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Amines as nucleophiles and their synthesis. L.O.: Understand the reaction of amines and ammonia with haloalkanes . Know one application of quaternary ammonium salts. Know how amines can be prepared from nitriles . Know how aromatic amines can be prepared from nitro compounds.
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Amines as nucleophiles and their synthesis L.O.: • Understand the reaction of amines and ammonia with haloalkanes. • Know one application of quaternary ammonium salts. • Know how amines can be prepared from nitriles. • Know how aromatic amines can be prepared from nitro compounds.
Consider the reaction between ethylamine and an excess of bromoethane • Write the mechanism for the nucleophilic reaction. • Consider what type of product you got. Would it react with an excess of bromoethane? • Write down all the possible products you may get in the reaction.
NUCLEOPHILIC SUBSTITUTION - X H NH3 + NH3 NH2 X C C NH2 C • Lone pair on N can attack + C of haloalkane. - NH4+X
NUCLEOPHILIC SUBSTITUTION • H on N is swapped by R from haloalkane. • Product also has lone pair on N so product reacts further. • To get mainly 1y amine use XS NH3. • To get mainly 4y ammonium salt use XS RX.
TASK • Give all the organic products formed when methylamine reacts with chloroethane. 2) Draw the mechanism for the formation of the tertiary amine in this reaction. 3) Draw all the organic products formed when diethylamine reacts with 2-bromopropane
NUCLEOPHILIC SUBSTITUTION methylamine + chloroethane (products & 1st mechanism)
NUCLEOPHILIC SUBSTITUTION diethylamine + 2-bromopropane (products)
NUCLEOPHILIC SUBSTITUTION 4y ammonium salts • Ones with some long chain alkyl groups can be used as cationic surfactants. • Often used in fabric softener as many fabrics have –ve charge on surface. e.g. (CH3)2N[(CH2)16CH3]2+ Cl-
MAKING ALIPHATIC 1y AMINES 1 – Reaction of NH3 with haloalkanes • Need to separate mixture of products. 2 – Reduction of nitrile compounds
Give two synthetic paths to making butylamine. Give equations for all reactions.
Butylamine a)CH3CH2CH2CH2Cl + 2 NH3→ CH3CH2CH2CH2NH2 + NH4Cl • CH3CH2CH2Cl + KCN → CH3CH2CH2CN + KCl • CH3CH2CH2CN + 4 [H] → CH3CH2CH2CH2NH2