HL Organic 1 - Types of Reactions
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Transcript of HL Organic 1 - Types of Reactions
HL ORGANIC CHEMISTRY20.1 – TYPES OF
REACTIONS
ISHCMC CHEMISTRY
Great, amazing, super resources
1. Chemguide – Organic2. MSJ Chem - HL organic3. Organic interactives
LEARNING OBJECTIVES 20.1
LEARNING OBJECTIVES 20.1
LEARNING OBJECTIVES 20.1
HL – TYPES OF REACTIONS Organic reactions are organised according to what
happens (the type of reaction) and how it happens (the mechanism of the reaction)
Different combinations of mechanisms and reaction type are described in this section
Many mechanisms describe the reactants according to electrophilic and nucleophilic behaviour
Nucleophiles and ElectrophilesNucleophiles
are electron rich and attack areas of electron deficiency.
They act as Lewis bases and donate a pair of electrons in forming a new covalent bond
Examples: OH-, H2O and CN-
Electrophiles are electron deficient & accept electron pairs
from a nucleophile They act as Lewis acids Examples: H+, Br+ and NO2
+
Nucleophiles and Electrophiles A convention used in depicting
mechanisms is the curly arrow – it shows the movement of an electron pair
The tail shows where the electron pair have come from and the arrow head showing where it is going
Video:
Thornley: SN1 and SN2
Nucleophilic substitution reactions: halogenoalkanes Shorthand used: SN – substitution nucleophilic
The polar carbon-halogen bond in halogenoalkanes means that the carbon atom is electron deficient and it is attacked by nucleophiles such as OH-
As a result, the carbon-halogen bond breaks and the halogen atom is released as a negative ion (halide)
This types of breakage where both the shared electrons go to one of the products is known as heterolytic fission
The halogen is known as the leaving group The exact mechanism of these reactions
depends on whether the halogenoalkane is primary, secondary or tertiary
Primary halogenoalkanes: SN2 mechanism
Primary halogenoalkanes have at least 2 hydrogen atoms attached to the carbon of the carbon-halogen bond, eg chloroethane
The overall reaction that occurs with NaOH is:
Primary halogenoalkanes: SN2 mechanism
As the hydrogen atoms are so small, the carbon atom is relatively open to attack by the nucleophile
An unstable transition state is formed in which the carbon is weakly bonded to the halogen & the nucleophile
The carbon-halogen bond then breaks heterolytically, releasing Cl- and forming the alcohol product
This is effectively a 1-step reaction with an unstable transition state
The mechanism depends on the concentration of both the halogenoalkane and the OH- ion, so it is known as a bimolecular reaction
Rate = k [halogenoalkane] [nucleophile] Therefore the full name of this is
described as:SN2: substitution nucleophilic
bimolecular
Primary halogenoalkanes: SN2 mechanism
Note: the nucleophile attacks the elctrophilic carbon atom on the opposite side from the leaving group which causes an inversion of the arrangement of the atoms around carbon
Primary halogenoalkanes: SN2 mechanism
The SN2 mechanism is favoured by polar aprotic solvents (those which are not able to form hydrogen bonds, although they may contain strong dipoles)
This means that they solvate the metal cation (eg Na+) rather than the nucleophile (OH-)
Suitable solvents include propanone (CH3)2CO and ethanentrile (CH3CN)
Tertiary halogenoalkane:SN1 mechanism
Tertiary halogenoalkanes have 3 alkyl groups attached to the carbon of the carbon-haologen bond, eg 2-chloro-2-methylpropane
The overall reaction that occurs with NaOH is:
Tertiary halogenoalkane:SN1 mechanism
Here the presence of the 3 alkyl groups around the carbon of the carbon-halogen bond causes what is called steric hindrance
Meaning that these bulky groups make it difficult for an incoming group to attack this carbon atom
Tertiary halogenoalkane:SN1 mechanism Instead the first step of the reaction involves the
halogenoalkane ionizing by breaking its C-halogen bond heterolytically
As the halide ion is detached, this leaves the carbon atom with a temporary positive charge, which is known as a carbocation intermediate
This is then attacked by the nucleophile in the second step of the reaction, leading to the formation of a new bond
Another reason which favours this mechanism in tertiary halogenoalkanes is that the carbocation is stabilized by the presence of the 3 alkyl groups, as each of these has an electron-donating or positive inductive effect (shown by the blue arrows)
This stabilizing effect helps the carbocation to persist for long enough for the second step to occur
Because the slow step of this reaction, the rate-determining step, is determined by the concentration only of the halogenoalkane, it is described as a unimolecular reaction
Rate = k [halogenoalkane] Therefore this reaction mechanism is
known as:SN1 substitution nucleophilic
unimoleclar
Tertiary halogenoalkane:SN1 mechanism The carbocation intermediate has a planar shape which
means that the nucleophile can attack from any position – this means that the SN1 reaction is not stereospecific (it can give rise to racemic mixtures – more on this later)
The influence of solvents is mostly a result of their stability to stabilize the carbocation intermediate – anything that stabilizes the intermediate will favour the reaction
Polar, protic solvents contain –OH or –NH and so are able to form hydrogen bonds
They are effective in stabilizing the positively charged intermediate by solvation involving ion-dipole interactions
Suitable solvents include water, alcohols & carboxylic acids
Secondary halogenoalkanes
It is not possible to be precise about the mechanism of nucleophilic substitution in secondary halogenoalkanes, as data show that they usually undergo a mixture of both SN1 and SN2 mechanisms
Comparison of the rates of nucleophilic substitution reactions
Comparison of the rates of nucleophilic substitution reactions In this section we will compare the rates of the nucleophilic substitution reactions by looking at:1. The effect of the mechanism2. The influence of the leaving group
(halogen)3. Choice of solvent
1. The effect of the mechanism Experimental data have shown that the
SN1 mechanism proceeds more quickly than the SN2
So substitution reactions occur more quickly with tertiary halogenoalkanes than with primary
Secondary halogenoalkanes show an intermediate rate of reaction as they use both mechanisms
When all other variables are kept constant, the relative rate of the 3 classes of halogenoalkanes is:
2. Influence of the leaving groupThere are 2 possible factors to consider:
a) The polarity of the carbon-halogen bond- As the electronegativity of the halogens decreases when going down the group, the carbon of the carbon-halogen bond becomes less electron deficient & so less vulnerable to nucleophilic attack
-
2. Influence of the leaving groupb) The strength of the carbon-halogen bond- Bond energy data shows that he carbon-
halogen bond decreases in strength from fluorine to iodine
- As the substitution reaction involves breaking this bond, we would expect the ease of breaking bonds to be:
- Reaction rate data indicate that b) wins this battle
- The relative rate of reaction of the different halogens in halogenoalkanes when all other variables is kept constant is:
2. Influence of the leaving group
3. Choice of solvent Recall that the SN1 mechanism is favoured by polar,
protic solvents while the SN2 mechanism is favoured by polar, aprotic solvents
So, overall, the fastest reactions will occur with tertiary iodoalkanes in polar, protic solvents
Nucleophilic substitution reactions of halogenoalkanes can be followed by the appearance of the halide ion
Silver nitrate solution added to the reaction mixture reacts with the halide ion, forming a precipitate of the silver halide, each with a distinct colour
Video:
Thornley - Rates of nucleophilic substitution
Summary of differences in the 2 mechanisms of nucleophilic substitution reactions
Electrophilic addition reactions: alkenes
As you hopefully know, alkenes are unsaturated molecules and are important in many synthesis pathways because they readily undergo addition reactions
To understand the mechanism of these reactions it is useful to focus first on some key features of the carbon-carbon double bond
Video:
thornley - reaction pathways
Electrophilic addition reactions:
alkenesThe nature of the double bond in alkenes is: The C atoms of the double bond are sp2
hybridized, forming a planar triangular shape (bond angle is 12o0)
This is a fairly open structure (easy for attack) The pi bond represents an area of electron
density above and below the plane of the bond axis – because electrons in the pi bond are less closely associated with the nuclei, it is weaker than the sigma bond and so breaks more easily during the addition reactions
The pi bond is attractive to electrophiles
Electrophilic addition reactions:
alkenes When this bond breaks, reactants attach at each carbon atom:
Reactions between these reagents and alkenes are known as electrophilic addition reactions
Examples include the addition of halogens and hydrogen halides to alkenes
In these reactions, the electrophile is produced through heterolytic fission
Electrophilic addition reactions:
ethene and bromine When ethene gas is bubbled through bromine at room temperature, the brown colour of the bromine fades as it reacts to form the saturated 1,2-dibromoethane
Youtube: Identifying alkenes using bromine
Electrophilic addition reactions:
ethene and bromineThe mechanism for the reaction is as follows: Bromine is a non-polar molecule, but as it
approaches the electron rich region of the alkene, it becomes polarized by electron repulsion
The bromine atom nearest the alkene’s double bond gains a 8+ charge and acts as the electrophile
The bromine molecule splits heterolytically, forming Br+ and Br-, and the initial attack on the ethene in which the pi bond breaks is carried out by the Br+
Electrophilic addition reactions:
ethene and bromine
This step is slow, resulting in an unstable carbocation intermediate in which the carbon atom has a share in only 6 outer electrons and carries an overall positive charge
This unstable species then reacts rapidly with the negative bromide ion, Br-, forming the product 1,2-dibromoethane
Electrophilic addition reactions:
ethene and bromine Overall the equation for the reaction is:
Similar reactions take place with other alkenes, eg propene
Electrophilic addition reactions:ethene and hydrogen bromide
When ethene gas is bubbled through a concentrated aqueous solution of hydrogen bromide, HBr, an addition reaction occurs fairly readily at room temperature, forming bromoethane
Electrophilic addition reactions:ethene and hydrogen bromide
The reaction is similar to ethene and bromine HBr as a polar molecule undergoes heterolytic
fission to form H+ and Br-, and the electrophile, H+, makes the initial attack on the alkene’s double bond
The unstable carbocation intermediate that forms from this step then reacts quickly with Br- to form the addition product
Electrophilic addition reactions:ethene and hydrogen bromide
The mechanism is as follows:
One piece of evidence that supports this mechanism is that the reaction is favoured by a polar solvent that facilitates the production of ions from heterolytic fission
The other hydrogen halides react similarly with alkenes – HI reacts more readily than HBr due to its weaker bond
Electrophilic addition reactions:
propene + hydrogen bromide (unsymmetric addition) When an unsymmetric alkene such as propene is
reacted with a hydrogen halide such as HBr, there are theoretically 2 different products that can form
These are isomers of each other and they result from 2 possible pathways through the electrophilic addition mechanism described previously
Electrophilic addition reactions:propene + hydrogen bromide(unsymmetric
addition)
The difference between these 2 depends on whether the attacking electrophile (H+ formed from heterolytic fission of HBr) is more likely to bond to the carbon labelled 2 or 1
So which is more likely? The answer comes from considering which pathway will give the most stable carbocation during the addition process
Electrophilic addition reactions:propene + hydrogen bromide(unsymmetric
addition) Recall that alkyl groups around a carbon stabilize it somewhat due to their positive inductive effects – meaning that they push electron density away from themselves and so lessen the density of the +ve charge
In a) below, the carbo cation is a primary carbocation and is stabilized by one one such positive inductive effect, whereas in b), a secondary carbocation forms in which there are two such effects and the stabilization is greater
Electrophilic addition reactions:propene + hydrogen bromide(unsymmetric
addition) Consequently, the more stable carbocation in b) will be more likely to persist and react with Br-, leading to 2-bromopropane as the main product of the reaction
Therefore the correct mechanism for the reaction is:
Markovnikov’s Rule We can predict such an outcome for any reaction
involving addition of a hydrogen halide to assymetric alkenes by using Markovnikov’s rule
This states: the hydrogen will attach to the carbon that is already bonded to the greater number of hydrogens
this is based on the fact that the mechanism that proceeds via the most stable carbocation will be favoured
In more general terms it can be stated as the more electropositive part of the reacting species bonds to the least highly substituted carbon atom in the alkene (the one with the smaller number of carbons attached)
Markovnikov’s Rule
Worked example:
Questions
Electrophilic substitution reactions: Benzene Despite its high saturation, benzene does not behave like
alkenes in its reactions Its unusual and highly stable aromatic ring determines that
substitution is favoured However, like alkenes it is attractive to electrophiles
because its ring is a region of electron density The delocalized electron cloud of pi electrons seeks electron
deficient species & forms a new bond as a H atom is lost The reactions are therefore electrophilic substitution
Electrophilic substitution reactions: Benzene The reaction has high activation energy and so
proceeds rather slowly – this is because the first step in the mechanism, in which an electron pair from the benzene is attracted to the electrophile, leads to a disruption of the symmetry of the delocalized pi system
This unstable carbocation intermediate that forms has both the entering atom or group and the leaving hydrogen temporarily bonded to the ring
E+ is used to represent an electrophile The incomplete circle inside the ring
shows its loss of symmetry, with the positive charge distributed over the bulk of the molecule
Loss of hydrogen ion, H+, from this intermediate leads to the electrically neutral substitution product as two electrons from the C-H bond move to regenerate the aromatic ring
Electrophilic substitution reactions: Benzene This product is more stable, as shown below
The reactions can be used with different electrophiles - eg nitration
Nitration of Benzene The nitration of benzene is the
substitution of –H by –NO2 to form nitrobenzene, C6H5NO2
The electrophile for the reaction is NO2+
This is generated by using nitrating mixture (a mixture of concentrated nitric and sulfuric acids)
The stronger of the acids, sulfuric protonates the nitric acid, which then loses a molecule of water to produce NO2
+
Nitration of Benzene
NO2+ is a strong electrophile & reacts with the pi
electrons of the benzene ring to form the carbocation intermediate
Loss of a proton from this leads to reformation of the arene ring in the product nitrobenzene, which appears as a yellow oil
The hyrogen ion released reacts with the base HSO4-
to reform sulfuric acid, H2SO4
Nitration of Benzene
Reduction reactions It is convenient in organic chem to analyse redox
reactions in terms of gain and loss of hydrogen (rather than electrons)
In the following reactions, reduction can be identified where hydrogen is gained by a reactant
Reduction of Carbonyl compounds We previously saw that the oxidation of alcohols
to carbonyl compounds (those that contain C=O) takes place in the presence of a suitable oxidizing agent and yields different products depending on the alcohol and the conditions Primary alcohol aldehyde carboxylic acid Secondary alcohol ketone
We can reverse these reactions using a suitable reducing agent:1. Sodium borohydride, NaBH4, in aqueous or alcoholic
solutions2. Lithium aluminium hydride, LiAlH4, in anhydrous
conditions, such as dry ether followed by aqueous acid
Reduction of Carbonyl compounds – NaBH4 and LiAlH4 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 it is not reactive enough to reduce the carboxylic acids so LiAlH4 must be used for this
[+H] represents reduction
The reaction cannot be stopped at the aldehyde as it reacts readily with LiAlH4
Reduction of Carbonyl compounds – NaBH4 and LiAlH4
Conditions for this reaction: heat with NaBH4
Conditions for this reaction: heat with LiAlH4 in dry ether
Reduction of nitrobenzene Nitrobenzene, C6H5NO2, can be converted
into phenylamine, C6H5NH2, in a 2 stage reduction
Questions
20.2 Synthetic routes
Learning Objectives
Summary of reactions of this chapter
Synthetic routes The development of new
organic compounds – from drugs to dyes, clothing to construction materials is vitally important
Organic chemists typically have to convert compounds from one form to another in sequence (often the product from one is the reactant in the next)
The series of discrete steps involved is known as a synthetic route
Retro-synthesis: Working backwards The desired product, known as the target
molecule, can serve as the starting point By studying its functional group, it can be
strategically broken down into progressively smaller fragments known as precursors
Each precursor then becomes the target for further analysis, eventually yielding a familiar molecule which the synthetic sequence can start
Worked Example
Worked Example
questions