Topic 4.5 Organic Chemistry

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4.5a(i) Functional groups

A functional group is a set of atoms joined together in a molecule which give a molecule its properties.  Each family of organic compounds has its own functional group.

Task 4.5a(i)
Draw full displayed structural formulae for the examples above.
Name and give a molecular formulae for one more example of each homologous series.

4.5a(ii) Structural isomers

All isomers are compounds with the same molecular formula. e.g. C₄H₁₀ or C₂H₆O.  Structural isomers have atoms arranged in different orders.  They have similar bp's. 
e.g. CH₃CH₂CH₂CH₃ (butane) and CH₃CHCH₃  (2-methylpropane)  or
                                                              |
                                                             CH₃
CH₃CH₂OH (ethanol)   and  CH₃OCH₃ (methoxymethane)
CH₃CH₂CH₂CH₂Br (1-bromobutane) and CH₃CH₂CHCH₃ (2-bromobutane)
                                                                                                   |
                                                                                                   Br
Task 4.5a(ii)
Make a model of pentane and then models of all of its structural isomers.   Name and draw them.
Make a model of propan-1-ol and then all of its structural isomers.  Name and draw them.
Draw the structural formulae of two isomers of a compound for each of the functional groups in turn.
Other learning tasks.

4.5a(iii) Stereoisomerism
    Stereoisomers have the same molecular formula and the same structural formula.  The same atoms are arranged in the same order but with different orientations in space.  Two types of stereoisomers are geometric and optical isomers.

4.5a(iv) Geometrical (cis-trans) Isomerism
    Geometric isomers exist for some compounds with C=C double bonds in their molecules.  No rotation is possible around the double bond so different compounds result if atoms or groups are in positions on the same side or opposite sides of a double bond.  For example  1,2-dichlorethene can have its two chlorine atoms on the same side (cis) or opposite sides (trans) of the double bond.

Cl     Cl         cis-1,2-dichloroethene   Cl        H   trans-1,2-dichloroethene
  \     /                                                      \      /
  C=C                                                      C=C
  /     \                                                      /      \
H      H                                                  H       Cl
Task 4.5(iv)
Make models of the two isomers above.
Draw structural formulae for cis-1,2-dibromopropene and trans-1,2-dibromopropene.
Name and draw geometric isomers for butenedioic acid.
Draw structural isomers for C₅H₁₀O and C₅H₁₂O.

4.5a(v) Optical Isomerism
    Optical isomers exist for compounds in which molecules have an asymmetric carbon atom.  That is a carbon atom bonded to 4 different atoms or groups.  These atoms or groups can be arranged in two ways in space.  One arrangement is the mirror image of the other.  The mirror image is non-superimposable on the first arrangement in the same way that your left and right hand cannot be superimposed..  This is called chirality.   The carbon atom at the centre is called a chiral centre.  The isomers are called enantiomers e.g.
2-hydroxypropanoic acid

       H     COOH |  HOOC     H
         \   /             |            \   /
          C              |             C
         /   \             |            /   \
       CH₃ OH      |        HO     CH₃
                     mirror
Task 4.5a(v)
Draw the above structures showing the tetrahedral structure and the bond angles.
Make, name and draw three more structures which will have optically active isomers and label their chiral centres.
Observe each of your models in a mirror and draw their enantiomers.

Functional group and chirality test.  (Needs Excel.)

4.5a(vi) Optical activity and racemic mixtures
    Optical isomers differ only in the extent to which they rotate the plane of polarised monochromatic light. 

Optical isomers exist in two forms called enantiomers  The dextrorotatory (+) form rotates light to the right (clockwise) but the laevorotatory (-) rotates light to the left.  A sample of an optically active substance may contain both optically active isomers.  An equimolar mixture does not rotate light at all as equal and opposite rotations cancel.  This optically inactive mixture is called the racemic mixture or racemate.
Task 4.5 (a)(vi) 
Draw  both enantiomers for two different substances and label the chiral centres.
Draw two enantiomers in a racemic mixture.
Draw the above diagram showing the plane of rotated light with a laevorotatory enantiomer.
Draw diagrams to show observations as light passes through first one then two pieces of polaroid material.  Write explanations of what is seen.
Draw the above diagram but show the effect of an optical isomer between the two pieces of polaroid.

4.5(b)(i) Grignard reagents

Reactions of halogenoalkanes with magnesium to form Grignard reagents
A Grignard reagent is prepared by refluxing an alkyl or aryl bromide or iodide, dissolved in a dry ether with Mg turnings.
R - Br + Mg ---------> R-Mg-Br
C₂H₅-Br + Mg --------> C₂H₅-Mg-Br
                                    ethyl magnesium bromide
reagent:  magnesium
conditions: halogenoalkane dissolved in dry ether and heat electrically to boil under reflux
Draw a diagram to show this reflux apparatus.
Write equations to show the preparation of other named grignard reagents.

Reactions of Grignard reagents
In each case the grignard reagent formed is dissolved in ether.  
It is then added to another organic reactant to form an intermediate. 
When the reaction is complete aqueous acid is added to bring about hydrolysis of the intermediate and form the final product .  
The mechanism for each reaction involves the grignard agent as a nucleophile with a carbon atom holding a partial negative charge.  
    H  H  
     |    |
H-C- C^D--Mg-I^D+
     |    |
    H  H
Draw some structural formulae to show the charge distribution on other grignard reagents.

Grignard reagents and water
Alkanes can be made with this reaction but no other organic reactant is needed.
R-Mg-I + H₂O ----> R-H + Mg(OH)I
reagent: dilute HCl
conditions: room temperature
Draw structural formulae to show the first step of the above reaction as a nucleophilic substitution.  Show how the charge distribution in each molecule makes this possible.

Grignard reagents and carbonyl compounds
 1. Grignard + methanal -----> primary alcohol
R-MgI + HCHO ------> R-CH₂-O-MgI
                                          (intermediate)
R-CH₂-O-MgI + H₂O -----> RCH₂OH + Mg(OH)I

2. Grignard + aldehyde ------> secondary alcohol
R-MgI + CH₃CHO ------> R-CH(CH₃)-O-MgI
                                           (intermediate)
R-CH(CH₃)-O-MgI + H₂O -----> RCH(CH₃)OH + Mg(OH)I

3. Grignard + ketones --------> tertiary alcohol
RMgBr + (CH₃)₂CO -----> RC(CH₃)₂OMgBr
                                            (intermediate)
 RC(CH₃)₂OMgBr + H₂O-----> RC(CH₃)₂OH + Mg(OH)Br

Grignard reagents and carbon dioxide
Grignard + carbon dioxide -----> carboxylic acid
R-MgI + CO₂ ----------> R-CO₂ MgI
R-CO₂ MgI + H₂O------> R-CO₂H + Mg(OH)I
Name the reagents and products for some specific examples of the above general reactions of grignard reagents.
For any example above show how the charge distribution in each molecule leads to the formation of the intermediate.
Draw a spider diagram to show grignard reagent formation and its use to make 5 possible products.
Draw full structural formulae for intermediates and products for actual examples.
How do you make (a) propan-1-ol (b) propan-2-ol (c) butan-2-ol (d) 2-methylpropan-1-ol (E) 2-methylpropan-2-ol.  Answers

4.5(b)(ii) Reactions of carboxylic acids

Carboxylic acids with alcohols
An ester is the product when acid and alcohol are heated under reflux with conc. H₂SO₄ catalyst.
CH₃ COOH + CH₃ CH₂OH -----> CH₃ COOCH₂CH₃ + H₂O
                                                 ester (sweet smell)
Task 4.5bii 
The formation of polyesters
Molecules with two carboxyl groups can join to molecules with two hydroxyl groups to form polyesters.  These polyesters are important for artificial fibres.
nHOOCCH₂(CH₂)₂CH₂COOH + nHOCH₂ (CH₂)₄CH₂OH ------>
monomer 1,6-hexandioic acid         monomer hexan-1,6-diol
                   -(OCCH₂(CH₂)₂CH₂COOCH₂ (CH₂)₄CH₂O)_n-
                         polymer polyhexylhexanoate

ethane-1,2-diol  H-O-CH2-CH2-O-H benzene-  1,4-dicarboxylic acid  H-O-COC6H4-CO-O-H

polyester terylene  H-O-(COC6H4-CO- O-CH2-CH2-O)n-H

stability to light resistance to abrasion high tensile strength low melting point

net curtains conveyor and drive belts ropes, safety belts permanent pleating

Redraw the full structures of the alcohols, acids and esters above showing all bonds.

Carboxylic acids with lithium aluminium hydride
Lithium aluminium hydride, lithium tetrahydridoaluminate(III) reduces acids to alcohols.  LiAlH₄ is dissolved in ether.  Water is added at the end of the reaction to destroy any remaining LiAlH₄.
CH₃ COOH + 4H -----> CH₃ CH₂OH + H₂O
How do HCOOH and C₃H₇COOH react LiAlH₄ is dissolved in ether?
How can you make propan-1-ol and pentan-1-ol from carboxylic acids?

Carboxylic acids with phosphorus pentachloride
CH₃COOH (l) + PCl₅ (s) -------------> CH₃COCl (l) + POCl₃ (l) + HCl (g)
                                                                         ethanoyl chloride (an acyl halide)
     

Carboxylic acids with sodium carbonate and sodium hydrogencarbonate
The acid is neutralised in both cases with the formation of carbon dioxide.  This can be used in analysis to detect carboxylic acids.
2CH₃COOH + Na₂CO₃ ---> 2CH₃COONa + H₂O + CO₂
CH₃COOH + NaHCO₃ ---> CH₃COONa + H₂O + CO₂
Write equations to show other examples of the carboxylic acid reactions above and name the organic reactants and products.
Draw displayed structural formulae for the products in the reactions above.

4.5(b)(iii) Esters
Hydrolysis of the product (ester) is possible if a slightly conc. sulphuric acid catalyst is used.
CH₃COOCH₂CH₃ + H₂O <-----> CH₃COOH + C₂H₅OH
This is a reversible reaction.  Starting from either side results in an equilibrium position being established.
The hydrolysis reaction above also takes place in the presence of an alkali such as sodium hydroxide.  In this case however the ethanoic acid formed reacts with sodium hydroxide to form sodium ethanoate.  The reaction now proceeds to completion as ethanoic acid is removed.
Task 4.5b(iii)
Write equations for the acid hydrolysis of ethyl methanoate and the alkaline hydrolysis of methyl ethanoate.
Describe and explain the differences in yield in these reactions.

4.5(b)(iv) Carbonyl compounds

with hydrogen cyanide (nucleophilic addition)
HCN made in situ from KCN and HCl.  Alkali is added to ensure a reasonable concentration of CN^- by shifting the following equilibrium to the left:
CN^- + H₂O <-----> HCN  +  OH^-
CH₃CHO + HCN ------> CH₃CH(OH)CN
ethanal                        2-hydroxypropanonitrile (a cyanohydrin)
This is a nucleophilic addition reaction with a two step mechanism.
                CH₃                          CH₃
            |                           |
    N=C^-   C=O   ------> N=C-C-O^-
                 |                                 |
                H                               H

              CH₃                                       CH₃
               |                                          |
     N=C-C-O^-    H⁺       ------>  N=C-C-O-H
               |                                               |
             H                                             H
Draw diagrams for mechanisms for reactions of propanone and methanal with HCN.

with 2,4-dinitrophenylhydrazine (condensation)
This is used as a test for the carbonyl group with a positive test giving a yellow/orange ppt of a hydrazone.
(CH₃)₂C=O  + NH₂-NH-C₆H₃(NO₂)₂ -----> (CH₃)₂C=N-NH-C₆H₃(NO₂)₂  + H₂O
propanone     2,4-dinitrophenylhydrazine    propanone-2,4-dinitrophenylhydrazone
This is a condensation reaction (note water molecule formed).  Hydrazones are crystalline solids which can be purified easily.  Their melting points are easy to find and are used to identify the original carbonyl compound.

The yellow orange 2,4-dinitrophenylhydrazine solution forms a yellow orange precipitate.

Draw full structural formulae for all reactants and products above.
Show how ethanal and butan-2-one react with 2,4-dinitrophenylhydrazine.

with Tollen's reagent (ammoniacal silver nitrate solution) (oxidation)
Aldehydes (but not ketones) reduce Ag⁺ to Ag giving a silver mirror in this test.  Drops of sodium hydroxide are added to silver nitrate solution followed by aqueous ammonia until the ppt. redissolves.  The aldehyde is warmed in a water bath with the reagent. (oxidation)
RCHO(aq) + Ag(NH₃)₂⁺(aq) + H₂O -----> RCOOH(aq) + Ag(s) +2NH₄⁺(aq)
 

with Fehling’s solution (oxidation)
Aldehydes (but not ketones) reduce Cu²⁺ to Cu⁺ giving a red brown precipitate of copper (I) oxide in this test.  Drops of the carbonyl compound are added to equal volumes of Fehling's solutions A and B.  The mixture is warmed in a water bath. (oxidation)
RCHO(aq) + 2Cu²⁺(aq) + 2H₂O(l) -----> RCOOH(aq) + 4H⁺(aq) +Cu₂O(s)

The blue Fehling's solutions forms a red/brown precipitate

with iodine in the presence of alkali[or potassium iodide and chlorate(I)] (oxidation)
Ethanal, ethanol or any methyl ketone can react in this reaction.  A yellow precipitate of iodoform which has an antiseptic smell is the result of a positive test.  There are 3 stages:
(i) oxidation
CH₃CH(OH)-  + I₂ ----> CH₃CO- +2H⁺ + 2I^-
(ii) substitution
CH₃CO- + 3I₂ -----> CI₃CO- + 3H⁺ + 3I^-
(iii) hydrolysis
CI₃CO- + OH^- -----> CHI₃ + -COO^-
                              iodoform
A similar reaction occurs with KI and ClO^- because ClO^- can oxidise I^- to I₂ and CH₃CH(OH)- to a methyl ketone.

A pale yellow precipitate of iodoform is formed.

Describe how Tollen's, Fehling's and iodine react with ethanal, propanone and ethanol.

with sodium tetrahydridoborate(III) (sodium borohydride) (reduction)
Aldehydes form primary alcohols and ketones form secondary alcohols with this reducing agent.  NaBH₄ in aquous solution is easier to use than LiAlH₄ in ether solution.   Lithium tetrahydridoaluminate(III) (lithium aluminium hydride) LiAlH₄ does bring about the same reaction.
4CH₃CHO + NaBH₄ -----> 4CH₃CH₂OH + NaH₂BO₃
Name the products of NaBH₄ with methanal, propanone and propanal.

4.5(b)(v) Ethanoyl chloride
nucleophilic substitutions

with water
CH₃COCl (l) + H₂O (l) -------------> CH₃COOH (l) + HCl (g)
                                                        ethanoic acid 

with alcohols
CH₃COCl + C₂H₅OH ------------> CH₃CO₂C₂H₅ + HCl
                                                    ethyl ethanoate

with ammonia
iii) CH₃COCl + NH₃ -----------> CH₃CONH₂ + HCl
                                                ethanamide

with primary amines
ii) CH₃COCl + 2RNH₂ ----------> CH₃CONHR + RNH₃Cl
                                                         an amide
Task 4.5b(v) Draw diagrams to show the electron distribution around CH₃COCl, H₂O, C₂H₅OH, NH₃ and RNH₂.

4.5(b)(vi) Primary amines
with hydrogen ions
Methylamine vapour reacts with HCl to form white crystalline solid (smoke), methylammonium chloride,
CH₃ NH₂ (g) + HCl(g) --------> CH₃ NH₃⁺ Cl ^- (s)
The solubility of aromatic primary amines in increased greatly by the addition of acid.
C₆H₅NH₂(l) + H⁺ (aq) ---> C₆H₅NH₃⁺ (aq)

with acid chlorides
CH₃COCl + 2RNH₂ ----------> CH₃CONHR₂ + RNH₃Cl
                                                         an amide
The formation of polyamides

In the laboratory the hexane-1,6-dioyl chloride can be used to make nylon as it reacts at a quicker more convenient rate.
Task 4.5(b)(vii) Draw the full structural formulae for the reactants and products in the table above.

4.5(b)(vii) Nitriles
Nitriles are reduced to amines when treated with reducing agents like LiAlH₄ in dry ether or sodium in ethanol.
R-C = N  + 4H  ------------------> RCH₂NH₂
On boiling with a dilute mineral acid or dilute alkali, nitriles e.g. ethanenitrile are hydrolysed to an acid or its salt.
CH₃-CN + HCl + 2H₂O -------------> CH₃COOH + NH₄Cl
or
CH₃-CN + NaOH + H₂O--------------> CH₃COONa + NH₃

Task 4.5biii Describe how the following can be made from a suitable nitrile:
Butylamine, propanoic acid and sodium pentanoate.

4.5(b)(viii) Amides
R-CONH₂ + Br₂ +4KOH -------------------> R-NH₂ + 2KBr + K₂CO₃ +2H₂O
Conditions liquid Br₂  and conc OH^-(aq) at room temperature.
reaction removes C atom from carbon chain

reaction with P₄O₁₀ (dehydrating agent)
CH₃CONH₂ ---------------------> CH₃-CN (l) + H₂O (l)
Reagent: P₄O₁₀ (s)
Conditions: heat
Product called ethanenitrile
Task 4.5bviii Describe how propyamine and propanenitrile can be made from suitable amides.
Describe how propanoic acid can be made from a suitable amide.

4.5(b)(ix) Amino acids
These compounds are vital building blocks for protein molecules in living organisms.  Each one contains both an amino group and a carboxyl group.
2-aminobutanoic acid  CH₃CH₂ CHCOOH
                                                  |
                                                 NH₂
reaction with acids 
like amines they behave like bases and react with acids 
CH₃CH₂ CH(NH₂)COOH (aq) + H⁺ (aq) ---> CH₃CH₂ CH(NH₃)⁺COOH (aq)
reaction with bases
like carboxylic acids they react with bases
CH₃CH₂CH(NH₂)COOH (aq) + OH^- (aq) ---> CH₃CH₂ CH(NH₂)COO^- (aq) + H₂O(l)
zwitterion ion structure
In solution the acidic hydrogen of the amino acid is lost and can attach itself to the nitrogen atom in the same molecule.  The result is called a zwitterion.
CH₃CH₂ CHCOO^-
                 |
                NH₃⁺

Both the zwitterion and the original amino acid structure can react with acids or bases so can act as buffers.  A buffer maintains the pH of a solution when small amounts of acid or alkali are added.
Task 4.5bix Draw full displayed structural formulae for the substances and ions shown above.
Write equations to show how 2-aminopropanoic acid and the resulting zwitterion react with small amounts of acid and alkali.