20.1 Introduction
20.1.1 Deduce
structural formulas for compounds containing up to six carbon atoms with one of
the following functional groups: amine, amide, ester and nitrile.
Amine
·
Amines are indicated by the suffix -amine or the prefix amino-.
·
Amines contain R-NH2 functional
group, nitrogen has one lone pair.
·
Amines contain hydrogen bonding.
Example: ethylamine or ethanamine
Example: ethylamine or ethanamine
Amide
·
Amides are indicated by the suffix -amide.
·
Amides contain R-CONH2 functional
group, nitrogen has one lone pair.
·
Amide group can only be at the end of a carbon
chain
Example: ethanamide
Ester
·
Esters are indicated by the suffix -oate.
·
Esters contain R-COOCH- functional group, one
oxygen forms a double bond with carbon and another oxygen forms to single bonds
and joins two carbons.
·
Esters are derived from carboxylic acid and
alcohol.
·
Their naming is composed of two words. For
example, ethyl butanoate. The first part comes from the alcohol, while the
second part comes from the carboxylix acid. Ethyl butanoate was created by
condensation reaction of ethanol and butanoic acid.
Example:
ethyl methanoate
Nitrile (cyanides)
·
Nitriles are indicated by the suffix -nitrile.
·
Nitriles contain R-CO functional group, where
the carbon forms a triple bond with the oxygen.
·
The shortest nitrile is ethanenitrile
Example: ethanonitril
20.1.2 Apply IUPAC rules for naming compounds containing up to six carbon atoms with one of the following functional groups: amine, amide, ester and nitrile.
Look at 20.1.1. Standard rules in naming these compounds
apply. We put number between the name of an alkyl group and the suffix or
prefix depending to which carbon the functional group is attached. Remember
that nitriles and amides cannot exist as methyl-suffix.
20.1 Nucleophilic substitution reactions
20.2.1 Explain why
the hydroxide ion is a better nucleophile than water.
OH- is a better nucleophile than H2O
because its negative charge is more dense. While water only forms hydrogen
bonding, hydroxide ion has an ionic charge.
Nucleophile activity:
CN- > OH- > NH3 > H2O
20.2.2 Describe and
explain how the rate of nucleophilic substitution in halogenoalkanes by the
hydroxide ion depends on the identity of the halogen.
The rate of nucleophilic substitution depends on the
identity of the halogen involced. The rate of nucleophilic substitution depends
on the bond strength, not on the electronegativity of species. In theory,
fluorine is the most electronegative halogen, so it will be repulsed the most
by the negative nucleophile. However, in practice the strength of the fluorine
bond is so high that flouroalkanes will not take part in nucleophilic
substitution.
The reactivity of halogenalkanes increases with decreasing
electronegativity and the halogen-carbon bond strength. Therefore, the
reactivity of halogenalkanes is:
RI > RBr > RCl
20.2.3 Describe and
explain how the rate of nucleophilic substitution in halogenoalkanes by the
hydroxide ion depends on whether the halogenoalkane is primary, secondary or
tertiary.
Tertiary halogenalkanes react through SN1. This
reaction is first order in relation to the halogenalkane and independent of the
nucleophile.
Primary halogenalkanes react through SN2. This
reaction is first order with respect to the halogenalkane and to the
nuclephile. Both species are involved in the rate-determining step.
Empirically, it was found that SN1 reactions are
quicker than SN2 reactions – the activation energy necessary to
create the carbocation intermediate (SN1) is lower than the
activation energy necessary to form the transition state of SN2.
The nucleophilic substitution of the secondary
halogenalkanes can proceed by both SN1 and SN2. Therefore,
the reactivity of halogenalkanes with OH- is:
Tertiary >
Secondary > Primary
20.2.4 Describe,
using equations, the substitution reactions of halogenoalkanes with ammonia and
potassium cyanide.
Reaction with ammonia has 5 steps:
1.
SN2: Ammonia approaches the carbon bonded to
a halogen from the opposite side of the halogen. Ammonia donates the carbon its
lone pair of electrons. Transition state is formed when ammonia and halogen are
in the same distance from the carbon.
2.
Ammodia bonds to
the carbon and repulses the halogen ion.
3.
The ammonia bonded to the carbon, which had lost its
halogen has a nitrogen inside (ammonia – NH3). This nitrogen bonded
to the carbon has a positive charge.
4.
A new ammonia nuclephile approaches the positive charge
in ammonia bonded to carbon. The new ammonia is actually approaching one of the
hydrogens bonded to the old ammonia.
5.
The new ammonia takes one of the old ammonia’s
electrons and creates ammonium ion (NH4+). The ammonia,
which had lost the hydrogen now has one unbounded pair of electrons – an
alkylamine is created.
Conditions: excess ammonia is added,
heat is added, the container is sealed so that ammonia gas does not escape
Reaction with cyanide has 3 steps:
1.
Cyanide (CN-) approaches the carbon bonded
to a halogen from the opposite side of the halogen. Cyanide donates the carbon
its lone pair of electrons.
2.
The transition state is formed, when halogen and
cyanide are in an equal state of repulsion.
3.
Nitrile is formed. Halogen ion stays in the system.
Conditions: heated
under reflux, alcoholic solvent to prevent OH- ions present in CN-
solution from reacting
20.2.5 Explain the
reactions of primary halogenalkanes with ammonia and potassium cyanide in terms
of the SN2 mechanism.
See explanation given at 20.2.5.
20.2.6 Describe,
using equations, the reduction of nitriles using hydrogen and a nickel catalyst.
A nitrile is
converted into an amine by reduction using hydrogen with a nickel catalyst.
Notice this is
analogous to hydrogenation of alkenes (converting an alkene into an alkane also
using hydrogen and nickel catalyst).
Significance of this process is that it adds another carbon
to the carbon chain.
20.3 Elimination reactions
20.3.1 Describe,
using equations, the elimination of HBr from bromoalkanes.
The elimination reactions transform bromoalkanes into
alkenes.
To eliminate HBr from bromoalkanes we use OH-
ions, possibly, from NaOH.
Normally, OH- reacts in an acidified solution of
Cr2O7-2 to create alcohols through the
nucleophilic substitution. However, in the elimination something very different
happens.
We use special condions:
we use alcoholic solvent and heat the system under the reflux.
Process:
1.
Thanks to the alcoholic solvent OH- acts as
a Bronsted-Lowry base a picks up one of the hydrogens (hydrogen proton) bonded
to a carbon, bonded to a carbon, which is bonded to a halogen.
2.
The electron pair from the hydrogen goes toward forming
a carbon-carbon double bond with the carbon bonded to the halogen.
3.
The double bond repulses the halogen, which leaves the
halogenalkane as a halogen ion.
OR
(Know
the first one)
20.3.2 Describe and explain the mechanism for the elimination of HBr
from bromoalkanes.
The reaction is elimination
reaction and creates alkenes. The reaction requires heat, alcoholic solvent and
reflux.
Also, see 20.3.1.
20.4 Condensation reactions
20.4.1 Describe, using equations, the reactions of alcohols with
carboxylic acids to form esters and state the uses of esters.
Alcohols react with carbocylic
acids to form esters. This is called a condensation reaction because water is
“condensated” as a result of it. In the reaction the acid loses a hydrogen and
alcohol loses the OH- group – these two form water. Then, the
single-bonded oxygen from the acid attaches to the carbon, which used to be
bonded to the OH- group.
Conditions:
sulphuric acid, heat
Use: Scent (they smell good), Solvents
20.4.2 Describe, using equations, the reactions of amines with
carboxylic acids.
The condensation reaction, when
water gets created in addition to the main products, can also happen with
amines and carboxylic acids. This creates amides.
In this reaction the carboxylic acid loses its –OH group and the amine loses
its –H. The lost atoms form H2O and the nitrogen bonds to the
carbon.
The product of this reaction is
secondary amide.
20.4.3 Deduce the structures of the polymers formed in the reactions of
alcohols with carboxylic acids
Combination of many alcohols and
carboxylix acid makes a polyester. Polyesters is composed of building blocks,
monomers. Monomers are the esters created from alcohol and carboxylic acid as
outlined in 20.4.1. The monomers can join into long structures called polymers.
Polymers are created by alcohols with fuctional groups on two sides and carboxylic
acids with two functional groups. !!!There are two functional groups on each
monomer creating the polymer!!!
Example: polyester formed from
ethane-1,2-diol and benzene-1,4-dicarboxylic acid
This is the repeating unit:
20.4.4 Deduce the structures of the polymers formed in the reactions of
amines with carboxylic acids.
Condensation reaction between
di-carboxylic acids and di-amines produces condensation polymers called
polyamides, the most common of it being Nylon 6,6 produced from hexanedioic acid and
1,6-diaminohexane. !!!There are two functional groups on each monomer
creating the polyamide!!! In this reaction the carboxylic acid loses its –OH
group and the amine loses its –H. The lost atoms form H2O and the
nitrogen bonds to the carbon. The functional group in the polymer is called
peptide / amide.
20.4.5 Outline the economic importance of condensation reactions.
Nylon is used in the manufacture
of climbing ropes, carpeting, gears and automobile fuel tanks. Nylon is very
durable, non-abrasive and easy to shape into different forms.
PET is plastic used to create
containers, used to carry food. Most bottles are made of PET.
20.5 Reaction Pathways
20.5.1 Deduce reaction pathways given the starting materials and the
product.
For example, the conversion of
1-bromopropane to 1-butylamine can be done in two stages: 1-bromopropane can be
reacted with potassium cyanide to form propanenitrile, which can then be
reduced by heating with hydrogen and a nickel catalyst.
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