Physical Organic and Amines

Question 1

sp3 amines

Answer 1

can invert, going through a sp2 transition state
at room temp - raceimisation 1E16-18 s-1
5-6 kcal mol-1 to do this (methane 110 kcal mol-1 -chiral centres)
ammonium salt can’t go through inversion

Question 2

ammonium salt conformation

Answer 2

ammonium salt can’t go through inversion
ammonium salts e.g. N(Ph)(Pr)(Me)(Et) are conformationally stable - chiral centre (4 R groups)
if one group was instead a H - stable as long as no trace of base. base would lead to exchange of proton and therefore raceimisation

Question 3

sp2 amines

Answer 3

need a driving force that is able to overcome the steric compression resulting from sp3 –> sp2
moving lone pair into p-orbital
e.g pyridine - electrons in p-orbital as they are needed for aromaticity so energetically favourable. sp2 lope of N facing out of ring perpendicular to pi-system makes it a good base
e.g. pyrrole: 4 carbons, each have an unhybridised p-orbital (4e-) Nitrogen adopts sp2 to donate lone pair in an unhybridised p-orbital - aromaticity driving force

Question 4

aniline structure/ hybridisation

Answer 4

PhNH2 can be sp3 with l.p. - no steric hindrance
sp2 - some aromaticity but steric hindrance from hydrogens which can be overcome
R groups on benzene leads to more steric bulk and therefore more sp3 character
depending on sp3 character - better or worse base
aniline is not a good base

Question 5

stability/ pKa of ammonium ions

Answer 5

NH4+ - RNH3+ - R2NH2+ - R3NH+
stability causes pKa to increase ( becoming more basic)
stability increases as no. R groups increase but peaks at R2NH2+
tertiary ammonium has steric hindrance - positive charge hard to solvate so pKa goes down
stability increases by induction (low no. R to high)
stability increases by solvation (small = easier to solvate)

Question 6

properties of amines

Answer 6

strong/pungent smelling

water solubility - small side chains means more soluble

Question 7

synthesis of amines

Answer 7

problem: POLYALKYLATION using an alkyl halide (?) and ammonia - the alkyl group pushes e- density onto the N making the target material more nucleophilic, resulting in a complex mixture
overcome by using excess ammonia (cheap and easy to remove)

Question 8

acidity of organic compounds

Answer 8

consider ionic component first
Factors that stabilise A- make AH a strong acid, lower pKa
Factors that stabilise +AH make it a weaker acid, higher pKa

Question 9

pKa equations

Answer 9

p = -log10
p[H+] = pH = -log[H+]
pKa = p([A-]/[AH]) + pH

Question 10

to measure pKa directly

Answer 10

pH meter in HA and H2O
add -OH in small aliquots and measure pH
measure pKa within limits of solvent being protonated (H2O) and solvent being deprotonated (H3O+)

Question 11

Measure pKa by comparison

Answer 11

‘AH ‘A- + H+. pKa’
2AH 2A- + H+ pKa2
k’/k2, p(k’/k2) = difference in pKa
mix ‘AH and 2AH - equilibrate and measure ratios (all requires equilibrium)

Question 12

Factors affecting the acidity of organic compounds

Answer 12

Induction
Solvation
Hybridisation
Aromaticity
Resonance 
Electrostatics ( I SHARE)

Question 13

Induction

Answer 13

EWG remove e density from anion - stabilisation
as EWG acidity increases, stabilises anion side and so pKa goes down
expect opposite for EDG

Question 14

Salvation

Answer 14

H2O: branching increases, pKa increases, solvation decreases
more ordered H2O - entropy lost so less stabilising
on going from water to organic solvent: most pKa increase

Question 15

Hybridisation

Answer 15

more s character therefore charge held closer too the nucleus
sp3 - 25% s, pKa (ethane) 50
sp - 50% s, pKa (ethyne) 25

Question 16

Aromaticity

Answer 16

if anion forms an aromatic system highly favoured

if it forms an anti-aromatic system, 4pi, highly disfavoured

Question 17

Resonance/delocalisation

Answer 17

delocalised charge favoured over localised charge

dissociation occurs more readily, therefore more acidic

Question 18

Electrostatics

Answer 18

if X is +vely charged then electrostatic stabilisation so favours dissociation

Question 19

Kinetic order

Answer 19

experimentally measured

dependency on concentration of component

Question 20

molecularity

Answer 20

no. molecules involved in RDS

elementary step: starting materials and product are connected by a single TS

Question 21

zero order

Answer 21

no dependency on conc of component
graph - straight line (conc vs time)
gradient = kobs M s-1
half life halves

Question 22

First order

Answer 22

as conc decreases rate decreases because of cnc dependency
half life is constant k = ln2/x (x= half life)
ln[A] = ln[A0] + kt
plot ln[A0]/ln[At], slope = kobs s-1
concentration doubles as rate doubles

Question 23

Second order

Answer 23

concentration dependency
1/[A]t = 1/[A]0 + kt
to obtain kobs plot 1/[A]t against t
gradient = kobs M-1 s-1

Question 24

Pseudo orders

Answer 24

arise when:
1. large excess of one reagent (kobs = k1 x [excess reagent]
(excess reagent conc is basically constant)
2. catalyst present (kobs = k1[cat])
3. phase transfer phenomena at play (kobs= k1[reagents])

Question 25

Linear free energy relationships

Answer 25

aim to quantify the relationship between structure and reactivity i.e. how does rate vary systematically within structure

Question 26

“Hammett” correlation

Answer 26

most widely used
RA –> RB
how does rate vary with R
Delta G affected by: steric factors and electronic factors
hammett considers both steric and electronic influence of R
but separated
On benzene, considered para and meta (orthodox is statically hindered)

Question 27

substituent constant (sigma)

Answer 27

independent of the reaction
how much push or pull X has on the aromatic ring, reaction centre
log(Ka x/ Ka H) = sigma when x=H sigma= 0
K = equilibrium constants
quantified scale for the electronic influence of K from EDG to EWG

Question 28

using substituent constant

Answer 28

To use these values:
1. Measure rates of reactions where X is varied from EDG –> EWG (5 points min.)
2. plot rates, log10(ka x/ka H) against sigma
3. is there a correlation?
no, reaction mechanism is changing as X is varied
yes, X is only affecting the rate and not the mechanism
4. measure the gradient - RHO (reaction constant)
dependent on the reaction
5. is rho +ve - EWG accelerates reaction
rho = 0 - X has no affect
rho -ve - EWG decelerates reaction
6. magnitude of rho tells us how sensitive the reaction is to change of X

Question 29

rho

Answer 29

reaction constant - dependent on reaction
gradient from plot of log10(ka x/ ka H) (rate constants) against sigma
sigma (substituent constant) is log10(Ka x/ Ka H)

Question 30

negative rho

Answer 30

EWG decelerates reaction
loss of electron density on approach to TS
EWG stabilises/slows down the reaction so is favoured

Question 31

how to distinguish between SN1 and SN2 using rho

Answer 31

predict whether rho is +ve, zero or -ve for SN1 and SN2

experimental value - tells you which one

Question 32

rate limiting step and rho

Answer 32

predict rho values for each step of reaction
experimental tells us which one is right (similar to SN1/2) because Hammett plot is using rate so RLS will affect rho
if rho is -ve RLS is loss of e- density. protonation is unlikely to be RLS, collapse of tetrahedral intermediate is likely to be RLS