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 Markinovs Rule Understand how to deduce the mechanism of nitration reaction of benzene Main Menu Electrophilic addition reactions: alkenes Alkenes unsaturated readiness to undergo addition reactions Carbon atoms of the double bond are sp 2 hybridized, forming a planar triangular shape (120 o ) 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 + 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 Carbocation intermediate (only shares 6 electrons) Bromine has bonded Main Menu 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 alkenes 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 Which is more likely? Need consider which will give the most stable carbocation during the addition process The product formed depends on whether the attacking electrophile (H + ) attacks carbon 1 or 2 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 Br - 2-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 Markovnikovs 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 Solutions 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: 14 of 36 Boardworks Ltd 2010 Electrophilic substitution reactions Main Menu Nitration of benzene Different electrophiles can introduce different functional groups into the ring Substitution of H by NO 2 to form nitrobenzene, C 6 H 5 NO 2 The electrophile for the reaction is NO 2 +, the nitronium ion NO 2 +, 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 16 of 36 Boardworks Ltd 2010 Nitration of benzene 17 of 36 Boardworks Ltd 2010 Nitrated arenes As well as being precursors to amines, nitrated arenes are also useful as explosives. TNT has a low melting point, and is fairly stable to shock and friction, making it safer to handle than other types of explosive. TNT (trinitrotoluene), or 2,4,6- trinitromethylbenzene, is formed by nitrating methylbenzene (also known as toluene) at a high temperature, leading to the substitution of three hydrogen atoms. Main Menu Reduction of carbonyl compounds Oxidation of alcohols leads to carbonyl groups (C=O group) Dependent on the alcohol and conditions: 1 o alcohol aldehyde carboxylic acid 2 o alcohol ketone We can reverse this by using a suitable reducing agent: 1. Sodium borohyride, NaBH 4, in aqueous or alcoholic solution 2. Lithium aluminium hydride, LiAlH 4, in anhydrous conditions Both produce the hydride ion (H - ) acts as a nucleophile on the electron deficient carbonyl carbon Main Menu NaBH 4 is the safer reagent but not reactive enough to reduce carboxylic acids, so LiAlH 4 must be used for this Examples of the reactions shown below: 1 o alcohol aldehyde carboxylic acid 2 o alcohol ketone Conditions: Heat with LiAlH 4 in dry ether. Reaction cannot be stopped at aldehyde as it reacts to readily Conditions: Heat with NaBH 4 (aq) [+H] represents reduction, just like [+O] represents oxidation Main Menu Reduction of nitrobenzene Nitrobenzene, C 6 H 5 NO 2, can be converted into phenylamine, C 6 H 5 NH 2, in a two stage reduction process. 1. Reaction is heated under reflux in a boiling water bath. The product C 6 H 5 NH 3 +, phenylammonium ions, is protonated by the acid conditions. 2. C 6 H 5 NH 3 + is reacted with NaOH, to remove the H + and form the product phenylamine, C 6 H 5 NH 2. Main Menu Summary of reaction mechanisms Main Menu Solutions