from carb acid till end ch 4. 4.5 carboxylic acids key ideasintended student learning 1.carboxylic...
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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 \
OHformic 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 3
2- 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 Amines
Key Ideas Intended Student LearningOwing 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 because of the ion–dipole attraction between the ions and water.
Explain why some drugs with amine groups are usually taken in the form of 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- RCOO- + H2O
+ NaOH
+ NaOH
• RCOOH + HCO 3- RCOO- + H2O + CO2
+ NaHCO3
• Carbonate ions:2RCOOH + CO 3
2- 2RCOO- + H2O + CO2
+ Na2CO3
With carbonates and hydrogencarbonatesIn 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 aminesAmines 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.
estersThe 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 pointsThe 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.
moleculetype boiling point (°C)
CH3COOCH2CH3 ester 77.1
CH3CH2CH2COOH 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 hydroxide- products 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 anti-bumping 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 pear-shaped 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 point- pure 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 LearningAn 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 LearningAmino 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 LearningEdible 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 LearningCarbohydrates 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.