Haloalkanes Alcohols Amines

Question 1

Amine tunnelling

Answer 1

Occurs for simple structures.
Rapid interconversion of pyramidal structures via planar sp2.
For rigid amines the isomers can be separated.

Question 2

Sn1 carbocation

Answer 2

Stabilisation increases rate
I~TS1
Can be stabilised by σconjugation

Question 3

Sn1 stereochemistry

Answer 3

Partial or complete racemisation
Poor leaving groups will linger.
Slight excess of inverted enantiomer

Question 4

Sn1 solvent

Answer 4

Polar protic
Stabilises carbocation like transition state
(High dielectric constant)

Question 5

Dielectric constant

Answer 5

Ability to screen charge.

More stable ions and a lower tendency to recombine.

Question 6

Sn2 transition state

Answer 6

Trigonal bipyramidal

Stabilised by π conjugation- donation of the ‘filled’ p orbital into adjacent π *

Question 7

Sn2 solvent

Answer 7

Polar aprotic
No preferential Nu- stabilisation
Naked anions with minimised solvation shell

Question 8

E2

Answer 8

Negative base
H and X antiperiplanar
Favours saytzev orientation

Question 9

Hofmann product

Answer 9

E2 
Least substituted 
Used to eliminate large groups 
CH3I + NaOH 
Gauche high in energy

Question 10

E1cb

Answer 10

Carbocation formation
Conjugated carbonyls
E isomer predominates

Question 11

Saytzev product

Answer 11

Most substituted

β hydrogen overlap with π* lowering energy

Question 12

E1

Answer 12

Produces saytzev Alkene (kinetic product)
H abstraction product determining
Regio and stereo selective
E isomer favoured

Question 13

E1 vs 2

Answer 13

Increasing carbocation stability favours E1
Increasing product stability favours E2
Conc. base forces E2

Question 14

Soft vs hard nucleophiles

Answer 14

Soft favour substitution

Hard favour elimination

Question 15

Hard bulky bases

Answer 15

Slow Sn2

Gabor Hofmann product as tales least hindered protons

Question 16

ΔG affect

Answer 16

Substitution ΔS is 0
Elimination ΔS > 0
High temp favours elimination

Question 17

Cyclohexane substitution

Answer 17

Soft nucleophile
Good leaving group
Axial H block attack on equatorial substituents

Question 18

Cyclohexane elimination

Answer 18

Must be axial but equatorial configuration favoured due to diaxial repulsion

Question 19

1,2-deghalogenation

Answer 19

NaI and acetone
To alkene + IBr and bromide
To I2 and 2 bromide

Question 20

Alkyne synthesis

Answer 20

LDA and vinyl halides

Alkene + Br2 then base to alkyne via E2

Question 21

C-L bond strength.

Answer 21

Affects leaving group ability.

Sp2 stronger due to higher s character, decreases dipole and electrons closer to carbon.

Question 22

L- stability

Answer 22

Affects leaving group ability
More stable better leaving group
decreasing pKa
larger anions have greater area of interaction and hydration spheres

Question 23

haloalkanes from alcohols

Answer 23

Via protonation of OH
primary via Sn2 - NaX and sulfuric acid
tertiary via Sn1- HX

Question 24

Methyl Aryl Ether cleavage

Answer 24

HI
produces aryl alcohol and CH3I
Attacks methyl as minimally hindered

Question 25

Alcohol to tosylate

Answer 25

TsCl in pyridine gives TsOR Methyl - benzene - SO2Cl

Question 26

Tosylate to iodoalkane

Answer 26

NaI in acetone | RI + NaOTs

Question 27

Bromoalkanes from alcohols

Answer 27

PBr3 goes to PBr2OH and RBr goes to 3RBr and H3PO3

Question 28

Iodoalkanes from alcohols

Answer 28

PI3 formed in situ

Question 29

Chloroalkanes | P

Answer 29

PCl5 goes to RCl + O=PCl3 + HCl goes to 4RCl +HCl + H3PO4

Question 30

Chloroalkanes | S

Answer 30

``` Thionyl Chloride Forms R3C-O-SOCl and HCl SnI maintains stereocentre + SO2 Sn2 with pyridine inverts Pyridine displaces remaining Cl ```

Question 31

Increasing nucleophilicity

Answer 31

Increasing basicity | Increasinf HNu pKa

Question 32

Soft Nucleophiles

Answer 32

Polarisable, high HOMO energy, react by orbital interaction. Substitution.

Question 33

Hard Nucleophiles

Answer 33

small with low HOMO energy, react by eldctrostatics

Question 34

Haloalkane + Ammonia

Answer 34

nucleophilic attack - Sn2 Lone pair basicity increases over alkylation to quaternary ammonium salt

Question 35

primary amines from haloalkanes | N

Answer 35

NaN3 to RN3 - no nucleophilic electrons LiAlH4 to RNH2 N3- -N=N+=N-

Question 36

Primary amines from haloalkanes | gabriel

Answer 36

potassium phthalimide and hydrozone

Question 37

Acyl chloride to amine

Answer 37

NaN3 N3 swaps Cl and loses nitrogen R-N=*=O and water zwitter ion decarboxylates

Question 38

ylid formation

Answer 38

RX and PPh3 then base | R- carbanion - P+Ph3

Question 39

EtX from haloalkane

Answer 39

(EtO)3P | Forms R-P and P=O bonds and EtX

Question 40

1 carbon homologation

Answer 40

haloalkane + CN- | cyano reduced to amine by LiAlH4

Question 41

alcohol + diazomethane

Answer 41

H2=N+=N- resonance to carbanion deprotonate acidic OH Phenoxide attack releases N2 Inserts CH2 into OH bond

Question 42

primary haloalkanes or tosylates to ethers

Answer 42

alcohol + KOH or NaH phenoxide attacks haloalkane can be intramolecular - high dilution or slow base addition

Question 43

conversion to I

Answer 43

conversion to good leaving group increases rate of subsequent reaction NaI + RX in acetone NaI is soluble