Medicinal chemistry

Question 1

Lipinski’s rule of 5

Answer 1

(all less than or equal to excpt MW)
<5 H bond donors (OH+NH)
<10 H bond acceptors (O + N)
Molecular weight <500
Logp <5

Question 2

Vebers Rules

Answer 2

10 or fewer rotatable bonds
polar surface area equal to or < 140 A^2
equal to or < 12 H bond donors and acceptors

Question 3

Partition coefficient (P) 
(hydrophobic effects)

Answer 3

P = conc. of drug in octanol/conc. of drug in aqueous solution
logP= log(drug)octanol - log(drug) water
hydrophobic = +ve
hydrophilic = -ve
log ~ 2- likley to get into brain

Question 4

Distribution coefficeint (D)
(hydrophobic effects)

Answer 4

logD is the partition coefficient at a particular pH
pH dependant
logD 7.4 = gives indication of lipophilicity of drug at the pH of blood plasma

Question 5

Hydrophobic parameter (pi)
(hydrophobic effects)

Answer 5

used to calculate logP:
pi(x) = logPx -LogPh
Ph is the partition coefficient for the standard compound. Px is the partition coefficient for the standard compound with the substituent 
pi>0 = hydrophobic
pi<0 = hydrophilic

Question 6

Hammet substituent constant (σ)

electronic effects

Answer 6

σx = log (Kx/Kh) = logKx-logKh
Kh is the ionisation constant of benzoic acid
Kx is the ionisation constant of the substituted benzoic acid
σx < 0 substituent is acting as an electron donor
σx> 0 substituent is acting as an EWG, since Kh

Question 7

Taft steric parameter (Es)

steric effects

Answer 7

Es = log (kRCOOMe/ kRCOOH) = logkRCOOMe - logkRCOOH

where k is the rate constant for ester hydrolysis

Question 8

Molar refractivity (MR)
(steric effects)

Answer 8

attempts to measure volume of given atom or group
MR = (n^2 -1)/(n^2 +2) x MW/d
n= index of refraction
MW = molecular weight 
d = density 
MW/d = molar volume 
First term reflects polarisability

Question 9

Verloops steric parameters

steric effects

Answer 9

computes steric substituent values from standard bond angles, van der Waals radii, bond lengths and conformations for substituent

Question 10

Hansch analysis

Answer 10

Based on the theory that drug action could be divided into two stages:

1. Transport of the drug to the site of action
2. Binding of the drug to the target site

Question 11

Options for drug optimisation

Answer 11

1) variation of alkyl group
2) hydrophobic interactions
3) Bulk
4) extension of structure
5) ring expansion/contraction
7) ring variation
8) Ring fusions
9) simplification
10) rigidification
11) Conformational blockers

Question 12

Isosteres of OH
isosteres of O
Isosteres of Br
Isostere of I

Answer 12

SH, NH2, CH3, F, Cl
S, NH, CH2
Br, iPr
I, tBu

Question 13

How to remove H bonding ability of OH

Answer 13

The O can act as a H bond acceptor and H as an H bond donor
Replace with Methyl ether (R-OCH3) which removes the proton that would have been a H bond donor. The O atom can still act as a H bond acceptor but not to the same extent as in hydroxyl as the Me hinders close apprach. Therefore the H bonding will be weakened but is not completely prevented
Replace with OAc prevents H bonding
Replace with ester as H bond donor is removed. O can still act as a H bond acceptor but is even more hindered by the acyl group. The carboxyl group also pulls electrons from the neighbouring O making it less effective as H bond acceptor.

Question 14

How to remove ionic and H bonding ability of amines

Answer 14

They are involved in hydrogen bonding wither as a H bond acceptor or donor. N atom has lone pair that can act as a H bond acceptor. primary and secoundary have N-H that can act as a H bond donor. Aromatic or heteroaromatic amines only react as H bond donors as lone pair interects with the ring.
In the body amine can be protonated and ionzised therefore cannot act as a H bond acceptor but can form strong bonds as H bond donor
convert into amide which stops N acting as a H bond acceptor, the lone pair will instead interact with the carbonyl. It prevents protonation
Additon of actyl to the N atom which will donate electron density into the carbonyl and therefore unable to take part in H-bonding or ionic bonding

Question 15

How to remove bonding ability from flat aromatic ring

Answer 15

Ar rings are planar, flat and hydrophobic and involved in van der Waals interactions with flat hydrophobic regions of the binding site
hydrogenation to cyclohexane reduces van der waals interactions and now the Ar ring is rigid and no loner flar. The binding region however may be narrow rather than planar surface which will not allow cyclohexane ring to fit as it is bulkier

Question 16

How to remove bonding ability of alkenes

Answer 16

alkenes are planar and hydrophobic and therefore interact with hydrophobic regions of the binding site through van der Waals.
Reduce by hydrogenation to alkane which is bulkier and cannot approach the region of the binding site as closely

Question 17

How to remove the binding abilty of ketones and aldehydes

Answer 17

Ketone is planar and can interact with the binding site through H bonding where the carbonyl O acts as a H bond acceptor. Leaving tow interactions from the two lone pairs on O. The carbonyl group has a dipole so dipole-dipole interactions wiht the binding site.
Reduce ketone to alcohol, changing geometry from planar to tetrahedral which will weaken any H bonding and dipole-dipole as it is no in a new orientation. Can also be reduced completely to alkane.
Aldehydes are susceptible to metabolic oxidation to carboxylic acids

Question 18

How to remove bonidng ability of amides

Answer 18

They interact through H bonding, it is planar and cannot rotate. The carbonyl O can act as a H bond acceptor and has the potential to form 2 H bonds. N cannot be a H bond acceptor as the lone pair interacts with the carbonyl. Primary and secondary amides have H that can act as a H bond donor.
replace with alkene or amine anologue to see if the amid is a H bond acceptor. Alkenes are planar cannot rotate and cannot act as a H bond donor or acceptor