HL ORGANIC CHEMISTRY20.1 – TYPES OF

REACTIONS

ISHCMC CHEMISTRY

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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