds travel along the plate at different rates

polar components interact with absorbant → travels slower

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12
Q

Rf value - TLC

A

the distance traveled by an individual component divided by the distance travelled by the solvent front

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13
Q

column (flash) chromatography

A

similar concept to TLC → used for bulk compounds

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14
Q

ion exchange chromatography

A

separates materials with varying charge states

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15
Q

when is ion exchange chromatography frequently used

A

separation of protein mixtures

(proteins within a mixture exist in a variety of charge states)

pI vs pH matters as well (pI > pH is positively charged)

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16
Q

high performance liquid chromatography

A

speed and efficiency of separation is increased (higher pressures mobile → stationary)

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17
Q

size exclusion chromatography

A

used to separate bulk materials based on molecular size

uses polymer beads with pores - allows for small molecules to go through, but not large ones

large ones go through a different (faster) pathway

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18
Q

affinity chromatography

A

used to purify proteins or nucleic acids from biochemical mixtures

target molecule gets trapped on the stationary phase

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19
Q

different ways of using affinity chromatography

A

protein linked beads (target protein binds)

magnetic beads (target protein binds)

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20
Q

what is an affinity tag

A

molecular tags that act as an antibody (binds to target protein)

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21
Q

gas chromatography

A

separation of mixtures based on their different volatilities

less volatile components spend more time dissolved in the liquid stationary phase

the more volatile components leave the column first

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22
Q

branching (melting & boiling points)

A

branching reduces attractive forces (van Der Waals) between molecules → lower melting and boiling points

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23
Q

molecular weight (melting & boiling points)

A

the greater the molecular weight → more SA → more areas for van Der Waals interactions → higher MP and BP

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24
Q

small hydrocarbons tend to be ___ at room temp.

A

gases

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25
Q

intermediate hydrocarbons tend to be ___ at room temp.

A

liquids

26
Q

large hydrocarbons tend to be ___ at room temp.

A

solids

27
Q

hydrogen bonding ___ MP and BP

A

increases

28
Q

which has a higher effect on MP and BPs? intra or inter H bonding

A

intermolecular hydrogen bonding

29
Q

what is distillation

A

raising the temp of a liquid → vapour → condensed back into the liquid phase in another container

30
Q

when is a simple distillation performed

A

when trace impurities need to be removed from a relatively pure compound

31
Q

what is fractional distillation

A

used when the difference in BPs of the mixture components is not large

top of the column = pure vapour (condensed + collected in its liquid phase)

32
Q

what is mass spectrometry

A

technique that can determine the mass of compounds in a sample

33
Q

what is UV spectroscopy used for

A

transition metals (promotions of their electrons)

conjugated organic systems

34
Q

IR spectroscopy

A

range of 2.5-20 um (causes bonds in organic molecules to become vibrationally excited)

35
Q

the higher the wavenumber..

A

the higher the frequency and greater the energy

36
Q

C=O stretch

A

1700 cm-1

very strong & intense

37
Q

if the C=O stretch is not present..

A

you can eliminate compounds that contain a carbonyl group

e.g. aldehydes, ketones, carboxylic acids, esters, amides, anhydrides etc.

38
Q

C=C stretch

A

1650 cm-1

39
Q

triple bond stretch

A

C to C

or

C to N

2260-2100 cm-1

40
Q

O-H stretch

A

strong and broad

3600-3200 cm-1

41
Q

why is the O-H stretch broad

A

due to H bonding

42
Q

C-H stretch

A

3300-2850 cm-1

43
Q

aliphatic C-H bonds wavenumber

A

less than 3000 cm-1

44
Q

NMR spectroscopy

A

light from the radio frequency range of the electromagnetic spectrum is used to induce energy absorptions

45
Q

equivalent hydrogens have..

A

identical electronic environments - identical locations in the NMR spectrum

46
Q

when are hydrogens considered equivalent

A

when they can be interchanged by a free rotation or symmetry operation

47
Q

when does spin spin splitting occur

A

when nonequivalent hydrogens interact with eachother

48
Q

degree of splitting

A

n + 1

where n = number of nonequivalent neighbouring protons

49
Q

what is an integration - NMR

A

measurement of the area under each absorption peak (resonance) → indicatives the number of protons for each peak

50
Q

what is a chemical shift - NMR

A

indicates the location of resonance (peaks) in the NMR spectrum

51
Q

less deshielded protons will appear..

A

upfield (right)

52
Q

more deshielded protons will appear..

A

downfield (left)

53
Q

factors involved in proton deshielding

A

the electronegativity of the neighbouring atoms

hybridization

acidity and hydrogen bonding

54
Q

if an electronegative atom is close to a proton.. (chemical shift)

A

it decrease the electron density near the iron → deshielding therefore downfield shift

55
Q

what is the hybridization effect

A

a result of the varying bond characteristics of carbon atoms connected to hydrogens

56
Q

the greater the s-orbital character of a C-H bond.. (chemical shift)

A

the less electron density on the H → more deshielded so downfield shift

57
Q

protons attached to heteroatoms are..

A

very deshielded

heteroatoms = O or N

58
Q

extreme example of a very large downfield shift

A

acidic protons on a carboxylic acid

59
Q

fractional distillation is useful for separating..

A

diastereomers

60
Q

how to resolve enantiomers

A

convert the enantiomers to diasteromeric salts with a chiral resolving agent (acid)

separate the salts

revert the salts back to the original compounds

61
Q

length of chain of sp2 hybridized atoms & UV

A

as the chain increases, the wavelength of light absorbed increases (redshifts)

62
Q

the colour we see..

A

is the complementary colour of the wavelength that is absorbed

Organic Chemistry - MCAT (4 decks)

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