Transcript Electrophilic Addition and Substitution

Title: Lesson 7 Electrophilic Addition and
Substitution
Learning Objectives:
– Understand how to deduce the mechanism of the electrophilic addition
reactions of alkenes with halogens/interhalogens (2 or more different
halogens in a molecule) and hydrogen halides.
– Know Markinov’s Rule
– Understand how to deduce the mechanism of nitration reaction of benzene
Electrophilic addition reactions: alkenes

Alkenes  unsaturated  readiness to undergo addition reactions

Carbon atoms of the double bond are sp2 hybridized, forming a planar triangular
shape (120o)

Open structure  easy to attack

Pi bond is an area of electron density above and below the plane of the bond axis


Pi bonds less associated with the nuclei  weaker than sigma bond
Pi bond is more attractive to electrophiles that are or have become electron
deficient (in presence of the pi bond)

Reactants can attach like this (when pi bond breaks):
Known as electrophilic addition reactions
Main Menu
Ethene + bromine
Mechanism for the reaction:

Bromine  non polar, but as it approaches the electron rich region of alkene it is
polarized by electron repulsion

Bromine nearest the alkenes double bond gains δ+  becomes electrophile

The bromine molecule splits heterolytically, forming Br+ and Br-

Initial attack on ethane pi bond is carried out by the Br+
Slow step
Bromine has bonded
Main Menu
Carbocation intermediate
(only shares 6 electrons)

The unstable carbocation reacts rapidly with the negative bromide ion, Br-,
forming the product

Overall equation is:
Main Menu
Ethene + hydrogen bromide
Reaction mechanism is similar to previous slide:
 HBr is polar, splits heterolytically to form H+ and Br H+ is the electrophile, which attacks the alkene’s double bond
 The unstable cation reacts quickly with Br- to form the addition product
The other hydrogen halides
react similarly with alkenes. (HI
reacts more readily than HBr
due to its weaker bond, HCl
reacts less readily due to its
stronger bond etc…)
Alkenes and Electrophilic Addition Mechanism Video
Main Menu
Propene + hydrogen bromide (unsymmetric addition)


When propene (unsymmetrical shape) reacts with HBr, 2 products can be
formed
They are isomers of each other and result from two possible pathways
through the electrophilic addition mechanism described before
The product formed
depends on whether
the attacking
electrophile (H+)
attacks carbon 1 or 2

Which is more likely? Need consider which will give the most stable
carbocation during the addition process…
Main Menu





The alkyl groups around a carbocation stabilize it somewhat due to their
positive inductive effects  they push electron density away from
themselves and lessen the positive charge
In (a) there is a primary carbocation (stabilized by one inductive effect)
In (b) there is a secondary carbocation (stabilized by two inductive effects)
(b) is more stable and more likely to persist and react with Br2-bromopropane will be the main product of the reaction
Inductive effect of alkyls
• Alkyls want to push electrons onto
the carbon atom
• If there are several groups pushing
from several sides of the carbon
atom there will be a spreading of
+ve charge and this will stabilize
the carbocation…
Main Menu
Markovnikov’s Rules



Can predict the outcome for any reaction using a hydrogen
halide to asymmetric alkenes
‘The hydrogen will attach to the carbon that is already
bonded to the greater number of hydrogens’
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)
Main Menu
Main Menu
Main Menu
Solutions
Main Menu
Main Menu
Electrophilic substitution mechanism

First step is slow  electron pair from benzene is attracted to the electrophile 
disruption of symmetry of delocalized pi system

Unstable carbocation intermediate has both the entering atom/group and the leaving
hydrogen temporarily bonded

Incomplete circle shows loss of symmetry, with positive charge distributed over the bulk
of the molecule

Loss of a 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

This product is more stable as shown on the diagram:
Main Menu
Electrophilic substitution reactions
14 of 36
© Boardworks Ltd 2010
Nitration of benzene

Different electrophiles can introduce different functional groups into the ring
Substitution of –H by –NO2 to form nitrobenzene, C6H5NO2

The electrophile for the reaction is NO2+, the nitronium ion

NO2+, is a strong electrophile and reacts with the pi electrons of the benzene
ring to form a carbocation intermediate
Loss of the proton (H+) leads to the reformation of the arene ring and the
product nitrobenzene


Main Menu
Nitration of benzene
16 of 36
© Boardworks Ltd 2010
Nitrated arenes
As well as being precursors to amines, nitrated arenes are
also useful as explosives.
TNT (trinitrotoluene), or 2,4,6trinitromethylbenzene, is
formed by nitrating
methylbenzene (also known as
toluene) at a high temperature,
leading to the substitution of
three hydrogen atoms.
TNT has a low melting point, and
is fairly stable to shock and friction,
making it safer to handle than
other types of explosive.
17 of 36
© Boardworks Ltd 2010
Reduction of carbonyl compounds


Oxidation of alcohols leads to carbonyl groups (C=O group)
Dependent on the alcohol and conditions:


1o alcohol  aldehyde  carboxylic acid
2o alcohol  ketone
We can reverse this by using a suitable reducing agent:
 1. Sodium borohyride, NaBH4, in aqueous or alcoholic solution
 2. Lithium aluminium hydride, LiAlH4, in anhydrous conditions

Both produce the hydride ion (H-)  acts as a nucleophile on the electron
deficient carbonyl carbon
Main Menu
1o alcohol  aldehyde  carboxylic acid
2o alcohol  ketone


NaBH4 is the safer reagent but not reactive enough to reduce carboxylic acids, so LiAlH4 must be
used for this
Examples of the reactions shown below:
Conditions:
Heat with LiAlH4
in dry ether.
Reaction cannot
be stopped at
aldehyde as it
reacts to readily
Conditions:
Heat with
NaBH4(aq)
Main Menu
[+H]
represents
reduction, just
like [+O]
represents
oxidation
Reduction of nitrobenzene

Nitrobenzene, C6H5NO2, can be converted into phenylamine,
C6H5NH2, in a two stage reduction process.
1. Reaction is heated under
reflux in a boiling water bath.
The product C6H5NH3+,
phenylammonium ions, is
protonated by the acid
conditions.
2. C6H5NH3+ is reacted
with NaOH, to remove
the H+ and form the
product phenylamine,
C6H5NH2.
Main Menu
Summary of reaction mechanisms
Main Menu
Main Menu
Solutions
Main Menu