Saturday, October 1, 2022

Special properties of carbon                                                           
Carbon has a number  of unique  properties  which influence  how it behaves  and how it bonds with other atoms:

  • Carbon (Figure 4.2) has four valence electrons ¬†which¬† means ¬†that each ¬†carbon atom ¬†can ¬†form a maximum ¬†of four bonds ¬†with¬† other ¬†atoms.¬† ¬†Because ¬†of the number ¬†of bonds ¬†that carbon ¬†can form with other ¬†atoms, ¬†organic ¬†compounds can be very complex.

‚Äď ¬†Carbon can form bonds with other carbon ¬†atoms to form single, double ¬†or triple covalent bonds.

‚Äď ¬†Carbon can also form bonds with other atoms like hydrogen, ¬†oxygen, nitro- gen and the halogens.

‚Äď ¬†Carbon can bond to form straight chain, branched, and cyclic molecules.

  • Because of its position on ¬†the ¬†periodic ¬†table, ¬†most¬† of the ¬†bonds ¬†that carbon forms with other atoms are covalent.¬† ¬†Think for example ¬†of a C ‚ąí C bond. ¬†The difference in electronegativity ¬†between ¬†the two atoms is zero,¬† so this is a pure covalent ¬†bond. ¬†In the case of a C ‚ąí H bond, ¬†the difference in electronegativity between ¬†carbon (2,5) and hydrogen (2,2) is so small that C ‚ąí H bonds are almost purely covalent. ¬†The result of this is that most organic compounds are non-polar. This affects some of the properties¬† of organic compounds.

Sources of carbon
The main  source  of the  carbon  in organic  compounds is carbon  dioxide  in the  at- mosphere.  Plants use sunlight to convert carbon dioxide and water (inorganic com- pounds) into sugar (an organic compound)  through the process of photosynthesis.


Plants are therefore  able  to make  their own  organic  compounds through  photosyn- thesis,  while  animals  feed on  plants  or plant  products  in order  to gain the  organic compounds that they need to survive.
Other important sources of carbon are fossil fuels such as coal, petroleum and natural gas.  This is because fossil fuels are themselves formed from the decaying remains of dead organisms (refer to Grade 11 for more information on fossil fuels).
Representing organic molecules
There are a number  of ways to represent  organic compounds. It is useful to know all of these so that you can recognise a molecule  regardless of how it is shown.  There are four main ways of representing  a compound in two dimensions  (on your page).  We will use the examples  of two molecules  called  2-methylpropane and butane  to help explain the difference between  each.

  • Structural formula

The structural formula of an organic compound shows every bond  between  ev- ery atom in the molecule.  Each bond  is represented by a line.  The structural formulae of 2-methylpropane and butane are shown in Figure 4.5.
Figure 4.5: The structural formula of (a) 2-methylpropane and (b) butane

Figure 4.6: Different ways of representing a carbon atom bonding to four hydrogen atoms.

  • Semi-structural formula

It is possible to understand the structure of an organic molecule without writing out all the carbon-hydrogen bonds.   This way of writing a structure is called  a semi-structural formula and is shown in Figure 4.7.

Figure 4.7: The semi-structural formulae of (a) 2-methylpropane and (b) butane.
Compare these semi-structural representations with the structural representations shown in Figure 4.5.

  • Condensed structural formula

It is also possible to represent a molecule without showing any bonds  between atoms at all. This is called a condensed structural formula  (Figure 4.8). As for a semi-structural representation, the carbon  atoms are grouped  with the hydrogen atoms bonded  directly to it.  The bonds  between  these  groups are not shown. Branched  or substituent  groups are shown  in brackets after the carbon  atom to which they are bonded.Figure 4.8: The condensed structural formulae of (a) 2-methylpropane and (b) butane.

  • Molecular formula

The molecular  formula of a compound shows how many atoms of each type are in a molecule. The number of each atom is written as a subscript after the atomic symbol. The molecular  formula of 2-methylpropane is: C4H10
This means that each molecule of 2-methylpropane consists of four carbon atoms and ten hydrogen atoms. The molecular formula of butane is also C4H10 . Molecular formula gives no structural information about the compound. Of course molecules  are not two-dimensional so shown below are a few examples  of different ways to represent  methane  (CH4 , Figure 4.9) and ethane  (C2H6 , Figure 4.10).

Figure 4.9: Different ways of representing  methane

Figure 4.10:  Different ways of representing  ethane.
 
Figure 4.11: (a) Two-dimensional and (b) three dimensional representation of butane
This means that butane can be represented in two dimensions  as shown in Figure 4.11 (a) but it actually looks more like the three-dimensional representation given in Figure 4.11 (b)
Exercise 4 ‚Äď 1:¬† Representing organic¬† compounds

  1. For each of the following,  give the structural formula  and  the molecular formula.
  2. a) CH3CH2CH3 b) CH3CH2CH(CH3)CH3  c) CH3CH3
  3. For each of the  following  organic  compounds, give the  condensed structural formula  and the molecular formula.

  1. Give two possible structural  formulae for the compound with a molecular  formula of C4H10 .

Functional groups                                                        
The way in which a compound will react is determined by a particular characteristic of a group of atoms and the way they are bonded ¬†(e.g. double C‚ąíC bond, C‚ąíOH group). This is called ¬†the¬† functional group.¬† ¬†This group ¬†is important ¬†in determining ¬†how ¬†a compound will react.¬† ¬†The same¬† functional ¬†group ¬†will undergo ¬†the same¬† or similar chemical ¬†reaction(s) regardless ¬†of the size of the molecule ¬†it is a part of. ¬†Molecules can have more than one functional group.
DEFINITION:  Functional  group
In organic  chemistry  a functional  group  is a specific group  of atoms (and the bonds between  them) that are responsible  for the characteristic  chemical  reactions  of those molecules.
In one group of organic compounds, called the hydrocarbons, the single, double  and triple bonds  between  carbon  atoms give rise to the alkanes,  alkenes  and alkynes, re- spectively.  The double  carbon-carbon bonds (in the alkenes) and triple carbon-carbon bonds (in the alkynes) are examples of functional groups.
In another  group of organic compounds, called the alcohols,  an oxygen and a hydro- gen atom are bonded  to each  other to form the functional  group  (in other words an alcohol  has an OH in it).  All alcohols  will contain  an oxygen and a hydrogen  atom bonded  together  in some  part of the  molecule.  Table 4.1  summarises  some  of the common  functional groups.  We will look at these in more detail later in this chapter.

Table 4.1: Some functional groups of organic compounds.

There are some important points to note as we discuss functional groups:

  • The beginning of a compound name (prefix) comes from the number of carbons in the longest chain:

  • The end of a compound name (suffix) comes from the functional group, e.g. an alkane has the suffix -ane. Refer to the examples in Table 4.1.

For more information on naming organic molecules  see Section 4.3.
Saturated and unsaturated structures   
Hydrocarbons  that  contain   only   single bonds  are called  saturated hydrocarbons because each carbon atom is bonded  to as many hydrogen  atoms as possible. Figure 4.12 shows  a molecule  of ethane,  which is a saturated hydrocarbon.

Figure 4.12:  A saturated hydrocarbon, ethane.

DEFINITION:  Saturated compounds
A saturated compound has no double  or triple bonds  (i.e.  they have  single bonds only). All carbon atoms are bonded  to four other atoms. Hydrocarbons that contain double or triple bonds are called unsaturated hydrocarbons because  they don’t contain as many hydrogen  atoms as possible.
DEFINITION:  Unsaturated compounds
An unsaturated compound contains double or triple bonds.  A carbon atom may there- fore be bonded  to only two or three other atoms.Figure 4.13 shows molecules  of ethene  and ethyne which are unsaturated hydrocarbons. If you compare  the number  of carbon and hydrogen  atoms in a molecule  of ethane  and a molecule  of ethene,  you will see that the number  of hydrogen  atoms in ethene  is less than the number  of hydrogen  atoms in ethane despite the fact that they both contain two carbon atoms. In order for an unsaturated hydrocarbon compound to become  saturated, one of the two (or three) bonds in a double  (or triple) bond has to be broken, and additional atoms added.

Figure 4.13:  Unsaturated  hydrocarbons:(a) ethene  and (b) ethyne

The hydrocarbons   
Let us first look at a group of organic compounds known as the hydrocarbons.
DEFINITION:  Hydrocarbon
An organic molecule  which contains only carbon  and hydrogen  atoms with no other functional groups besides single, double  or triple carbon-carbon bonds.
The hydrocarbons that we are going to look at are called aliphatic  compounds.  The aliphatic compounds are divided into acyclic compounds (chain structures) and cyclic compounds (ring structures).  The chain  structures  are further divided  into structures that contain  only single bonds  (alkanes), those that contain  at least one double  bond (alkenes) and those that contain at least one triple bond (alkynes).
Cyclic compounds (which will not be covered  in this book) include  structures  such as a cyclopentane ring, which  is found in insulating  foam and in appliances  such as fridges and freezers. Figure 4.14 summarises the classification of the hydrocarbons.

Figure 4.14:  The classification of the aliphatic hydrocarbons.

We will now look at each of the acyclic, aliphatic hydrocarbon groups in more detail.
The alkanes
The alkanes are hydrocarbons that only contain single covalent bonds between their carbon atoms. This means that they are saturated compounds and are quite unreactive. The simplest alkane has only one carbon atom and is called methane. This molecule is shown in Figure 4.15.

Figure 4.15:  The (a) structural and (b) molecular formula representations of methane
The second  alkane  in the series has two carbon  atoms and is called  ethane.  This is shown in Figure 4.16.

Figure 4.16:  The (a) structural,  (b) condensed structural and (c) molecular  formula representations of ethane. (d) An atomic model of ethane
The third alkane  in the series has three  carbon  atoms and is called  propane (Figure 4.2).

When you look at the molecular  formula for each of the alkanes, you should notice a pattern developing.  For each carbon atom that is added to the molecule, two hydrogen atoms are added.  In other words, each molecule  differs from the one before it by CH2 This is called a homologous series.
DEFINITION:  Homologous series
A homologous series is a series of compounds with the same  general  formula.   All molecules  in this series will contain the same functional groups
The  general   formula  is  similar  to  both the molecular  formula and the condensed structural  formula.   The functional  group is written as it would be in the condensed structural  formula (to make  it more  obvious),  while  the  rest  of the  atoms  in  the compound are written in the same style as the molecular  formula.  The alkanes have the general  formula:  CnH2n+2 .

  • The alkanes are the most important source of fuel in the world and¬† are used extensively in the chemical¬† industry.
  • Alkanes that contain four or less car- bon atoms are gases (e.g. methane and ethane).
  • Others¬† are liquid fuels (e.g.¬† octane, an important component of petrol).

Figure 4.18:  (a) Methane  gas bubbles  burning and  (b) propane  (under  high  pressure)  being transported  by truck

Figure 4.19:  Liquid fuels that contain octane are kept in tanks at petrol stations.

The alkenes
In the alkenes there must be at least one double bond between  two carbon atoms. This means that they are unsaturated and are more reactive than the alkanes.  The simplest alkene is ethene  (also known as ethylene), which is shown in Figure 4.20.

Figure 4.20:  The (a) structural,  (b) condensed structural and (c) molecular  formula representa- tions of ethene.  (d) An atomic model of ethene
As with the alkanes, the alkenes also form a homologous series. They have the general formula:   CnH2n .   The second  alkene  in the  series  would  therefore  be  C3H6 .   This molecule  is known as propene  (Figure 4.21).

Figure 4.21:  The (a) structural,  (b) condensed structural and (c) molecular  formula representations of propene.
There can be more than one double  bond  in an alkene as shown in Figure 4.22.  The naming of these compounds is covered  in Section 4.3, IUPAC naming and formulae.

Figure 4.22:  The structural representations of (a) pent-1-ene and (b) pent-1,3-diene.
The alkenes  are  more  reactive  than  the  alkanes  because  they  are  unsaturated.   As with the alkanes, compounds that have four or less carbon  atoms are gases at room temperature. Those with five or more carbon atoms are liquids.
The alkenes have a variety of uses:

  • For example, ¬†ethene ¬†is a chemical compound used in plants to stimulate the ripening of fruits and the opening¬† of flowers

      Figure 4.23:  (a) Unripe (green) and ripe (yellow) bananas and (b) a flowering plant.

  • Propene is an important compound in the petrochemicals industry. ¬†It is used to make polypropylene (see Section 4.7 for more information) and is also used as a fuel gas for other industrial processes.


 
 
Figure   4.24:     A  lamp   made   of  polypropylene. Propene  is used to make polypropylene.
 
…………………………………………………………………………………………………………
The alkynes
In the alkynes there must be at least one triple bond between  two of the carbon atoms.They are unsaturated compounds and are therefore more reactive than alkanes. Their general  formula  is CnH2n-2 . For example  but-1-yne has the molecular  formula C4H6 . The simplest  alkyne is ethyne  (Figure 4.25),  also known  as acetylene.   Many of the alkynes are used to synthesise other chemical products

Figure 4.25:  The (a) structural,  (b) condensed structural  and  (c) molecular  representations of ethyne (acetylene). (d) An atomic model of ethyne

Remember  that organic molecules  do not need to be straight chains.  They can have branched  groups as well, as shown in Figure 4.26.

 
Figure 4.26:  A methyl branched  group on carbon 2 of butane (2-methylbutane).
 
A summary of the relative reactivity and the homologous series that occur in the hydrocarbons is given in Table 4.2.

Table 4.2: A summary of the homologous  series of the hydrocarbons.

Experiment:  Saturated vs.  unsaturated compounds
Aim:
To study the effect of bromine  water and potassium  permanganate on saturated  and unsaturated compounds.
Apparatus: WARNING!
Liquid bromine  (required to make bromine  water) is a highly volatile,  corrosive  and toxic compound.   Please  handle  with  care:   wear  the  appropriate safety  clothing including gloves, labcoat, safety glasses and mask.  Work in a fumehood. If you do not have the apparatus to handle liquid bromine  safely, use potassium  permanganate only

  • cyclohexane, cyclohexene, bromine water ¬†(Br2(aq)), potassium ¬†permanganate (KMnO4¬†) in an acidic solution
  • 4 glass containers (test tubes/beakers/shallow basins), two A4 sheets of paper
  • 2 plastic pipettes

Method:

  1. Label one piece of paper A and the other piece of paper B.
  2. Place 20 ml of cyclohexane into a container and place the container on paper A.
  3. Place 20 ml of cyclohexane into a container and place the container on paper B.
  4. Repeat steps 2 and 3 with cyclohexene.
  5. Take 12 ml of bromine water and add it to the beaker of cyclohexane on paper A. Observe any colour chang
  6. Repeat step 5 with the beaker of cyclohexene on paper A.
  7. Take 12 ml of KMnO4   and  add  it to the  beaker  of cyclohexane on  paper  B. Observe  any colour changes.
  8. Repeat step 7 with the beaker of cyclohexene on paper B.

Results:Record your results in the table. Cyclohexane  is an alkane, cyclohexene is an alkene.

Questions:

  • Which of these compounds (cyclohexane, cyclohexene) is saturated ¬†and which is unsaturated?
  • What colour changes did you observe with the alkane compound?
  • What colour changes did you observe with the alkene compound?
  • Can you suggest a reason for the differences?

Discussion and conclusion:
Bromine  water and  KMnO4  both  have  intense  colours.   Cyclohexane  is a saturated, colourless  liquid.   When  bromine  water  and  KMnO4  are added  to the cyclohexane there  is no  reaction  and  the  solution  becomes  the  colour  of the  bromine  water  or KMnO4 .
Cyclohexene is also a colourless  liquid,  but it is unsaturated.  This results in a reaction with bromine  water  and  with KMnO4 .  Cyclohexene will form  a bromoalkane with  bromine  water.   Bromoalkanes  are colourless  liquids  and  the  solution  will be colourless  Рliquid bromine  is decolourised by cyclohexene. Similarly KMnO4  will be decolourised by the cyclohexene
Exercise 4 ¬†‚Äď ¬†2: ¬†The hydrocarbons

  1. Answer these questions on the hydrocarbons.

a) What is the difference between the alkanes, alkenes and alkynes?

b) Give the general formula for the alkynes

c) Of the alkanes, alkenes and alkynes which is:

i.  saturated  ii.  unsaturated

d) Which series is the least reactive? Explain why.

2. Draw the structural formulae for:

a) CHCCH3          b) CH3CH2CH3            c) CH2CHCH3


 

  1. Fill in the table:

 
……………………………………………………………………………………………………………….
The alcohols

An alcohol is any organic com- pound ¬†where ¬†there¬† is a hydroxyl functional¬† ¬†group¬†¬† (‚ąíOH) ¬†bound to a carbon atom.¬† ¬†The general formula¬† ¬†for ¬†a ¬†simple ¬†alcohol¬†¬† is CnH2n+1¬†OH.

Figure 4.27:  The (a) structural,  (b) condensed structural and (c) molecular  formula representations of methanol

The   simplest    and    most    commonly used alcohols are methanol and   ethanol (Figures  4.27   and 4.28).

Figure 4.28:  The (a) structural,  (b) condensed structural and (c) molecular  formula representa- tions of ethanol.  (d) An atomic model of ethanol.
There are three possible  types of carbon  atoms Рprimary,  secondary  and tertiary.  A primary  carbon  is attached  to only one  other  carbon  atom.   A secondary  carbon  is attached  to two other carbon  atoms, while a tertiary carbon  atom is attached  to three other carbon atoms.
There can be a functional group attached to these different types of carbon atom. When a hydroxyl (‚ąíOH) functional ¬†group is attached ¬†to a primary carbon ¬†atom it is called a primary ¬†alcohol.¬† ¬†For a secondary alcohol ¬†the hydroxyl is bonded ¬†to a secondary¬†carbon atom. ¬†When ¬†the hydroxyl is bonded ¬†to a tertiary carbon ¬†atom it is a tertiary alcohol. Examples are given below.

 

  1. Primary Alcohols (C atom at the end of a-OH (hydroxyl group) is bonded to a carbon atom that is bonded to only one other carbon atom butan-1-ol
  2. Secondary  Alcohols (C atom inside a chain)-OH group is bonded to a carbon  atom that is bonded to two other carbon  atoms butan-2-ol
  3. Tertiary Alcohols (needs a branched chain)-OH group is bonded to a carbon  atom that is bonded to three other carbon  atoms 2-methylpropan-2-ol

Figure 4.29:  A (a) primary (butan-1-ol), (b) secondary  (butan-2-ol) and (c) tertiary (2-methylpropan-2-ol) alcohol
The alcohols have a number  of different uses:

  • methylated spirits is ethanol with methanol added
  • all alcohols are toxic, but in low concentrations ethanol can be used in alcoholic drinks
  • ethanol is the only alcohol used in alcoholic drinks
  • ethanol is used as an industrial solvent
  • methanol and ethanol can both ¬†be used ¬†as a fuel and they burn ¬†more ¬†cleanly than petrol or diesel.
  • ethanol is used as a solvent in medical drugs, perfumes and plant essences
  • ethanol is an antiseptic

Exercise 4 ¬†‚Äď ¬†3: ¬†The alcohols
Give the structural and condensed structural formula for the following alcohols.  State, with reasons, whether  the compound is a primary, secondary,  or tertiary alcohol. (Note: a black ball represents a carbon atom, a white ball represents a hydrogen atom, and a red ball represents  an oxygen atom)

……………………………………………………………………………………………………………….
Alkyl halides   
Alkyl  halides   are  hydrocarbons  with one hydrogen atom replaced by a halogen  atom  (F, Cl, Br, I). The alkyl is due  to the  fact  that a  hydrocarbon  branched   group has  the suffix -yl and  is one  of the three hydrocarbons:

alkanes, alkenes  or alkynes.  These  alkyl groups contain one or more halo- gen  atoms,   which  leads  to  the name  alkyl halides.    Our  focus will be on the alkane alkyl halides also known as the haloalkanes  (or halogenoalkanes) (see Table 4.1).
Figure 4.30:  Representations  of a halomethane where X can be F, Cl, Br or I: (a) structural, (b) molecular formula, (c) 3-D line drawing, (d) 3-D ball and stick model and e)3-D space-filling model.

Figure 4.31:  Representations  of 2-halopropane where  X can be F, Cl, Br or I: (a) structural,  (b)condensed structural, (c) molecular  formula and (d) a ball and stick model.

 
Note that the halogen atom is called a substituent.
 

Figure 4.32:  A fluorine atom as a substituent on carbon 2 of butane (2-fluorobutane).

Remember  the  branched  chain  shown  in Figure 4.26.   That branched  chain  is also called a substituent.
DEFINITION:  Substituent
A substituent  is an atom or group of atoms bonded  to a carbon  chain.  This can be an inorganic atom (e.g. halogen) or an alkyl group that is shorter than the main group.
An organic compound is always named in accordance with the longest chain of carbon atoms that contains the functional group. If the substituent is an alkyl group it is known as a branched  chain. Some uses of haloalkanes  include:

  • in fire extinguishers
  • as aerosol propellants
  • in refrigeration
  • generating foamed plastics
  • solvents in dry cleaning processes (not actually dry, but no water is required)


Extension                    Chloroform
Haloalkanes  can contain more than one halogen atom. Chloromethanes are substances that can be used as anaesthetics during operations.  One example is trichloromethane, also known as chloroform (Figure 4.33).

Figure 4.33:  The (a) structural and (b) molecular formula representations of trichloromethane (chloroform).

Exercise 4 ¬†‚Äď ¬†4: ¬†Haloalkanes

  1. Answer these questions on the haloalkanes.
  2. a) Give the general formula for the haloalkanes with only one halogen atom
  3. b) Are haloalkanes saturated compounds?
  4. Draw the structural formulae for:
  5. a) CH2 (Br)CH2CH3
  6. b) CH3 CH(Cl)CH2CH3
  7. c) CH2 (F)CH3

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Carbonyl-containing compounds                                   
The carbonyl group consists of a carbon atom that is joined to an oxygen by a double  bond (see Figure 4.34) In Figure 4.34 R’ and R are used to represent  the rest of the atoms in the  molecule. For example R could represent  an alkyl chain, or a hydrogen  atom

Aldehydes and ketones If the functional  group  is on the end of the carbon chain, the organic com- pound  is called  an  aldehyde (Figure 4.35 (a)). Being at the end of the chain means that R’ or R represents a hydro- gen  atom.   The simplest  aldehyde  is methanal
The aldehyde  containing  4 carbon  atoms, butanal,  is illustrated in Figure 4.2.  In this example R represents  H and R’ represents  CH3CH2CH2
 

Some uses of aldehydes  include:

  • in resins (over 6 million tons of formaldehyde are produced per year)
  • in the production of plasticisers and alcohols used in detergents
  • in perfumes and flavourants

If the carbonyl  group is in the middle  of the carbon  chain,  the compound is called a ketone  (Figure 4.35 (b)). Being in the middle of the chain means that R’ and R cannot represent H. The simplest ketone is propanone (also known as acetone,  the compound in nail varnish remover), which contains three carbon atoms. The ketone containing  4 carbon atoms, butanone, is illustrated in Figure 4.2.

Some uses of ketones include:

  • as solvents
  • in the production of polymers
  • in the production of pharmaceuticals

The general  formula  for both the aldehydes  and ketones  can be written as: CnH2nO. This means that they cannot be told apart from their general formula alone.  There are more complex general formulas that allow aldehydes  and ketones to be distinguished, but they are not covered  in this book.
Carboxylic  acids
Carboxylic acids are organic acids that are characterised by having a carboxyl group, written as ‚ąíCOOH. In a carboxyl group a carbon atom is double-bonded to an oxygen atom (carbonyl ¬†group),¬† and ¬†it is also bonded ¬†to a hydroxyl ¬†group ¬†(R).¬† The simplest carboxylic acid, methanoic acid, is shown ¬†in Figure 4.38 and ethanoic ¬†acid is shown in Figure 4.2.

Carboxylic acids are widespread in nature.  Methanoic acid (also known as formic acid) has the formula HCOOH  and is found in insect stings. Ethanoic acid (CH3COOH), or acetic acid, is the main component of vinegar. More complex organic acids also have a variety of different functions. Benzoic acid for example, is used as a food preservative. Carboxylic acids have the general  formula:  CnH2n+1COOH
Ethanoic  acid  can  be  produced through the oxidation of ethanol upon exposure  to the oxygen in air. This is why wine that is left too  long can  taste acidic.   Wine  can easily  go  sour  if exposed  to  the  oxygen molecules  (O2 ) in the air, especially  if the weather is warm.

The oxidation  of ethanol  to ethanoic  acid can also be seen in the reaction  of ethanol with potassium dichromate:
2(Cr2O7 )2‚ąí¬†(aq)+3C2H5OH(aq)+16H+¬†(aq) ‚Üí 4Cr3+(aq)+3CH3COOH(aq)+11H2O(`) The colour change that occurs is shown in the image below and the following video:

Case study:  Breathalysers
Read the following extract taken from HowStuffWorks (12/08/13):   The Breathalyzer device contains:

  • A system to sample the breath of the suspect
  • Two glass vials containing the chemical reaction¬† mixture
  • A system of photocells connected to a meter to measure the ¬†color change associated ¬†¬†with the chemical ¬†reaction

To measure  alcohol,  a suspect breathes  into the device.  The breath  sam- ple is bubbled  in one  vial through  a mixture of sulfuric acid,  potassium dichromate, silver nitrate and water.  The principle  of the measurement is based on the following chemical  reaction:
2K2Cr2O7(aq) + 3CH3CH2OH(aq) + 8H2SO4(aq) ‚Üí
2Cr2(SO4 )3(aq) + 2K2SO4  + 3CH3COOH(aq) + 11H2O(`) In this reaction:

  1. The sulfuric acid removes the alcohol from the air into a liquid solu- tion.
  2. The  alcohol    reacts    with   potassium    dichromate   to   produce: chromium  sulfate, potassium  sulfate, acetic acid, water.

The silver nitrate is a catalyst,  a substance  that makes a reaction  go faster without  participating  in it.  The sulfuric acid, in addition  to removing the alcohol  from the air, also might provide  the acidic  condition  needed for this reaction.
During this reaction, ¬†the reddish-orange dichromate ion changes ¬†color to the ¬†green ¬†chromium ¬†ion when ¬†it reacts ¬†with ¬†the ¬†alcohol; ¬†the ¬†degree ¬†of the color change ¬†is directly related ¬†to the level of alcohol ¬†in the expelled air. To determine the amount ¬†of alcohol¬† in that air, the reacted ¬†mixture is compared to a vial of unreacted mixture in the photocell system,¬† which produces ¬†an electric ¬†current that causes the needle¬† in the meter to move from its resting place. ¬†The operator then rotates a knob to bring the needle back to the resting place ¬†and reads the level of alcohol ¬†from the knob ‚Äď the more the operator ¬†must turn the knob to return it to rest, the greater the level of alcohol.
Break into groups of three or four.  Research breathalysers  and then report your infor- mation to the class.
Make sure to cover the following areas:

  • The effect of alcohol on the body
  • The effect of alcohol on reaction times
  • The origins of the breathalyser
  • The term mouth alcohol and its effect on breathalyser tests.

…………………………………………………………………………………………………….
Esters
When  an alcohol  reacts with a carboxylic  acid, an ester  is formed.  Most esters have a characteristic  smell.  In the reaction  a molecule  of water is removed  from the two compounds and a new bond  is formed between  what remains of the alcohol  and the carboxylic  acid.   A catalyst  is required  in this reaction,  in this case  it must  be  an inorganic acid (e.g. H2SO4 ). An example is shown in Figure 4.42.

The esterification  process  with methanol  and  methanoic acid  is shown  with atomic models  in Figure 4.43.  Esters have the general  formula: CnH2nO2 . This general formula can also be applied  to carboxylic  acids, but the more complex  general  formula for esters alone is not covered  in this book.
Some common  uses for esters are:

  • in cosmetics and beauty ¬†products ¬†because ¬†they ¬†typically ¬†have ¬†a fruity smell, making them good as artificial flavourants and scents
  • in nail varnish remover and model plane glue
  • as solvents for non-water soluble compounds (e.g. oils, resins) because the ester of a specific carboxylic acid will be less water soluble than the carboxylic acid
  • as plasticisers because esters can make a compound less brittle, and more flexible

Exercise 4 ¬†‚Äď ¬†5: ¬†Carbonyl ¬†compounds

  1. Answer these questions on carbonyl compounds.
  2. a) What other functional group does a carboxylic acid have in addition  to a carbonyl group?
  3. b) What is the main difference between aldehydes and ketones c) What two reactants are required  to make an ester?
  4. d) How is ethanoic acid produced?
  5. Draw the structural formulae for each of the following compounds. What series does each compound belong to?                                                                                             a) CH3COCH3     b) CH3CH2COOH    c) CH3CH2CHO  d) CH3COOCH3

Exercise 4 ¬†‚Äď ¬†6: ¬†Functional ¬†groups

a) Complete the table by identifying the functional group of each com b) Give the structural representation of the compounds represented by condensed  structural formulae.

3. A chemical reaction takes place and ethyl methanoate is formed.

a) Identify the homologous series to which ethyl methanoate belongs?            b) Name the two types of reactants used to produce this compound in a chem-ical reaction.                                                                                                c) Give the structural formula of ethyl methanoate (HCOOCH2CH3 ).

4. The following reaction takes place: CH3CHCH2 (g) + H2(g) ‚Üí CH3CH2CH3 (g)

a) Give the name of the homologous series of the organic compound in the react                                                                                                            b) What is the name of the homologous series of the product?                          c) Which compound in the reaction is a saturated hydrocarbon?

……………………………………………………………………………………………………….
ISOMERS
It is possible  for two organic  compounds to have the same molecular  formula but a different  structural  formula.  Look at the two organic  compounds that are shown  in Figure 4.44 for example

If you were to count  the
number  of carbon  and hydrogen  atoms in each  compound, you would  find that they are the same.  They both have the same molecular formula C4 H10 , but their structure is different and so are their properties.  Such compounds are called isomers.
DEFINITION:  Isomer
In chemistry, isomers are molecules with the same molecular formula but different structural formula. Isomers are molecules with the same molecular formula and often (though not always) with the same kinds of chemical  bonds  between  atoms, but with the atoms arranged differently.
The isomers shown in Figure 4.44 differ only in the location of the carbon atoms. The functional groups are the same but butane ¬†has all four carbons ¬†in one ¬†chain, ¬†while 2-methylpropane has three carbons in the longest chain and a methyl group attached to the second carbon in the chain. ¬†It is also possible to have positional isomers (Figure 4.45). ¬†In this case the ‚ąíOH functional ¬†group can be on different carbon ¬†atoms,¬† for example carbon 1 for pentan-1-ol, ¬†on¬† carbon ¬†2 for pentan-2-ol ¬†or on¬† carbon ¬†3 for pentan-3-ol

Figure 4.45:  The structural and condensed structural representations of the isomers (a) pentan-1-ol, (b) pentan-2-ol and (c) pentan-3-ol.

Positional  isomers are also found in esters.  If one ester was made  from ethanol  and hexanoic  acid and another  was made  from hexanol  and ethanoic  acid the two esters produced are isomers (see Figure 4.46).

It is important  to note that molecules  need  not have the same functional groups to be isomers.  For example propanone (commonly known as acetone) and propanal  (Figure 4.47) have the same molecular  formula (C3H6O) but different functional  groups and properties  (Table 4.3). These types of isomers are functional isomers.

Heptanoic  acid and butyl propanoate (Figure 4.48) are another  example  of functional isomers (containing  different functional  groups).  They both have the same molecular formula of C7 H14 O2  but have different functional groups and different properties (Table 4.4).

Exercise 4 ¬†‚Äď ¬†7: ¬†Isomers

  1. Match the organic compound in Column A with its isomer in Column B:

What is IUPAC naming?
In order to give compounds a name, certain rules must be followed.  When naming organic compounds, the IUPAC (International Union of Pure and Applied Chemistry) nomenclature (naming scheme) is used.  This is to give consistency to the names.   It also enables every compound to have a unique name, which is not possible with the common names used (for example in industry). We will first look at some of the steps that need to be followed when naming a compound, and then try to apply these rules to some specific examples. A good general rule to follow is to start at the  end  (the suffix) and  work  backwards (from right to left) in the name.

  1. Recognise the functional group in the compound.   This will determine  the suffix of the name (see Table 4.5).
  2. Find  the   longest   continuous  carbon chain that contains the functional group (it won’t always be a straight chain) and count  the  number  of carbon  atoms  in this chain. This  number   will  determine   the  pre- fix (the beginning) of the  compound’s name (see Table 4.6).

Table   4.5:     The   suffix  associated                                                                               with various functional groups.
3. Number  the   carbons   in  the   longest carbon  chain  (Important:    If  the molecule   is  not  an  alkane  (i.e.    has a functional group) you need to start numbering  so that the functional  group is on the carbon  with the lowest possi- ble number).   Start with the carbon  at the end closest to the functional group.
4. Look for any branched groups:

  • The branched ¬†groups must be listed before the name of the main chain in alphabetical order (ignoring di/tri/tetra)Name them by counting th
    e number of carbon atoms in the branched group and referring to Table 4.6, these groups will all end in -yl.
  • Note the position of the group on the main carbon chain.
  • If there is more than one of the same type of branched ¬†group then both numbers ¬†must be listed (e.g.¬† 2,4 -) and one of the prefixes listed in Table 4.7 must be used. Important: I
    f the molecule  is an alkane the branched  group must be on the carbon with the lowest possible number.
  • The branched ¬†groups must be listed before the name of the main chain in alphabetical ¬†order (ignoring di/tri/tetra).

 
 Table 4.6: The prefix of a compound’s name is determined                                               by  the  number   of  carbon atoms in the longest chain                                                     that contains the functional group.
If there are no branched  groups this step can be ignored.

Table 4.7: Prefixes for multiple substituents with the same name.  These apply to multiple functional groups as well.

  1. For the alkyl halides  the  halogen  atom  is treated  in  much  the  same  way  as branched  groups:
  • To name them take the name of the halogen atom (e.g. iodine) and replace the -ine with -o (e.g. iodo).
  • Give ¬†the ¬†halogen¬† ¬†atom ¬†a ¬†number to¬† show ¬†its position ¬†on ¬†the ¬†carbon chain.¬†¬† ¬†If there is more than one halogen atom the numbers should be listed ¬†and ¬†a¬† prefix should ¬†be ¬†used (e.g. ¬†3,4-diiodo-¬† or 1,2,2-trichloro-). See Table 4.7 for a list of the prefixes

 
                                                                       Table  4.8:    Naming  halogen   atoms                                                                             in organic molecules

  • The halogen atoms must be listed before the name ¬†of the main ¬†chain ¬†in alphabetical ¬†order (ignore di/tri/tetra).

If there are no halogen atoms this step can be ignored.
6. Combine the elements of the name into a single word in the following order:

‚ÄĘ branched groups/halogen atoms in alphabetical order (ignoring prefixes)
‚ÄĘ prefix of main chain
‚ÄĘ name ending according to the functional group and its position on the longest carbon chain.

Naming hydrocarbons  
Naming alkanes

 
The suffix for an alkane is -ane.
Worked  example  1:  Naming the alkanes
QUESTION
Give the IUPAC name for the following compound:
Note: The numbers  attached  to the carbon atoms would not normally be shown.
The carbon atoms have been numbered to help you to name the compound.
SOLUTION
Step 1: Identify the functional group
The compound is a hydrocarbon with single bonds between  the carbon atoms. It is an alkane and will have a suffix of -ane.
Step 2: Find the longest carbon  chain
There are four carbon atoms in the longest chain.  The prefix of the compound will be but-.
Step 3: Number  the carbon  atoms in the longest chain
The numbering  has been done for you here.
Step  4:  Look for  any  branched group,  name  them  and  give their  position  on  the carbon  chain
There are no branched  groups in this compound.
Step 5: Combine the elements  of the name into a single word
The name of the compound is butane.
Worked  example  2:  Naming the alkanes
QUESTION

 
Give the IUPAC name for the following compound
SOLUTION  
Step 1: Identify the functional group
The compound is a hydrocarbon with single bonds between the carbon atoms. It is an alkane and will have the suffix -ane.
Step 2: Find the longest carbon chain
There are three carbon atoms in the longest chain. The prefix for this compound is prop-.
Step 3: Number the carbons in the carbon chain
If we start at the carbon on the left, we can number the atoms as shown in red (left). If we start at the carbon on the right, we can number the atoms as shown in blue (right).

Step 4: Look for any branched groups, name them and give their position on the carbon chain.There is a branched group attached to the second carbon atom. In this case the methyl group is on carbon 2 regardless of which side you number the longest chain from.
This group has the formula CH3 , which is methane without a hydrogen atom. However, because it is not part of the main chain, it is given the suffix -yl (i.e. methyl). The position of the methyl group comes just before its name (see the next step).
Step 5: Combine the elements of the compound’s name into a single word in the order of branched group; prefix; name ending according to the functional group
The compound’s name is 2-methylpropane.
Worked example 3: Naming the alkanes
QUESTION
Draw the semi-structural structural and condensed structural formula for the organic compound 2,2,4-trimethylhexane
SOLUTION
Step 1: Identify the functional group
The name ends in -ane therefore the compound is an alkane.
Step 2: Determine the number of carbon atoms in the longest chain
The longest chain has the prefix hex-.

There are therefore 6 carbon atoms in the longest chain.
Step 3: Look for any branched groups and place them on the structure

The compound is 2,2,4-trimethylhexane. Therefore there are three branched groups. Two on carbon 2 and one on carbon 4.
 
 
Step 4: Combine this informationand add the hydrogen  atoms

Carbon atoms can have four single bonds. Therefore wherever a carbon atom has less than  four bonds  draw in hydrogen  atoms until there are four bonds
Step 5: Condense the structural formula
First condense the main chain: CH3CCH2CHCH2CH3
Then add the side chains (in brackets) on the relevant carbon atoms:
       CH3C(CH3)2CH2CH(CH3)CH2CH3
1. Give the structural formula for each of the following alkanes

a) Octane b) Propane c) 2-methylpropane d) 3-ethylpentane

2. Give the IUPAC name for each of the following alkanes:

a) CH3 CH2 CH(CH3 )CH2 CH3 b) CH3 CH(CH3 )CH2 CH(CH3 )CH3
c) CH3 CH2 CH2 CH2 CH2 CH3 d) CH3 CH3


Naming alkenes
The suffix for an alkene is -ene.
Worked example 4: Naming the alkenes
QUESTION
Give the IUPAC name for the following compound:

SOLUTION
Step 1: Identify the functional group
The compound has a double carbon-carbon bond and is an alkene. It will have the suffix -ene.
Step 2: Find the longest carbon chain containing the functional group
The functional group is a double bond, so the longest chain must contain the double bond. There are four carbon atoms in the longest chain and so the prefix for this compound will be but-.
Step 3: Number the carbon atoms
Remember that the carbon atoms must be numbered so that the functional group is at the lowest numbered carbon atom possible. In this case, it doesn’t matter whether we number the carbons from the left to right, or from the right to left. The double bond will still fall between the second and third carbon atoms.
Step 4: Look for any branched groups, name them and give their position on the carbon chain
There are no branched groups in this molecule.
Step 5: Combine the elements of the name into a single word in the following order: branched groups; prefix; name ending according to the functional group and its position along the longest carbon chain.The name of this compound is but-2-ene or 2-butene.
Worked example 5: Naming the alkenes
QUESTION
Draw the structural and molecular formula for the organic compound
3-methylbut-1-ene
SOLUTION
Step 1: Identify the functional group
The suffix -ene means that this compound is an alkene and there must be a double bond in the molecule. The number 1 immediately before the suffix means that the double bond must be at the first carbon in the chain (but-1-ene).
Step 2: Determine the number of carbon atoms in the longest chain containing the functional group
The prefix for the compound is but- so there must be four carbons in the longest chain containing the double bond.
C=¬†C‚ÄĒC‚ÄĒC

Step 4: Combine this information and add the hydrogen atoms

Step 5: Reduce the structural formula  to the molecular formula
There are 5 carbon atoms and 10 hydrogen atoms so the molecular  formula is C5H10 . (Remember that there is no structural information given by the molecular  formula)
Worked  example  6:  Naming the alkenes
QUESTION
Give the IUPAC name for the following compound:

SOLUTION
Step 1: Identify the functional group
The compound is an alkene and will have the suffix -ene. There is a double bond between the first and second carbons and also between the third and fourth carbons. The organic compound therefore contains ’1,3-diene’.
Step 2: Find the longest carbon chain containing the functional group, and number the carbon atoms
Remember that the main carbon chain must contain both the double bonds. There are four carbon atoms in the longest chain containing the double bonds and so the prefix for this compound will be but-. The carbon atoms are already numbered 1 to 4 in the diagram.
Step 3: Look for any branched groups, name them and give their position on the carbon chain
There is an ethyl group on the second carbon.
Note that if we had numbered from the right to left the suffix would still have been 1,3-diene, however the ethyl group would have been on the third carbon. So we had to number left to right.
Step 4: Combine the elements of the name into a single word in the following order: branched groups; prefix; name ending according to the functional group and its position along the longest carbon chain
The name of this compound is 2-ethylbut-1,3-diene.
Exercise 4 ‚Äď 9: Naming alkenes
1. Give the IUPAC name for each of the following alkenes:

2. Give the structural formula for each of the following alkenes:
a) ethene b) hex-1-ene c) hept-3-ene d) 4-ethyloct-3-ene
Naming alkynes
The suffix for an alkyne is -yne.
Worked example 7: Naming the alkynes
Give the IUPAC name for the following compound:

SOLUTION H
Step 1: Identify the functional group
There is a triple bond between two of the carbon atoms, so this compound is an alkyne. The suffix will be -yne.
Step 2: Find the longest carbon chain containing the functional group
The functional group is a triple bond, so the longest chain must contain the triple bond. There are six carbon atoms in the longest chain. The prefix of the compound’s name will be hex-.
Step 3: Number the carbons in the longest chain
In this example, you will need to number the carbons from right to left so that the triple bond is between carbon atoms with the lowest numbers (the suffix for the compound will therefore be -2-yne).

Step 4: Look for any branched groups, name them and assign the number of the carbon atom to which the group is attached
There is a methyl (CH3 ) group attached to the fifth carbon (remember we have num- bered the carbon atoms from right to left).
Step 5: Combine the elements of the name into a single word in the following order: branched groups; prefix; name ending according to the functional group and its position along the longest carbon chain
If we follow this order, the name of the compound is 5-methylhex-2-yne.
Worked example 8: Naming the alkynes
QUESTION
Give the IUPAC name for the following compound:

SOLUTION
Step 1: Identify the functional group
There are two triple bonds. The suffix will therefore be -diyne.
Step 2: Find the longest carbon chain containing the functional group
The functional group is a triple bond, so the longest chain m
 
ust contain all triple bonds. The longest carbon chain contains seven carbon atoms, therefore the prefix will be hept-.
Step 3: Number the carbons in the longest chain

Numbering from left to right(shown in red) the Ô¨Ārst triple bondis on carbon 1 and the second is on carbon 5. The sufÔ¨Āx will there-fore be -1,5-diyne.
(Numbering from right to left (shown in blue) will give the sufÔ¨Āx -2,6-diyne, and is incorrect).
Step 4: Look for any branched groupsThere are no branched groups for this molecule.
Step 5: Combine the elements of the name into a single word in the following order: branched groups; preÔ¨Āx; name ending according to the functional group and its position along the longest carbon chain
The name of the compound is hept-1,5-diyne
Worked example 9: Naming the alkynes
QUESTION
Draw the structural and condensed structural formula for the organic compound
6-methylhept-3-yne
SOLUTION
Step 1: Identify the functional group
The suffix -3-yne means that this compound is an alkyne and there must be a triple bond located on carbon number 3.
Step 2: Determine the number of carbon atoms in the longest chain containing the functional group

The prefix for the compound is hept- so there must be seven carbons in the longest chain.
Step 3: Look for any branched groups
 
There is a methyl group located on carbon number 6.
Step 4: Combine this information and add the hydrogen atoms

Step 5: Condense the structural formula
First condense the main chain: CH3CH2CCCH2CHCH3
Then add the side chains (in brackets) on the relevant carbon atoms:
CH3CH2CCCH2CH(CH3)CH3
Exercise 4 ‚Äď 10: Naming alkynes
1. Give the structural formula for each of the following alkynes:
a) ethyne b) pent-1-yne c) 5-methylhept-3-yne
2. Give the IUPAC names for the following alkynes:

 

 
Activity: Building hydrocarbons
An example of ethane, ethene and ethyne built with atomic model kits is given in the picture below:

1. Using atomic model kits, build the molecules of methane, propane, butane, pentane and octane. If you don’t have atomic model kits, jelly tots (or playdough) and toothpicks will work just as well. Use one colour jelly tot for the carbon atoms and one for the hydrogen atoms.
‚ÄĘ Remember that carbon atoms should have four bonds and hydrogen atoms can have only one. You should see that all these compounds have a similar formula, remember they all have the general formula Cn H2n+2 .
‚ÄĘ What is the name of the homologous series that all these molecules belong to?
2. Build the molecules of prop-1-ene, but-1-ene, pent-1-ene and oct-1-ene. Use two toothpicks to represent a double bond. You should see that all these compounds have a similar formula. Remember, they all have the general formula Cn H2n .
‚ÄĘ Try placing the double bond at different positions within the molecule.
Does this make any difference to the total number of carbon and hydrogen atoms in the molecule?
‚ÄĘ What is the name of the homologous series that all these molecules belongto?
3. Build the molecules of prop-1-yne, but-1-yne, pent-1-yne and oct-1-yne. Use three toothpicks to represent a triple bond. You should see that all these com- pounds have a similar formula. Remember, they are all alkynes.
‚ÄĘ Try placing the triple bond at different positions within the molecule. Does this make any difference to the total number of carbon and hydrogen atoms in the molecule?
‚ÄĘ What is the general formula for the alkynes?
 
Exercise 4 ‚Äď 11: Naming hydrocarbons
1. Give the structural formula for each of the following compounds:
a) oct-2-ene b) 3-methylhexane c) 4-ethyl-4-methylhept-2-yne d) hept-3-yne e) pentane f) 2-methylbut-1-ene g) propyne
2. Give the IUPAC name for each of the following organic compounds:

 
Naming alkyl halides
 
All the same rules apply when naming the alkyl halides as for naming the hydrocar- bons. We will only be dealing with the haloalkanes (i.e. there are no other functional groups). The halogen atom is treated in the same way as a branched group.

 
Worked example 10: Naming the haloalkanes
QUESTION
 
Give the IUPAC name for the following compound:
SOLUTION
Step 1: Identify the functional group
There is a halogen atom and no other functional group. This compound is therefore a haloalkane, and will have the suffix -ane.
Step 2: Find the longest carbon chain containing the functional group
There are three carbons in the longest chain containing the halogen atom. The prefix is prop-.
Step 3: Number the carbon atoms in the longest chain
You need to number the carbon atoms so that the halogen atom is on the carbon atom
with the lowest number. In this case you can number from either side.
Step 4: Name the halogen atom and assign the number for the carbon atom it is attached to The halogen is a chlorine atom. It is attached to carbon number 2 and so will have the name 2-chloro.
Step 5: Look for branched groups
There are no branched groups in this compound.
Step 6: Combine the elements of the name into a single word in the following order:
halogen atoms; prefix; name ending according to functional group
The name of the compound is 2-chloropropane.
Worked example 11: Naming the haloalkanes
QUESTION

Give the IUPAC name for the following compound:
SOLUTION
Step 1: Identify the functional group
There are three halogen atoms and no other functional groups. This compound is therefore a haloalkane, and will have the suffix -ane.
Step 2: Find the longest carbon chain containing the functional group
There are four carbons in the longest chain containing all the halogen atoms. The prefix for this compound will be but-.
Step 3: Number the carbon atoms in the longest chain
You need to number the carbon atoms so that the halogen atoms are on the carbon atoms with the lowest numbers. You must number from left to right here so that one halogen atom is on carbon 1 and two halogen atoms are on carbon 3.
Worked example 12: Naming the haloalkanes
QUESTION
Draw the structural and condensed structural formula for the organic compound
2-iodo-3-methylpentane
SOLUTION
Step 1: Identify the functional group
This compound has the suffix -ane, but also contains a halogen atom. It is therefore a haloalkane. Note that the methyl and iodo are written in alphabetical order.
Step 2: Find the longest carbon chain containing the functional group
The prefix is pent- therefore there are 5 carbons in the longest chain.

Step 3: Place the halogen atom(s) and any branched groups
 
 
There is an iodine atom on the second carbon atom, and a methyl branched group on the third  carbon atom.
Step 4: Combine this information to and add the hydrogen atoms

Step 5: Condense the structural formula
First condense the main chain: CH3CHCHCH2CH3
Then add the side chains and halogen atoms (in brackets) on the relevant carbon atoms:
CH3 CH(I)CH(CH3)CH2 CH3
Exercise 4 ‚Äď 12: Naming haloalkanes
1. Give the structural representation for the following haloalkanes:
a) 2-chlorobutane b) 1-bromopropane c) 2,3-difluoropentane
2. Give the IUPAC name for the following haloalkanes:
 

The rules used to name the alcohols are similar to those already discussed for the hydrocarbons. The suffix of an alcohol is ‚Äďol (see Table 4.5).
Worked example 13: Naming the alcohols
QUESTION
Give the IUPAC name for the
following organic compound
SOLUTION
Step 1: Identify the functional group

The compound has an ‚ąíOH (hydroxyl) functional group and is therefore an alcohol. The compound will have the suffix¬†-ol.
Step 2: Find the longest carbon chain containing the functional group
There are three carbon atoms in the longest chain that contains the functional group.
The prefix for this compound will be prop-. As there are only single bonds between the carbon atoms, the prefix includes an to become propan-.
Step 3: Number the carbons in the carbon chain
In this case, it doesn’t matter whether you start numbering from the left or right. The hydroxyl group will still be attached to second carbon atom (-2-ol).
Step 4: Look for branched groups
There are no branched groups in this compound.
Step 5: Combine the elements of the compound’s name into a single word in the order of branched groups; prefix; name ending according to the functional group The compound’s name is propan-2-ol or 2-propanol.
Worked example 14: Naming the alcohols
QUESTION
Give the IUPAC name for the

following compound:
SOLUTION
Step 1: Identify the functional groupThe compound has an ‚ąíOH (hydroxyl) functional group and is therefore an alcohol. There are two hydroxyl groups in the compound, so the suffix will be -diol.
Step 2: Find the longest carbon chain that contains the functional group
There are four carbon atoms in the longest chain that contains the functional group
(but-) and only single bonds (an-). The prefix for this compound will be butan-.
Step 3: Number the carbons in the carbon chain

There are two hydroxyl groups attached to the main chain. If we number as shown in red (on the left) they are attached to the first and second carbon atoms. If we number as shown in blue (on the right) they are attached to the third and fourth carbon atoms.
The functional groups should have the lowest numbers possible. Therefore the red numbering is correct. The hydroxyl groups are attached to the first and second carbon atoms (1,2-diol).
Step 4: Look for branched groups
There are no branched groups in this compound.
Step 5: Combine the elements of the compound’s name into a single word in the order of branched groups; prefix; name ending according to the functional group The compound’s name is butan-1,2-diol
Worked example 15: Naming the alcohols
QUESTION
Draw the structural and condensed structural representations for the organic com- pound 4-ethyloctan-2,5-diol
SOLUTION
Step 1: Identify the functional group
The compound has the suffix -ol. It is therefore an alcohol.
Step 2: Find the longest carbon chain that contains the functional group
The prefix is oct- therefore there are 8 carbons in the longest chain containing the functional group.
Step 3: Place the functional group as well as any branched groups
There is one ‚ąíOH attached to carbon 2 and one attached to carbon 5. There is also an ethyl (‚ąíCH2 CH3 ) branched group attached to carbon 4.

Step 4: Combine this information and add the hydrogen atoms

Step 5: Condense the structural formula
First condense the main chain: CH3 CHCH2 CHCHCH2 CH2 CH3
Then add the side chains and alcohol functional groups (in brackets) on the relevant carbon atoms: CH3 CH(OH)CH2 CH(CH2 CH3 )CH(OH)CH2 CH2 CH3
Activity:  Building alcohols
The structural  representation of butan-1-ol  built using an atomic  model  kit is given below:

 
Using atomic model kits, build the molecules of methanol, ethanol, propan-1-ol, pentan-1-ol and octan-1-ol. If you don’t have an atomic model kit remember that you can use jelly tots (or playdough) and toothpicks. Use different colour jelly tots to represent the different atoms.
 

  • You should see that all these compounds have a similar formula. Remember, they belong to the homologous series of the alcohols. What is the general formula for this series?
  • Try placing the hydroxyl group at different positions within the molecule. Does this make any difference to the total number of carbon, hydrogen and oxygen atoms in the molecule?

¬†Exercise 4 ‚Äď 13: Naming alcohols
1. Give the structural representation of each of the following organic compounds:
a) pentan-3-ol b) butan-2,3-diol c) 2-methylpropan-1-ol
2. Give the IUPAC name for each of the following:

 
Naming carbonyl compounds
A carbonyl group consists of a carbon atom that is bonded to an oxygen atom through a double bond (C=O). There are many different functional groups that contain a car- bonyl group.
Naming aldehydes
If the carbonyl group is on the end of the carbon chain, the organic compound is called an aldehyde. An aldehyde has the suffix -al.
Worked example 16: Naming aldehydes
QUESTION
Give the IUPAC name and molecular formula for the following organic compound

SOLUTION
Step 1: Identify the functional group
The compound has a C=O (carbonyl) group and no other functional groups. It is therefore either an aldehyde or a ketone. The carbonyl group is on the last (terminal) carbon in the main chain so the compound is an aldehyde. It will have the suffix -al.
Step 2: Find the longest carbon chain containing the functional group
There are three carbons in the longest chain that contains the functional group. The prefix for this compound will be prop-. As there are only single bonds between the carbon atoms, the prefix becomes propan-.
Step 3: Number the carbon atoms in the carbon chain
The carbon atoms will be numbered so that the carbon atom of the aldehyde group has the lowest number possible. In this case that is from right to left.
Step 4: Look for any branched groups
There are no branched groups in this compound.
Step 5: Combine the elements of the compound’s name into a single word in the order of branched groups; prefix; name ending according to the functional group The compound’s name is propanal (there is no need to say propan-1-al as by definition
all aldehydes are -1-al).
Step 6: Reduce the structural representation to the molecular formula
There are 3 carbon atoms, 6 hydrogen atoms and 1 oxygen atom so the molecular formula is C3H6O.
(Remember that there is no structural information given by the molecular formula)
Naming ketones
If the carbonyl group is in the middle of the carbon chain, the compound is called a
ketone. A ketone has the suffix -one.
TIP: Ketone is pronounced keytone.  Therefore propanone is pronounced propanown.
FACT: Note that butanone can only bebutan-2-one.  If the carbonyl group were on carbon 1 it would be an aldehyde,  while if it were on carbon 3 we would simply count from the other side of the molecule. The 2 is still required for IUPAC naming however.
Naming ketones
If the carbonyl group is in the middle of the carbon chain, the compound is called a
ketone. A ketone has the suffix -one.
Worked example 17: Naming ketones
QUESTION
Give the IUPAC name and molecular formula for the following compound:

SOLUTION
Step 1: Identify the functional group
The compound has a C=O (carbonyl) group and no other functional groups. It is therefore either an aldehyde or a ketone. The carbonyl group is not at the end of the chain. Therefore the compound is a keton
 
e and the suffix will be -one.
Step 2: Find the longest carbon chain containing the functional group
There are four carbons in the longest chain that contains the functional group, and only single carbon-carbon bonds. The prefix for this compound will be butan-.
Step 3: Number the carbon atoms in the carbon chain
The carbon atoms will be numbered from left to right so that the carbon atom of the ketone group has the lowest number possible (-2-one).
Step 4: Look for any branched groups
There are no branched groups in this compound.
 
Step 5: Combine the elements of the compound’s name into a single word in the order of branched groups; prefix, name ending according to the functional group The compound’s name is butan-2-one or 2-butanone.
Step 6: Reduce the structural representation to the molecular formula
There are 4 carbon atoms, 8 hydrogen atoms and 1 oxygen atom so the molecular formula is C4H8O. (Remember that there is no structural information given by the molecular formula)
Worked example 18: Naming carbonyl compounds
QUESTION
Give the IUPAC name for the following compound
SOLUTION

Step 1: Identify the functional group
The compound has a C=O (carbonyl) group and no other functional groups. It is therefore either an aldehyde or a ketone. The carbonyl group is not at the end of the chain. Therefore the compound is a ketone and the suffix will be -one.
 
Step 2: Find the longest carbon chain that contains the functional group
The longest carbon chain that contains the functional group has four carbon atoms in it, and only single bonds. The prefix for this compound will be butan-.
Step 3: Number the carbon atoms in the carbon chain
The carbon atoms will be numbered from left to right so that the carbon atom of the ketone functional group has the lowest possible number. The suffix will be -2-one.
Step 4: Look for any branched groups
There is a branched group on carbon 3. This group has only one carbon atom. The branched group is attached to the third carbon atom (3-methyl).
Step 5: Combine the elements of the compound’s name into a single word in the order of branched groups; prefix; name ending according to the functional group The compound’s name is 3-methylbutan-2-one.
Worked example 19: Naming carbonyl compounds
QUESTION
Draw the structural and condensed structural representations for the organic com- pound 3-methylpentanal.
SOLUTION
Step 1: Identify the functional group
The compound has the suffix -al. It is therefore an aldehyde and has a C=O (carbonyl)
group on the first carbon atom.
Step 2: Find the longest carbon chain that contains the functional group
The prefix is pent- so there are 5 carbon atoms in the longest chain.
C‚ÄĒC‚ÄĒC‚ÄĒC‚ÄĒC
Step 3: Place the functional group as well as any branched groups
There is a C=O (carbonyl) group at the first carbon atom and a methyl group attached to the third carbon atom (3-methyl).

Step 4: Combine this information and add the hydrogen atoms
 

Step 5: Condense the structural representation
First condense the main chain including the carbonyl group oxygen atom:                  CHOCH2CHCH2CH3
Then add the side chains on the relevant carbon atoms: CHOCH2CH(CH3)CH2CH3
Worked example 20: Naming carbonyl compounds
QUESTION
Draw the structural and condensed structural representations for the organic com- pound 3-methylpentan-2-one.
SOLUTION
Step 1: Identify the functional group
The compound has the suffix -one. It is therefore a ketone and has a C=O (carbonyl)
group. This group cannot be on the first carbon atom.
Step 2: Find the longest carbon chain that contains the functional group
The prefix is pent- therefore there are 5 carbon atoms in the longest chain.
C‚ÄĒC‚ÄĒC‚ÄĒC‚ÄĒC
Step 3: Place the functional group as well as any branched groups
There is a C=O (carbonyl) group at the second carbon atom (-2-one) and a methyl group attached to the third carbon atom (3-methyl).

Step 4: Combine this information and add the hydrogen atoms

Step 5: Condense the structural representation
First condense the main chain including the carbonyl group oxygen atom: CH3 COCHCH2 CH3
Then add the side chains on the relevant carbon atoms: CH3 COCH(CH3 )CH2 CH3
Exercise 4 ‚Äď 14: Naming aldehydes and ketones
1. Give the IUPAC name for each of the following compounds:

2. Give the structural representation for the following:

a) ethanal b) propanone c) heptan-3-one

 
Naming carboxylic acids
Carboxylic acids are characterised by having a carboxyl group, which has the formula
‚ąíCOOH. In a carboxyl group a carbon atom is double-bonded to an oxygen atom (a carbonyl group), and is also bonded to a hydroxyl (alcohol) group. The IUPAC suffix for a carboxylic acid is -oic acid.
Worked example 21: Naming carboxylic acids
QUESTION
Give the IUPAC name and molecular formula for the following compound:

SOLUTION
Step 1: Identify the functional group
The compound has a ‚ąíCOOH group and is therefore a carboxylic acid. The suffix will be -oic acid.
Step 2: Find the longest carbon chain that contains the functional group
There are five carbon atoms in the longest chain that contains the functional group, and only single bonds between carbon atoms. The prefix for this compound is pentan-.
Step 3: Number the carbon atoms in the carbon chain
The carbon atoms will be numbered from right to left so that the carboxylic acid func- tional group has the lowest numbered carbon atom.
Step 4: Look for any branched groups
There are no branched groups in this compound.
Step 5: Combine the elements of the compound’s name into a single word in the order of branched groups; prefix, name ending according to the functional group The compound’s name is pentanoic acid.
Step 6: Reduce the structural representation to the molecular formula
There are 5 carbon atoms, 10 hydrogen atoms and 2 oxygen atoms so the molecular formula is C5H10O2 .
Worked example 22: Naming carboxylic acids
QUESTION
Draw the structural and condensed structural representations for the organic com- pound 2-ethylhexanoic acid.
SOLUTION
Step 1: Identify the functional group
The compound has the suffix -oic acid. It is therefore a carboxylic acid and has a
‚ąíCOOH group. This group can only be on carbon 1 (at the end of the carbon chain).
Step 2: Find the longest carbon chain that contains the functional group
The prefix is hex- therefore there are 6 carbons in the longest chain.
¬† ¬† ¬† ¬† ¬†C‚ÄĒC‚ÄĒC‚ÄĒC‚ÄĒC‚ÄĒC
Step 3: Place the functional group as well as any branched groups
There is a C=O (carbonyl) group at the second carbon atom (-2-one) and a methyl group attached to the third carbon atom (3-methyl).

Step 4: Combine this information and add the hydrogen atoms

Step 5: Condense the structural representation
First condense the main chain including the carbonyl group oxygen atom: CH3COCHCH2CH3
Then add the side chains on the relevant carbon atoms: CH3COCH(CH3)CH2CH3
Activity: Building carboxylic acids
Using atomic model kits build the molecules of methanoic acid, ethanoic acid, bu- tanoic acid, pentanoic acid and octanoic acid. If you don’t have an atomic model kit remember that you can use jelly tots (or playdough) and toothpicks. Use different colour jelly tots to represent the different atoms and two toothpicks to represent double bonds.
‚ÄĘ You should see that all these compounds have a similar formula. Remember, they belong to the carboxylic acid homologous series. What is the general formula for this series?
‚ÄĘ Remember that carbon must have four bonds, oxygen must have two bonds and hydrogen can only have one bond. Thinking about this fact, is it possible to have the carboxylic acid (‚ąíCOOH) group in a position other than the last (or first) carbon atom?
Exercise 4 ‚Äď 15: Naming carboxylic acids
1. Give the structural representation for the following:
a) pentanoic acid b) 4-ethyl-7-methyloctanoic acid c) 4,4-diethylheptanoic acid
2. Give the IUPAC name for each of the following
 


Naming esters
When an alcohol reacts with a carboxylic acid an ester is formed. A new bond is formed between the oxygen atom of the hydroxyl group and the carbonyl carbon atom of the carboxylic acid. The suffix for an ester is -oate.

Although the part of the ester from the alcohol (ethyl) is on the right, and the part from the carboxylic acid (butan-) is on the left in Figure 4.51, when naming the structure the part of the name from the alcohol is written first (ethyl butanoate). Don’t forget to count the carbon atom in the carbonyl group when determining the number of carbon atoms in the chain

Figure 4.51: The (a) structural, (b) semi-structural structural and (c) condensed structural repre- sentations of ethyl butanoate.
Worked example 23: Naming esters
QUESTION
Give the IUPAC name for the following
compound:
SOLUTION
Step 1: Identify the functional group
There is a ‚ąíC=O (carbonyl) group as well as an oxygen atom bonded to the carbon atom of the carbonyl and another carbon atom. This is therefore an ester and the suffix is -oate.

Step 2: Determine which part is from the alcohol and which is from the carboxylic acid An ester is a carboxylic acid derivative.
Divide the molecule in two with the car-bonyl group on one side and the oxygen
bonded to two carbon atoms on the other

The part containing the oxygen atom
bonded to two different carbon atoms was
formed from the alcohol and is on the left
here. The part containing the carbonyl
group was formed from the carboxylic acid and is on the right here
Step 3: Number the carbon atoms on the carbon chains
There is only one carbon atom in the left-hand chain (from the alcohol). Therefore this will be methyl. There are three carbon atoms in the right-hand chain (from the carboxylic acid) therefore the prefix will be propan-.
Step 4: Combine the elements of the compound’s name into a single word in the order of chain from the alcohol; prefix (from chain containing carbonyl functional group), name ending according to functional group
The compound’s name is methyl propanoate.
Worked example 24: Naming esters
QUESTION
Draw the structural and condensed structural representations for the organic com- pound ethyl hexanoate.
SOLUTION
Step 1: Identify the functional group
The compound has the suffix -oate. It is therefore an ester and has a ‚ąíC=O (carbonyl) group as well as an oxygen atom bonded to the carbon atom of the carbonyl and another carbon atom.

Step 2: Determine which part is from the alcohol and which is from the carboxylic acid
The ethyl tells us that there are two carbon atoms in the part of the chain from the alcohol. The prefix hex- tells us that there are six carbon atoms from the part of the chain from the carboxylic acid.
Step 3: Place the functional group as well as any branched groups
The oxygen atom bonded to two different carbon atoms is located between the two sections. The ‚ąíC=O (carbonyl) group is located at the first carbon atom of the carboxylic acid chain.
Step 4: Combine this information and add the hydrogen atoms

Step 5: Condense the structural representation
Condense the part of the compound that came from the carboxylic acid first, so start from the right here: CH3 CH2 CH2 CH2 CH2 COO
The first O is for the ‚ąíC=O, the second is from the ‚ąíO‚ąí. Now condense the part that came from the alcohol, starting from the ‚ąíO‚ąí:
CH3 CH2 CH2 CH2 CH2 COOCH2 CH3
Exercise 4 ‚Äď 16: Naming esters
1. Give the IUPAC name for each of the following compounds:
a) CH3COOCH2CH3

2. Give the structural representations for the following esters:

a) heptyl propanoate b) methyl octanoate c) hexyl pentanoate

Worked example 25: Naming carbonyl compounds
QUESTION
Give the IUPAC name for the following compound: CH3 CH2 CH(CH3 )CH2 CH2 COOH
(Remember that the side groups are shown in brackets after the carbon atom to which they are attached)
SOLUTION
Step 1: Draw the structural representation

Step 2: Identify the functional group
The compound has a ‚ąíCOOH functional group. It is therefore a carboxylic acid and the suffix is -oic acid.
Step 3: Find the longest carbon chain containing the functional group
There are six carbon atoms in the longest chain containing the functional group. The prefix for this compound is hexan-.
Step 4: Number the carbon atoms in the carbon chain
The carbon atoms should be numbered from right to left so that the carboxylic acid functional group has the lowest numbered carbon atom.
Step 5: Look for any branched groups, name them and give their position on the carbon chain
There is a branched group attached to the fourth carbon atom. This group has only one carbon atom and is therefore a methyl group (4-methyl).
Step 6: Combine the elements of the compound’s name into a single word in the order of branched groups; prefix; name ending according to the functional group The compound’s name is 4-methylhexanoic acid.
Worked example 26: Naming carbonyl compounds
QUESTION
Give the IUPAC name for the following compound: CH3COOCH2CH2CH2CH3
SOLUTION
Step 1: Identify the functional group
The compound has a ‚ąíCOO functional group. It is therefore an ester and the suffix
will be -oate. This can also be shown as:

Step 2: Determine which part is from the
 
alcohol and which is from the carboxylic acid
The left half of the compound contains the carbonyl group and is therefore from the carboxylic acid. The right half of the compound must be from the alcohol.

Step 3: Determine the number of carbon atoms in each chain
There are 2 carbon atoms in the part that contains the C=O group (from the carboyxlic acid), therefore the prefix is ethan-. There are 4 carbon atoms in part of the chain without the carbonyl group (from the alcohol), which is therefore butyl.

Step 4: Look for any branched groups, name them and give their position on the carbon chain
There are no branched groups.
 
Step 5: Combine the elements of the compound’s name into a single word in the order of name of chain from the alcohol; prefix (from chain containing carbonyl functional group); suffix
The compound’s name is butyl ethanoate.
Exercise 4 ‚Äď 17: Naming carbonyl compounds
1. Give the structural representation for the following compounds:. a) 3-methylpentanal b) butyl pentanoate
c) 2-methylbutanoic acid d) octan-4-one
2. Give the IUPAC name for each of the following compounds:
                           c) CH3 CH2 CH2 CHO d) HCOOH
Activity: Building molecules
Using atomic model kits build molecules of butane, but-1-ene, but-1-yne, butan-1-ol, butanoic acid, butyl butanoate, butan-2-one, butanal. If you don’t have an atomic model kit remember that you can use jelly tots (or playdough) and toothpicks. Use dif- ferent colour jelly tots to represent the different atoms and extra toothpicks to represent double and triple bonds.
‚ÄĘ Identify the functional group in each of these molecules. Move the functional group around the molecule. You should find that you can do so with but-1-ene, but-1-yne and butan-1-ol. It will not be possible with butanoic acid and there is no functional group to move in butane.
‚ÄĘ Can you see that if you move the carbonyl group in butan-2-one you will either still get butan-2-one or you will get butanal? Similarly, when moving the carbonyl group in butanal you can only get butanal or butan-2-one.
‚ÄĘ Build a model of the molecule below:

Compare this to butyl butanoate. What differences are there between these two molecules? What similarities are there? What is the name of the new molecule?
‚ÄĘ Continue by building different compounds for your friends. Make them identify the functional group, the number of carbon atoms and therefore the name of the compound
Exercise 4 ‚Äď 18: IUPAC naming
1. Study the table below and answer the questions that follow:

a) Match the compounds in column A with the correct functional group in column B. For example methane is an alkane: A1, B1.
b) Match the compounds in column A with the correct number of carbon atoms in column C. For example methane has one carbon atom in its longest chain: A1, C1.

2. Match the structural representation in column A with the condensed structural representation (column B) and IUPAC name (column C).

 

3. Fill in the gaps in the table below:

   

 

 

Physical properties  and intermolecular forces      
The types of intermolecular forces that occur  in a substance  will affect its physical properties,  such as its phase,  melting point and boiling point.  You should remember from the  kinetic  theory  of matter  (see Grade  10),  that  the  phase  of a substance  is determined by how strong the forces are between  its particles.  The weaker the forces, the more  likely the substance  is to exist as a gas.  This is because  the particles  are able to move far apart since they are not held together very strongly.  If the forces are very strong, the particles are held closely together in a solid structure.  Remember  also that the temperature of a material affects the energy of its particles.  The more energy the particles have, the more likely they are to be able to overcome  the forces that are holding them together.  This can cause a change in phase.
Figure 4.52 shows the three phases of water. Note that we are showing two-dimensional figures when in reality these are three-dimensional.

Figure 4.52:  Submicroscopic diagrams  of the three  phases  of water.   (Photos by Iaszlo-photo, Fields of View and Capt’ Gorgeous  on Flickr)
The effects of intermolecular forces
The forces between  molecules  that bind them together  are known  as intermolecular forces.   Intermolecular  forces allow  us to determine  which  substances  are likely to dissolve in other  substances,  and what the melting and boiling  points of substances are. Without intermolecular forces holding molecules  together we would not exist.
DEFINITION:  Intermolecular forces
Intermolecular  forces are forces that act between  molecules.
Remember  from Grade  11 that a dipole  molecule  is a molecule  that has its charge unevenly  distributed.   One  end  of the molecule  is slightly positive  and  the other  is slightly negative.  An overview  of the different types of intermolecular forces that are discussed in this chapter are given below:

  1. Dipole-dipole forces

When  one dipole  molecule  comes  into contact  with another  dipole  molecule, the positive pole of the one molecule  will be attracted to the negative pole of the other, and the molecules  will be held together in this way.
One special case of this is hydrogen  bonding:

  • Hydrogen bonds

As the name implies, this type of intermolecular bond involves a hydrogen atom.  When a molecule  contains a hydrogen  atom covalently bonded  to a small, highly electronegative atom (e.g. O, N or F) this type of intermolecular force can  occur.   The highly electronegative atom  on  one  molecule attracts the hydrogen  atom on a nearby molecule (see Figure 4.53).

TIP: Do not confuse hydrogen  bonds with intramolecular covalent bonds. Hydrogen  bonding  is an example of a scientist naming something,  believing it to be one thing when in fact it was another. In this case the strength of the hydrogen  bonds misled scientists into thinking this was an intramolecular bond when it is really just a strong intermolecular force
FACT: Dipole-induced-dipole intermolecular forces are also sometimes called London forces or dispersion  forces.

  1. van der Waals forces
  • Induced-dipole forces

In non-polar  molecules  the electronic  charge is usually evenly distributed but it is possible  that at a particular  moment  in time, the electrons  might not be evenly distributed (remember that the electrons are always moving in their orbitals). The molecule  will have a temporary  dipole.  When this hap- pens, molecules  that are next to each other attract each other very weakly.

  • Dipole-induced-dipole forces

These forces exist between  dipoles  and non-polar  molecules.  The dipole induces  a dipole  in the non-polar  molecule  leading to a weak, short lived force which holds the compounds together.
In this chapter  we  will focus  on  the  effects  of van  der  Waals  forces  and  hydrogen bonding  on the physical properties  of organic molecules.
Viscosity
Viscosity is the  resistance  to flow of a liquid.   Think how  easy  it is to pour  water compared to syrup or honey.  The water flows much faster than the syrup or honey.
DEFINITION:  Viscosity
Viscosity is a measure  of how  much  a liquid  resists flowing.   i.e.   The higher  the viscosity, the more viscous a substance  is.

You can  see  this if you  take a cylinder filled  with  water  and  a  cylinder  filled with glycerol (propane-1,2,3-triol).  Drop a small metal ball into each cylinder and note how easy it is for the ball to fall to the bottom (Figure 4.55).  In the glycerol the ball falls slowly, while in the water it falls faster.

Figure 4.55:  The higher the viscosity (red) the slower the ball moves through the liquid
As implied by the definition, substances  with stronger intermolecular forces are more viscous  than  substances  with weaker  intermolecular forces.   The stronger  the  inter- molecular  forces the more the substance  will resist flowing.  The greater the internal friction the more a substance  will slow down an object moving through it.
Activity:  Resistance  to flow
Take a sheet of glass at least 10 by 15 cm in size.  Using a water-proof marker draw a straight line across the width of the glass about 2 cm from each end.
Place the glass flat on top of two pencils,  then carefully put a drop of water on one end of the line. Leave at least 2 cm space next to the water drop, and carefully place a drop of alcohol.  Repeat this with a drop of oil and a drop of syrup.
Slowly and carefully remove  the pencil  from the end  opposite  the drops (make sure you don’t tilt the glass to either side in the process).

  • Which drop moves fastest and reaches the end line first?
  • Which drop moves slowest?

The fastest moving substance has the least resistance to flow, and therefore has the least viscosity (is the least viscous).  The slowest moving substance  has the most resistance to flow, and therefore has the most viscosity (is the most viscous).
Density
DEFINITION:  Density
Density is a measure of the mass per unit of volume.
The solid phase  is often the most dense  phase  (water is one noteworthy  exception  to this). This can be explained  by the strong intermolecular forces found in a solid. These forces pull the  molecules  together,  which  results in more  molecules  in one  unit of volume  than in the liquid or gas phases.   The more  molecules  in a unit volume  the denser, and heavier that volume of the substance  will be.
Density can be used to separate  different liquids, with the more dense  liquid settling to the bottom of the container,  while the less dense liquid floats on top. If you throw a leaf into a river or pond  the leaf will float. If you instead throw a rock (with the same surface area and volume) into the river or pond  the rock will sink.  This is due to the different densities of the two substances:  rocks are more dense than water while leaves are less dense than water.
Melting and boiling points
Intermolecular  forces affect the boiling and melting points of substances.  Substances with weak intermolecular forces will have low melting and boiling points as less energy (heat) is needed to overcome  these forces. Those with strong intermolecular forces will have high melting and boiling points as more  energy (heat) is required  to overcome these  forces.  When  the temperature of a substance  is raised  beyond  it’s melting  or boiling point the intermolecular forces are not weakened. Rather, the molecules  have enough  energy to overcome  those forces.
FACT:An ether is a compound that contains two alkyl chains (e.g. methyl, ethyl) joined at an angle by an oxygen atom (‚ąíO‚ąí).
Table 4.9:  Relationship  between  intermolecular forces  and  melting  point,  boiling  point  and physical state.
As the  intermolecular forces  increase  (from top  to bottom in Table 4.9) the melting and boiling points  increase.  The stronger the intermolecular forces the more likely a substance is to be a liquid or a solid at room temperature

 


Step 2: What are the intermolecular forces that each molecule  will experience?

  • propane has only single carbon-carbon bonds and no other functional group. It will therefore have induced-dipole forces only.
  • butanoic acid has the carboxylic ¬†acid¬† functional ¬†group.¬† ¬†It can therefore ¬†form hydrogen ¬†bonds.
  • bromoethane has the highly electronegative bromine atom. ¬†This means ¬†that it will form dipole-dipole interactions ¬†with neighbouring molecules.
  • diethyl ether will have induced-dipole forces, however due to the flexible nature of the molecule is can also have dipole-induced-dipole (stronger van der Waals interactions) and dipole-dipole forces.

Step 3: Do these forces make sense with the melting and boiling points provided? Propane  has the lowest melting and boiling points and the weakest interactions.  The next lowest melting and boiling points are for bromoethane and diethyl ether,  which
both  have  dipole-dipole interactions,  the  next  strongest  intermolecular forces.   The
highest melting and boiling  points are for butanoic  acid which  has strong hydrogen bonds.  Therefore these forces do make sense.
Step 4: Calculate  the molecular mass of these molecules

  • propane – C3H8 , molecular ¬†mass = 44,08 g.mol‚ąí1
  • butanoic acid – C4H8 O2 , molecular ¬†mass = 88,08 g.mol‚ąí1
  • bromoethane – C2H5Br, molecular mass = 108,95¬†g.mol‚ąí1
  • diethyl ether – C4H10 O, molecular mass = 74,10¬† g.mol‚ąí1

Step 5: What will the phase of each compound be at 25‚ó¶C ?
To determine  the phase of a molecule  at 25◦C look at the melting and boiling points:

  • The melting and boiling points of propane are both below 25‚ó¶C , therefore the molecule ¬†will be a gas at 25‚ó¶C .

‚ÄĘ The melting points of butanoic ¬†acid, ¬†bromoethane and ¬†dimethyl ¬†ether ¬†are be- low 25‚ó¶C , however ¬†the boiling points of all three molecules ¬†are above 25‚ó¶C . Therefore these molecules ¬†will be liquids at 25‚ó¶C .

Experiment:  Investigation of boiling and melting  points
Aim:
To investigate the relationship  between  boiling points and intermolecular forces
Apparatus:

  • butan-1-ol (CH2(OH)CH2CH2CH3 ), propanoic acid (CH3CH2COOH) ¬†and ethyl methanoate (HCOOCH2CH3 ), cooking oil
  • three test tubes, a beaker, a thermometer, a hot plate

Method:
WARNING!
Ethyl methanoate can irritate  your eyes, skin, nose and lungs. Keep open flames away from your experiment and make sure you work in a well ventilated area.

  1. Label the test tubes 1, 2 and 3.  Place20  ml  of butan-1-ol  into  test  tube  1, 20 ml of propanoic  acid into test tube 2, and 20 ml of ethyl methanoate into test tube 3.
  2. Half-fill the beaker with cooking oil and place it on the hot plate. three test tubes in the beaker.

 

  1. Make a  note  of  the  temperature when each substance  starts to boil.

Results:
Fill in the gaps in the table below.  Do the values you obtained  match those reported in literature?

Draw  the  structural  representation of ethyl  methanoate,  butan-1-ol  and  propanoic acid.
Discussion and conclusion:
You should have found that the ethyl methanoate boiled first, then the
butan-1-ol and then the propanoic  acid. Ethyl methanoate has some dipole-dipole interactions,  but cannot form a hydrogen bond.  The alcohol (butan-1-ol) can form hydrogen  bonds and so has a higher boiling point.  This strong intermolecular force needs more energy to break and so the boiling point is higher.

Figure 4.57:  A carboxylic acid hydrogen  bonding dimer
For propanoic  acid hydrogen bonds form between  the carbonyl group on one acid and the hydroxyl group on another.  This means that each molecule  of propanoic  acid can be part of two hydrogen  bonds (this is called dimerisation,  see Figure 4.57) and so the boiling point is even higher for propanoic  acid than for butan-1-ol.
Flammability and vapour  pressure
Flammability is a measure  of how easy it would be for a substance  to catch alight and burn.   The flash point  of a substance  is the lowest temperature that is likely to form a gaseous mixture you could  set alight.  If a liquid has a low enough  flash point it is considered flammable (able to be ignited easily) while those with higher flash points are considered nonflammable. A substance  that is classified as nonflammable can still be forced to burn, but it will not ignite easily.
When  a substance  is in the liquid or solid state there will be some molecules  in the gas state. These molecules  have enough energy to overcome  the intermolecular forces holding the majority of the substance in the liquid or solid phase.  These gas molecules exert a pressure on the liquid or solid (and the container) and that pressure is the vapour pressure of that compound. The weaker the intermolecular forces within a substance the higher the vapour pressure will be. Compounds with higher vapour pressures have lower flash points and are therefore more flammable.
DEFINITION:  Vapour pressure
The pressure  exerted  (at a specific temperature)  on  a solid or liquid  compound by molecules  of that compound that are in the gas phase.

Table 4.10:  Relationship between  intermolecular forces and the flammability of a substance.
As the intermolecular forces increase (from top to bottom in Table 4.10) you can see a decrease  in the vapour pressure.  This corresponds  with an increase  in the flash point temperature and a decrease  in the flammability of the substance.  Figure 4.58 shows a few examples

Figure 4.58:  The vapour pressure of (a) water, (b) ethanol  and (c) propanone at 20 ◦ C. (Images by Duncan  Watson)
Exercise 4 ¬†‚Äď ¬†19: ¬†Types of intermolecular forces

  1. Use your knowledge of different  types of intermolecular forces  to explain  the following statements:
  1. The boiling point of hex-1-ene  is much  lower  than  the  boiling  point  of propanoic acid.
  2. Water evaporates  more slowly than propanone.

2.

  1. Which of the compounds listed in the table are gases at room temperature?
  2. Name the main type of intermolecular forces for A, B and C.
  3. Account for the difference in boiling point between A and B.
  4. Account for the difference in boiling point between B and C.
  5. Draw the structural representations of A, B and C.

3. a) Which container (A or B) has the compound with higher vapour pressure in it? Explain your answer.

b) Draw the condensed structural formula for each of these compounds:

i. propanoic acid ii. butan-1-ol

c) Propanoic acid has a vapour pressure of 0,32 kPa at 20 ◦ C  Butan-1-ol has a vapour pressure of 0,64 kPa at 20 ◦ C Explain the difference in vapour pressure.
TIP: Vanilla has a sweet smell. The smell of ethene  and dimethyl ether cling in your nose though, and the smell of ether can even stay on your skin up to 24 hours
FACT: In humans,  ethanol reduces the secretion of a hormone  called antidiuretic  hormone (ADH). The role of ADH is to control the amount of water that the body retains. When this hormone  is not secreted  in the right quantities,  it can cause dehydration because  too much water is lost from the body in the urine (ethanol is a diuretic). This is why people who drink too much alcohol can become dehydrated, and experience symptoms such as headaches, dry mouth, and lethargy. Part of the reason for the headaches is that dehydration causes the brain to shrink away from the skull slightly.
———————————————————————-
Physical properties  and functional groups  
Compounds that contain very similar atoms can have very different properties depending on how those atoms are arranged.  This is especially true when they have different functional groups.  Table 4.11 shows some properties  of different functional groups.   

Table 4.11:  Some properties of compounds with different functional groups.
Listed in the  table  are the  common  smells and  other  physical  properties  found for common  functional groups.  Only one representative  example from each homologous series is provided.  This does not mean that all compounds in that series have exactly the same properties.   For example,  short-chain  and long-chain  alkanes  are generally odourless, while those with moderate chain length (approximately 6 Р12 carbon atoms) smell like petrol (Table 4.15).
Some specific properties  of a few functional  groups  will now  be discussed  in more detail.
Physical properties of the alcohols
The hydroxyl group (‚ąíOH) affects the solubility of the alcohols
DEFINITION:  Solubility
Solubility is a measure  of the ability of a substance  (solid, liquid or gas) to dissolve in another  substance.   The amount  of the substance  than can dissolve is the measure  of its solubility.


Table  4.12:   Melting and  boiling  points  of haloalkanes  with  increasing  numbers  of chlorine atoms.
Properties of carbonyl  compounds
Carboxylic  acids are weak  acids,  in other words they only dissociate  partially.  Why does the carboxyl group have acidic properties?  In the carboxyl group, the hydrogen tends to separate itself (dissociate) from the oxygen atom. In other words, the carboxyl group becomes  a source  of positively-charged  hydrogen  ions (H+ ). This is shown  in Figure 4.60.

Figure 4.60:  The dissociation  of a carboxylic acid.
The carboxylic  acid functional  group is soluble in water.  However,  as the number  of carbon  atoms in the attached  carbon  chain increases  the solubility decreases.  This is discussed in greater detail in the next section.

Remember that
carboxylic acids form
hydrogen bonding
dimers (the formation is
called dimerisation) as
shown in Figure 4.61.
Figure 4.61: Ethanoic acid forming a carboxylic acid hydrogen bonding dimer.

Table 4.13: The melting and boiling points of similar organic compounds that can form different numbers  of hydrogen  bonds

Physical properties of ketones Hydrogen bonds are stronger than the van der Waals forces found in ketones.  There- fore  compounds  with  functional  groups that  can  form hydrogen  bonds  are  more likely to be soluble in water.  This applies to aldehydes  as well as to ketones.
 
 
Exercise 4 ¬†‚Äď ¬†20: ¬†Physical properties and functional groups

  1. Refer to the data table below which shows the melting point and boiling point for a number  of organic compounds with different functional groups.

  1. a) At room temperature (appro 25‚ó¶C), which of the organic compounds in the table are:

i. gases  ii.  liquids

b) Look at an alkane, alkene, alcohol and carboxylic acid with the same num- ber of carbon atoms:

  1. How do their melting and boiling points compare?
  2. Explain why their melting points and boiling points are different?

 
Physical properties  and chain length
Remember that the alkanes are a group of organic compounds that contain carbon and hydrogen  atoms bonded  together.   The carbon  atoms link together  to form chains  of varying lengths.  We have already mentioned that the alkanes are relatively unreactive because  of their stable C-C and C-H bonds.  The boiling points and melting points of these molecules  are determined by their molecular  structure and their surface area.

Figure 4.63:  Some alkanes,  from top left to bottom  right:  methane, ethane,  propane, butane, hexane,  octane and icosane.

The more carbon atoms there are in an alkane, the greater the surface area available for intermolecular interactions.
 
 
 
Figure 4.64: van der Waals intermolecular forces increase as chain length increases.  (Image by Duncan  Watson)
This increase  in intermolecular attractions leads to higher melting and boiling points. This is shown in Table 4.14.


Table 4.15 shows some other properties of the alkanes that also vary with chain length.

Table 4.15:  Properties of some of the alkanes.
Density  increases  with  increasing  molecular  size.    Note  that  compounds that  are gaseous are much less dense than compounds that are liquid or solid.
Remember that the flash point of a volatile molecule is the lowest temperature at which that molecule  can form a vapour  mixture with air and be ignited.  The flash point in- creases with increasing chain length meaning that the longer chains are less flammable (Table 4.10) although they can still be ignited.
The change  in physical properties  due to
chain length does not only apply to the hy- drocarbons. The solubility of ketones in water decreases  as the chain length increases. The longer the chain length, the stronger the intermolecular interactions  between  the ketone molecules.
This means that more energy is required  to overcome  those interactions  and the molecule  is therefore less soluble in water. Long chains can also fold around the polar carbonyl group and  stop water molecules  from bonding  with it.

 
 
This same idea can be applied to all the compounds with
water-soluble functional groups. The longer the chain becomes, the less soluble the compound isin water
 
 
 
Figure 4.66:  The water solubility of a ketone                                                                   decreases as the chain length increases.
Exercise 4 ¬†‚Äď ¬†21: ¬†Physical properties and chain ¬†length

  1. Refer to the table below which gives information about a number  of carboxylic acids, and then answer the questions that follow.


a) Fill in the missing spaces in the table by writing the formula, common name or IUPAC
b) Draw the structural representation of butyric c) Give the molecular formula for caprylic acid.
d) i. Draw a graph to show the relationship between  molecular mass (on the x-axis) and boiling point (on the y-axis) ii. Describe the trend you see.  iii.  Suggest a reason for this trend.
2.  Refer to the data table below  which shows the melting point and boiling point for a number  of organic compounds with different functional groups.

a) At room temperature (approx. 25 ◦ C), which of the organic compounds in the table are:             i. gases ii. liquids
b) In the alkanes:
i. Describe what happens to the melting point and boiling point as the number of carbon atoms in the compound increases.
ii. Explain why this is the case.
3. Fill in the table below. Under boiling point put 1 for the compound with the lowest boiling point, 2 for the next lowest boiling point until you get to 6 for the compound with the highest boiling point. Do not use specific boiling point values, but rather use your knowledge of intermolecular forces.

4. Draw the structural representations for each of the following compounds and answer the question that follows.
a) but-2-yne b) hex-2-yne c) pent-2-yne
d) Which of these compounds will have the highest viscosity?


Physical properties  and branched  groups         
In a straight chain  molecule, the carbon  atoms  are connected to at most two other carbon  atoms.  However,  in a branched  molecule  some carbon  atoms are connected to three or four other carbon atoms. This is the same principle that was discussed with primary, secondary  and tertiary alcohols.
Straight chains  always have higher boiling  points than the equivalent  molecule  with branched  chains.   This is because  the  molecules  with  straight chains  have  a larger surface area that allows close contact.  Branched chains have a lower boiling point due to a smaller area of contact.  The molecules  are more compact and cannot get too close together, resulting in fewer places for the van der Waals forces to act.


 
Shown in Figure 4.68 are three isomers that all have the molecular formula C5 H12 . They are all alkanes with the only difference being branched groups leading to longer or shorter main chains
 
Figure 4.68:  Three isomers of C5 H12 : (a) pentane,  (b) 2-methylbutane and (c) 2,2-dimethylpropane.
As shown in Table 4.16 the properties  of these three compounds are significantly different. The melting points have significant differences and the boiling points steadily decrease with an increasing number  of branched  groups.

Table 4.16:  Properties of compounds with different numbers  of branched  groups.
Although the  densities  of all three  compounds are very similar,  there  is a decrease in density with the increase  of branched  groups.  The vapour  pressure increases  with increasing branched  groups and the flash point of 2,2-dimethylpropane is significantly higher than those of pentane  and 2-methylbutane.
It is interesting  that the common  trend of melting and boiling points is not followed here Р2,2-dimethylpropane has the highest melting point, lowest boiling point and is the least flammable  although  all three compounds are very flammable.  Symmetrical molecules (such as 2,2-dimethylpropane) tend to have higher melting points than similar molecules with less symmetry  due to their packing in the solid state. Once  in the liquid state they follow the normal trends.
Exercise 4 ¬†‚Äď ¬†22: ¬†Physical properties and branched chains
             a) Give the IUPAC name for each compound
b) Explain the difference in the melting points
2. There are five isomers with the molecular formula C6 H14 .

a) Draw the structural representations of:
2-methylpentane and 2,2-dimethylbutane (two of the isomers)
b) What are the names of the other three isomers?
c) Draw the semi-structural representations of these three molecules.
d) The melting points of these three isomers are: ‚ąí118, ‚ąí95 and ‚ąí130 ‚ó¶ C.
Assign the correct melting points to the correct isomer. Give a reason for your answers.

Exercise 4 ¬†‚Äď ¬†23: ¬†Physical properties of organic ¬†compounds

  1. The table shows  data  collected  for four  organic  compounds (A РC)  during  a practical investigation.


a) Is compound A saturated or unsaturated? Give a reason for your answer. b) To which homologous series does compound B belong?
c) Write down the IUPAC name for each of the following compounds:
i. B ii. C
d) Refer to intermolecular forces to explain the difference in boiling points between compounds A and C.
e) Which one of compounds B or C will have the highest vapour pressure at a specific temperature? Give a reason for your answer.
2. Give the IUPAC names for the following compounds and answer the questions that follow.

Alkanes as fossil fuels      
Fossil fuels are fuels formed by the natural process of the decomposition of organisms under heat and pressure.  They contain a high percentage  of carbon  and include  fuels such as coal, petrol, and natural gases. They are also a non-renewable.
DEFINITION:  Hydrocarbon cracking
Hydrocarbon cracking is the process  of breaking carbon-carbon bonds  in long-chain hydrocarbons to form simpler, shorter-chain hydrocarbons.
Hydrocarbon cracking  is an important  industrial process.  Through this process long, bulky alkanes  are  broken  up  into  smaller  compounds.   These  compounds include shorter alkanes  and alkenes.   A few examples  of cracking hydrocarbons are given in Figure 4.69

Figure 4.69:  The cracking of various hydrocarbons to produce alkanes and alkenes.
There are two types of hydrocarbon cracking.   Thermal cracking occurs  under  high pressures and temperatures without a catalyst, catalytic  cracking occurs at lower pressures and temperatures in the presence  of a catalyst. This process is a common source of shorter (more useful) alkanes as well as unsaturated alkenes.  The alkanes are then used in combustion processes.
It is possible  to separate  the  products  of hydrocarbon cracking  and  obtain  specific products  from a crude  oil mix through  a process  called  fractional  distillation.  This is done  using a fractionating  column.   Crude oil evaporates  when  heated  to 700◦ C. The gas bubbles  through a tray that is kept at a certain temperature. The alkanes and alkenes that condense at that temperature will then condense in the tray. For example if the tray is kept at 170◦ C the product will be paraffin oil (Figure 4.70).

Figure 4.70:  One layer of a fractionating column.  (Image by Duncan  Watson)
A fractionating column has a series of these trays (Figure 4.71), each at a constant temperature.  This means that many compounds can be separated from the crude oil mix. The crude oil is heated to 700◦ C and the gas of the crude oil is passed through the column.  Bitumen for tar roads is collected at the bottom of the fractionating column. These are all compounds with more than 70 carbon atoms. The temperature decreases as you move up the column.  As the gases rise, compounds with different length carbon chains condense until only the chains with 1 Р4 carbon atoms are collected at the top of the column.  These are used for liquid petroleum gas.

Note that this fits what we learned previous section. More carbon atoms in the chain, the greater the intermolecular forces and the ref higher the boiling point.  That me these molecules condense at higher temperatures.

 
 
 

Figure 4.71:  An example of fractionating column

For more information on the process of fractional distillation have a look at this animation:
Combustion of alkanes
Alkanes are our most important fossil fuels. The combustion (burning) of alkanes (also known as oxidation) is highly exothermic.
DEFINITION:  Combustion
In a combustion reaction a substance  reacts with an oxidising agent (e.g. oxygen), and heat and light are released.
In the  complete  combustion reaction  of alkanes,  carbon  dioxide  (CO2 )  and  water (H2O) are released  along with energy.   Fossil fuels are burnt  for the energy  they release. The general reaction for the combustion of an alkane as a fossil fuel is given in Figure 4.72

Worked  example  28:  Balancing equations
QUESTION
Balance the following equation:  C4H10(g) + O2(g) → CO2(g) + H2O(g)
SOLUTION
Step 1: Balance the carbon  atoms
There are 4 carbon  atoms on the left. There is 1 carbon  atom on the right. Add a 4 in front of the CO2  molecule  on the right: C4H10(g) + O2(g) → CO2(g) + H2O(g)
Step 2: Balance the hydrogen  atoms
There are 10 hydrogen  atoms on the left.  There are 2 hydrogen  atoms on the right. Add a 5 in front of the H2O molecule  on the right:
C4H10(g) + O2(g) ‚Üí 4CO2(g) + 5H2O(g)
Step 3: Balance the oxygen atoms
There are 2 oxygen atoms on the left. There are 13 oxygen atoms on the right (4 x 2 in the CO2  and 5 in the H2O). Divide the number  of O atoms on the right by 2 to get  13/2  ,this is the number  of O2  molecules  required  on the left:
C4H10(g) + 13/2 O2(g) ‚Üí 4CO2(g) + 5H2O(g)
This is acceptable but it is better for all numbers  to be whole numbers.
Step 4: Make sure all numbers are whole numbers
There is 13/2 in front of the O2  while all other numbers  are whole numbers.  So multiply the entire equation  by 2:
2C4H10(g) + 13O2(g) ‚Üí 8CO2(g) + 10H2O(g)
Worked  example  29:  Balancing equations
QUESTION
Balance the equation  for the complete  combustion of heptane
SOLUTION
Step 1: Write the unbalanced equation
The molecular  formula  for heptane  is C7H16 .   Combustion  always  involves  oxygen (O2). The complete  combustion of an alkane always produces  carbon  dioxide  (CO2 ) and water (H2O):
C7H16(l) + O2(g) ‚Üí CO2(g) + H2O(g)
Step 2: Balance the carbon  atoms
There are 7 carbon  atoms on the left. There is 1 carbon  atom on the right. Add a 7 in front of the CO2 molecule  on the right:
Step 3: Balance the hydrogen  atoms
There are 16 hydrogen  atoms on the left.  There are 2 hydrogen  atoms on the right.
Add an 8 in front of the H2O molecule  on the right:
C7H16(l) + O2(g) ‚Üí 7CO2(g) + 8H2O(g)
Step 4: Balance the oxygen atoms
There are 2 oxygen atoms on the left. There are 22 oxygen atoms on the right (7 x 2 in the CO2 and 8 in the H2O ). Divide the number  of O atoms on the right by 2 to get 11, this is the number  of O2 molecules  required  on the left:
C7H16(l) + O2(g) ‚Üí 7CO2(g) + 8H2O(g)
Exercise 4 ¬†‚Äď ¬†24: ¬†Alkanes as fossil fuels

  1. Balance the following complete combustion equations:

a) C3H8(g) + O2(g) ‚Üí CO2(g) + H2O(g)

b) C7H16 (`) + O2 (g) ‚Üí CO2 (g) + H2O(g)

c) C2H6 (g) + O2 (g) ‚Üí CO2 (g) + H2O(g)

  1. Write the balanced equation  for the complete  combustion of:

a) octane b) pentane             c) hexane              d) butane

  1. Is the combustion of alkanes exothermic or endothermic?  What does that mean?

——————————————————————————-
Esters
Production of esters
As was discussed  earlier in this chapter  one  way to form  an ester is through  the re- action  of an alcohol  and a carboxylic  acid.  This process  is called  an acid-catalysed condensation or esterification of a carboxylic acid.

          Figure 4.74:  The acid-catalysed condensation of a carboxylic acid to form an ester.
In the general form above an alcohol  (red) and a carboxylic acid (orange) combine  to form an ester and water.  A specific example  is given in Figure 4.75 for the formation of butyl propanoate and water.

Figure 4.75:  The esterification of butanol and propanoic  acid to form butyl propanoate, water is also formed in this reaction.
This reaction can also be written as:

A more general example is:

alcohol   +  carboxylic  acid   →   ester      +   water

It is important  to be able to identify what ester a specific alcohol  and carboxylic acid will form. Remember  that the first part of the ester name takes its prefix from the alco- hol with the suffix -yl. The second  part of the ester takes its prefix from the carboxylic acid with the ester suffix -oate
Worked  example  30:  Determining ester  names
QUESTION
What is the name of the ester that will form from hexanol  and propanoic acid
SOLUTION
Step 1: Which compound forms the first part of the ester name and which forms the second  part of the ester name?
The alcohol forms the first part of the ester name and takes the suffix -yl. The carboxylic acid forms the second part of the ester name and takes the suffix -oate.
Step 2: Determine the first part of the ester name
The alcohol is hexanol, therefore there are 6 carbons and this will be hexyl.
Step 3: Determine the second  part of the ester name
The carboxylic  acid is propanoic acid, therefore  there are 3 carbons  and this will be propanoate.
Step 4: Combine the first and second  parts of the ester name
The ester will be hexyl propanoate.
It is also important  to be able to determine  which  compounds were used to form an ester.
Worked  example  31:  Determining starting  materials  of esters
QUESTION
What compounds did the ester octyl heptanoate come from?
SOLUTION
Step 1: What types of compounds are used to form esters?
Esters are formed from alcohols  (which become  the first part of the ester name) and carboxylic acids (which become  the second part of the ester name).
Step 2: Determine the prefix for the alcohol
The first part of the ester name  comes  from the alcohol  (-ol). Therefore the prefix is oct-.
Step 3: Determine the prefix for the carboxylic  acid
The second part of the ester name comes from the carboxylic acid (-oic acid). Therefore the prefix is hept-.
Step 4: Determine the compounds use to form the ester O
ctyl heptanoate was formed from octanol  and heptanoic acid.
A few examples of esters are given in Table 4.17.

Table 4.17:  The uses of some esters.

The following experiment  will help you to prepare esters. Use what you have learned in this section to answer the questions that follow.
Experiment:  Preparation of esters
Aim:
To prepare and identify esters.
Apparatus:

  • Methanol (CH3OH), ethanol (CH3CH2(OH)), pentan-1-ol (CH3CH2CH2CH2CH2(OH)), methanoic acid (HCOOH), ethanoic ¬†acid (CH3COOH), sulfuric acid (H2SO4 )
  • Marble chips (or small, clean stones)
  • Five tall test tubes or beakers, a water ¬†bath, ¬†a hot-plate ¬†(or bunsen ¬†burner), ¬†a thermometer, rubber bands, paper towel, five bowls of cold water

Method:
WARNING!
Concentrated acids  can  cause  serious  burns.   We  suggest  using  gloves and  safety glasses whenever you work with an acid.  Always add the acid to the water  and avoid sniffing the acid.
Remember  that  all alcohols  are toxic,  methanol is particularly toxic and can cause blindness,  coma or death.  Handle  all chemicals  with care.

  1. Place the marble chips (or clean  stones) in the test tubes and label them A РE.
  2. Add 4 ml methanoic acid to test tubes A and B.

3. Add 4 ml ethanoic acid to test tubes C, D and E.
 

  1. Add 5 ml methanol to test tubes A and A
  2. Add 5 ml ethanol to test tubes B and D.

6.  Add 5 ml pentanol  to test tube E

  1. Slowly add 2  ml  H2 SO4   to  each  test tube.
  2. Soak the paper towel in cold water and attach it to the sides near the top of each test tube with  a rubber  band  (do  not close the test tube, just wrap the paper around  the sides near the top).
  3. Heat the water bath to 60 ◦C (using the hot-plate or bunsen  burner)  and  place the test tubes in it for 10-15 min.
  4. After 10-15 min cool each test tube in cold water. Label the bowls of cold water A РE and  pour  the contents  of test tube A into bowl A, etc.
  5. Observe the surface of the water and note the smell in each bowl.

What is the purpose of the wet paper towel?
What did you observe on top of the water in each bowl at the end?
Fill in the details in a table like this one

Discussion and Conclusion:
An ester is the product  of an acid-catalysed  condensation between  an alcohol  and a carboxylic  acid.  Esters have identifiable  aromas (like the fragrant smell and odour  of fruit) and are used in perfumes.  If you didn’t smell anything it is likely that your water bath was set at too high a temperature and the ester degraded.
Esters are less water soluble than the carboxylic acid they were formed from and appear as an oily substance  on water.  Before being cooled  however  some of the ester would have been a vapour, so the wet paper towel would help to prevent a loss of the product.
Refer to Table 4.17 and see if the smells match those listed there.  Remember  to waft the smell towards you using your hand, do not sniff the fumes directly!
Exercise 4 ¬†‚Äď ¬†25: ¬†Esters

  1. Give the IUPAC name for the product in the esterification of ethanoic acid with:

a) methanol b) octanol              c) hexanol             d) propanol

2. What is another name for the type of reaction in the question above?
3. Give the IUPAC name for the product in the reaction of butanol with:

a) ethanoic acid  b) pentanoic  acid c) heptanoic  acid  d) methanoic acid

4. Fill in the missing reactant or part of the product name in the reactions below:

a) octanol +_ _ _ _ _ ‚Üí_ _ _ _ _ exanoate

b) _ _ _ _ _ _  propanoic  acid → hexyl_ _ _ _ _ _

c)_ _ _ _ _ _ _ + butanol ‚Üí_ _ _ _ _ pentanoate