Distinguishing between aldehydes and ketones. Adehydes and ketones can be structural isomers of each...
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Transcript of Distinguishing between aldehydes and ketones. Adehydes and ketones can be structural isomers of each...
Distinguishing between aldehydes and ketones
Adehydes and ketones can be structural isomers of each other. Aldehydes are produced by the oxidation of a primary alcohol and have the C=O on end carbon.
Ketones are produced by the oxidation of a secondary alcohol and have the C=O on a carbon atom in the middle of the carbon chain.
Aldehydes can be further oxidised to carboxylic acids, while ketones are not oxidised further.
The tests we use to distinguish between aldehydes and ketones all involve oxidising the aldehyde but not the ketone.
While acidified dichromate or permanganate will distinguish between aldehydes and ketones, they are strong oxidising agents which will also change colour in the presence of alcohol or other reagents.
Oxidising agents which oxidise aldehydes are:
• Tollen’s reagent• Benedict solution• Fehling’s solution
The ‘silver mirror’.
You’re more likely to get a mirror with a very clean test tube.
Tollen’s reagent
Tollen’s reagent is [Ag(NH3)2]+ which, when reduced, forms Ag(s). It must be freshly prepared.
Silver nitrate solution
A few drops of NaOH to form a precipitate.
Add ammonia solution till the precipitate dissolves.
Add a few drops of the aldehyde or ketone, shake, and warm gently.
The ketone remains colourless, the aldehyde will react.
If you are lucky you will get a ‘silver mirror’ as elemental silver forms on the inside of the test tube.
Less spectacular, but just as valid is the formation of a grey or black precipitate, also of elemental silver.
A grey precipitate of silver.
Tollens’ Test Tollens reagent is a complex of Ag(NH3)2
+ .
When heated with an aldehyde a redox reaction occurs producing a silver mirror on the inner surface of the test tube.
The aldehyde is oxidised to a carboxylic acid.
The reduction half-equation is
• Ag+(aq) + e Ag(s)
•If Tollens’ reagent is heated with a ketone or an alcohol no reaction occurs.
Tollens’ Test
The overall reaction is :
RCHO + 2Ag(NH3)2OH RCOONH4 + 2Ag + H2O + 3NH3
Silver mirror
Task – write the balanced redox reactions for the oxidation of propan-1-ol to propanal using Cr207
2- /H+ and give all colour changes.
CH3CH2CH2OH
Cr2072-
CH3CH2COH + 2H+
+ 2e-
2Cr3+
+ 7H20
+ 14H+
+ 6e-
(X3) 3
3
+ 6H+
+ 6e-
3CH3CH2CH2OH + Cr2072- + 2Cr3+
+ 7H20
+ 8H+
3CH3CH2COH
Full balanced redox equation
Benedict solution
Benedict solution is an alkaline solution of Cu2+, complexed with citrate ions to keep it in solution. It is a mild oxidising agent which is reduced to Cu+.
In the alkaline solution the Cu+ is in the form of Cu2O which is a brick-red precipitate which is a positive test for an aldehyde.
Take about 2 mL of Benedict solution in each of two test tubes.
Add a few drops of aldehyde to one tube, and ketone to the other tube, and shake to mix.
Heat the mixture by putting the tubes in hot water. Shake several times to mix.
A reaction has occurred in the left hand (aldehyde) tube, but not in the right hand (ketone) tube.
If you wait long enough you will see the red-brown precipitate of Cu2O form.
Benedicts and Fehlings
Aldehydes reduce the copper (II) ions in both Fehlings and Benedicts solution to form a reddish brown copper (I) oxide which is precipitated
The reaction is :
RCHO + 2Cu 2+ + 2OH- RCOO- + Cu2O + 3H+
Brick red precipitate
Fehling’s solution
Like Benedict solution, Fehling’s contains alkaline Cu2+, but Fehling’s uses potassium tartrate to complex the copper. The mixture is freshly prepared:
Pour a little Fehling’s A solution into each test tube.
Add the ‘B’ solution until a precipitate forms.
Keep adding ‘B’ solution until the precipitate has redissolved and the solution is a clear, dark blue.
A reaction occurs in the aldehyde tube as Cu2O forms. No reaction occurs in the ketone tube.
Add a few drops of aldehyde and ketone to separate tubes, shake, and heat in a beaker of hot water.
Glucose is an aldehyde and will form a mirror with Tollens and will also give a positive test with Benedicts and Fehlings solutions
Do you remember doing this test for sugars in yr10 ?
In all of these reactions (Tollen’s, Benedict and Fehling’s), the aldehyde is oxidised to the carboxylic acid while no reaction occurs to the ketone.
Task – write the balanced redox reactions for the oxidation of the oxidation of propan-2-ol to propanone using Mn04
- /H+ and give all colour changes.
CH3CHOHCH3
Mn04-
CH3COCH3 + 2H+
+ 2e-
Mn2+
+ 4H20
+ 8H+ + 5e-
(X5) 5
5
+ 10H+
+ 10e-
5CH3CHOHCH3 + 2Mn04- +
2Mn2+ +8H20 +
6H+ 5CH3COCH3
Full balanced redox equation
(X2)
2MnO4-
+ 16H+ + 10e- +
8H20 2Mn2+
Producing an Aldehyde from a Primary Alcohol
When forming the aldehyde (ethanal) from ethanol the alcohol and dichromate must be added to the hot concentrated H2SO4
We use a distillation technique to collect a volatile product
Producing an Aldehyde from a Primary Alcohol using distillation
Dropping funnel with K2Cr2O7 and ethanol
Distillation flask with hot H2SO4
The alcohol/dichromate mixture is added to the acid and only the first 2-3 mls of distillate is collected – why?
To make sure that the immediate production of ethanal with a lower boiling point of 21 deg C was vapourised and collected quickly and was not allowed to be converted to ethanoic acid.
The alcohol/dichromate mixture is added to the acid and only the first 2-3 mls of distillate is collected – why?
To make sure a reaction goes to completion one such as:
•Primary alcohol is completely oxidised to form a carboxylic acid
•A secondary alcohol is oxidised to form a ketone
You would use a reflux arrangement
Making Aspirin – using methylsalicylate
methylsalicylate
Acetic anhydride
Aspirin Carboxylic acid
Carboxylic Acids O
OH C R
Named with -oic on the end
They are all organic acids which are weak acids
Reactions of Carboxylic Acids
Carboxylic acids undergo 4 types of substitution reactions (of the –OH)
Reactions of Carboxylic Acids
1. Forming acid chlorides from carboxylic acids – reagents are PCl5,PCl3 or SOCl2
acid chlorides are named at the end by the –oyl group
Functional group of the acid chloride
O
Cl C R
Reactions of Acid Chlorides2. Forming esters from acid chloride – reagents are a primary alcohol
O
O C R
R’
From acyl chloride
From alcohol
+ HCl
CH3COCl + CH3OH CH3COOCH3 + HClethanoyl chloride + methanol methyl ethanoate + hydrogen chloride
Ester
Reactions of Acyl Chlorides
3. acyl chlorides form amides – reagent ammonia and heat
O
NH2
C RFunctional group of the amide
CH3COCl + 2NH3 CH3CONH2 + NH4Clethanoyl chloride + ammonia ethanamide + ammonium
chloride
4. Acyl chlorides forming N - substituted amides – reagent amine O
NH C R N substituted amide
Reactions of Acyl Chlorides
R’
O
Cl C CH3 + NH2 CH3
O C CH3 +
HClCH3NH
Ethanoyl chloride + aminomethane N – methyl ethanamide
+ hydrochloric acid
Acyl chlorides react with water to form acidic solutions
Reactions of Acyl Chlorides
O
Cl C CH3 H
HO
O C CH3
HO+ HCl
ethanoyl chloride + water ethanoic acid + hydrogen chloride
Carboxylic acids react with PCl3, PCl5 or SOCl2 (not HCl) to form Acyl chloridesby substituting the –OH for a Cl.
Making Acyl Chlorides
O
OH C CH3
O C CH3
Cl
Ethanoic acid + PCl5 ethanoyl chloride
PCl5
Lysergic acid diethylamide (LSD)
ESTERS
C
O
O
CH3
CH3
Carboxylic acids react with alcohols, in the presence of conc sulfuric acid as a catalyst, to form esters. The reagents are heated together to bring about a reaction. Any excess acid is neutralised by the addition of sodium carbonate. If ethanoic acid is reacted with methanol the ester, methyl ethanoate is formed.
+ CH3OH H3O
+
heat
C
O
O
CH3
CH3
+ H2OC
OH
O
CH3
Hydrolysis of estersThe hydrolysis of an ester in aqueous solution results in the break up of the ester and the formation of an alcohol and the carboxylic acid or carboxylate ion
*(depending on the pH of the solution). Hydrolysis in acid produces the alcohol + carboxylic acid
CH3CH2COOCH3 + CH3CH2COOH + CH3OH
H2O / H+
methyl propanoate propanoic acid methanol
Hydrolysis of esters
Hydrolysis in NaOH soln gives alcohol + the sodium salt of the carboxylic acid.
CH3CH2COOCH3 + NaOH CH3CH2COONa+ +
CH3OH
methyl propanoate methanol
sodium propanoate
O CH3C
O
OH
hydrolysing an ester (methyl salicylate)
Methyl salicylate
Write the products if we hydrolyse it in acid ie H2O/H+
O HC
O
OH
hydrolysing an ester (methyl salicylate)
Salicylic acid
if we hydrolyse it in acid ie H2O/H+ the products are
+ CH3OH
methanol
O – Na + C
O
OH
Write the products if we hydrolyse it in alkaline conditions ie H2O/NaOH
sodium salicylate
+ CH3OH
methanol
Turn to the last page in the bookletEster Hydrolysis read the method (the ester is methyl salicylate)
Change the method as follows:Weigh out 4.8 grams of NaOH place in boiling flask then add 20mls of water (careful! it may get very hot)
Measure out and add 5mls of oil of wintergreen (methyl salycilate) to flask
Place boiling chips in boiling flask and reflux carefully for 30 mins
Reflux – heating mixture without losing volatile substances
Distillation – using the deferring boiling points of substances to separate them
Write equations using structural formulae for the formation of the following esters
a) ethyl methanoate from ethanol and methanoic acid
b) butyl propanoate from butan-1-ol and propanoic acid
complete the following scheme giving all Formula
Ethyl ethanoate
Formula?
Ethanoyl chloride
Formula?
Ethanoic acid
CH3COOH
CH3COCl
CH3COOC2H5
Reagents?C2H5OH
C2H5OH
+ Conc H2SO4
Reagents?
H+ CH3COOH + C2H5OH
OH-
C2H5NH2
/ ethanol
salt + ethanol
CH3COO- + C2H5OH
ethanamide
CH3CONH2
NH3 / ethanol
N-ethylethanamideCH3CONH2C2H5
Fats and oils Fats and oils (lipids) are all triesters made from glycerol (propane-1,2,3-triol) and three long chain carboxylic acids (fatty acids) as shown below.
Glycerol is an example of a”triol” which has three -OH groups present. Each of these can form an ester link with a different carboxylic acid, for example the fat called stearin.
C
C
C OH
OH
OH
H
H
H
H
H + R1COOH
+ R3COOH
+ R2COOH
CH2
CH
CH2
O
O
O
C
O
C
C
O
O
R3
R2
R1
Glycerol3 fatty acids
+ 3H2O
Fat or oil
CH2
CH
CH2
O
O
O
C
O
C
C
O
O
R3
R2
R1C
C
C OH
OH
OH
H
H
H
H
H
SoapThe three ester links present in these molecules can be broken (or hydrolysed) by heating with sodium hydroxide solution.This releases the original glycerol molecule plus the sodium salts of the long chain fatty acids which are soaps. This “saponification” process is shown in the diagram below.
CH2
CH
CH2
O
O
O
C
O
C
C
O
O
R3
R2
R1
+ R1COONa+
+ R
2COONa+
+ R
3COONa+
C
C
C OH
OH
OH
H
H
H
H
HHeat
3 soap molecules
+ 3NaOH
CCH2
CH2
CH2CH2
CHCH
CH2CH
CHCH2
CH2CH2
CH2CH2
CH3OH
O
Soaps work because the tail of the molecule is a long non-polar hydrocarbon chain (from the fatty acid) which readily dissolves grease and dirt (as “like dissolves like”). Then the ionic carboxylate ion readily dissolves in water (which is also polar) and is able to carry away the grease with it in the rinse water.
non-polar hydrocarbon tail (dissolves grease)
polar carboxylate head (dissolves in water)
bestchoice.net.nz is an excellent revision website run by the Auckland University-To get on to this site you must use the following details – go to new user on main page register using the following ID: 919
Password: compound
Go on and try the organic section – lasts years class went nuts on this site
Log on using your own username and the supplied password and ID
Log on and give it a go! There is also a scholarship section-
This is your homework Tim!
I will log on and see your progress –
Go on and try the organic section – lasts years class went nuts on this site
bestchoice.net.nz is an excellent revision website run by the Auckland University-To get on to this site you must use the following details – go to new user on main page register using the following ID: 919
Password: compound
Go on and try the organic section – lasts years class went nuts on this site
Log on using your own username and the supplied password and ID