Download From carb acid till end ch 4

From carb acid till end ch 4
4.5
Carboxylic Acids
Key Ideas
Intended Student Learning
1. Carboxylic acids can be produced by
the oxidation of aldehydes or primary
alcohols.
1. Identify the aldehyde or primary
alcohol from which a carboxylic acid
could be produced by oxidation, given
its structural formula.
2. Carboxylic acids are weak acids and, to
a small extent, ionise in water.
1. Write an equation for the ionisation of
a carboxylic acid in water.
3. Carboxylic acids react with bases to
form ionic carboxylate salts.
1. Write equations for the reactions of
carboxylic acids with hydroxides,
carbonates, and hydrogen carbonates,
and describe changes that accompany
these reactions.
4. The salts of sodium and potassium
carboxylates are soluble in water
because of the ion–dipole attraction
between the ions and water.
1. Explain why some drugs with carboxyl
groups are usually taken in the form of
their salts.
Preparation of carboxylic acid
• How can it be prepared??
• Hint Primary alcohol
• Complete Q4.18 page 262
H H
O
| | //
H-C-C-C
| | \
H H
OH
propanoic acid
(from propan + oic acid)
O
//
H-C
\
OH
formic acid
methanoic acid
Ionisation of carboxylic acids in
water
• Ionisation – molecular substance react
with water to form ionic products.
• Equilibrium process
• Carboxylic acids are weak acids- if soluble
in water, partially ionise in water to form
hydronium and carboxylate ions.
2. Write an equation for the ionisation of a
carboxylic acid in water.
• General formula to show ionisation of
carboxylic acids in water??
• RCOOH + H2O
H3 O+ + RCOO-
Neutralisation of carboxylic acids
• Can be neutralised in three ways: by
hydroxide ion, carbonate ion and
hydrogencarbonate ions.
• Three equations are important page 262
• Last paragraph is important
• Hydroxide ions:
RCOOH + OH-
RCOO- + H2O
• Carbonate ions:
2RCOOH + CO 32-
2RCOO- + H2O + CO2
• Hydrogencarbonate ions:
• RCOOH + HCO 3-
RCOO- + H2O + CO2
The solubility of carboxylate salts
• Potassium and sodium carboxylates are soluble in water.
Why?? Strong ion dipole bonds that form between the
negatively charged carboxylate ions and polar water
molecules
• Fig 4.8 important page 262
• Complete Q4.19 pg 263
Drugs with carboxyl groups as part
of their molecular structure
• Pain relieving drugs such as aspirin and iburofen has a
carboxyl group as part of its compound.
• Long non polar chains- therefore not soluble in water.
So how do we make it water
soluble??
• In tablet form, these drugs are mixed with
sodium hydrogencarbonate. So when it is
mixed in water, carboxylic acid is
converted to water soluble carboxylate
ions. Co2 is given off during dissolving
process.
• Onc in stomach it reacts with HCl and is
converted back to carboxylic acid.
• Complete Q4.20 page 263
• Complete Q4.21 page 264
4.6
Key Ideas
Amines
Intended Student Learning
Owing to the presence of an unbonded
electron pair, amines are able to act as
bases and accept H+ ions.
Draw the structural formula of the
protonated form of an amine, given the
structural formula of its molecular
form, and vice versa.
Amines are classified as primary,
secondary, or tertiary.
Identify an amino group in an amine as
primary, secondary, or tertiary, given
the structural formula.
The salts of amines are soluble in water Explain why some drugs with amine
because of the ion–dipole attraction
groups are usually taken in the form of
between the ions and water.
their salts.
Ammonia is the simplest example of a functional group called amines. Just as there are
primary, secondary, and tertiary alcohols, there are primary, secondary, and tertiary
amines depending on the number of hydrogen atoms replaced.
Similarity!!
• Unbonded pair of electrons on the nitrogen
atom. Because of this molecules can
accept one proton each from an acid.
Therefore Amines are bases. When an
amine molecule accepts a proton in this
way, a positively charged substituted
ammonium ion is formed and this is called
the protonated form of the amine.
Owing to the presence of an unbonded
electron pair, amines are able to act as
bases and accept H+ ions.
Draw the structural formula of the
protonated form of an amine, given the
structural formula of its molecular
form, and vice versa.
• Complete Q 4.22 page 265
Solubility of protonated amines
• Protonated amines are soluble in water.
• This is because of the strong ion-dipole
bonds that form between the positively
charged protonated amine ions and polar
water molecules.
• Amines that are insoluble in water are
reacted with acid to form water soluble
protonated form of the amine Fig 4.9 pg
266
The salts of amines are soluble in water because of the
ion–dipole attraction between the ions and water.
Dissolving process
Drugs with amine groups as
part of their molecular structure
• Amines are insoluble
• Drugs are usually administered in the
protonated form because it is water
soluble
• Note: only tertiary amine group is
protonated.
• Complete Q4.23 page 267
Questions
• RCOOH + OH-
+ NaOH
RCOO- + H2O
+ NaOH
• RCOOH + HCO 3-
RCOO- + H2O + CO2
+ NaHCO3
• Carbonate ions:
2RCOOH + CO 32-
2RCOO- + H2O + CO2
+ Na2CO3
With carbonates and hydrogencarbonates
In both of these cases, a salt is formed together with carbon dioxide and water. Both
are most easily represented by ionic equations.
For carbonates:
. . . and for hydrogencarbonates:
•If you pour some dilute ethanoic acid onto some white sodium carbonate or
sodium hydrogencarbonate crystals, there is an immediate fizzing as carbon
dioxide is produced. You end up with a colourless solution of sodium ethanoate.
With sodium carbonate, the full equation is:
•. . . and for sodium hydrogencarbonate:
With amines
Amines are compounds in which one or more of the hydrogen atoms in an
ammonia molecule have been replaced by a hydrocarbon group such as an alkyl
group. For simplicity, we'll just look at compounds where only one of the hydrogen
atoms has been replaced. These are called primary amines.
The small amines are very similar indeed to ammonia in many ways. For example,
they smell very much like ammonia and are just as soluble in water. Because all
you have done to an ammonia molecule is swap a hydrogen for an alkyl group, the
lone pair is still there on the nitrogen atom.
That means that they will react with acids (including carboxylic acids) in just
the same way as ammonia does.
For example, ethanoic acid reacts with methylamine to produce a colourless
solution of the salt methylammonium ethanoate.
Explain why some drugs with amine groups are
usually taken in the form of their salts
Polarity of carboxylic acid molecule
Shown above is the hydrogen bonded dimer structure that is found for carboxylic acid
molecules in a non-polar solvent like chloroform. The hydrogen bonds are shown in
yellow and occur between the hydrogen atom of one molecule, and the carbonyl
oxygen atom of the other. Note in the electrostatic potential surface models below how
this is just partial negative charge (carbonyl oxygen atom, red color) making an
electrostatic bond with partial positive charge (hydrogen atoms, blue color). Carboxylic
acids can "stick together" via hydrogen bonding, so they have relatively high boiling
points compared to other types of molecules with similar molecular weights.
Using the scheme below show the
charges in acetic acid (ethanoic)
H-bond - DIMERS
H-bond
Identify the aldehyde or primary alcohol from which a carboxylic acid could be
produced by oxidation, given its structural formula.
Write equations for the reactions of carboxylic acids with hydroxides,
carbonates, and hydrogen carbonates, and describe changes
that accompany these reactions.
1. Weak acids:
2. Reaction with metals:
3. Dilute ethanoic acid with sodium hydroxide solution - a colourless solution
containing sodium ethanoate. The only sign that a change has happened is
that the temperature of the mixture will have increased.
However, you would notice the difference if you used a slower reaction - for example
with calcium carbonate in the form of a marble chip. With ethanoic acid, you would
eventually produce a colourless solution of calcium ethanoate.
In this case, the marble chip would react noticeably more slowly with ethanoic acid than
with hydrochloric acid.
esters
The boiling points of esters are lower
than those of isomeric acids because
of the absence of hydrogen bonding
between molecules of the ester.
Predict and explain the boiling points of
esters in comparison with those of
isomeric acids.
An ester can be produced by a
condensation reaction between an alcohol
and a carboxylic acid.
Draw the structural formula of the ester
that could be produced by the
condensation reaction between an alcohol
and a carboxylic acid, given their
structural formulae, and write an equation
for the reaction.
The production of an ester from the
reaction of an alcohol and a carboxylic
acid is slow at 25°C.
Explain the use of heating under reflux
and the presence of a trace of concentrated
sulfuric acid in the laboratory production
of esters.
Esters may be hydrolysed under acidic or
alkaline conditions.
Identify the products of hydrolysis of an
ester, given its structural formula.
• Ester can be regarded as two alkyl groups
linked by an ester functional group.
Notice that the acid is named by counting up the total number
of carbon atoms in the chain - including the one in the -COOH
group. So, for example, CH3CH2COOH is propanoic acid, and
CH3CH2COO is the propanoate group.
Boiling points
The small esters have boiling points which are similar to those of aldehydes and
ketones with the same number of carbon atoms.
Like aldehydes and ketones, they are polar molecules and so have dipole-dipole
interactions as well as van der Waals dispersion forces. However, they don't form
hydrogen bonds, and so their boiling points aren't anything like as high as an
acid with the same number of carbon atoms.
Note: boiling points are lower than those of alcohols and carboxylic acids of similar
relative molecular mass.
molecule
CH3COOCH2CH3
CH3CH2CH2COOH
type
boiling point (°C)
ester
77.1
carboxylic acid
164
Solubility of ester
• Most esters are essentially insoluble in
water
• Water solubility decreases markedly with
increasing molecular mass.
Preparation of esters
• Any ester forming reactions are called
esterifications.
An ester can be produced by a condensation reaction between an
alcohol and a carboxylic acid, under reflux Catalyst is concentrated
sulfuric acid. Water is the other product
Reflux assembly page 269
• Reactions are slow
• Extended period of heating
• Reflux= process whereby reactants and products are boiled for a
long time.
• Reactant and product vapour pass into condenser where they are
condensed to liquid drops then fall back into the reaction mixture.
• Reflux process allows extended heating of the reaction mixture with
minimum loss of reactants and products by evaporation.
• Reaction is reversible and equilibrium is established in the reaction
vessel during reflux
• Equilibrium mixture contains similar quantities of both reactants and
products
• Yield of ester can be increased by using an excess of alcohol in the
reaction mixture to drive the reaction towards the products
Explain the use of heating under reflux
and the presence of a trace of
concentrated sulfuric acid in the
laboratory production of esters
.
• Complete Q4.24- Q4.25 pg 269-270
Hydrolysis of ester
• Esters undergo hydrolysis when refluxed
with aqueous acid or base.
• Reverse of esterification- water is
consumed as a reactant in a hydrolysis
reaction
Refluxed under acidic condition
(Hydrolysis)
• Acidic condition- products are carboxylic
acid and alcohol. Reaction is catalysed by
the acid.
• Large excess of water, position of
equilibrium favours the formation of
carboxylic acid and alcohol
Refluxed under basic condition
(Hydrolysis)
• In basic condition – sodium hydroxideproducts of hydrolysis are carboxylate salt
and alcohol
Esters may be hydrolysed under
acidic or alkaline conditions.
• Complete Q4.26 page 271
• Complete Q4,27 page 272
Practical Notes
Reflux- 1st step
• Many organic reactions are quite slow and
need heating to achieve a reasonable
reaction rate. However, most organic
chemicals are quite volatile, and if heated
they will evaporate and be lost. The
solution to this problem is to heat the
reaction mixture under reflux.
• This involves having the reaction mixture in a
flask which is attached to a vertical, open Liebig
condenser.
• Never attempt to stopper the Liebig
condenser. (This is quite a popular idea among
students "to stop the vapour escaping". If you
attempt to heat sealed glass apparatus it may
explode!
• There should be no problem with vapour
escaping - as it hits the cold surface of the
condenser it will condense and drip back in to
the flask.
• The flow of cold water through the condenser should be
slow.
• Most students seem to think that a fast flow will be more
efficient in cooling, but it makes little difference, and a
fast flow usually leads to floods!
• If vapour is escaping from the top of your condenser
reduce your rate of heating.
• Using a gauze over the Bunsen allows the mixture to boil
more smoothly. It is a good idea to use a few antibumping granules in the boiling liquid.
• A granules are added to a liquid before heating and
they encourage smooth boiling
• In reflux set up, the flask must never be
more than half-full.
• Heating should be controlled so that
solution is constantly simmering not
violently boiling.
Common mistakes during
refluxing
1. Water entering the condenser at the
wrong end. It should enter at the bottom.
If it enters at the top there are likely to be
air pockets which result in poor
condensation
2. Too fast a flow of water. This may result
in flooding.
3. Loose or inadequate clamping. This
may lead to later glassware breakage. Both
the pear-shaped flask and the Liebig
condenser should be clamped. The pearshaped flask should be clamped firmly and
just below the glass lip at the top. The Liebig
condenser can be lightly clamped - its main
purpose is to prevent the weight of the
rubber tubing from twisting it over
4. Quickfit not sealing. Always check just
prior to heating. The tightening of the clamps
often leads to the Quickfit joint coming apart.
Vapour will escape from your apparatus and
yield will be reduced. There is a safety
hazard from organic vapour escaping into
the lab. The vapour may also catch fire
resulting in another hazard.
5. Failure to lightly grease the Quickfit
joints. This leads to the apparatus being
difficult to pull apart after use.
6. Excessive heating. This may cause the
mixture to froth up the Liebig
condenser. Remember that the Bunsen
burner can be turned down at the gas tap
2nd step- Separation procedure!!
• As the reaction is in equilibrium, after the reflux all
reactants, products and catalysts are present together in
the pear shaped flask.
• As ester in not water soluble- mixture is washed with
water, water soluble liquids in mixture will collect with
water and form two layers in separating funnel.
• In separating funnel, two layers with different densities,
both are colourless. Hard to work out aqueous layer.
• To work out aqueous layer-drop of water is added and
see if water is incorporated or sinks to second layer.
The less dense organic layer floats on top of the more
dense aqueous layer.
• The stop cock on the separating funnel is opened to
allow the more dense aqueous layer to be run off,
leaving the less dense organic layer containing the ester
in the separating funnel.
• Calcium carbonate (CaCO3), or a similar neutralizing
agent, is added to the mixture to react with any
unreacted acid present in the mixture.
• calcium carbonate + acid → water soluble salt + carbon
dioxide gas + water
• Final washing with water
• This removes any traces of water-soluble
materials (such as the salts formed in the
last washing) from the separating funnel.
• Stand over CaCl2
• CaCl2 is a dehydrating agent and so
removes any traces of water trapped in the
ester layer. There was so much water that
it formed a new layer.
7. No tripod and gauze. The flask can be
directly heated if the Bunsen is held and
played across the flask so as not to overheat
it. However, if the Bunsen is static the flask
will overheat leading to bumping and
charring. The gauze spreads the heat and
reduces the probability of overheating. A few
anti-bumping granules also encourage
smooth boiling. The use of an electric
heating mantle would be a safer alternative.
Distillation-
3rd
part
• Distillation
• Distillation allows the separation of liquids
that have different boiling points as each
liquid has its own boiling point and the
temperature only rises when that liquid
has been completely boiled off. The ester
had the highest boiling point and so was
collected last.
Each liquid in the mixture has a different
boiling point and the boiling point of ester is
approx 126. Distillation process will
continue, fluids will be collected, discarded
until temp reaches approx 126. At this time,
new beaker is used to collect distillate until
temperature begins to rise at such pointpure ester
• The flask must never be more than half-full
and anti-bumping granules should always
be used.
• Distilallation should be at a rate that no
more than 2 drops per second of distillate
is obtained.
• Temperature should be continuously
monitored.
Position of thermometer
• The placement of thermometer bulb is just
below the level of the sidearm of the
distillation head. If the bulb were placed
higher than this position, it would not be in
the vapour path and consequently would
show an error of low reading for the boiling
point.
Few more things to note!!
• Remember boiling chips!- promotes even
heating of the solution….NO SUPER
HEATING!!!
Excessive heat will lead to some solution
entering the condenser and therefore
contaminating the distilate
• Condenser should be at an angle!!
• Why> to ensure vapour that has
condensed will actually flow into the
collecting flask and not back into the pear
shaped flask
• When removing hose!!
• Remove the hose from the tap and allow
the water in the condenser to run into the
sink. Place one end of the condenser in a
beaker of hot water to soak and soften the
hose. Reverse the condenser to remove
the other hose
4.8 Amides
Key Ideas
Intended Student Learning
An amide can be produced by a
condensation reaction between an
amine and a carboxylic acid.
Draw the structural formula of the
amide that could be produced by the
condensation reaction between an
amine and a carboxylic acid, given
their structural formulae.
Amides may be hydrolysed under
acidic or alkaline conditions.
Identify the products of hydrolysis
of an amide, given its structural
formula.
Production
• Condensation reaction - spot the differences
Hydrolysis
Identify
• Products of hydrolysis of the amide, given
its structural formulae
4.9 Proteins
Key Ideas
Intended Student Learning
Amino acids contain a carboxyl group and an amino
group.
Determine whether or not a compound is an amino
acid, given its structural formula.
Amino acids can self-ionise to produce an ion.
Draw the structural formula of the product formed
when an amino acid self-ionises.
Proteins are large molecules in which amide groups
link monomer units. In proteins the amide group is
called a ‘peptide link’ or a ‘peptide bond’.
Identify the amide group and deduce the structural
formula(e) of the monomer(s), given the structural
formula of a section of a protein.
Proteins are polyamides consisting of covalently
bonded long chains of amino acid units.
Write the general formula of amino acids and
recognise their structural formulae.
Proteins have sites that allow hydrogen bonding
between sections of chains and between the chain
and water.
Identify where hydrogen bonding can occur between
protein chains or between the chain and water, given
the structural formula of a section of the chain.
The biological function of a protein is a consequence
of its unique spatial arrangement.
Explain why the biological function of a protein (e.g.
an enzyme) is altered if its spatial arrangement is
altered.
Changes in pH and temperature disrupt the
secondary interactions, and hence the spatial
arrangements, of a protein chain.
Explain why proteins are sensitive to changes in pH
and temperature.
4.10 Triglycerides
Key Ideas
Intended Student Learning
Edible oils and fats are esters of
propane-1,2,3-trio (glycerol) and various
carboxylic acids. The carboxylic acids are
unbranched and usually contain an even
number of carbon atoms between twelve and
twenty.
Draw the structural formula of an edible oil
or fat, given the structural formula(e) of the
carboxylic acid(s) from which it is derived.
Triglycerides can be hydrolysed to produce
propane-1,2,3-triol and various carboxylic
acids.
Identify the alcohol and acid(s) from which a
triglyceride is derived, given its structural
formula.
Edible oils are liquids at 25°C and are
commonly obtained from plants and fish.
Edible fats are solids at 25°C and are
commonly obtained from land animals.
Identify the most likely source of a
triglyceride, given its state at 25°C.
Most liquid triglycerides contain a larger
proportion of unsaturated carbon chains than
solid triglycerides contain.
Describe and explain the use of a solution of
bromine or iodine to determine the degree of
unsaturation of a compound. Draw the
structural formula of the reaction product.
Liquid triglycerides can be converted into
triglycerides of higher melting point by a
process that involves the addition of
hydrogen under pressure and at increased
temperature, in the presence of a catalyst.
Explain the role of pressure, temperature,
and a catalyst in the hydrogenation of liquid
triglycerides.
4.11 Carbohydrates
Key Ideas
Intended Student Learning
Carbohydrates are naturally occurring sugars
and their polymers. They usually have the
general formula CxH2yOy. They are defined
more precisely as either polyhydroxy aldehydes
or polyhydroxy ketones, or their polymers.
Given its structural formula, determine the
molecular formula of an organic compound, and
whether or not it is a carbohydrate.
Carbohydrates can be classified as
monosaccharides, disaccharides, or
polysaccharides.
Write molecular formulae for glucose, and for
disaccharides and polysaccharides based on
glucose monomers.
Polysaccharides are produced by the
condensation of many monosaccharide units
linked in chains by covalent bonds.
Identify the repeating unit and draw the
structural formula of the monomer, given the
structural formula of a section of a
polysaccharide derived from one monomer.
Glucose molecules can occur in either a chain
form or a ring form. There is equilibrium
between the two structures. In the chain form an
aldehyde group is present.
Explain the ability of glucose to react as an
aldehyde when in chain form but not when in
ring form.
Many simple carbohydrates are soluble in water,
whereas polysaccharides are insoluble in water.
Explain the differences in solubility in water of
simple carbohydrates and polysaccharides in
terms of the size of the molecules and the
number of hydroxyl groups.
Reflux and distillation Procedure
• Larger esters tend to form more slowly. In
these cases, it may be necessary to heat
the reaction mixture under reflux for some
time to produce an equilibrium mixture.
The ester can be separated from the
carboxylic acid, alcohol, water and
sulphuric acid in the mixture by fractional
distillation.