Brainscape's Knowledge GenomeTM

Browse over 1 million classes created by top students, professors, publishers, and experts.


See full index

Chemistry for Biomedicine > Terms > Flashcards

Terms Flashcards

You may prefer our related Brainscape-certified flashcards:
Periodic Table flashcards Chemistry 101 flashcards AP® Chemistry flashcards Organic Chemistry 101 flashcards
1
Q

Conformation Isomers (Conformers)

A

Arise from free rotation about a σ-bond. Vary in stability due to differences in steric interactions. Staggered conformers are more stable than eclipsed conformers, while anti arrangements are more stable than gauche.

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
2
Q

Cycloalkanes

A

Have a cyclic structure and vary in stability and strain energies based on sized and bond angles. Rings prevent free rotation about single bonds. Therefore substituent isomers that do not interconvert exist

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
3
Q

What is the most stable conformation of cycloalkanes?

A

Chair conformation, which results in axial and equatorial positions (ring-flips can exchange these). Equatorial substituents result in more stable conformations due to reduces steric interactions.

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
4
Q

Enantiomers

A

Molecules that are non superimposable mirror images - chiral.

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
5
Q

Racemic Mixture

A

1:1 mixture of enantiomers

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
6
Q

How can enantiomers be separated?

A

By converting them into diastereomers (different 3D shapes), separating them, then retransforming them into enantiomers

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
7
Q

Stereochemistry of Alkenes

A

Stereoisomers due to lack of rotation about the C=C bond.

- diastereomers - non mirror image stereoisomers, named using E/Z nomenclature

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
8
Q

Addition Reactions

A

Overall replacement of a weak π-bonds with a strong σ-bond

  • hydrogenation, halogenation, hydrohalogenation, hydration
  • σ-bond is a nucleophile, supplies e-, produces a carbocation intermediate
  • rate increases with acidity
How well did you know this?
1
Not at all
2
3
4
5
Perfectly
9
Q

Stability of Carbocations

A

3° > 2° > 1°

stability increases with variety of substituents

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
10
Q

Markovnikov’s Rule

A

Predicts the regiochemistry of HX addition to unsymmetrically substituted alkenes
H will preferentially end up on the side that has more Hs, while X will end up on the side with more substituents.

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
11
Q

Alkene addition reactions with no nucleophile

A

Can result in polymers

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
12
Q

Aromaticity requirements

A

Cyclic
conjugated
planar
4n + 2 π-electrons (n=integer)

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
13
Q

Nomenclature for disubstituted benznes

A

Ortho (1,2-)
Meta (1,3-)
Para (1,4-)

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
14
Q

Nucleophilic Substitution Reaction

A
  • replacement of a leaving group by a nucleophile
  • nucleophile attacks carbon of an organic molecule and displaces a leaving group, which carries away its bonding e-
  • Two mechanisms
How well did you know this?
1
Not at all
2
3
4
5
Perfectly
15
Q

First Order Nucleophilic Substitution (SN1)

A 
Two distinct steps:
- ionisation
- nucleophilic attack
Reaction proceeds via an intermediate and rate depends only on substrate concentration. Results in a racemic mixture.
Ease parallels carbocation stability.
How well did you know this?
1
Not at all
2
3
4
5
Perfectly
16
Q

Second Order Nucleophilic Substitution (SN2)

A

Making/Breaking of bonds is concerted. There is no intermediate and rate depends on both substrate and nucleophile concentrations. Able to interconvert enantiomers.
Ease is the reverse of carbocation stability

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
17
Q

Elimination Reactions

A

The loss of an atom or group of atoms resulting in the formation of a multiple bond (opposite of addition reactions). Require reagents that are strong bases and relatively weak leaving groups. Two distinct mechanisms.

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
18
Q

First Order Elimination (E1)

A

Two distinct steps:
- ionisation
- deprotonation
Proceeds via an intermediate with a rate dependent only on substrate concentration

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
19
Q

Second Order Elimination (E2)

A

One step with simultaneous making/breaking of bonds. Rate dependent on concentration of substrate and base.

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
20
Q

Regiochemistry of Elimination

A

Often more than one product possible (placement of double bond and rotation around it). The transition state for E2 reactions dictates the stereochemical outcome. Newman projections are needed.

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
21
Q

Zaitsev’s Rule

A

In the elimination of H-X from an alkyl halide, the more highly substituted alkene product predominates. (less sterically crowded)

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
22
Q

Carbonyl Functional Group

A

Carbonyl bond strongly polarised - electrophilic at the carbon and can undergo nucleophilic addition reactions.
Most carbonyl reactions involve the breaking of π-bonds

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
23
Q

Formation of Aldehydes and Ketones

A

Can be formed via oxidation of primary and secondary alcohols.
1° alcohols oxidise to aldehydes and then to carboxylic acids
2° alcohols oxidise to ketones
3° alcohols are not oxidised

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
24
Q

Jones Reagent

A

Oxidising agent (Cr)3/H2SO4)

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
25
Q

Reduction of Carbonyls

A

Addition of hydride leads to an alkoxide, protonation of the alkoxide affords an alcohol

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
26
Q

Common reducing agents

A

(H- sources)
LiAlH4 - Lithium aluminium hydride
NABH4 - Sodium borhydride

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
27
Q

Reduction of Carboxylic Acids

A

Lithium aluminium hydride will reduce carboxylic acids directly to alcohols. It is not possible to stop at the intermediate aldehyde as these are more reactive than the acid.
Sodium borohydride does not reduce carboxylic acids

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
28
Q

Nucleophilic Addition Reactions with CArbonyls

A

Addition of water affords hydrates - creates an acid base equilibrium
Addition of alcohols (catalysed by acids) affords hemiacetals; SN2 substitution then affords acetals

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
29
Q

Carboxyl Group

A

Highly polarised and undergoes nucleophilic substitution through an addition/elimination mechanism

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
30
Q

Nucleophilic Substitution of Esters

A

Nucleophilic substitution of esters by hydride results in reduction to primary alcohols

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
31
Q

Preparation of Acid Chlorides

A

Acid Chlorides can be prepared from carboxylic acids and thionyl chloride (SOCl2).

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
32
Q

Preparation of Esters

A

Acid Chlorides react readily with alcohols to yield esters. Nucleophilic Substitution via addition/elimination mechanism.

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
33
Q

Preparation of Amides

A

Amides can be prepared via nucleophilic substitution (addition/elimination mechanism) of acid chlorides with primary or secondary amines. Tertiary amines are unreactive.

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
34
Q

Reaction of Anhydrides

A

Carboxylic anhydrides have similar reactivity to acid chlorides.
- can produce esters via substitution with alcohols
- can produce amides by substitution with amines.
One equivalent of carboxylic acid is released.

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
35
Q

Carbohydrates

A

Polyhydroxy aldehydes and ketones of general formula Cn(H2O)m
Energy storage material that also plays important roles in the social behaviour of cells and organisms.

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
36
Q

Stereochemistry of Carbohydrates

A 
Indicated by (D)- or (L)- descriptors. 
(L)- indicates highest numbered stereogonic sentre has hydroxyl on the left.
How well did you know this?
1
Not at all
2
3
4
5
Perfectly
37
Q

Nomenclature of Carbohydrates

A

Named according to:

  • whether they are an aldehyde or a ketone
  • number of carbons
  • suffix ‘ose’
  • D or L descriptor
How well did you know this?
1
Not at all
2
3
4
5
Perfectly
38
Q

Chemistry of Sugars

A

Dominated by intramolecular interaction of the alcohol and carbonyl functional groups

  • aldehyde –> hemiacetal of aldehyde –> acetal of an aldehyde
  • ketone –> hemiacetal of ketone –> acetal of ketone
How well did you know this?
1
Not at all
2
3
4
5
Perfectly
39
Q

Cyclisation of Aldoses

A

Aldoses are able to form cyclic hemiacetals. Hydroxide on second last carbon binds to carbonyl - addition of alcohol to carbonyl group).
This produces a new stereocentre called the anomeric centre.

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
40
Q

Anomeric Isomers

A

Stereoisomer found in carbohydrate chemistry. Able to interconvert - mutarotation.

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
41
Q

Cyclisation of Ketoses

A

Ketoses are able to form cyclic hemiacetals via addition of last hydroxyl group to oxygen.

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
42
Q

Pyranose and Furanose

A

Used to denote 6 and 5 membered rings.

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
43
Q

α- and β-anomers

A

For sugars of the D-series, in a standard Haworth projection, α-anomers have the anomeric hydroxyl pointing down while β-anomers have it pointing up.
This is the opposite for L-sugars

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
44
Q

Reduction of Carbohydrates

A

The carbonyl group is readily reduced as the ring forms are in equilibrium with the open chain forms

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
45
Q

Tautomers

A

Two molecules with the same molecular formula but different connectivity - constitutional isomers, in other words - which can interconvert in a rapid equilibrium

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
46
Q

Oxidation of Carbohydrates

A

Can be used as a diagnostic method (reducing/non-reducing sugars).
Both aldoses and ketoses react because of tautomerism, if a sugar cannot mutarotate it is anon-reducing sugar (glycocsides)

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
47
Q

Glycosides

A

A molecule (acetal) in which a sugar is bound to another functional group via a glycosidic bond. Produced from hemiacetal monosaccharides upon treatment with an alcohol and catalytic acid. Do not mutarotate and are not reducing sugars.

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
48
Q

Maltose

A

Formed from joining the anomeric centre of one D-glucose and the 4-position of another (α-D-glucapyranosyl-1,4-β-glucapyranose) A reducing Sugar.

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
49
Q

Sucrose

A

Formed from joining the anomeric centres of D-glucose and D-fructose. As both anomeric centres are acetals, sucrose is a non reducing sugar. (α-D-glucapyranosyl-1,1-β-fructafuranose)

50
Q

Cellulose

A

Polymer of D-glucose

β-linkage between 1 and 4 positions

51
Q

Starch

A

Comprised of amylose and amylopectin

α-linkage

52
Q

Amylose

A

Exists in a helical conformation

Core of helix can form characteristic inclusion complexes

53
Q

Amylopectin

A

linear chains of 24-30 α-1,4-linked D-glucopyranose interupted by α-1,6-branch points

54
Q

Nucleosides

A

N-glycosides formed from a sugar and a nucleobase

55
Q

Nucleotides

A

Formed from nucleosides by phosphorylation
Oligo and polynucleotides form the structure of DNA
Base pairing allows the double helix to form - need donor to form hydrogen bond

56
Q

Keto-enol tautomerism

A

Each of the bases in DNA can appear as different tautomers, which are in equilibrium.
The keto form of each base normally present in DNA - critical for correctly matching the base pairs.
The imino and enol forms are rare.
The wrong tautomer of one base results in a mispairing

57
Q

Amino Acids

A

Contain a basic amino group an acidic carboxyl group
20 naturally occurring α-amino acids, which can be classified according to their side chain
All are chiral except glycine.
Take (L)- and (D)- descriptors (proteins derived exclusively from L)

58
Q

Amide Bonds

A

Form between -NH2 and of one amino acid and -CO2H of the next
Can be broken to establish amino acid sequence - acid hydrolysis (can be non-selectively or selectively broken, involves nucleophilic substitution via addition/elimination)

59
Q

Synthesis of Proteins

A

Peptides can be synthesized chemically using nucleophilic substitution reactions of the carboxyl group.
Requires protecting the carboxylic acid of one amino acid and the amine of the other (and sometimes the R group)

60
Q

Steps to make a peptide bond

A
  1. Amino group of one amino acid 1 is protected by the BOC derivative (via substitution)
  2. Carboxyl end of amino acid 2 is protected as the methyl ester
  3. The two protected amino acids are coupled using DDC
  4. The BOC protecting group is removed by acid treatment
  5. The methyl ester is removed by basic hydrolyisis
61
Q

The Nernst Equation

A

Allows for calculation of cell potential under non-standard conditions

62
Q

Concentration Cells

A

At unequal concentrations, current will flow until the concentration in each beaker is equal. At equilibrium, cell potential = 0V

63
Q

The Stability Field of Water

A

Species with a potential more positive than the top line will liberate Oxygen from water - oxidation
Species with a potential more negative than the bottom line will liberate hydrogen from water - reduction.

64
Q

Transition Metal Cation

A

Have strong attractions in water leading to metal water interactions approaching strength of covalent bonds - thus results in acid-base interactions.
A strong transition metal acid can bind up to 6OH2 ligands - an aqua complex, discrete and soluble

65
Q

Bio availability

A

Requires concentration of at least 10^-6M

66
Q

Availability of Fe

A

Ferric Ion soluble and available under acidic conditions, but not at the surface of natural fresh water bodies.
Iron enters the food chain in deep waters via the metabolisms of anaerobic bacteria.

67
Q

Ligands

A

Natural molecules or anions that have a lone pair of electrons which is donated towards a metal ion - donor atoms

68
Q

Coordination Number

A

The number of donor atoms bound to the metal ion. This determines geometric shape.

69
Q

Charge on Metal Complexes

A
  • If ligands are all neutral, the charge on the complex will equal the charge on the metal ion - the complex is cationic.
  • If the total charge on the ligands equals the charge on the metal ion then the complex then the complex is neutral.
  • if the total charge on the ligands exceeds the charge on the metal ion then the complex is anionic
70
Q

Ligand name for water

A

aqua

71
Q

Ligand name for ammonia

A

Ammine

72
Q

Ligand name for carbon monoxide

A

carbonyl

73
Q

Ligand name for Nitric Oxide

A

Nitrosyl

74
Q

Ligand name for hydroxide

A

Hydroxido

75
Q

Ligand name for oxide

A

oxido

76
Q

Ligand name for Cyanide

A

cyanido

77
Q

Ligand name for carbonate

A

Carbonato

78
Q

Coordination Isomerism

A

Compound has the same formula but the formula of the complex ion is different
(structural)

79
Q

Linkage Isomerism

A

The complex has the same formula but a ligand binds through different donor atoms
(Structural)

80
Q

Complex Ion

A

A metal plus its ligands

81
Q

Bidentate

A

A ligand that can form two bonds to a metal ion (e.g. oxalato

82
Q

Chelate

A

A ligand that can bind to a metal ion through more than one donor atom to form a ring (e.g. catechol, ethylendiamine)

83
Q

Geometric Isomerism

A

Occurs when atoms or groups of atoms can assume different positions in a complex (cis/trans, fac/mer)
(stereoisomerism)

84
Q

Optical Isomers

A

Non-superimposible mirror images (chiral)

85
Q

Enterobactin

A

A compound produced by aerobic bacteria that can extract iron from rust.
A cyclic ester with amide links to three catechol-type ligands

86
Q

Catechol

A

A dianionic chelating ligand. Upon deprotonation can chelate metal ions.

87
Q

Siderophores

A

Chelating molecules secreted by microorganisms that are able to bind iron ions very strongly.

88
Q

Transferrin

A

An iron transport protein in the blood

  • two identical subunits
  • Fe ligands: 2 tyrosine, 1 histidine and 1 aspartic acid
  • coordination environment completed by an anion (e.g. carbonate)
  • drop in pH causes release of Fe2+
  • acts as an iron scavanger
89
Q

Ferritin

A

An iron storage protein

  • rust covered with protein
  • 24 subunits
  • Fe ions enter through channels
90
Q

Haem Group

A

An Fe2+ at the centre of a porphyrin ligand. Binds oxygen in oxygen carrying proteins.

91
Q

Haemogloin

A

An oxygen binding protein α2,β2 with 4 haem units.
Exhibits cooperativity - binding of oxygen to one haem group causes conformational change that increases affinity of other haem groups (histidine is moved up)

92
Q

The Bohr Effect

A

Low pH decreases haemoglobin’s affinity for oxygen

93
Q

Zinc in Biological systems

A

Catalyses acid-base reactions

Zinc enzymes can act as OH- reagents - allow it to act as a base in biological systems

94
Q

Carbonic Anhydrase

A

Catalyses formation of bicarbonate from carbon dioxide and water, and the reverse reaction.

  • active site: zinc bound to three histidines
  • binds water first, loses proton, nucleophilic O attacks C of carbon dioxide, bicarbonate formed
  • bicarbonate can displace the water ligand, but not the coordinated OH-
95
Q

Internal Energy (U)

A

The sum of the energies for all of the individual particles in a sample of matter.

96
Q

Enthalpy (H)

A

A function related to the heat adsorbed or evolved by a chemical system.

97
Q

Entropy (S)

A

A measure of the number of ways energy is distributed throughout a chemical system.

98
Q

Gibbs Energy (G)

A

The energy from a reaction that is available to do work.

99
Q

Heat (q)

A

A transfer of thermal energy due to a temperature difference.

100
Q

First Law of Thermodynamics

A

The only way we can change the total energy of a system is through the transfer of heat or work to the surroundings.
Energy cannot be created or destroyed

101
Q

Reaction Enthalpy

A

The sum of the enthalpies of any sequence of reactions into which the overall reaction may be divided.

102
Q

Standard Enthalpy of Formation

A

The enthalpy change when 1mol of substance is formed at 10^5Pa and specified T from its elements in their standard states.

103
Q

Second Law of Thermodynamics

A

The entropy of the universe increases in the course of a spontaneous change. Thus spontaneous processes tend to disperse energy.

104
Q

Boltzmann Equation

A

S = klnW (W - microstates into which energy can be dispersed)

105
Q

Spontaneous Process

A

A process is spontaneous if the change in Gibbs energy is less than 0 at constant p and T

106
Q

Chemical Equilibrium

A

At equilibrium the rates of the forward and reverse reactions are equal and there is no net change to the overall reaction mixture.
Equilibrium concentrations are independent of direction of approach
Change in Gibbs Energy = 0, T=change in H/change in S can be used to find when a non-spontaneous process becomes spontaneous.
At equilibrium deltaG = -RTlnK

107
Q

Reaction Quotient

A

An expression for systems that are not necessarily at equilibrium.

108
Q

How do we derive rate laws for reactions?

A

Rate laws must always be derived experimentally using the method of initial rates.

109
Q

What is Collision Theory?

A

The rate of reaction is proportional to the number of effective collisions per second among the reactant molecules. This requires sufficient energy and correct orientation.

110
Q

How can Ea be determined?

A

Using the Arrhenius Equation:

  • graphically: when ln(k) is plotted against 1/T the slope is -Ea/R
  • using two temperatures: ln(k1) - ln (k2)
111
Q

What is a homogenous catalyst?

A

A catalyst in the same phase as the reactants. Usually reacts with the reactants.

112
Q

What is a heterogenous catalyst?

A

A catalyst in a different phase to the reactants. Commonly a solid and promotes reactions on surface through adsorption of one or more reactants.

113
Q

How are Enzyme Kinetics modelled?

A

Using the Michaelis-Menten Model.

114
Q

How do we derive rate laws for reactions?

A

Rate laws must always be derived experimentally using the method of initial rates.

115
Q

What is Collision Theory?

A

The rate of reaction is proportional to the number of effective collisions per second among the reactant molecules. This requires sufficient energy and correct orientation.

116
Q

How can Ea be determined?

A

Using the Arrhenius Equation:

  • graphically: when ln(k) is plotted against 1/T the slope is -Ea/R
  • using two temperatures: ln(k1) - ln (k2)
117
Q

What is a homogenous catalyst?

A

A catalyst in the same phase as the reactants. Usually reacts with the reactants.

118
Q

What is a heterogenous catalyst?

A

A catalyst in a different phase to the reactants. Commonly a solid and promotes reactions on surface through adsorption of one or more reactants.

119
Q

How are Enzyme Kinetics modelled?

A

Using the Michaelis-Menten Model.

120
Q

What are some common oxidising agents?

A 
Hydrogen Peroxide
Potassium Permaganate
Chromium  Trioxide
Potassium Dichromate
Jones Reagent
121
Q

What are some common reducing agents?

A

Lithium aluminium hydride

sodium borohydride

Chemistry for Biomedicine (2 decks)

Brainscape helps you reach your goals faster, through stronger study habits.
© 2024 Bold Learning Solutions. Terms and Conditions