Organics HL

Question 1

Nucleophiles

Answer 1

electron rich
attack areas of electron deficiency
act as lewis bases and donate a pair of electrons

Question 2

Electrophiles

Answer 2

electron deficient
accept electron pairs from a nucleophile
act as lewis acids

Question 3

Nucleophilic substitution (halogenoalkanes)

Answer 3

haloalkane + sodium hydroxide -> alcohol
R-X + OH- -> R-OH + X-
two mechanisms SN1 and SN2

Question 4

Strength of nucleophiles

Answer 4

CN- > OH- > NH3 > H2O

Question 5

SN2 mechanism

Answer 5

SN2 = substitution nucleophilic bimolecular
1 step process
primary halogenoalkanes (and secondary)
as 1o have at least two small H atoms attached they are open to attack by nucleophile
An unstable transition state is formed in which the carbon is weakly bonded to both the halogen and the nucleophile
the C-halogen bond then breaks heterolytically
as this mechanism is dependent on the concentration of both the halogenoalkane and the OH- it is bimolecular
rate = k [halogenoalkane] [nucleophile]
the nucleophile attacks the the C on the opposite side from the leaving group, which causes an inversion of the arrangement of the atoms
this is a result of different spatial arrangements around tetrahedrally bonded C atoms
described as stereospecific because the 3D arrangement of reactants determines the 3D configuration of the products
favoured by polar, aprotic solvents (not able to form H bonds as they do not contain -OH or -NH bonds)

Question 6

SN1 mechanism

Answer 6

(123 bang)
SN1 = substitution nucleophilic unimolecular
2 step process
tertiary halogenoalkanes (and secondary)
3o halogenoalkanes have three alkyl groups which causes steric hindrance meaning that these bulky groups make it difficult for incoming group
first step involves the halogenoalkane ionising by breaking its C-halogen bond heterolytically. As the halide ion is detached, the C is left with a temporary +ve charge (carbocation intermediate)
second step the carbocation is attacked by the nucleophile
as both of the shared e- go to one of the products the bond breakage is heterolytic fission
rate determining step: spontaneous dissociation of leaving group
transition state: formation of carbocation
very fast step: reaction of nucleophile and carbocation
as the rate of this reaction is determined by the concentration of the halogenoalkane it is unimolecular
rate = k [halogenoalkane]
not stereospecific and can give rise to racemic mixture
favoured by polar, protic solvents (contain -OH or -NH and so are able to form H bonds)

Question 7

Influence of the leaving group on rate of nucleophilic substitution

Answer 7

strength of C-halogen bond
carbon to halogen bond decreases in strength from F to I
therefore ease of breaking bond is as follows:
C-I > C-Br > C-Cl > C-F

Question 8

Protic solvent

Answer 8

solvent that can make H bonds (donate a proton)
e.g H2O

Question 9

Aprotic solvent

Answer 9

cannot make H bonds (cannot donate a proton) e.g methoxymethane

Question 10

Solvation

Answer 10

solvent surrounding an atom/molecule/ion in a cage of solvent
reduces reactivity (stabilises)

Question 11

Nature of double bond in alkenes

Answer 11

C atoms of the double bond are sp2 hybridised forming a trigonal planar shape (120 degrees)
open structure that makes it easy for incoming groups to attack
the pi bond is weaker and is more easily broken
because it is an area of electron density it is attracted to electrophiles
therefore can undergo electrophilic addition reactions
e.g. halogen, hydrogen halides

Question 12

Electrophilic addition
ethene + bromine

Answer 12

C2H4 + Br2 -> CH2BrCH2Br (1,2-dibromoethane)
as Br2 approaches the e rich region it becomes polar, one of the C-C bond breaks, the pair of e- attaches to the slightly positive Br end of Br-Br. The Br-Br bond breaks heterolytically to form a bromide ion and a carbocation is formed
the bromide ion behaves as a nucleophile and attacks the carbocation

Question 13

Electrophilic addition
ethene + hydrogen bromide

Answer 13

HBr + C2H4 -> CH3CH2Br (bromoethane)
as the HBr nears the alkene, one of the C-C bonds breaks, the pair of e- attaches to the slightly positive H end of H-Br. The H-Br bond breaks heterolytically to form a bromide ion. A carbocation is formed. The bromide ion behaves as a nucleophile and attacks the carbocation.

Question 14

Electrophilic addition
Propene + hydrogen bromide

Answer 14

-> either 1-bromopropane or 2-bromopropane
when an unsymmetrical alkene is reacted with a hydrogen halide there are two dif. products that can form
we can predict the outcome by using Markovnikov’s rule which states that the hydrogen will attach to the carbon that is already bonded to the greater number of hydrogens
therefore the major product is 2-bromoethane

Question 15

Markovnikov’s rule

Answer 15

Used for electrophilic addition reactions involving unsymmetrical alkenes
the hydrogen will attach to the carbon that it already bonded to the greater number of hydrogens

Question 16

Interhalogens electrophilic addition

Answer 16

2 or more halogens bonded together
The more elctronegative halogen attaches to the centre carbon as this allows for a more stable spread of the density of positive charge

Question 17

Electrophilic substitution of benzene

Answer 17

called the nitration of benzene
substitution of -H by -NO2 to form nitrobenzene (C6H5NO2)
the electrophile is NO2+ (nitronium ion)
generated by a nitrating mixture (concentrated nitric acid and concentrated sulfuric acid at 50 degrees)
the stronger acid (H2SO4) protonates the nitric acid to form NO2+
NO2+ reacts with the pi electrons of the benzene ring to form the carbocation intermediate
loss of a proton lead to reformation of the arene ring in the product nitrobenzene
the H+ released reacts with the base HSO4- to reform H2SO4
C6H6 + HNO3 -> (catalyst H2SO4 at 50 degrees) C6H5NO2 + H2O

Question 18

Reduction of carbonyl compounds

Answer 18

reverse the oxidation reactions using a suitable reducing agent
1. sodium borohydride (NaBH4) aq or alcoholic
2. lithium aluminium hydride (LiAlH4) in anhydrous conditions such as dry ether followed by aqueous acid
both reagents produce the hydride ion (H-) which acts as a nucleophile on the electron deficient carbonyl carbon
NaBH4 tends to be the safer reagent but is not reactive enough to reduce carboxylic acids so LiAlH4 must be used for this
[H+] represents reduction

Question 19

Examples of reduction of carbonyl compounds

Answer 19

ethanoic acid -> [H+] ethanal -> [H+] ethanol
CH3COOH -> CH3CHO -> CH3CH2OH
carbox. acid -> aldehyde -> primary alcohol
conditions: heat with LiAlH4 in dry ether

pronanone -> propan-2-ol
(CH3)2CO -> [H+] (CH3)2CHOH
ketone -> secondary alcohol
conditions: heat with NaBH4 (aq)

Question 20

Reduction of nitrobenzene

Answer 20

nitrobenzene can be converted to penylamine (C6H5NH2) in a two stage reduction process
1. nitrobenzene is reacted with a mixture of Sn and conc. HCl, heated under reflux in a boiling water bath the product C6H5NH3+, phenylammonium ions is protonated because of acidic conditions
2. C6H5NH3+ is reacted with NaOH to remove H+ and forms phenylamine (C6H5NH2)

Question 21

Retrosynthesis

Answer 21

desired product (target molecule) can serve as a starting point for thinking
can be broken down into precursors
target molecule => precursor 1 => precursor 2 => starting material
retrosynthesis analysis describes this systematic backward approach led to synthesis of thousands of molecules

Question 22

Isomerism

Answer 22

compounds with same molecular formula but different arrangements of the atoms

Question 23

Structural isomerism

Answer 23

atoms and functional groups attached in different ways

Question 24

Stereoisomerism

Answer 24

different spatial arrangements of atoms in molecules

Question 25

Conformational isomerism

Answer 25

can be interconverted by free rotation about sigma bonds

Question 26

Cis - trans and E/Z isomerism

Answer 26

exist where there is restricted rotation around atoms

Question 27

Optical isomerism

Answer 27

chirality exists where there is an asymmetric carbon atom

Question 28

Double bonded molecules isomers

Answer 28

free rotation of the double bond is not possible as it would push the p orbitals out of position and the pi bond would break

Question 29

Cyclic molecules isomers can be isomers because …

Answer 29

the ring of carbon atoms restricts rotation

Question 30

Cis

Answer 30

isomer that has the same groups on the same side of the double bond

Question 31

Trans

Answer 31

isomer that has the same groups on opposite sides of the double bond

Question 32

E/Z isomers

Answer 32

occurs when the carbon atoms of the double bond are bonded to more than two different substituents
as all the groups are different there are no ‘same groups’
therefore Cahn-Ingold-Prelog rules of priority are used
Z is where the two highest priority groups are on the same side
E is where the two highest priority groups are on the opposite side

Question 33

Cahn - Ingold - Prelog rules of priority

Answer 33

1 - look at the atom bonded to the C of the double bond. The atom with the higher atomic number has the higher priority
2 - if the atoms are the same, then the longer hydrocarbon chains have higher priority
3. if the two highest priority are on the same side then its Z and if they are on the opposite sides then its E

Question 34

Optical isomers

Answer 34

C atom attached to four different atoms or groups is known as chiral
known as asymmetric or as a stereocentre
arranged tetrahedrally (109.5 degrees)
can be arranged in two dif. 3D config. which are mirror images of each other
refers to the ways in which the isomers interact with plane polarised light
they are said to be chiral molecules and have no plane of symmetry
are molecules that are non-superimposable mirror images of each other
the two non-superimposable forms are known as enantiomers
many molecules have more than one chiral centre and so can give rise to different configs. at each position
when molecules have dif. configs. at one or more, but not all, chiral centres they are known as diastereomers and are not mirror images

Question 35

Racemic mixture

Answer 35

a mixture containing equal amounts of the two enantiomers

Question 36

Chiral

Answer 36

a carbon atom attached to four different atoms or groups

Question 37

Diastereomers

Answer 37

molecules that have different configurations at one or more, but not all, chiral centres
not mirror images

Question 38

Properties of enantiomers and diastereomers

Answer 38

diastereomers usually differ from each other in their physical and chemical properties
enantiomers have identical physical and chemical properties with two exceptions (optical activity and reactivity with other chiral molecules)

Question 39

Optical activity enantiomers

Answer 39

when a beam of plane polarised light passes through a solution of optical isomers, they rotate the plane of polarisation
separate solutions of enantiomers at the same concentration, rotate plane polarised light in equal amounts but opposite directions (optically active)
a racemic mixture does not rotate the light (optically inactive)

Question 40

Reactivity of enantiomers with other chiral molecules

Answer 40

when a racemic mixture is reacted with a single enantiomer of another chiral compound the two components of the mixture react to form different products (with different chemical and physical properties and so can be separated easily)
therefore, this is a means by which the two enantiomers can be separated from a racemic mixture in a process called resolution