Aldehyde_ketone_acylchlorides.ppt - CashmereChemistry

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Aldehyde_ketone_acylchlorides.ppt - CashmereChemistry Powered By Docstoc
					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

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    A few drops of    Add ammonia solution
 solution          NaOH to form a    till the precipitate
                   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 Cr2072- /H+
and give all colour changes.

(X3) 3 CH3CH2CH2OH                      6H     6e-
                          3 CH3CH2COH + 2H++ + 2e-

      Cr2072- + 14H+ + 6e-           2Cr3+ + 7H20

 Full balanced redox equation
3CH3CH2CH2OH + Cr2072-+ 8H+     3CH3CH2COH + 2Cr3++ 7H20
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
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
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     Add the ‘B’ solution until
solution into each test       a precipitate forms.
Keep adding ‘B’ solution
until the precipitate has
redissolved and the
solution is a clear, dark
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
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.

(X5) 5 CH3CHOHCH3                        2H + 2e-
                          5 CH3COCH3 + +10H+++ 10e-

(X2) 2MnO4- ++16H+ + 10e-
      Mn04-    8H+ + 5e-             Mn     8H
                                    2Mn2+ + 4H20

 Full balanced redox equation
5CH3CHOHCH3 + 2Mn04-+ 6H+       5CH3COCH3 + 2Mn2++8H20
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
Producing an Aldehyde from a Primary Alcohol
using distillation

 Dropping funnel with K2Cr2O7
 and ethanol

           flask with hot

The alcohol/dichromate mixture is added to the acid and
only the first 2-3 mls of distillate is collected – why?
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.
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


              Acetic               Aspirin   Carboxylic
              anhydride                      acid
       Carboxylic Acids       R   C

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
               R   C
       Reactions of Acid Chlorides
 2. Forming esters from acid chloride – reagents are
 a primary alcohol
      From acyl chloride            Ester

                    R      C                 +   HCl
                               O     R’ From alcohol

CH3COCl + CH3OH                     CH3COOCH3 + HCl

ethanoyl chloride + methanol       methyl ethanoate + hydrogen
      Reactions of Acyl Chlorides
 3. acyl chlorides form amides – reagent
 ammonia and heat
          Functional                            O
          group of the              R       C

CH3COCl + 2NH3                CH3CONH2 + NH4Cl
ethanoyl chloride + ammonia    ethanamide       +    ammonium
      Reactions of Acyl Chlorides
4. Acyl chlorides forming
N - substituted amides – reagent amine
 N substituted amide                R       C
                                                NH   R’

           O                                O
CH3    C        + NH2 CH3     CH3       C             + HCl
           Cl                               NH CH3
Ethanoyl chloride + aminomethane    N – methyl ethanamide
                                        + hydrochloric acid
      Reactions of Acyl Chlorides

 Acyl chlorides react with water to form
 acidic solutions

            O                          O
CH3     C          O   H    CH3    C            + HCl
            Cl H                       O    H

ethanoyl chloride + water   ethanoic acid   + hydrogen
        Making Acyl Chlorides
Carboxylic acids react with PCl3, PCl5 or
SOCl2 (not HCl) to form Acyl chlorides
by substituting the –OH for a Cl.

                 O      PCl5               O
      CH3    C                  CH3    C
                 OH                        Cl

 Ethanoic acid + PCl5          ethanoyl chloride
Lysergic acid diethylamide (LSD)

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
If ethanoic acid is reacted with methanol the ester,
methyl ethanoate is formed.
                      H3C   C
           O                    O

H3C    C        + CH3OH                    H3C   C             + H2O

           OH                       heat             O   CH3
Hydrolysis of esters
The 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 + H2O / H+            CH3CH2COOH       + CH3OH

methyl propanoate                propanoic acid       methanol
Hydrolysis of esters

  Hydrolysis in NaOH soln gives alcohol + the sodium salt of
the carboxylic acid.

 CH3CH2COOCH3 + NaOH         CH CH COO Na
                                    3   2
                                                 +   + CH3OH

 methyl propanoate               sodium                methanol
hydrolysing an
ester (methyl salicylate)

                     Methyl salicylate
        C O    CH3

 Write the products if we hydrolyse
 it in acid ie H2O/H+
hydrolysing an ester (methyl salicylate)
 if we hydrolyse it in acid ie H2O/H+
 the products are

          C O     H   +   CH3OH


 Salicylic acid
Write the products if we hydrolyse
it in alkaline conditions ie

         C O – Na   +   +   CH3OH

sodium salicylate           methanol
Turn to the last page in the booklet
Ester 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
Ethanoyl chloride                  H+     CH3COOH + C2H5OH
                              NH3 / ethanol

                    Ethyl ethanoate OH-       salt   + ethanol

                    Formula? 2H5
                    CH3COOC               CH3COO- + C2H5OH
+ Conc H2SO4

  Ethanoic acid
  CH3COOH                    C2H5NH2      N-ethylethanamide
                             / ethanol    CH3CONH2C2H5
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.

      H                                               O

   H C       OH   +     R1COOH            H2C    O   C       R1
   H C       OH   +     R2COOH                                   + 3H2O
                                          HC     O    C      R2
                  +     R3COOH                        O
   H C       OH
                                          H2C    O    C      R3
                      3 fatty acids             Fat or oil
The 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.

                       O                            H
          O                       H

H2C O         OR       C   R1                           OH + R1COONa+
H C       C                     H C   OH
 2                 1
                                                  H C
          O                     H C   OH
 HC   O   C       R2   O        H C   OH

H2C       C
              OR C 3
                           R2+ 3NaOH
                                                  H C   OH + R COONa+
              O        C   R3
                                                  H C   OH + R 3COONa+
                                                          3 soap molecules
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

HO        CH2         CH2         CH        CH2        CH         CH2         CH2         CH3
      C         CH2         CH2        CH         CH        CH2         CH2         CH2


polar carboxylate                       non-polar hydrocarbon
head (dissolves in                      tail (dissolves grease)
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