Chemistry Form 4 Notes : ORGANIC CHEMISTRY 2 – ALKANOLS (Alcohols)

ORGANIC CHEMISTRY 2

ALKANOLS (Alcohols)

(A) INTRODUCTION.

Alkanols belong to a homologous series of organic compounds with a general formula C_nH_2n
+1 OH and thus -OH as the functional group .The 1^st ten alkanols include

Alkanols like Hydrocarbons( alkanes/alkenes/alkynes) form a homologous series where:

(i)general name is derived from the alkane name then ending with “-ol”

(ii)the members have –OH as the fuctional group

(iii)they have the same general formula represented by R-OH where R is an alkyl group.

(iv) each member differ by –CH₂ group from the next/previous.

(v)they show a similar and gradual change in their physical properties e.g. boiling and melting points.

(vi)they show similar and gradual change in their chemical properties.

B. ISOMERS OF ALKANOLS.

Alkanols exhibit both structural and position isomerism. The isomers are named by using the following basic guidelines:

(i)Like alkanes , identify the longest carbon chain to be the parent name.

(ii)Identify the position of the -OH functional group to give it the smallest /lowest position.

(iii) Identify the type and position of the side branches.

Practice examples of isomers of alkanols

(i)Isomers of propanol C₃H₇OH

CH₃CH₂CH₂OH – Propan-1-ol

OH

CH₃CHCH₃ – Propan-2-ol

Propan-2-ol and Propan-1-ol are position isomers because only the position of the –OH functional group changes.

(ii)Isomers of Butanol C₄H₉OH

CH₃ CH₂ CH₃ CH₂ OH Butan-1-ol

CH₃ CH₂ CH CH₃

  OH Butan-2-ol

CH₃

CH₃ CH₃ CH₃

OH 2-methylpropan-2-ol

Butan-2-ol and Butan-1-ol are position isomers because only the position of the -OH functional group changes.

2-methylpropan-2-ol is both a structural and position isomers because both the position of the functional group and the arrangement of the atoms in the molecule changes.

(iii)Isomers of Pentanol C₅H₁₁OH

CH₃ CH₂ CH₂CH₂CH₂ OH Pentan-1-ol (Position isomer)

CH₃ CH₂ CH CH₃

  OH Pentan-2-ol (Position isomer)

CH₃ CH₂ CH CH₂ CH₃

  OH Pentan-3-ol (Position isomer)

CH₃

CH₃ CH₂ CH₂ C CH₃

  OH 2-methylbutan-2-ol (Position /structural isomer)

CH₃

CH₃ CH₂ CH₂ C CHOH

  CH₃ 2,2-dimethylbutan-1-ol (Position /structural isomer)

CH₃

CH₃ CH₂ CH
C CH₃

  CH₃ OH 2,3-dimethylbutan-1-ol (Position /structural isomer)

(iv)1,2-dichloropropan-2-ol

CClH₂ CCl CH₃

OH

(v)1,2-dichloropropan-1-ol

CClH₂ CHCl CH₂

OH

(vi) Ethan1,2-diol

H H

 HOCH₂CH₂OH   H-O – C – C – O-H

H H

(vii) Propan1,2,3-triol H
OH H

HOCH₂CHOHCH₂OH H-O – C- C – C – O-H

  H H H

C. LABORATORY PREPARATION OF ALKANOLS.

For decades the world over, people have been fermenting grapes juice, sugar, carbohydrates and starch to produce ethanol as a social drug for relaxation.

In large amount, drinking of ethanol by mammals /human beings causes mental and physical lack of coordination.

Prolonged intake of ethanol causes permanent mental and physical lack of coordination because it damages vital organs like the liver.

Fermentation is the reaction where sugar is converted to alcohol/alkanol using biological catalyst/enzymes in yeast.

It involves three processes:

(i)Conversion of starch to maltose using the enzyme diastase.

(C₆H₁₀O₅)n (s) + H₂O(l) –diastase enzyme –> C₁₂H₂₂O₁₁(aq)

  (Starch)  (Maltose)

(ii)Hydrolysis of Maltose to glucose using the enzyme maltase.

C₁₂H₂₂O₁₁(aq)+ H₂O(l) — maltase enzyme –>2 C₆H₁₂O₆(aq)

 (Maltose) (glucose)

(iii)Conversion of glucose to ethanol and carbon(IV)oxide gas using the enzyme zymase.

C₆H₁₂O₆(aq) — zymase enzyme –> 2 C₂H₅OH(aq) + 2CO₂(g)

(glucose) (Ethanol)

At concentration greater than 15% by volume, the ethanol produced kills the yeast enzyme stopping the reaction.

To increases the concentration, fractional distillation is done to produce spirits (e.g. Brandy=40% ethanol).

Methanol is much more poisonous /toxic than ethanol.

Taken large quantity in small quantity it causes instant blindness and liver, killing the consumer victim within hours.

School laboratory preparation of ethanol from fermentation of glucose

Measure 100cm3 of pure water into a conical flask.

Add about five spatula end full of glucose.

Stir the mixture to dissolve.

Add about one spatula end full of yeast.

Set up the apparatus as below.

Preserve the mixture for about three days.

D.PHYSICAL AND CHEMICAL PROPERTIES OF ALKANOLS

Use the prepared sample above for the following experiments that shows the characteristic properties of alkanols

1. Role of yeast
   

Yeast is a single cell fungus which contains the enzyme maltase and zymase that catalyse the fermentation process.

1. Observations in lime water.
   

A white precipitate is formed that dissolve to a colourless solution later. Lime water/Calcium hydroxide reacts with carbon(IV)0xide produced during the fermentation to form insoluble calcium carbonate and water.

More carbon (IV)0xide produced during fermentation react with the insoluble calcium carbonate and water to form soluble calcium hydrogen carbonate.

Ca(OH)₂(aq)  +  CO₂ (g)  ->  CaCO₃(s)

H₂O(l) +  CO₂ (g) +  CaCO₃(s) ->  Ca(HCO₃) ₂ (aq)

(c)Effects on litmus paper

Experiment

Take the prepared sample and test with both blue and red litmus papers.

Repeat the same with pure ethanol and methylated spirit.

Sample Observation table

Explanation

Alkanols are neutral compounds/solution that have characteristic sweet smell and taste.

They have no effect on both blue and red litmus papers.

(d)Solubility in water.

Experiment

Place about 5cm3 of prepared sample into a clean test tube Add equal amount of distilled water.

Repeat the same with pure ethanol and methylated spirit.

Observation

No layers formed between the two liquids.

Explanation

Ethanol is miscible in water.Both ethanol and water are polar compounds .

The solubility of alkanols decrease with increase in the alkyl chain/molecular mass.

The alkyl group is insoluble in water while –OH functional group is soluble in water.

As the molecular chain becomes longer ,the effect of the alkyl group increases as the effect of the functional group decreases.

e)Melting/boiling point.

Experiment

Place pure ethanol in a long boiling tube .Determine its boiling point.

Observation

Pure ethanol has a boiling point of 78^oC at sea level/one atmosphere pressure.

Explanation

The melting and boiling point of alkanols increase with increase in molecular chain/mass .

This is because the intermolecular/van-der-waals forces of attraction between the molecules increase.

More heat energy is thus required to weaken the longer chain during melting and break during boiling.

f)Density

Density of alkanols increase with increase in the intermolecular/van-der-waals forces of attraction between the molecule, making it very close to each other.

This reduces the volume occupied by the molecule and thus increase the their mass per unit volume (density).

Summary table showing the trend in physical properties of alkanols

g)Burning

Experiment

Place the prepared sample in a watch glass. Ignite. Repeat with pure ethanol and methylated spirit.

Observation/Explanation

Fermentation produce ethanol with a lot of water(about a ratio of 1:3)which prevent the alcohol from igniting.

Pure ethanol and methylated spirit easily catch fire / highly flammable.

They burn with an almost colourless non-sooty/non-smoky blue flame to form carbon(IV) oxide (in excess air/oxygen)or carbon(II) oxide (limited air) and water.

Ethanol is thus a saturated compound like alkanes.

Chemica equation

C₂ H₅OH(l) +  3O₂ (g) -> 3H₂O(l) +  2CO₂ (g) ( excess air)

C₂ H₅OH(l) +  2O₂ (g) -> 3H₂O(l) +  2CO (g) ( limited air)

2CH₃OH(l) +  3O₂ (g) -> 4H₂O(l) +  2CO₂ (g) ( excess air)

2 CH₃OH(l) +  2O₂ (g) -> 4H₂O(l) +  2CO (g) ( limited air)

2C₃ H₇OH(l) +  9O₂ (g) -> 8H₂O(l) +  6CO₂ (g) ( excess air)

C₃ H₇OH(l) +  3O₂ (g) -> 4H₂O(l) +  3CO (g) ( limited air)

2C₄ H₉OH(l) +  13O₂ (g) -> 20H₂O(l) +  8CO₂ (g) ( excess air)

C₄ H₉OH(l) +  3O₂ (g) -> 4H₂O(l) +  3CO (g) ( limited air)

Due to its flammability, ethanol is used;

1. as a fuel in spirit lamps
2. as gasohol when blended with gasoline

(h)Formation of alkoxides

Experiment

Cut a very small piece of sodium. Put it in a beaker containing about 20cm3 of the prepared sample in a beaker.

Test the products with litmus papers. Repeat with absolute ethanol and methylated spirit.

Sample observations

Explanations

Sodium/potassium reacts slowly with alkanols to form basic solution called alkoxides and producing hydrogen gas.

If the alkanol has some water the metals react faster with the water to form soluble hydroxides/alkalis i.e.

Sodium  +  Alkanol   ->  Sodium alkoxides   + Hydrogen gas

Potassium   +  Alkanol  ->   Potassium alkoxides + Hydrogen gas

Sodium  +  Water   ->  Sodium hydroxides   + Hydrogen gas

Potassium   +  Water  ->   Potassium hydroxides + Hydrogen gas

Examples

1.Sodium metal reacts with ethanol to form sodium ethoxide

Sodium metal reacts with water to form sodium Hydroxide

2CH₃CH₂OH(l)  +   2Na(s)   ->   2CH₃CH₂ONa (aq) + H₂ (s)

2H₂O(l)  +   2Na(s)   ->   2NaOH (aq) + H₂ (s)

2.Potassium metal reacts with ethanol to form Potassium ethoxide

Potassium metal reacts with water to form Potassium Hydroxide

2CH₃CH₂OH(l)  +   2K(s)   ->   2CH₃CH₂OK (aq) + H₂ (s)

2H₂O(l)  +   2K(s)   ->   2KOH (aq) + H₂ (s)

3.Sodium metal reacts with propanol to form sodium propoxide

Sodium metal reacts with water to form sodium Hydroxide

2CH₃CH₂ CH₂OH(l)  +   2Na(s) -> 2CH₃CH₂ CH₂ONa (aq) + H₂ (s)

2H₂O(l)  +   2Na(s)   ->   2NaOH (aq) + H₂ (s)

4.Potassium metal reacts with propanol to form Potassium propoxide

Potassium metal reacts with water to form Potassium Hydroxide

2CH₃CH₂ CH₂OH(l)  +   2K(s)   -> 2CH₃CH₂ CH₂OK (aq) + H₂ (s)

2H₂O(l)  +   2K(s)   ->   2KOH (aq) + H₂ (s)

5.Sodium metal reacts with butanol to form sodium butoxide

Sodium metal reacts with water to form sodium Hydroxide

2CH₃CH₂ CH₂ CH₂OH(l) + 2Na(s) -> 2CH₃CH₂ CH₂ CH₂ONa (aq) + H₂ (s)

2H₂O(l)  +   2Na(s)   ->   2NaOH (aq) + H₂ (s)

6.Sodium metal reacts with pentanol to form sodium pentoxide

Sodium metal reacts with water to form sodium Hydroxide

2CH₃CH₂ CH₂ CH₂ CH₂OH(l)+2Na(s) -> 2CH₃CH₂ CH₂ CH₂ CH₂ONa (aq) + H₂ (s)

2H₂O(l)  +   2Na(s)   ->   2NaOH (aq) + H₂ (s)

(i)Formation of Esters/Esterification

Experiment

Place 2cm3 of ethanol in a boiling tube.

Add equal amount of ethanoic acid.To the mixture add carefully 2drops of concentrated sulphuric(VI)acid.

Warm/Heat gently.

Pour the mixture into a beaker containing about 50cm3 of cold water.

Smell the products.

Repeat with methanol

Sample observations

Explanation

Alkanols react with alkanoic acids to form a group of homologous series of sweet smelling compounds called esters and water. This reaction is catalyzed by concentrated sulphuric(VI)acid in the laboratory.

Alkanol  + Alkanoic acid –Conc. H₂SO₄-> Ester + water

Naturally esterification is catalyzed by sunlight. Each ester has a characteristic smell derived from the many possible combinations of alkanols and alkanoic acids that create a variety of known natural(mostly in fruits) and synthetic(mostly in juices) esters .

Esters derive their names from the alkanol first then alkanoic acids. The alkanol “becomes” an alkyl group and the alkanoic acid “becomes” alkanoate hence alkylalkanoate. e.g.

 Ethanol  +  Ethanoic acid  ->  Ethylethanoate + Water

 Ethanol  +  Propanoic acid  ->  Ethylpropanoate + Water

 Ethanol  +  Methanoic acid  ->  Ethylmethanoate + Water

 Ethanol  +  butanoic acid  ->  Ethylbutanoate + Water

 Propanol  +  Ethanoic acid  ->  Propylethanoate  + Water

 Methanol  +  Ethanoic acid  ->  Methyethanoate  + Water

 Methanol  +  Decanoic acid  ->  Methyldecanoate  + Water

 Decanol  +  Methanoic acid  ->  Decylmethanoate  + Water

During the formation of the ester, the “O” joining the alkanol and alkanoic acid comes from the alkanol.

 R₁ -COOH + R₂ –OH -> R₁ -COO –R₂ + H₂O

e.g.

1. Ethanol reacts with ethanoic acid to form the ester ethyl ethanoate and water.

 Ethanol  + Ethanoic acid  –Conc. H₂SO₄ –>Ethylethanoate + Water

C₂H₅OH (l)  + CH₃COOH(l) –Conc. H₂SO₄ –> CH₃COO C₂H₅(aq)
+H₂O(l)

CH₃CH₂OH (l)+ CH₃COOH(l) –Conc. H₂SO₄ –> CH₃COOCH₂CH₃(aq)
+H₂O(l)

2. Ethanol reacts with propanoic acid to form the ester ethylpropanoate and water.

 Ethanol  + Propanoic acid  –Conc. H₂SO₄ –>Ethylethanoate + Water

C₂H₅OH (l)+ CH₃ CH₂COOH(l) –Conc. H₂SO₄ –>CH₃CH₂COO C₂H₅(aq)
+H₂O(l)

CH₃CH₂OH (l)+ CH₃ CH₂COOH(l) –Conc. H₂SO₄ –> CH₃ CH₂COOCH₂CH₃(aq)
+H₂O(l)

3. Methanol reacts with ethanoic acid to form the ester methyl ethanoate and water.

 Methanol  + Ethanoic acid  –Conc. H₂SO₄ –>Methylethanoate + Water

CH₃OH (l)  + CH₃COOH(l) –Conc. H₂SO₄ –> CH₃COO CH₃(aq)
+H₂O(l)

4. Methanol reacts with propanoic acid to form the ester methyl propanoate and water.

 Methanol  + propanoic acid  –Conc. H₂SO₄ –>Methylpropanoate + Water

CH₃OH (l)+ CH₃ CH₂COOH(l) –Conc. H₂SO₄ –> CH₃ CH₂COO CH₃(aq)
+H₂O(l)

5. Propanol reacts with propanoic acid to form the ester propylpropanoate and water.

 Propanol  + Propanoic acid  –Conc. H₂SO₄ –>Ethylethanoate + Water

C₃H₇OH (l)+ CH₃ CH₂COOH(l) –Conc. H₂SO₄ –>CH₃CH₂COO C₃H₇(aq)
+H₂O(l)

CH₃CH₂ CH₂OH (l)+ CH₃ CH₂COOH(l) –Conc. H₂SO₄ –> CH₃ CH₂COOCH₂ CH₂CH₃(aq)
+H₂O(l)

(j)Oxidation

Experiment

Place 5cm3 of absolute ethanol in a test tube.Add three drops of acidified potassium manganate(VII).Shake thoroughly for one minute/warm.Test the solution mixture using pH paper. Repeat by adding acidified potassium dichromate(VII).

Sample observation table

Explanation

Both acidified KMnO₄ and K₂Cr₂O₇ are oxidizing agents(add oxygen to other compounds. They oxidize alkanols to a group of homologous series called alkanals then further oxidize them to alkanoic acids.The oxidizing agents are themselves reduced hence changing their colour:

 (i) Purple KMnO₄ is reduced to colourless Mn²⁺

 (ii)Orange K₂Cr₂O₇is reduced to green Cr³⁺

The pH of alkanoic acids show they have few H⁺ because they are weak acids i.e

Alkanol  +   [O]  ->   Alkanal  +  [O]  ->  alkanoic acid

NB The [O] comes from the oxidizing agents acidified KMnO₄ or K₂Cr₂O₇

Examples

1.When ethanol is warmed with three drops of acidified KMnO₄ there is decolorization of KMnO₄

Ethanol + [O] -> Ethanal + [O] -> Ethanoic acid

CH₃CH₂OH
+ [O] -> CH₃CH₂O
+ [O] -> CH₃COOH

2.When methanol is warmed with three drops of acidified K₂Cr₂O₇ ,the orange colour of acidified K₂Cr₂O₇ changes to green.

methanol + [O] -> methanal + [O] -> methanoic acid

CH₃OH
+ [O] -> CH₃O
+ [O] -> HCOOH

3.When propanol is warmed with three drops of acidified K₂Cr₂O₇ ,the orange colour of acidified K₂Cr₂O₇ changes to green.

Propanol + [O] -> Propanal + [O] -> Propanoic acid

CH₃CH₂ CH₂OH
+ [O] -> CH₃CH₂ CH₂O
+ [O] -> CH₃ CH₂COOH

4.When butanol is warmed with three drops of acidified K₂Cr₂O₇ ,the orange colour of acidified K₂Cr₂O₇ changes to green.

Butanol + [O] -> Butanal + [O] -> Butanoic acid

CH₃CH₂ CH₂ CH₂OH
+ [O] ->CH₃CH₂ CH₂CH₂O
+[O] -> CH₃ CH₂COOH

Air slowly oxidizes ethanol to dilute ethanoic acid commonly called vinegar. If beer is not tightly corked, a lot of carbon(IV)oxide escapes and there is slow oxidation of the beer making it “flat”.

(k)Hydrolysis /Hydration and Dehydration

I. Hydrolysis/Hydration is the reaction of a compound/substance with water.

Alkenes react with water vapour/steam at high temperatures and high pressures in presence of phosphoric acid catalyst to form alkanols.i.e.

Alkenes  +  Water  – H₃PO₄ catalyst->   Alkanol

Examples

(i)Ethene is mixed with steam over a phosphoric acid catalyst at 300^oC temperature and 60 atmosphere pressure to form ethanol

Ethene + water —60 atm/300^oC/ H₃PO₄ –> Ethanol

H₂C =CH₂ (g) + H₂O(l)
–60 atm/300^oC/ H₃PO₄ –> CH₃ CH₂OH(l)

This is the main method of producing large quantities of ethanol instead of fermentation

(ii) Propene + water —60 atm/300^oC/ H₃PO₄ –> Propanol

CH₃C =CH₂ (g) + H₂O(l)
–60 atm/300^oC/ H₃PO₄ –> CH₃ CH₂ CH₂OH(l)

(iii) Butene + water —60 atm/300^oC/ H₃PO₄ –> Butanol

CH₃ CH₂ C=CH₂ (g) + H₂O(l)
–60 atm/300^oC/ H₃PO₄ –> CH₃ CH₂ CH₂ CH₂OH(l)

II. Dehydration is the process which concentrated sulphuric(VI)acid (dehydrating agent) removes water from a compound/substances.

Concentrated sulphuric(VI)acid dehydrates alkanols to the corresponding alkenes at about 180^oC. i.e

Alkanol –Conc. H₂ SO₄/180^oC–> Alkene + Water

Examples

1. At 180^oC and in presence of Concentrated sulphuric(VI)acid, ethanol undergoes dehydration to form ethene.

Ethanol —180^oC/ H₂SO₄ –> Ethene + Water

CH₃ CH₂OH(l)
–180^oC/ H₂SO₄ –> H₂C =CH₂ (g) + H₂O(l)

2. Propanol undergoes dehydration to form propene.

Propanol —180^oC/ H₂SO₄ –> Propene + Water

CH₃ CH₂ CH₂OH(l)
–180^oC/ H₂SO₄ –> CH₃CH =CH₂ (g) + H₂O(l)

3. Butanol undergoes dehydration to form Butene.

Butanol —180^oC/ H₂SO₄ –> Butene + Water

CH₃ CH₂ CH₂CH₂OH(l)
–180^oC/ H₂SO₄ –> CH₃ CH₂C =CH₂ (g) + H₂O(l)

3. Pentanol undergoes dehydration to form Pentene.

Pentanol —180^oC/ H₂SO₄ –> Pentene + Water

CH₃ CH₂ CH₂ CH₂ CH₂OH(l)–180^oC/ H₂SO₄–>CH₃ CH₂ CH₂C =CH₂ (g)+H₂O(l)

(l)Similarities of alkanols with Hydrocarbons

I. Similarity with alkanes

Both alkanols and alkanes burn with a blue non-sooty flame to form carbon(IV)oxide(in excess air/oxygen)/carbon(II)oxide(in limited air) and water. This shows they are saturated with high C:H ratio. e.g.

Both ethanol and ethane ignite and burns in air with a blue non-sooty flame to form carbon(IV)oxide(in excess air/oxygen)/carbon(II)oxide(in limited air) and water.

CH₂ CH₂OH(l) + 3O₂(g)
-Excess air-> 2CO₂ (g)
+ 3H₂ O(l)

CH₂ CH₂OH(l) + 2O₂(g)
-Limited air-> 2CO
(g)
+ 3H₂ O(l)

CH₃ CH₃(g) + 3O₂(g)
-Excess air-> 2CO₂ (g)
+ 3H₂ O(l)

2CH₃ CH₃(g) + 5O₂(g)
-Limited air-> 4CO
(g)
+ 6H₂ O(l)

II. Similarity with alkenes/alkynes

Both alkanols(R-OH) and alkenes/alkynes(with = C = C = double and – C = C- triple ) bond:

 (i)decolorize acidified KMnO₄

 (ii)turns Orange acidified K₂Cr₂O₇ to green.

Alkanols(R-OH) are oxidized to alkanals(R-O) ant then alkanoic acids(R-OOH).

Alkenes are oxidized to alkanols with duo/double functional groups.

Examples

1.When ethanol is warmed with three drops of acidified K₂Cr₂O₇ the orange of acidified K₂Cr₂O₇ turns to green. Ethanol is oxidized to ethanol and then to ethanoic acid.

Ethanol + [O] -> Ethanal + [O] -> Ethanoic acid

CH₃CH₂OH
+ [O] -> CH₃CH₂O
+ [O] -> CH₃COOH

2.When ethene is bubbled in a test tube containing acidified K₂Cr₂O₇ ,the orange of acidified K₂Cr₂O₇ turns to green. Ethene is oxidized to ethan-1,2-diol.

Ethene   + [O]  ->  Ethan-1,2-diol.

H₂C=CH₂ + [O] ->  HOCH₂ -CH₂OH

III. Differences with alkenes/alkynes

Alkanols do not decolorize bromine and chlorine water.

Alkenes decolorizes bromine and chlorine water to form halogenoalkanols

Example

When ethene is bubbled in a test tube containing bromine water,the bromine water is decolorized. Ethene is oxidized to bromoethanol.

Ethene   + Bromine water  ->  Bromoethanol.

H₂C=CH₂ + HOBr ->  BrCH₂ -CH₂OH

IV. Differences in melting and boiling point with Hydrocarbons

Alkanos have higher melting point than the corresponding hydrocarbon (alkane/alkene/alkyne)

This is because most alkanols exist as dimer.A dimer is a molecule made up of two other molecules joined usually by van-der-waals forces/hydrogen bond or dative bonding.

Two alkanol molecules form a dimer joined by hydrogen bonding.

Example

In Ethanol the oxygen atom attracts/pulls the shared electrons in the covalent bond more to itself than Hydrogen.

This creates a partial negative charge (^δ-) on oxygen and partial positive charge(^δ+) on hydrogen.

Two ethanol molecules attract each other at the partial charges through Hydrogen bonding forming a dimmer.

 H  H   H

H  C  C  O H  H  

 H  H   H   O  C  C  H

H  H

Dimerization of alkanols means more energy is needed to break/weaken the Hydrogen bonds before breaking/weakening the intermolecular forces joining the molecules of all organic compounds during boiling/melting.

E.USES OF SOME ALKANOLS

(a)Methanol is used as industrial alcohol and making methylated spirit

(b)Ethanol is used:

 1. as alcohol in alcoholic drinks e.g Beer, wines and spirits.

 2.as antiseptic to wash woulds

 3.in manufacture of vanishes, ink ,glue and paint because it is volatile and thus easily evaporate

 4.as a fuel when blended with petrol to make gasohol.

B.ALKANOIC ACIDS (Carboxylic acids)

(A) INTRODUCTION.

Alkanoic acids belong to a homologous series of organic compounds with a general formula C_nH_2n
+1 COOH and thus -COOH as the functional group .The 1^st ten alkanoic acids include:

Alkanoic acids like alkanols /alkanes/alkenes/alkynes form a homologous series where:

(i)the general name of an alkanoic acids is derived from the alkane name then ending with “–oic” acid as the table above shows.

(ii) the members have R-COOH/R C-O-H as the functional group.

O

(iii)they have the same general formula represented by R-COOH where R is an alkyl group.

(iv)each member differ by –CH₂– group from the next/previous.

(v)they show a similar and gradual change in their physical properties e.g. boiling and melting point.

(vi)they show similar and gradual change in their chemical properties.

(vii) since they are acids they show similar properties with mineral acids.

(B) ISOMERS OF ALKANOIC ACIDS.

lkanoic acids exhibit both structural and position isomerism. The isomers are named by using the following basic guidelines

(i)Like alkanes. identify the longest carbon chain to be the parent name.

(ii)Identify the position of the -C-O-H functional group to give it the smallest

O

/lowest position.

(iii)Identify the type and position of the side group branches.

Practice examples on isomers of alkanoic acids

1.Isomers of butanoic acid C₃H₇COOH

CH₃ CH₂ CH₂ COOH

Butan-1-oic acid

 CH₃

H₂C C COOH 2-methylpropan-1-oic acid

2-methylpropan-1-oic acid and Butan-1-oic acid are structural isomers because the position of the functional group does not change but the arrangement of the atoms in the molecule does.

2.Isomers of pentanoic acid C₄H₉COOH

CH₃CH₂CH₂CH₂ COOH pentan-1-oic acid

CH₃

    CH₃CH₂CH COOH 2-methylbutan-1-oic acid

CH₃

H₃C C COOH   2,2-dimethylpropan-1-oic acid

CH₃

3.Ethan-1,2-dioic acid

O O

HOOC- COOH //  H – O – C – C – O – H

4.Propan-1,3-dioic acid

O H O

HOOC- CH₂COOH //  H – O – C – C – C – O – H

H

5.Butan-1,4-dioic acid

O H H O

HOOC CH₂ CH₂ COOH   H- O – C – C – C – C –O – H

H H

6.2,2-dichloroethan-1,2-dioic acid

HOOCCHCl₂ Cl

H – O – C – C – Cl

O H

(C) LABORATORY AND INDUSTRIAL PREPARATIONOF ALKANOIC ACIDS.

In a school laboratory, alkanoic acids can be prepared by adding an oxidizing agent (H⁺/KMnO₄ or H⁺/K₂Cr₂O₇)to the corresponding alkanol then warming.

The oxidation converts the alkanol first to an alkanal the alkanoic acid.

NB Acidified KMnO₄ is a stronger oxidizing agent than acidified K₂Cr₂O₇

General equation:

R- CH₂ – OH  + [O] –H⁺/KMnO₄–>  R- CH –O  +  H₂O(l)

(alkanol)  (alkanal)

R- CH – O  + [O]  –H⁺/KMnO₄–>  R- C –OOH

(alkanal)  (alkanoic acid)

Examples

1.Ethanol on warming in acidified KMnO₄ is oxidized to ethanal then ethanoic acid .

CH₃– CH₂ – OH  + [O] –H⁺/KMnO₄–>  CH₃– CH –O  +  H₂O(l)

(ethanol)  (ethanal)

CH₃– CH – O  + [O]  –H⁺/KMnO₄–>  CH₃– C –OOH

(ethanal)  (ethanoic acid)

2Propanol on warming in acidified KMnO₄ is oxidized to propanal then propanoic acid

CH₃– CH₂ CH₂ – OH  + [O]  –H⁺/KMnO₄–>  CH₃– CH₂ CH –O  +  H₂O(l)

(propanol) (propanal)

CH₃– CH – O  + [O]  –H⁺/KMnO₄–>  CH₃– C –OOH

(propanal)  (propanoic acid)

Industrially,large scale manufacture of alkanoic acid like ethanoic acid is obtained from:

(a)Alkenes reacting with steam at high temperatures and pressure in presence of phosphoric(V)acid catalyst and undergo hydrolysis to form alkanols. i.e.

Alkenes + Steam/water  — H₂PO₄ Catalyst–> Alkanol

The alkanol is then oxidized by air at 5 atmosphere pressure with Manganese (II)sulphate(VI) catalyst to form the alkanoic acid.

Alkanol + Air  — MnSO₄ Catalyst/5 atm pressure–> Alkanoic acid

Example

Ethene is mixed with steam over a phosphoric(V)acid catalyst,300oC temperature and 60 atmosphere pressure to form ethanol.

CH₂=CH₂  + H₂O -> CH₃ CH₂OH

(Ethene) (Ethanol)

This is the industrial large scale method of manufacturing ethanol

Ethanol is then oxidized by air at 5 atmosphere pressure with Manganese (II)sulphate(VI) catalyst to form the ethanoic acid.

CH₃ CH₂OH  +   [O] — MnSO₄ Catalyst/5 atm pressure–> CH₃ COOH

(Ethanol) (Ethanoic acid)

(b)Alkynes react with liquid water at high temperatures and pressure in presence of Mercury(II)sulphate(VI)catalyst and 30% concentrated sulphuric(VI)acid to form alkanals.

Alkyne  +  Water — Mercury(II)sulphate(VI)catalyst–> Alkanal

The alkanal is then oxidized by air at 5 atmosphere pressure with Manganese (II) sulphate(VI) catalyst to form the alkanoic acid.  

Alkanal  + air/oxygen — Manganese(II)sulphate(VI)catalyst–> Alkanoic acid

Example

Ethyne react with liquid water at high temperature and pressure with Mercury (II) sulphate (VI)catalyst and 30% concentrated sulphuric(VI)acid to form ethanal.

CH = CH  + H₂O –HgSO₄–> CH₃ CH₂O

(Ethyne) (Ethanal)

This is another industrial large scale method of manufacturing ethanol from large quantities of ethyne found in natural gas.

Ethanal is then oxidized by air at 5 atmosphere pressure with Manganese (II)sulphate(VI) catalyst to form the ethanoic acid.

CH₃ CH₂O  +   [O] — MnSO₄ Catalyst/5 atm pressure–> CH₃ COOH

(Ethanal) (Oxygen from air) (Ethanoic acid)

(D) PHYSICAL AND CHEMICAL PROPERTIES OF ALKANOIC ACIDS.

I.Physical properties of alkanoic acids

The table below shows some physical properties of alkanoic acids

From the table note the following:

1. Melting and boiling point decrease as the carbon chain increases due to increase in intermolecular forces of attraction between the molecules requiring more energy to separate the molecules.
2. The density decreases as the carbon chain increases as the intermolecular forces of attraction increases between the molecules making the molecule very close reducing their volume in unit mass.
3. Solubility decreases as the carbon chain increases as the soluble –COOH end is shielded by increasing insoluble alkyl/hydrocarbon chain.
4. Like alkanols ,alkanoic acids exist as dimmers due to the hydrogen bonds within the molecule. i.e..

   

II Chemical properties of alkanoic acids

The following experiments shows the main chemical properties of ethanoic (alkanoic) acid.

 (a)Effect on litmus papers

Experiment

Dip both blue and red litmus papers in ethanoic acid. Repeat with a solution of succinic acid, citric acid, oxalic acid, tartaric acid and dilute nitric(V)acid.

Sample observations

Explanation

All acidic solutions contains H⁺/H₃O⁺(aq) ions. The H⁺ /H₃O⁺ (aq) ions is responsible for turning blue litmus paper/solution to red

(b)pH

Experiment

Place 2cm3 of ethaoic acid in a test tube. Add 2 drops of universal indicator solution and determine its pH. Repeat with a solution of succinic acid, citric acid, oxalic acid, tartaric acid and dilute sulphuric (VI)acid.

Sample observations

Explanations

Alkanoic acids are weak acids that partially/partly dissociate to release few H⁺ ions in solution. The pH of their solution is thus 4/5/6 showing they form weakly acidic solutions when dissolved in water.

All alkanoic acid dissociate to releases the “H” at the functional group in -COOH to form the alkanoate ion; –COO^–

Mineral acids(Sulphuric(VI)acid, Nitric(V)acid and Hydrochloric acid) are strong acids that wholly/fully dissociate to release many H⁺ ions in solution. The pH of their solution is thus 1/2/3 showing they form strongly acidic solutions when dissolved in water.i.e

Examples

1. CH₃COOH(aq) CH₃COO^–(aq)   +   H⁺(aq)

   (ethanoic acid) (ethanoate ion) (few H⁺ ion)

    

2. CH₃ CH₂COOH(aq) CH₃ CH₂COO^–(aq) +   H⁺(aq)

(propanoic acid)  (propanoate ion)   (few H⁺ ion)

1. CH₃ CH₂ CH₂COOH(aq) CH₃ CH₂ CH₂COO^–(aq) + H⁺(aq)

(Butanoic acid)   (butanoate ion)   (few H⁺ ion)

1. HOOH(aq)   HOO^–(aq)   +   H⁺(aq)

   (methanoic acid) (methanoate ion) (few H⁺ ion)

2. H₂ SO₄ (aq)  SO₄^2- (aq)   +   2H⁺(aq)

   (sulphuric(VI) acid) (sulphate(VI) ion)   (many H⁺ ion)

3. HNO₃ (aq)  NO₃^– (aq)   +   H⁺(aq)

   (nitric(V) acid) (nitrate(V) ion)   (many H⁺ ion)

    (c)Reaction with metals

Experiment

Place about 4cm3 of ethanoic acid in a test tube. Put about 1cm length of polished magnesium ribbon. Test any gas produced using a burning splint. Repeat with a solution of succinic acid, citric acid, oxalic acid, tartaric acid and dilute sulphuric (VI) acid.

Sample observations

Explanation

Metals higher in the reactivity series displace the hydrogen in all acids to evolve/produce hydrogen gas and form a salt. Alkanoic acids react with metals with metals to form alkanoates salt and produce/evolve hydrogen gas .Hydrogen extinguishes a burning splint with a pop sound/explosion. Only the “H”in the functional group -COOH is /are displaced and not in the alkyl hydrocarbon chain.

Alkanoic acid + Metal -> Alkanoate + Hydrogen gas. i.e.

Examples

1. For a monovalent metal with monobasic acid

2R – COOH + 2M -> 2R- COOM + 2H₂(g)

2.For a divalent metal with monobasic acid

2R – COOH + M -> (R- COO) ₂M + H₂(g)

3.For a divalent metal with dibasic acid

HOOC-R-COOH+ M -> MOOC-R-COOM + H₂(g)

4.For a monovalent metal with dibasic acid

HOOC-R-COOH+ 2M -> MOOC-R-COOM + H₂(g)

5 For mineral acids

(i)Sulphuric(VI)acid is a dibasic acid

 H₂ SO₄ (aq) + 2M -> M₂ SO₄ (aq) + H₂(g)

 H₂ SO₄ (aq) + M -> MSO₄ (aq) + H₂(g)

(ii)Nitric(V) and hydrochloric acid are monobasic acid

 HNO₃ (aq) + 2M -> 2MNO₃ (aq) + H₂(g)

 HNO₃ (aq) + M -> M(NO₃ ) ₂ (aq) + H₂(g)

Examples

1.Sodium reacts with ethanoic acid to form sodium ethanoate and produce. hydrogen gas.

Caution: This reaction is explosive.

CH₃COOH (aq) + Na(s) -> CH₃COONa (aq) + H₂(g)

 (Ethanoic acid) (Sodium ethanoate)

2.Calcium reacts with ethanoic acid to form calcium ethanoate and produce. hydrogen gas.

2CH₃COOH (aq) + Ca(s) -> (CH₃COO) ₂Ca (aq) + H₂(g)

 (Ethanoic acid) (Calcium ethanoate)

3.Sodium reacts with ethan-1,2-dioic acid to form sodium ethan-1,2-dioate and produce. hydrogen gas.

HOOC-COOH+ 2Na -> NaOOC – COONa + H₂(g)

(ethan-1,2-dioic acid) (sodium ethan-1,2-dioate)

Commercial name of ethan-1,2-dioic acid is oxalic acid. The salt is sodium oxalate.

4.Magnesium reacts with ethan-1,2-dioic acid to form magnesium ethan-1,2-dioate and produce. hydrogen gas.

HOOC-R-COOH+ Mg -> ( OOC – COO)
Mg + H₂(g)

(ethan-1,2-dioic acid) (magnesium ethan-1,2-dioate)

5.Magnesium reacts with

(i)Sulphuric(VI)acid to form Magnesium sulphate(VI)

 H₂ SO₄ (aq) + Mg -> MgSO₄ (aq) + H₂(g)

(ii)Nitric(V) and hydrochloric acid are monobasic acid

 2HNO₃ (aq) + Mg -> M(NO₃ ) ₂ (aq) + H₂(g)

(d)Reaction with hydrogen carbonates and carbonates

Experiment

Place about 3cm3 of ethanoic acid in a test tube. Add about 0.5g/ ½ spatula end full of sodium hydrogen carbonate/sodium carbonate. Test the gas produced using lime water. Repeat with a solution of succinic acid, citric acid, oxalic acid, tartaric acid and dilute sulphuric (VI) acid.

Sample observations

All acids react with hydrogen carbonate/carbonate to form salt ,water and evolve/produce bubbles of carbon(IV)oxide and water.

Carbon(IV)oxide forms a white precipitate when bubbled in lime water/extinguishes a burning splint.

Alkanoic acids react with hydrogen carbonate/carbonate to form alkanoates ,water and evolve/produce bubbles of carbon(IV)oxide and water.

Alkanoic acid + hydrogen carbonate -> alkanoate + water + carbon(IV)oxide

Alkanoic acid + carbonate -> alkanoate + water + carbon(IV)oxide

Examples

1. Sodium hydrogen carbonate reacts with ethanoic acid to form sodium ethanoate ,water and carbon(IV)oxide gas.

CH₃COOH (aq) + NaHCO₃ (s) -> CH₃COONa (aq) + H₂O(l) + CO₂ (g)

 (Ethanoic acid) (Sodium ethanoate)

2.Sodium carbonate reacts with ethanoic acid to form sodium ethanoate ,water and carbon(IV)oxide gas.

2CH₃COOH (aq) + Na₂CO₃ (s) -> 2CH₃COONa (aq) + H₂O(l) + CO₂ (g)

 (Ethanoic acid) (Sodium ethanoate)

3.Sodium carbonate reacts with ethan-1,2-dioic acid to form sodium ethanoate ,water and carbon(IV)oxide gas.

HOOC-COOH+ Na₂CO₃ (s) -> NaOOC – COONa + H₂O(l) + CO₂ (g)

(ethan-1,2-dioic acid) (sodium ethan-1,2-dioate)

4.Sodium hydrogen carbonate reacts with ethan-1,2-dioic acid to form sodium ethanoate ,water and carbon(IV)oxide gas.

HOOC-COOH+ 2NaHCO₃ (s) -> NaOOC – COONa + H₂O(l) + 2CO₂ (g)

(ethan-1,2-dioic acid) (sodium ethan-1,2-dioate)

(e)Esterification

Experiment

Place 4cm3 of ethanol acid in a boiling tube.

Add equal volume of ethanoic acid. To the mixture, add 2 drops of concentrated sulphuric(VI)acid carefully. Warm/heat gently on Bunsen flame.

Pour the mixture into a beaker containing 50cm3 of water. Smell the products. Repeat with a solution of succinic acid, citric acid, oxalic acid, tartaric acid and dilute sulphuric (VI) acid.

Sample observations

Explanation

Alkanols react with alkanoic acid to form the sweet smelling homologous series of esters and water.The reaction is catalysed by concentrated sulphuric(VI)acid in the laboratory but naturally by sunlight /heat.Each ester has a characteristic smell derived from the many possible combinations of alkanols and alkanoic acids.

 Alkanol  +  Alkanoic acids  ->  Ester   +  water

Esters derive their names from the alkanol first then alkanoic acids. The alkanol “becomes” an alkyl group and the alkanoic acid “becomes” alkanoate hence alkylalkanoate. e.g.

 Ethanol  +  Ethanoic acid  ->  Ethylethanoate + Water

 Ethanol  +  Propanoic acid  ->  Ethylpropanoate + Water

 Ethanol  +  Methanoic acid  ->  Ethylmethanoate + Water

 Ethanol  +  butanoic acid  ->  Ethylbutanoate + Water

 Propanol  +  Ethanoic acid  ->  Propylethanoate  + Water

 Methanol  +  Ethanoic acid  ->  Methyethanoate  + Water

 Methanol  +  Decanoic acid  ->  Methyldecanoate  + Water

 Decanol  +  Methanoic acid  ->  Decylmethanoate  + Water

During the formation of the ester, the “O” joining the alkanol and alkanoic acid comes from the alkanol.

 R₁ -COOH + R₂ –OH -> R₁ -COO –R₂ + H₂O

Examples

1. Ethanol reacts with ethanoic acid to form the ester ethyl ethanoate and water.

 Ethanol  + Ethanoic acid  –Conc. H₂SO₄ –>Ethylethanoate + Water

C₂H₅OH (l)  + CH₃COOH(l) –Conc. H₂SO₄ –> CH₃COO C₂H₅(aq)
+H₂O(l)

CH₃CH₂OH (l)+ CH₃COOH(l) –Conc. H₂SO₄ –> CH₃COOCH₂CH₃(aq)
+H₂O(l)

2. Ethanol reacts with propanoic acid to form the ester ethylpropanoate and water.

 Ethanol  + Propanoic acid  –Conc. H₂SO₄ –>Ethylethanoate + Water

C₂H₅OH (l)+ CH₃ CH₂COOH(l) –Conc. H₂SO₄ –>CH₃CH₂COO C₂H₅(aq)
+H₂O(l)

CH₃CH₂OH (l)+ CH₃ CH₂COOH(l) –Conc. H₂SO₄ –>

CH₃ CH₂COOCH₂CH₃(aq)
+H₂O(l)

3. Methanol reacts with ethanoic acid to form the ester methyl ethanoate and water.

 Methanol  + Ethanoic acid  –Conc. H₂SO₄ –>Methylethanoate + Water

CH₃OH (l)  + CH₃COOH(l) –Conc. H₂SO₄ –> CH₃COO CH₃(aq)
+H₂O(l)

4. Methanol reacts with propanoic acid to form the ester methyl propanoate and water.

 Methanol  + propanoic acid  –Conc. H₂SO₄ –>Methylpropanoate + Water

CH₃OH (l)+ CH₃ CH₂COOH(l) –Conc. H₂SO₄ –> CH₃ CH₂COO CH₃(aq)
+H₂O(l)

5. Propanol reacts with propanoic acid to form the ester propylpropanoate and water.

 Propanol  + Propanoic acid  –Conc. H₂SO₄ –>Ethylethanoate + Water

C₃H₇OH (l)+ CH₃ CH₂COOH(l) –Conc. H₂SO₄ –>CH₃CH₂COO C₃H₇(aq)
+H₂O(l)

CH₃CH₂ CH₂OH (l)+ CH₃ CH₂COOH(l) –Conc. H₂SO₄ –> CH₃ CH₂COOCH₂ CH₂CH₃(aq)
+H₂O(l)

C. DETERGENTS

Detergents are cleaning agents that improve the cleaning power /properties of water.A detergent therefore should be able to:

 (i)dissolve substances which water cannot e.g grease ,oil, fat

 (ii)be washed away after cleaning.

There are two types of detergents:

 (a)Soapy detergents

 (b)Soapless detergents

1. SOAPY DETERGENTS
   

Soapy detergents usually called soap is long chain salt of organic alkanoic acids.Common soap is sodium octadecanoate .It is derived from reacting concentrated sodium hydroxide solution with octadecanoic acid(18 carbon alkanoic acid) i.e.

Sodium hydroxide + octadecanoic acid -> Sodium octadecanoate + water

NaOH(aq) + CH₃ (CH₂) ₁₆ COOH(aq)  -> CH₃ (CH₂) ₁₆ COO ^– Na⁺ (aq) +H₂ O(l)

Commonly ,soap can thus be represented

R-
COO ^– Na⁺
where;

R is a long chain alkyl group and -COO ^– Na⁺ is the alkanoate ion.

In a school laboratory and at industrial and domestic level,soap is made by reacting concentrated sodium hydroxide solution with esters from (animal) fat and oil. The process of making soap is called saponification. During saponification ,the ester is hydrolyzed by the alkali to form sodium salt /soap and glycerol/propan-1,2,3-triol is produced.

Fat/oil(ester)+sodium/potassium hydroxide->sodium/potassium salt(soap)+ glycerol

Fats/Oils are esters with fatty acids and glycerol parts in their structure;

C₁₇H₃₅COOCH₂

C₁₇H₃₅COOCH

C₁₇H₃₅COOCH₂

When boiled with concentrated sodium hydroxide solution NaOH;

 (i)NaOH ionizes/dissociates into Na⁺ and OH^– ions

 (ii)fat/oil split into three C₁₇H₃₅COO^– and one CH₂ CH
CH₂

(iii) the three Na⁺ combine with the three C₁₇H₃₅COO^– to form the salt C₁₇H₃₅COO^– Na⁺

(iv)the three OH^–ions combine with the CH₂ CH
CH₂ to form an alkanol with three functional groups CH₂ OH CH OH
CH₂ OH(propan-1,2,3-triol)

C₁₇H₃₅COOCH₂ CH₂OH

C₁₇H₃₅COOCH +NaOH -> 3 C₁₇H₃₅COO^– Na⁺ + CHOH

C₁₇H₃₅COOCH₂   CH₂OH

Ester Alkali  Soap  glycerol

Generally:

C_nH_2n+1COOCH₂ CH₂OH

C_nH_2n+1COOCH +NaOH -> 3 C_nH_2n+1COO^– Na⁺ +  CHOH

C_nH_2n+1COOCH₂   CH₂OH

Ester Alkali  Soap  glycerol

R – COOCH₂ CH₂OH

R – COOCH +NaOH -> 3R-COO^– Na⁺ +   CHOH

R- COOCH₂   CH₂OH

Ester Alkali  Soap  glycerol

During this process a little sodium chloride is added to precipitate the soap by reducing its solubility. This is called salting out.

The soap is then added colouring agents ,perfumes and herbs of choice.

 School laboratory preparation of soap

Place about 40 g of fatty (animal fat)beef/meat in 100cm3 beaker .Add about 15cm3 of 4.0M sodium hydroxide solution. Boil the mixture for about 15minutes.Stir the mixture .Add about 5.0cm3 of distilled water as you boil to make up for evaporation. Boil for about another 15minutes.Add about four spatula end full of pure sodium chloride crystals. Continue stirring for another five minutes. Allow to cool. Filter of /decant and wash off the residue with distilled water .Transfer the clean residue into a dry beaker. Preserve.

The action of soap

Soapy detergents:

 (i)act by reducing the surface tension of water by forming a thin layer on top of the water.

 (ii)is made of a non-polar alkyl /hydrocarbon tail and a polar -COO^–Na⁺ head. The non-polar alkyl /hydrocarbon tail is hydrophobic (water hating) and thus does not dissolve in water .It dissolves in non-polar solvent like grease, oil and fat. The polar -COO^–Na⁺ head is hydrophilic (water loving)and thus dissolve in water. When washing with soapy detergent, the non-polar tail of the soapy detergent surround/dissolve in the dirt on the garment /grease/oil while the polar head dissolve in water.

Through mechanical agitation/stirring/sqeezing/rubbing/beating/kneading, some grease is dislodged/lifted of the surface of the garment. It is immediately surrounded by more soap molecules It float and spread in the water as tiny droplets that scatter light in form of emulsion making the water cloudy and shinny. It is removed from the garment by rinsing with fresh water. The repulsion of the soap head prevent /ensure the droplets do not mix. Once removed, the dirt molecules cannot be redeposited back because it is surrounded by soap molecules.

Advantages and disadvantages of using soapy detergents

Soapy detergents are biodegradable. They are acted upon by bacteria and rot. They thus do not cause environmental pollution.

Soapy detergents have the disadvantage in that:

 (i)they are made from fat and oils which are better eaten as food than make soap.

 (ii)forms an insoluble precipitate with hard water called scum. Scum is insoluble calcium octadecanoate and Magnesium octadecanoate formed when soap reacts with Ca²⁺ and Mg²⁺ present in hard water.

Chemical equation

2C₁₇H₃₅COO^– Na⁺ (aq)
+ Ca²⁺(aq) ->  (C₁₇H₃₅COO^– )Ca²⁺ (s) +
2Na⁺(aq)

(insoluble Calcium octadecanote/scum)

2C₁₇H₃₅COO^– Na⁺ (aq)
+ Mg²⁺(aq) ->  (C₁₇H₃₅COO^– )Mg²⁺ (s) +
2Na⁺(aq)

(insoluble Magnesium octadecanote/scum)

This causes wastage of soap.

Potassium soaps are better than Sodium soap. Potassium is more expensive than sodium and thus its soap is also more expensive.

(b)SOAPLESS DETERGENTS

Soapless detergent usually called detergent is a long chain salt fromed from by-products of fractional distillation of crude oil.Commonly used soaps include:

 (i)washing agents

 (ii)toothpaste

 (iii)emulsifiers/wetting agents/shampoo

Soapless detergents are derived from reacting:

 (i)concentrated sulphuric(VI)acid with a long chain alkanol e.g. Octadecanol(18 carbon alkanol) to form alkyl hydrogen sulphate(VI)

Alkanol + Conc sulphuric(VI)acid -> alkyl hydrogen sulphate(VI) + Water

R –OH + H₂SO₄ -> R –O-SO₃H + H₂O

 (ii)the alkyl hydrogen sulphate(VI) is then neutralized with sodium/potassium hydroxide to form sodium/potassium alkyl hydrogen sulphate(VI)

Sodium/potassium alkyl hydrogen sulphate(VI) is the soapless detergent.

alkyl hydrogen + Potassium/sodium -> Sodium/potassium + Water

sulphate(VI) hydroxide alkyl hydrogen sulphate(VI)

R –O-SO₃H + NaOH -> R –O-SO₃^– Na⁺ + H₂O

Example

Step I : Reaction of Octadecanol with Conc.H₂SO₄

C₁₇H₃₅CH₂OH
(aq) + H₂SO₄ -> C₁₇H₃₅CH₂–O- SO₃^– H⁺
(aq) + H₂O
(l)

octadecanol + sulphuric(VI)acid -> Octadecyl hydrogen sulphate(VI) + water

Step II: Neutralization by an alkali

C₁₇H₃₅CH₂–O- SO₃^– H⁺
(aq) + NaOH
-> C₁₇H₃₅CH₂–O- SO₃^– Na⁺
(aq) + H₂O
(l)

Octadecyl hydrogen + sodium/potassium ->  sodium/potassium octadecyl+Water

sulphate(VI) hydroxide hydrogen sulphate(VI)

School laboratory preparation of soapless detergent

Place about 20g of olive oil in a 100cm3 beaker. Put it in a trough containing ice cold water.

Add dropwise carefully 18M concentrated sulphuric(VI)acid stirring continuously into the olive oil until the oil turns brown.Add 30cm3 of 6M sodium hydroxide solution.Stir.This is a soapless detergent.

The action of soapless detergents

The action of soapless detergents is similar to that of soapy detergents.The soapless detergents contain the hydrophilic head and a long hydrophobic tail. i.e.

vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv-COO^–Na⁺

vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv-O-SO₃^– Na⁺

(long hydrophobic /non-polar alkyl tail)   (hydrophilic/polar/ionic head)

The tail dissolves in fat/grease/oil while the ionic/polar/ionic head dissolves in water.

The tail stick to the dirt which is removed by the attraction of water molecules and the polar/ionic/hydrophilic head by mechanical agitation /squeezing/kneading/ beating/rubbing/scrubbing/scatching.

The suspended dirt is then surrounded by detergent molecules and repulsion of the anion head preventing the dirt from sticking on the material garment.

The tiny droplets of dirt emulsion makes the water cloudy. On rinsing the cloudy emulsion is washed away.

Advantages and disadvantages of using soapless detergents

Soapless detergents are non-biodegradable unlike soapy detergents.

They persist in water during sewage treatment by causing foaming in rivers ,lakes and streams leading to marine /aquatic death.

Soapless detergents have the advantage in that they:

 (i)do not form scum with hard water.

 (ii)are cheap to manufacture/buying

(iii)are made from petroleum products but soapis made from fats/oil for human consumption.

Sample revision questions

1. Study the scheme below  

(a)Identify the process

 Saponification

(b)Fats and oils are esters. Write the formula of the a common structure of ester

C₁₇H₃₅COOCH₂

C₁₇H₃₅COOCH

C₁₇H₃₅COOCH₂

(c)Write a balanced equation for the reaction taking place during boiling

C₁₇H₃₅COOCH₂ CH₂OH

C₁₇H₃₅COOCH +3NaOH -> 3 C₁₇H₃₅COO^– Na⁺ +  CHOH

C₁₇H₃₅COOCH₂   CH₂OH

Ester Alkali   Soap glycerol

(d)Give the IUPAC name of:

 (i)Residue X

Potassium octadecanoate

 (ii)Filtrate Y

Propan-1,2,3-triol

(e)Give one use of fitrate Y

 Making paint

(f)What is the function of sodium chloride

 To reduce the solubility of the soap hence helping in precipitating it out

(g)Explain how residue X helps in washing.

 Has a non-polar hydrophobic tail that dissolves in dirt/grease /oil/fat

 Has a polar /ionic hydrophilic head that dissolves in water.

 From mechanical agitation,the dirt is plucked out of the garment and surrounded by the tail end preventing it from being deposited back on the garment.

(h)State one:

 (i)advantage of continued use of residue X on the environment

Is biodegradable and thus do not pollute the environment

 (ii)disadvantage of using residue X

Uses fat/oil during preparation/manufacture which are better used for human consumption.

(i)Residue X was added dropwise to some water.The number of drops used before lather forms is as in the table below.

(i)State and explain which sample of water is:

 I. Soft

Sample B .Very little soap is used and no effect on amount of soap even on boiling/heating.

 II. Permanent hard

Sample C . A lot of soap is used and no effect on amount of soap even on boiling/heating. Boiling does not remove permanent hardness of water.

 III. Temporary hard

Sample A . A lot of soap is used before boiling. Very little soap is used on boiling/heating. Boiling remove temporary hardness of water.

(ii)Write the equation for the reaction at water sample C.

Chemical equation

2C₁₇H₃₅COO^– K⁺ (aq)
+ CaSO₄(aq) ->  (C₁₇H₃₅COO^– )Ca²⁺ (s) +
K₂SO₄(aq)

(insoluble Calcium octadecanote/scum)

Ionic equation

2C₁₇H₃₅COO^– K⁺ (aq)
+ Ca²⁺(aq) ->  (C₁₇H₃₅COO^– )Ca²⁺ (s) +
2K⁺(aq)

(insoluble Calcium octadecanote/scum)

Chemical equation

2C₁₇H₃₅COO^– K⁺ (aq)
+ MgSO₄(aq) ->  (C₁₇H₃₅COO^– )Mg²⁺ (s) +
K₂SO₄(aq)

(insoluble Calcium octadecanote/scum)

Ionic equation

2C₁₇H₃₅COO^– K⁺ (aq)
+ Mg²⁺(aq) ->  (C₁₇H₃₅COO^– )Mg²⁺ (s) +
2K⁺(aq)

(insoluble Magnesium octadecanote/scum)

(iii)Write the equation for the reaction at water sample A before boiling.

Chemical equation

2C₁₇H₃₅COO^– K⁺ (aq)
+ Ca(HCO₃)(aq) ->(C₁₇H₃₅COO^– )Ca²⁺ (s) +
2KHCO₃ (aq)

(insoluble Calcium octadecanote/scum)

Ionic equation

2C₁₇H₃₅COO^– K⁺ (aq)
+ Ca²⁺(aq) ->  (C₁₇H₃₅COO^– )Ca²⁺ (s) +

2K⁺(aq)

(insoluble Calcium octadecanote/scum)

Chemical equation

2C₁₇H₃₅COO^– K⁺ (aq)
+ Mg(HCO₃)(aq) ->(C₁₇H₃₅COO^– )Mg²⁺ (s) +
2KHCO₃ (aq)

(insoluble Calcium octadecanote/scum)

Ionic equation

2C₁₇H₃₅COO^– K⁺ (aq)
+ Mg²⁺(aq) ->  (C₁₇H₃₅COO^– )Mg²⁺ (s) +

2K⁺(aq)

(insoluble Magnesium octadecanote/scum)

(iv)Explain how water becomes hard

Natural or rain water flowing /passing through rocks containing calcium (chalk, gypsum, limestone)and magnesium compounds (dolomite)dissolve them to form soluble Ca²⁺ and Mg²⁺ ions that causes water hardness.

(v)State two useful benefits of hard water

 -Used in bone and teeth formation

 -Coral polyps use hard water to form coral reefs

 -Snails use hard water to make their shells

2.Study the scheme below and use it to answer the questions that follow.

(a)Identify :

 (i)brown solid A

Alkyl hydrogen sulphate(VI)

 (ii)substance B

Sodium alkyl hydrogen sulphate(VI)

(b)Write a general formula of:

 (i)Substance A.

  O

R-O-S O₃
H
//
R- O – S – O – H

O

(ii)Substance B   O

 R-O-S O₃
^– Na⁺  R- O – S – O ^– Na⁺

  O

(c)State one

(i) advantage of continued use of substance B

 –Does not form scum with hard water

 -Is cheap to make

 -Does not use food for human as a raw material.

(ii)disadvantage of continued use of substance B.

 Is non-biodegradable therefore do not pollute the environment

(d)Explain the action of B during washing.

 Has a non-polar hydrocarbon long tail that dissolves in dirt/grease/oil/fat.

 Has a polar/ionic hydrophilic head that dissolves in water

Through mechanical agitation the dirt is plucked /removed from the garment and surrounded by the tail end preventing it from being deposited back on the garment.

(e) Ethene was substituted for olive oil in the above process. Write the equation and name of the new products A and B.

Product A

Ethene  +  Sulphuric(VI)acid ->  Ethyl hydrogen sulphate(VI)

H₂C=CH₂ +   H₂SO₄ –>  H₃C – CH₂ –O-SO₃H

Product B

Ethyl hydrogen sulphate(VI) + sodium hydroxide -> sodium Ethyl + Water

hydrogen sulphate(VI)

H₃C – CH₂ –O-SO₃H + NaOH ->  H₃C – CH₂ –O-SO₃^–Na⁺
+ H₂O  

(f)Ethanol can also undergo similar reactions forming new products A and B.Show this using a chemical equation.

Product A

Ethanol  +  Sulphuric(VI)acid ->Ethyl hydrogen sulphate(VI) + water

H₃C-CH₂OH  +   H₂SO₄ –>  H₃C – CH₂ –O-SO₃H + H₂O

Product B

Ethyl hydrogen sulphate(VI) + sodium hydroxide -> sodium Ethyl + Water

hydrogen sulphate(VI)

H₃C – CH₂ –O-SO₃H + NaOH ->  H₃C – CH₂ –O-SO₃^–Na⁺
+ H₂O

3.Below is part of a detergent

 H₃C – (CH₂ )₁₆ – O – SO₃ ^– K ⁺

(a)Write the formular of the polar and non-polar end

 Polar end

H₃C – (CH₂ )₁₆ –

 Non-polar end

– O – SO₃ ^– K ⁺

(b)Is the molecule a soapy or saopless detergent?

 Soapless detergent

(c)State one advantage of using the above detergent

 -does not form scum with hard water

 -is cheap to manufacture

4.The structure of a detergent is

 H H H H H H H H H H H H H

H- C- C- C-C- C- C- C- C- C- C -C- C- -C- COO^–Na⁺

 H H H H H H H H H H H H H

a)  Write the molecular formula of the detergent. (1mk)

CH₃(CH₂)₁₂COO^–Na⁺

b)  What type of detergent is represented by the formula? (1mk)

Soapy detergent

c)  When this type of detergent is used to wash linen in hard water, spots (marks) are left on the linen. Write the formula of the substance responsible for the spots (CH₃(CH₂)₁₂COO^–)₂Ca²⁺ / CH₃(CH₂)₁₂COO^–)₂Mg²⁺

D. POLYMERS AND FIBRES

Polymers and fibres are giant molecules of organic compounds. Polymers and fibres are formed when small molecules called monomers join together to form large molecules called polymers at high temperatures and pressures. This process is called polymerization.

Polymers and fibres are either:

 (a)Natural polymers and fibres

 (b)Synthetic polymers and fibres

Natural polymers and fibres are found in living things(plants and animals) Natural polymers/fibres include:

 -proteins/polypeptides making amino acids in animals

 -cellulose that make cotton,wool,paper and silk

 -Starch that come from glucose

 -Fats and oils

-Rubber from latex in rubber trees.

Synthetic polymers and fibres are man-made. They include:

 -polyethene

 -polychloroethene

 -polyphenylethene(polystyrene)

 -Terylene(Dacron)

 -Nylon-6,6

 -Perspex(artificial glass)

Synthetic polymers and fibres have the following characteristic advantages over natural polymers

1. They are light and portable

2. They are easy to manufacture.

3. They can easily be molded into shape of choice.

4. They are resistant to corrosion, water, air , acids, bases and salts.

5. They are comparatively cheap, affordable, colourful and aesthetic

Synthetic polymers and fibres however have the following disadvantages over natural polymers

1. They are non-biodegradable and hence cause environmental pollution during disposal
2. They give out highly poisonous gases when burnt like chlorine/carbon(II)oxide
3. Some on burning produce Carbon(IV)oxide. Carbon(IV)oxide is a green house gas that cause global warming.
4. Compared to some metals, they are poor conductors of heat,electricity and have lower tensile strength.

To reduce environmental pollution from synthetic polymers and fibres, the followitn methods of disposal should be used:

1.Recycling: Once produced all synthetic polymers and fibres should be recycled to a new product. This prevents accumulation of the synthetic polymers and fibres in the environment.

2.Production of biodegradable synthetic polymers and fibres that rot away.

There are two types of polymerization:

 (a)addition polymerization

 (b)condensation polymerization

(a)addition polymerization

Addition polymerization is the process where a small unsaturated monomer (alkene ) molecule join together to form a large saturated molecule. Only alkenes undergo addition polymerization.

Addition polymers are named from the alkene/monomer making the polymer and adding the prefix “poly” before the name of monomer to form a polyalkene

During addition polymerization

(i)the double bond in alkenes break

(ii)free radicals are formed

(iii)the free radicals collide with each other and join to form a larger molecule. The more collisions the larger the molecule.

Examples of addition polymerization

1.Formation of Polyethene

Polyethene is an addition polymer formed when ethene molecule/monomer join together to form a large molecule/polymer at high temperatures and pressure.

During polymerization:

(i)many molecules are brought nearer to each other by the high pressure(which reduces the volume occupied by reacting paticles)

 H H H H    H H H H

C = C  +   C = C  +  C = C  +  C = C + …

H H H H    H H H H

Ethene  +  Ethene  +  Ethene  +  Ethene + …

(ii)the double bond joining the ethane molecule break to free readicals

H H H H    H H H H

•C – C•  +  •C – C•  + •C – C•  + •C – C• + …

H H H H    H H H H

Ethene radical + Ethene radical + Ethene radical  + Ethene radical + …

(iii)the free radicals collide with each other and join to form a larger molecule

H H H H H H H H lone pair of electrons

•C – C – C – C – C – C – C – C• + …

H H H H H H H H

Lone pair of electrons can be used to join more monomers to form longer polyethene.

Polyethene molecule can be represented as:

H H H H H H H H extension of

molecule/polymer

– C – C – C – C – C – C – C – C- + …

H H H H H H H H

Since the molecule is a repetition of one monomer, then the polymer is:

H H

( C – C )n

H H

Where n is the number of monomers in the polymer. The number of monomers in the polymer can be determined from the molar mass of the polymer and monomer from the relationship:

Number of monomers/repeating units in monomer = Molar mass polymer

Molar mass monomer  

Examples

Polythene has a molar mass of 4760.Calculate the number of ethene molecules in the polymer(C=12.0, H=1.0 )

Number of monomers/repeating units in polyomer = Molar mass polymer

Molar mass monomer

=> Molar mass ethene (C₂H₄ )= 28 Molar mass polyethene = 4760

Substituting 4760 = 170 ethene molecules

28

The commercial name of polyethene is polythene. It is an elastic, tough, transparent and durable plastic. Polythene is used:

 (i)in making plastic bag

 (ii)bowls and plastic bags

 (iii)packaging materials

2.Formation of Polychlorethene

Polychloroethene is an addition polymer formed when chloroethene molecule/monomer join together to form a large molecule/polymer at high temperatures and pressure.

During polymerization:

(i)many molecules are brought nearer to each other by the high pressure(which reduces the volume occupied by reacting particles)

 H H H H    H H H H

C = C  +   C = C  +  C = C  +  C = C + …

H Cl H Cl    H Cl   H Cl

chloroethene  + chloroethene  + chloroethene  + chloroethene + …

(ii)the double bond joining the chloroethene molecule break to free radicals

H H H H    H H H H

•C – C•  +  •C – C•  + •C – C•  + •C – C• + …

H Cl H Cl    H Cl   H Cl

(iii)the free radicals collide with each other and join to form a larger molecule

H H H H H H H H lone pair of electrons

•C – C – C – C – C – C – C – C• + …

H Cl H Cl H Cl H Cl

Lone pair of electrons can be used to join more monomers to form longer polychloroethene.

Polychloroethene molecule can be represented as:

H H H H H H H H extension of

molecule/polymer

– C – C – C – C – C – C – C – C- + …

H Cl H Cl H Cl H Cl

Since the molecule is a repetition of one monomer, then the polymer is:

H H

( C – C )n

H Cl

Examples

Polychlorothene has a molar mass of 4760.Calculate the number of chlorethene molecules in the polymer(C=12.0, H=1.0,Cl=35.5 )

Number of monomers/repeating units in monomer = Molar mass polymer

Molar mass monomer

=> Molar mass ethene (C₂H₃Cl )= 62.5 Molar mass polyethene = 4760

Substituting   4760   = 77.16 => 77 polychloroethene molecules(whole number)

  62.5

The commercial name of polychloroethene is polyvinylchloride(PVC). It is a tough, non-transparent and durable plastic. PVC is used:

 (i)in making plastic rope

 (ii)water pipes

 (iii)crates and boxes

3.Formation of Polyphenylethene

Polyphenylethene is an addition polymer formed when phenylethene molecule/monomer join together to form a large molecule/polymer at high temperatures and pressure.

During polymerization:

(i)many molecules are brought nearer to each other by the high pressure(which reduces the volume occupied by reacting particles)

 H H H H    H H H H

C = C  +   C = C  +  C = C  +  C = C + …

H C₆H₅ H C₆H₅    H C₆H₅   H C₆H₅

phenylethene  + phenylethene  + phenylethene  + phenylethene + …

(ii)the double bond joining the phenylethene molecule break to free radicals

H H H H    H H H H

•C – C•  +  •C – C•  + •C – C•  + •C – C• + …

H C₆H₅ H C₆H₅    H C₆H₅   H C₆H₅

(iii)the free radicals collide with each other and join to form a larger molecule

H   H   H   H H H H H lone pair of electrons

• C – C – C – C – C – C – C – C • + …

H C₆H₅ H   C₆H₅ H C₆H₅ H C₆H₅

Lone pair of electrons can be used to join more monomers to form longer polyphenylethene.

Polyphenylethene molecule can be represented as:

H   H   H   H H H H H

– C – C – C – C – C – C – C – C –

H C₆H₅ H   C₆H₅ H C₆H₅ H C₆H₅

Since the molecule is a repetition of one monomer, then the polymer is:

H H

( C – C )n

H C₆H₅

 Examples

Polyphenylthene has a molar mass of 4760.Calculate the number of phenylethene molecules in the polymer(C=12.0, H=1.0, )

Number of monomers/repeating units in monomer = Molar mass polymer

Molar mass monomer

=> Molar mass ethene (C₈H₈ )= 104 Molar mass polyethene = 4760

Substituting   4760   = 45.7692 =>45 polyphenylethene molecules(whole number)

  104

The commercial name of polyphenylethene is polystyrene. It is a very light durable plastic. Polystyrene is used:

 (i)in making packaging material for carrying delicate items like computers, radion,calculators.

 (ii)ceiling tiles

 (iii)clothe linings

4.Formation of Polypropene

Polypropene is an addition polymer formed when propene molecule/monomer join together to form a large molecule/polymer at high temperatures and pressure.

During polymerization:

(i)many molecules are brought nearer to each other by the high pressure(which reduces the volume occupied by reacting particles)

 H H H H    H H H H

C = C  +   C = C  +  C = C  +  C = C + …

H CH₃ H CH₃    H CH₃   H CH₃

propene  + propene  + propene  + propene + …

(ii)the double bond joining the phenylethene molecule break to free radicals

H H H H    H H H H

•C – C•  +  •C – C•  + •C – C•  + •C – C• + …

H CH₃ H CH₃    H CH₃   H CH₃

(iii)the free radicals collide with each other and join to form a larger molecule

H   H   H   H H H H H lone pair of electrons

• C – C – C – C – C – C – C – C • + …

H CH₃ H   CH₃ H CH₃ H CH₃

Lone pair of electrons can be used to join more monomers to form longer propene.

propene molecule can be represented as:

H   H   H   H H H H H

– C – C – C – C – C – C – C – C –

H CH₃ H   CH₃ H CH₃ H CH₃

Since the molecule is a repetition of one monomer, then the polymer is:

H H

( C – C )n

H CH₃

 Examples

Polypropene has a molar mass of 4760.Calculate the number of propene molecules in the polymer(C=12.0, H=1.0, )

Number of monomers/repeating units in monomer = Molar mass polymer

Molar mass monomer

=> Molar mass propene (C₃H₈ )= 44 Molar mass polyethene = 4760

Substituting   4760   = 108.1818 =>108 propene molecules(whole number)

  44

The commercial name of polyphenylethene is polystyrene. It is a very light durable plastic. Polystyrene is used:

 (i)in making packaging material for carrying delicate items like computers, radion,calculators.

 (ii)ceiling tiles

 (iii)clothe linings

5.Formation of Polytetrafluorothene

Polytetrafluorothene is an addition polymer formed when tetrafluoroethene molecule/monomer join together to form a large molecule/polymer at high temperatures and pressure.

During polymerization:

(i)many molecules are brought nearer to each other by the high pressure(which reduces the volume occupied by reacting particles)

 F F   F F    F F F F

C = C  +   C = C  +  C = C  +  C = C + …

F F F F    F F   F F

tetrafluoroethene  + tetrafluoroethene+ tetrafluoroethene+ tetrafluoroethene + …

(ii)the double bond joining the tetrafluoroethene molecule break to free radicals

F F F F     F F F F

•C – C•  +  •C – C•  + •C – C•  + •C – C• + …

F F F F   F F   F F

(iii)the free radicals collide with each other and join to form a larger molecule

F F F F F F F F lone pair of electrons

•C – C – C – C – C – C – C – C• + …

F F F F
F F F F

Lone pair of electrons can be used to join more monomers to form longer polytetrafluoroethene.

polytetrafluoroethene molecule can be represented as:

F F F F F F F F extension of

molecule/polymer

– C – C – C – C – C – C – C – C- + …

F F F F F F F F

Since the molecule is a repetition of one monomer, then the polymer is:

F F

( C – C )n

F F

Examples

Polytetrafluorothene has a molar mass of 4760.Calculate the number of tetrafluoroethene molecules in the polymer(C=12.0, ,F=19 )

Number of monomers/repeating units in monomer = Molar mass polymer

Molar mass monomer

=> Molar mass ethene (C₂F₄ )= 62.5 Molar mass polyethene = 4760

Substituting   4760   = 77.16 => 77 polychloroethene molecules(whole number)

  62.5

The commercial name of polytetrafluorethene(P.T.F.E) is Teflon(P.T.F.E). It is a tough, non-transparent and durable plastic. PVC is used:

 (i)in making plastic rope

 (ii)water pipes

 (iii)crates and boxes

5.Formation of rubber from Latex

Natural rubber is obtained from rubber trees.

During harvesting an incision is made on the rubber tree to produce a milky white substance called latex.

Latex is a mixture of rubber and lots of water.

The latex is then added an acid to coagulate the rubber.

Natural rubber is a polymer of 2-methylbut-1,3-diene

H CH₃ H   H

CH₂=C (CH₃) CH = CH₂ H – C = C – C = C – H

During natural polymerization to rubber, one double C=C bond break to self add to another molecule. The double bond remaining move to carbon “2” thus;

H CH₃ H   H H CH₃ H   H

– C – C = C – C – C – C = C – C –

H   H   H   H

Generally the structure of rubber is thus;

  H CH₃ H   H

  -(- C – C = C – C -)_n–

H H

Pure rubber is soft and sticky. It is used to make erasers, car tyres. Most of it is vulcanized. Vulcanization is the process of heating rubber with sulphur to make it harder/tougher.

During vulcanization the sulphur atoms form a cross link between chains of rubber molecules/polymers. This decreases the number of C=C double bonds in the polymer.

H CH₃ H   H H CH₃ H   H

– C – C – C – C – C – C – C – C –

H S H   H S   H  

H CH₃ S   H H CH₃ S   H

– C – C – C – C – C – C – C – C –

H H H   H H   H  

Vulcanized rubber is used to make tyres, shoes and valves.

6.Formation of synthetic rubber

Synthetic rubber is able to resist action of oil,abrasion and organic solvents which rubber cannot.

Common synthetic rubber is a polymer of 2-chlorobut-1,3-diene

H Cl H   H

CH₂=C (Cl CH = CH₂ H – C = C – C = C – H

During polymerization to synthetic rubber, one double C=C bond is broken to self add to another molecule. The double bond remaining move to carbon “2” thus;

H Cl H   H H Cl H   H

– C – C = C – C – C – C = C – C –

H   H   H   H

Generally the structure of rubber is thus;

  H Cl H   H

  -(- C – C = C – C -)_n–

H H

Rubber is thus strengthened through vulcanization and manufacture of synthetic rubber.

(b)Condensation polymerization

Condensation polymerization is the process where two or more small monomers join together to form a larger molecule by elimination/removal of a simple molecule. (usually water).

Condensation polymers acquire a different name from the monomers because the two monomers are two different compounds

During condensation polymerization:

(i)the two monomers are brought together by high pressure to reduce distance between them.

(ii)monomers realign themselves at the functional group.

(iii)from each functional group an element is removed so as to form simple molecule (of usually H₂O/HCl)

(iv)the two monomers join without the simple molecule of H₂O/HCl

Examples of condensation polymerization

1.Formation of Nylon-6,6

Method 1: Nylon-6,6 can be made from the condensation polymerization of hexan-1,6-dioic acid with hexan-1,6-diamine.Amines are a group of homologous series with a general formula R-NH₂ and thus -NH₂ as the functional
group.

During the formation of Nylon-6,6:

(i)the two monomers are brought together by high pressure to reduce distance between them and realign themselves at the functional groups.

  O O H H

H- O – C – (CH₂ ) ₄ – C – O – H + H –N – (CH₂) ₆ – N – H

(iii)from each functional group an element is removed so as to form a molecule of H₂O and the two monomers join at the linkage .

O O H H

H- O – C – (CH₂ ) ₄ – C – N – (CH₂) ₆ – N – H + H ₂O

.

Polymer bond linkage

Nylon-6,6 derive its name from the two monomers each with six carbon chain

Method 2: Nylon-6,6 can be made from the condensation polymerization of hexan-1,6-dioyl dichloride with hexan-1,6-diamine.

Hexan-1,6-dioyl dichloride belong to a group of homologous series with a general formula R-OCl
and thus -OCl
as the functional
group.

The R-OCl is formed when the “OH” in R-OOH/alkanoic acid is replaced by Cl/chlorine/Halogen

During the formation of Nylon-6,6:

(i)the two monomers are brought together by high pressure to reduce distance between them and realign themselves at the functional groups.

  O O H H

Cl – C – (CH₂ ) ₄ – C – Cl + H –N – (CH₂) ₆ – N – H

(iii)from each functional group an element is removed so as to form a molecule of HCl and the two monomers join at the linkage .

O O H H

Cl – C – (CH₂ ) ₄ – C – N – (CH₂) ₆ – N – H + HCl

.

Polymer bond linkage

The two monomers each has six carbon chain hence the name “nylon-6,6”

The commercial name of Nylon-6,6 is Nylon It is a a tough, elastic and durable plastic. It is used to make clothes, plastic ropes and carpets.

2.Formation of Terylene

Method 1: Terylene can be made from the condensation polymerization of ethan-1,2-diol with benzene-1,4-dicarboxylic acid.

Benzene-1,4-dicarboxylic acid a group of homologous series with a general formula R-COOH
where R is a ring of six carbon atom called Benzene ring .The functional
group is -COOH.

During the formation of Terylene:

(i)the two monomers are brought together by high pressure to reduce distance between them and realign themselves at the functional groups.

  O O

H- O – C – C₆H₅ – C – O – H + H –O – CH₂ CH₂ – O – H

(iii)from each functional group an element is removed so as to form a molecule of H₂O and the two monomers join at the linkage .

O O

H- O – C – C₆H₅ – C – O – (CH₂) ₆ – N – H + H ₂O

.

Polymer bond linkage of terylene

Method 2: Terylene can be made from the condensation polymerization of benzene-1,4-dioyl dichloride with ethan-1,2-diol.

Benzene-1,4-dioyl dichloride belong to a group of homologous series with a general formula R-OCl
and thus -OCl
as the functional
group and R as a benzene ring.

The R-OCl is formed when the “OH” in R-OOH is replaced by Cl/chlorine/Halogen

During the formation of Terylene

(i)the two monomers are brought together by high pressure to reduce distance between them and realign themselves at the functional groups.

  O O

Cl – C – C₅H₅ – C – Cl + H –O – CH₂ CH₂ – O – H

(iii)from each functional group an element is removed so as to form a molecule of HCl and the two monomers join at the linkage .

O O

Cl – C – C₅H₅ – C – O – CH₂ CH₂ – O – H + HCl

.

Polymer bond linkage of terylene

The commercial name of terylene is Polyester /polyster It is a a tough, elastic and durable plastic. It is used to make clothes, plastic ropes and sails and plastic model kits.

Practice questions Organic chemistry

1. A student mixed equal volumes of Ethanol and butanoic acid. He added a few drops of concentrated Sulphuric (VI) acid and warmed the mixture

(i) Name and write the formula of the main products   Name………………………………….

Formula……………………………………..

(ii) Which homologous series does the product named in (i) above belong?

2. The structure of the monomer phenyl ethene is given below:-

a) Give the structure of the polymer formed when four of the monomers are added together

b) Give the name of the polymer formed in (a) above  

3. Explain the environmental effects of burning plastics in air as a disposal method 4.  Write chemical equation to represent the effect of heat on ammonium carbonate

5.  Sodium octadecanoate has a chemical formula CH₃(CH₂)₆ COO^–Na⁺, which is used as soap.

 Explain why a lot of soap is needed when washing with hard water

6.  A natural polymer is made up of the monomer:

 (a) Write the structural formula of the repeat unit of the polymer  (b) When 5.0 x 10^-5 moles of the polymer were hydrolysed, 0.515g of the monomer were obtained.

Determine the number of the monomer molecules in this polymer.  

(C = 12; H = 1; N = 14; O =16)

7.  The formula below represents active ingredients of two cleansing agents A and B

 Which one of the cleansing agents would be suitable to be used in water containing magnesium hydrogen carbonate? Explain

 (a) Give the name of the monomer and draw its structures    (b) Identify the type of polymerization that takes place    (c) State one advantage of synthetic polymers

9.  Ethanol and Pentane are miscible liquids. Explain how water can be used to separate a mixture of ethanol and pentane  

10.

 (a) What is absolute ethanol?

(i) Plot a graph of volume of oxygen gas against time     (ii) Determine the rate of reaction at time 156 seconds     (iii) From the graph, find the time taken for 18cm³ of oxygen to be produced   (iv) Write a chemical equation to show how hydrogen peroxide decomposes in the presence of manganese (IV) Oxide

(b) The diagram below shows how a Le’clanche (Dry cell) appears:-

(i) What is the function of MnO₂ in the cell above?

 (ii) Write the equation of a reaction that occurs at the cathode

 (iii) Calculate the mass of Zinc that is consumed when a current of 0.1amperes flows through the above cell for 30minutes (1F =96500c Zn =65)  

12.  (a) Give the IUPAC names of the following compounds:

  (i) CH₃COOCH₂CH₃ *

(ii)  

(b) The structure below shows some reactions starting with ethanol. Study it and answer

the questions that follow:

(i) Write the formula of the organic compounds P and S

  (ii) Name the type of reaction, the reagent(s) and condition for the reactions in the following steps :-

(I) Step I  

   (II) Step II  

(III) Step III  

  (iii) Name reagent R  ……………………………………………………………   (iv) Draw the structural formula of T and give its name

(v) (I) Name compound U………………………………………………………..

(II) If the relative molecular mass of U is 42000, determine the value of n (C=12, H=1)

 (c) State why C₂H₄ burns with a more smoky flame than C₂H₆  13. a) State two factors that affect the properties of a polymer  b) Name the compound with the formula below : CH₃CH₂CH₂ONa

 c) Study the scheme below and use it to answer the questions that follow:-

i) Name the following compounds:-

I. Product T ………………………… II. K ……… ii) State one common physical property of substance G

iii) State the type of reaction that occurred in step J

iv) Give one use of substance K

v) Write an equation for the combustion of compound P vi) Explain how compounds CH₃CH₂COOH and CH₃CH₂CH₂OH can be distinguished chemically

vii) If a polymer K has relative molecular mass of 12,600, calculate the value of n (H=1 C =12)

14.  Study the scheme given below and answer the questions that follow:-

(a) (i) Name compound P ……………………………………………………………………  

(ii) Write an equation for the reaction between CH₃CH₂COOH and Na₂CO₃   (b) State one use of polymer Q

(c) Name one oxidising agent that can be used in step II   …………………………………..

(d) A sample of polymer Q is found to have a molecular mass of 4200. Determine the number of

monomers in the polymer  (H = 1, C = 12)

(e) Name the type of reaction in step I …………………………………………………………..

(f) State one industrial application of step III  

(g)State how burning can be used to distinguish between propane and propyne. Explain your

answer

(h) 1000cm³ of ethene (C₂H₄) burnt in oxygen to produce Carbon (II) Oxide and water vapour.

Calculate the minimum volume of air needed for the complete combustion of ethene

(Air contains 20% by volume of oxygen)  

15.  (a) Study the schematic diagram below and answer the questions that follow:-

 (i) Identify the following:

Substance Q ………………………………………………………………………………………………..  

Substance R…………………………………………………………………………………………………  

 Gas P…………………………………………………………………………………………………………..  

  (ii) Name:

 Step 1…………………………………………………………………………………….

 Step 4…………………………………………………………………………………….

  (iii) Draw the structural formula of the major product of step 5  (iv) State the condition and reagent in step 3  16.  Study the flow chart below and answer the questions that follow

(a) (i) Name the following organic compounds:

 M……………………………………………………………..……..

 L…………………………………………………………………..   (ii) Name the process in step:

Step 2 ………………………………………………………….….

Step 4 ………………………………………………………….…

(iii) Identify the reagent P and Q

 (iv) Write an equation for the reaction between CH₃CH₂CH₂OH and sodium 17.  a) Give the names of the following compounds:

 i) CH₃CH₂CH₂CH₂OH ……………………………………………………………………

 ii) CH₃CH₂COOH  …………………………………………………………………

 iii) CH₃C – O- CH₂CH₃ ……………………………………………………………………

18.   Study the scheme given below and answer the questions that follow;

 i) Name the reagents used in:

Step I: ………………………………………………………………………

Step II ……………………………………………………………………  

Step III  ………………………………………………………………………

 ii) Write an equation to show products formed for the complete combustion of CH = CH  

 iii) Explain one disadvantage of continued use of items made from the compound formed in step III

19.   A hydrated salt has the following composition by mass. Iron 20.2 %, oxygen 23.0%, sulphur 11.5%, water 45.3%

 i) Determine the formula of the hydrated salt (Fe=56, S=32, O=16, H=11)  ii) 6.95g of the hydrated salt in c(i) above were dissolved in distilled water and the total volume made to 250cm³ of solution. Calculate the concentration of the resulting salt solution in moles per litre. (Given that the molecula mass of the salt is 278)

20. Write an equation to show products formed for the complete combustion of CH = CH  

 iii) Explain one disadvantage of continued use of items made from the compound formed in step III

21.  Give the IUPAC name for each of the following organic compounds;

  i) CH₃ – CH – CH₂ – CH₃

  OH

ii)CH₃ – CH – CH₂ – CH₂ – CH₃

C₂H₅

  iii)CH₃COOCH₂CH₂CH₃

22.  The structure below represents a cleansing agent.

  O

R – S – O^–Na⁺

O

a) State the type of cleansing agent represented above b) State one advantage and one disadvantage of using the above cleansing agent.

23.  The structure below shows part of polymer .Use it to answer the questions that follow.

CH₃ CH₃ CH₃





― CH – CH₂ – CH- CH₂ – CH – CH₂ ―

 a) Derive the structure of the monomer  

 b) Name the type of polymerization represented above  

24.  The flow chart below represents a series of reactions starting with ethanoic acid:-

(a) Identify substances A and B

 (b) Name the process I

25.  a) Write an equation showing how ammonium nitrate may be prepared starting with ammonia gas

 (b) Calculate the maximum mass of ammonium nitrate that can be prepared using 5.3kg of ammonia (H=1, N=14, O=16)

26.  (a) What is meant by the term, esterification?

 (b) Draw the structural formulae of two compounds that may be reacted to form ethylpropanoate  

27.  (a) Draw the structure of pentanoic acid  

 (b) Draw the structure and give the name of the organic compound formed when ethanol reacts with pentanoic acid in presence of concentrated sulphuric acid

28.  The scheme below shows some reactions starting with ethanol. Study it and answer the questions

 that follow:-

 (i) Name and draw the structure of substance Q  

(ii) Give the names of the reactions that take place in steps 2 and 4 (iii) What reagent is necessary for reaction that takes place in step 3

29.  Substances A and B are represented by the formulae ROH and RCOOH respectively.

They belong to two different homologous series of organic compounds. If both A and B

react with potassium metal:

(a) Name the common product produced by both  (b) State the observation made when each of the samples A and B are reacted with sodium hydrogen carbonate

  (i) A

  (ii) B

30.  Below are structures of particles. Use it to answer questions that follow. In each case only electrons in the outermost energy level are shown

key

P = Proton

N = Neutron

X = Electron

(a) Identify the particle which is an anion  

31.  Plastics and rubber are extensively used to cover electrical wires.

 (a) What term is used to describe plastic and rubbers used in this way? (b) Explain why plastics and rubbers are used this way

32.  The scheme below represents the manufacture of a cleaning agent X

(a) Draw the structure of X and state the type of cleaning agent to which X belong  (b) State one disadvantage of using X as a cleaning agent  

33.  Y grams of a radioactive isotope take 120days to decay to 3.5grams. The half-life period of the isotope is 20days

 (a) Find the initial mass of the isotope  

 (b) Give one application of radioactivity in agriculture  

34.  The structure below represents a polymer. Study and answer the questions that follow:-

 (i) Name the polymer above……………………………………………………………………….  (ii) Determine the value of n if giant molecule had relative molecular mass of 4956  

35. RCOO^–Na⁺ and RCH₂OSO₃^–Na⁺ are two types of cleansing agents;

i) Name the class of cleansing agents to which each belongs  

ii) Which one of these agents in (i) above would be more suitable when washing with water from the Indian ocean. Explain  

iii) Both sulphur (IV) oxide and chlorine are used bleaching agents. Explain the difference in their bleaching properties

36. The formula given below represents a portion of a polymer

(a) Give the name of the polymer

(b) Draw the structure of the monomer used to manufacture the polymer