Organics HL Flashcards
Nucleophiles
electron rich
attack areas of electron deficiency
act as lewis bases and donate a pair of electrons
Electrophiles
electron deficient
accept electron pairs from a nucleophile
act as lewis acids
Nucleophilic substitution (halogenoalkanes)
haloalkane + sodium hydroxide -> alcohol
R-X + OH- -> R-OH + X-
two mechanisms SN1 and SN2
Strength of nucleophiles
CN- > OH- > NH3 > H2O
SN2 mechanism
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)
SN1 mechanism
(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)
Influence of the leaving group on rate of nucleophilic substitution
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
Protic solvent
solvent that can make H bonds (donate a proton)
e.g H2O
Aprotic solvent
cannot make H bonds (cannot donate a proton) e.g methoxymethane
Solvation
solvent surrounding an atom/molecule/ion in a cage of solvent
reduces reactivity (stabilises)
Nature of double bond in alkenes
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
Electrophilic addition
ethene + bromine
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
Electrophilic addition
ethene + hydrogen bromide
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.
Electrophilic addition
Propene + hydrogen bromide
-> 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
Markovnikov’s rule
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
Interhalogens electrophilic addition
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
Electrophilic substitution of benzene
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
Reduction of carbonyl compounds
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
Examples of reduction of carbonyl compounds
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)
Reduction of nitrobenzene
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)
Retrosynthesis
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
Isomerism
compounds with same molecular formula but different arrangements of the atoms
Structural isomerism
atoms and functional groups attached in different ways
Stereoisomerism
different spatial arrangements of atoms in molecules