Lecture 3

Question 1

hybridisation of nitrogen in primary, secondary and tertiary amines vs quaternary ammonium salts.

Answer 1

• BOTH n hybrid orbitals = 4
• BOTH sp3 hybridisation
• AMINES are trigonal pyramidal
  -AMMONIUM SALT is tetrahedral
• AMINES bond angle = 107
  -AMMONIUM SALT = 109.5
• AMINES = 1 lone pair of e
• AMMONIUM SALT = no lone pairs

Question 2

equilibrium equation

Answer 2

keq = Ka reactant acid / Ka reactant product

at equilibrium weaker acid is generated

Question 3

what does the keq value indicate?

Answer 3

• keq > 0 = favourable reaction
• keq<0 = unfavourable reaction

Question 4

what can pka tables be used to predict?

Answer 4

ionisation

Question 5

what affects the avaliability of the nitrogens lone pair?

Answer 5

EWG
EDG

Question 6

what affects the aqueous solubility of a compound containing N?

Answer 6

the number of hydrocarbon substituents bonded to the N

Question 7

what affects the steric hindrance at the lone pair?

Answer 7

the number and size of substituents

Question 8

what does the geometry affect?

Answer 8

protonation

Question 9

what does the steric hindrance to the lone pair of electrons cause?

Answer 9

decrease in basicity.

Question 10

compare the pka of an anilinium ion to a methylammonium ion.

Answer 10

the pKa of anilinium ion is 4.63
whereas
pKa of methyl ammonium is 10.66

Question 11

why are arylamines (aniline) less basic than alkylamines?

Answer 11

• the nitrogen’s lone pair of electrons are delocalised by interaction with aromatic ring pi electron system
• less available to bond with H+

Question 12

arylamines are stabilized how?

Answer 12

with respect to alkylamines because of five resonance structures which can be drawn.

Question 13

what do EWG and EDG on the aromatic ring do?

Answer 13

increase or decrease basicity of the substituted analine.

Question 14

sp2 hybridised nitrogen properties

Answer 14

• n of hybrid orbitals =3
• sp2 hybridised
• geometry = trigonal planar
• bond angles = 120
• the lone pair can be in an sp2 hybrid orbital or in a non-hybridised p orbital.

Question 15

explain lone pair in unhybridised sp2 orbital.

Answer 15

-pi excessive
- N is electron donating group due to resonance effect (donation of lone pairs)
- lone pair is part of pi cloud of aromatic electrons
- lone pair cannot be protonated

Question 16

explain lone pair in hybridised sp2 orbital.

Answer 16

• pi deficient
• N is electron withdrawing group due to inductive effect. (electronegativity)
• lone pair not part of pi cloud of aromatic electrons
• lone pair can be protonated.

Question 17

EWG and EDG effect on heterocycles

Answer 17

electron density
hence their basicity

the effect of EWG and EDG is linked to their position on the ring (especially if the ring is aromatic)

Question 18

the case of conjugation/aromaticity on heterocycles.

Answer 18

substituted heterocycles can exist in different tautomeric forms.

Question 19

what is guanadine?

Answer 19

nitrogen rich functional group characteristic of the proteinogenic amino acid L-arginine.

Question 20

characteristics of guanidine.

Answer 20

• strong base due to resonance delocalisation of the positive charge present on its conjugate acid (guanidinium cation)
• guanidine is flat and all notrogen atoms are sp2 hybridised
• high pka = guanidine derivatives can mimic inorganic metal cations
• can be used as anti-hypertensives due to blockage of ion channels.

Question 21

what are free amine bases?

Answer 21

uncharged organic molecules - not very water soluable

Question 22

what are quaternity ammonium salts?

Answer 22

charged organic molecules - relatively water soluable

Question 23

common quaternary ammonium salts

Answer 23

• hydrochlorides
• hydrobromides
• carboxylates (succinates, tartrates, citrates)
• methanesulphonates (mesylates)

Question 24

selection criteria for quaternary ammonium salts

Answer 24

• solubility
• pH of resulting solution
• strength of ionic couple
• ease of formulation

Question 25

imide groups mean….

Answer 25

weak acid