S5: Chemistry

         Key unit competency

The learner should be able to relate the physical and chemical
properties of halogenoalkanes to their reactivity and their uses

Learning objectives
• Define halogenoalkanes and homologous series.
• Explain the reactivity of halogenoalkanes.
• Explain the physical properties of halogenoalkanes.
• Describe preparation methods for halogenoalkanes. 
• Explain different mechanisms in halogenoalkanes.
• Explain the uses and dangers associated with halogenoalkanes.
• Draw displayed structural formulae of halogenoalkanes and give names using 
   IUPAC system.
• Classify halogenoalkanes according to developed formula as primary, 
   secondary and tertiary.
• Write reaction mechanisms of halogenoalkanes as SN1, SN2, E1 and E2.
• Test for the presence of halogenoalkanes in a given sample organic compound.
• Appreciate the uses and dangers of halogenoalkanes in everyday life.

• Develop the awareness in protecting the environment.

Introductory activity

Look at the pictures below and answer the following questions.
Record your answers and discuss them.

a. Observe carefully pictures 4.1 and 4.2 and suggest the similarity 
    between them.
b. Observe carefully pictures 4.1 and 4.2 and suggest the difference 
    between them.
Substances which are used in the pictures belong to the same homologous series. 
They may be obtained from the reaction between alkanes and halogens.

What homologous series do these substances belong to?

         

4.1. Definition and nomenclature of halogenoalkane

      Activity 4.1

1. Look at the following and answer the questions that follow.

               

Questions:

i. Which structures do represent halogenoalkanes?
ii. What are the similarities between the selected structures?

iii. From your answers above deduce the general formula for alkanes.

      1. Definition

Halogenoalkanes compounds are compounds in which the halogen atoms like 
chlorine, bromine, iodine or fluorine are attached to a hydrocarbon chain. When 
the halogen atom is attached to a hydrocarbon chain the compound is called a 

halogenoalkane or haloalkane or an alkyl halide. 

Halogenoalkanes contain halogen atom(s) attached to the  hybridised carbon 
atom of an alkyl group. 

2. Nomenclature of halogenoalkanes
Halogenoalkanes are organic compounds that contain a halogen atom: F, Cl, Br, I. 
They are named using the prefixes fluoro-, chloro-, bromo- and iodo-.

Numbers are used if necessary to indicate the position of the halogen atom in the 

molecule.

         

If the molecule contains more than one halogen atom of the same kind, the prefixes 
di-, tri-, tetra-, etc… are used. 

Examples:

Checking up 4.1

1. Name these compounds
                 

2. Write the structural formulae for the following compounds:
a. 1,2-dibromo-3-chloropropane

b. 1,1,2-trichloro-1,2,2-trifluoroethane

4.2. Classification and isomerism

Activity 4.2

Consider the following compounds and based on the carbon atom attached to 

the halogen atom, classify them.

                              

Do research to find an appropriate name for each class

4.2.1. Classification of halogenoalkanes

  There are three types of halogenoalkanes:

     

A primary halogenoalkane has a halogen atom attached to the ended carbon atom 
of the chain. A secondary halogenoalkane has a halogen atom attached to a carbon 
bonded to two other carbon atoms while a tertiary halogenoalkane has a halogen 

atom attached to a carbon bonded to three other carbon atoms.

4.2.2. Isomerism

Halogenoalkanes exhibit both chain and position isomerism.

 Example: Molecular formula
a. Chain isomerism: This arises due to arrangement of carbon atoms in chains of 

different size.

              

b. Position isomerism: This arises due to the different positions
 taken by the halogen atom on the same carbon chain.

The following compounds are position isomers:and
 because the atoms of bromine are on different positions
of the chain. 
Hence, all isomers of the compound with molecular formula
are the following.
Checking Up 4.2:
1. How many positional isomers possess the chlorobromopropane,

      Enumerate those able to form optical isomers. 

    Draw their structuralformulae.

    2. Illustrate the structural formulas of : 

         a. 1,1,2-trichloropropane

        a. 2-chloro-2-methylpropane

4.3. Physical properties of halogenoalkanes

       Activity 4.3

 1. Consider the following substances:

Sodium chloride, potassium bromide, hexane, pentane, trichlomethane, 
terachloromethane.

Mix a sample of each compound (1g for solids, 2ml for liquids) with 10ml 
of water.
2. Record your observations.
3. Write down your conclusions.
4. Based on the physical state and the nature of chemical bonding, predict 
the increasing order in the boiling points of the compounds above.

5. Write down your conclusions. 

1. Volatility

Volatility is a property that shows if a substance transforms easily or not into vapour 
or gaseous form. This property depends on the nature of the bonds that make up 
the molecule of the substance. Generally non polar covalent compounds are more 
volatile than polar covalent compounds. We know that halogens when bonded to 

other atoms form polar bonds because they possess high electronegativities: 

     F =  
The more the difference of electronegativities of the atoms that form the bond, 
the more polar is the bond. This explains the high polarity of C-F bond with an 
electronegativity difference of 1.5, and the low polarity of C-Cl and C-Br bonds where 

the electronegativity differences are 0.5 and 0.3 respectively. 

The presence of polarity or charge distribution results into more attraction between 
polar molecules called dipole-dipole attraction forces, one type of Van der Waals 

forces, as shown below:

     

The dashed line represents the attraction forces between the polar molecules or 

dipoles. 

Therefore, more energy must be supplied to separate polar molecules and this 
explains why melting and boiling temperatures of fluoroalkanes and chloroalkanes 
are higher than those of alkanes of similar molecular mass. 

As we have already learnt, molecules of organic halogen compounds are generally 
polar. Due to the greater polarity as well as higher molecular mass as compared 
to the parent hydrocarbons, the intermolecular forces of attraction (dipole-dipole 
and Van der Waals) are stronger in the halogen derivatives. That is why the boiling 
points of chlorides, bromides and iodides are considerably higher than those of the 

hydrocarbons of comparable molecular mass (Table 4.1).

Table 4.1: Comparison of boiling points of some halogenoalkanes

             

Chloromethane, bromomethane, chloroethane and some chlorofluoromethanes 
are gases at room temperature. Higher members are liquids or solids.
 
The attractions get stronger as the molecules get bigger in size. The pattern of 
variation of boiling points of different halides is depicted in Figure 4.1. For the same 
alkyl group, the boiling points of alkyl halides increase in the order: RF <RCl < RBr, < 
RI This is because with the increase in size and mass of halogen atom, the magnitude 
of Van der Waal forces increases.

2. Solubility
The solubility is the capacity of a substance to dissolve in a given solvent; in chemistry 
the most common solvent we refer to is water. It is a result of the interaction between 
the molecules of the substance, a solute, and the molecules of the solvent.

Polar molecules can interact with water molecules, but the attractive forces set 
up between water molecules and molecules concerned are not as strong as the 
hydrogen bonds present in water. Halogenoalkanes therefore, although they 
dissolve more than alkanes, are only slightly soluble in water.

3. State
The state of matter is the physical appearance of that matter: solid, liquid and 
gaseous. 

Chloromethane, bromomethane, chloroethane and chloroethene are colourless 
gases at room temperature and pressure. The higher members are colourless liquids 
with a sweet pleasant smell.

4. Density
The density is a measure of the quantity of matter by volume unit. Cotton wool is 
less dense than sand because if you compare the quantity of matter cotton wool 
and sand contained in for instance you find that there more matter
 in sand than in cotton wool.

The density of halogenoalkanes increases in the order RCl < RBr < RI, since the 
atomic weight of halogens increases in order Cl < Br < I. Iodo, bromo and polychloro 

derivatives are denser than water but chloro derivatives are less dense than water.

Checking up 4.3

1. Arrange each set of compounds below in order of increasing boiling 
     points and explain why.
     a. Bromomethane, tribromomethane, chloromethane, 
          dibromomethane.
     b. 1-chloropropane, 2-chloro-2-methylpropane, 1-chlorobutane.
2. Explain the origin of the difference between the boiling temperatures of 

    the following compounds:

               

4.4. Preparation methods of halogenoalkanes

  1. From alkenes and alkynes

  Activity 4.4.1

           1. Give the product for each of the following chemical reaction.

                                   

         2. Identify the class of the products of the reactions above.

    Halogenoalkanes can be prepared by a reaction of alkenes or alkynes with:

i. hydrogen halides 
   Addition of hydrogen halide to alkenes, gives alkyl halides as the products. The 

   orientation in the addition reaction is described by Markovnikov’s rule (see alkenes). 

                              

                

               ii. Halogens

                  

               

             2. From alcohols

When ethanol reacts with potassium bromide in the presence of concentrated 
sulphuric acid, bromoethane is formed. The reactions that took place in flask A are 

the following.

            

  In this reaction the hydroxyl group –OH is replaced with a bromine atom.
Halogenoalkanes are also obtained from alcohols using other reagents such as 

phosphorus halides. 

                                

                        

           3. From alkanes

          Activity 4.4.2

Observe the set up below and answer to the following questions.
Record your answers and discuss. 
                       
                         Picture 4.2: preparation of halogenoalkanes

a. Observe carefully the picture and suggest the product of the reaction in 
     flask A.
b. What is the role of: 
i. Concentrated sulphuric acid? 

ii. Water in the picture above?

Direct halogenation of alkanes in the presence of ultraviolet light
gives alkyl halides and a hydrogen halide.

Example:

4. From aldehydes or ketones

                 

                

Checking up 4.4:

1. Complete the following chemical reactions :

             

2. Give the reagents and conditions needed to make the following 
     compounds from 1-bromopropane:

a) propan-1-ol, b) propene.

4.5. Chemical properties

         Activity 4.5.1 

To investigate some reactions of halogenoalkanes
To of dilute sodium hydroxide solution in a test tube, 
add 5 drops of 1-bromobutane and gently warm the mixture.
Carefully smell the product. 
Neutralize the solution with dilute nitric acid. Acidify the solution by adding 5 
more drops of nitric acid.
Then add 5 drops of silver nitrate and observe. Write down your observations.
Write the equation of the reactions that take place. What is the role of sodium 
hydroxide in this experiment?  
When 1-bromobutane reacts with dilute sodium hydroxide solution, a product 
with a sweet alcoholic smell is formed. That indicates that an alcohol is formed. 
The formation of a pale yellow precipitate on addition of silver nitrate indicates 
the presence of bromide ions. That means the carbon-bromine bond has been 
heterolytically broken (the bromine atom takes the whole bonding electron pair). In 
other words the bromine atom has been replaced by hydroxide ions. Thus, sodium 
hydroxide provides the OH- ion which replaces bromine atom which leaves as a 
bromide ion. As OH is a nucleophile (Lewis base) this reaction is called nucleophilic 

substitution.

       

The order of reactivity for the same alkyl group is such that iodides > bromides > 

chlorides >fluorides.”>”, “are more reactive than “

            

The greater the electronegativity of the halogen, the greater the separation of charges 
on the carbon and the halogen atoms, hence the stronger the bond. Therefore the 
reaction is fastest with Iodoalkane because iodine is less electronegative compared 
to bromine and Cl. Hence it will have weak C-I bond unlike that of C-Cl which will be 
very strong due to the strong electronegativity of the chlorine atom. Hence bond 

energies below are due to the above reason.

      

  Because the carbon atom attached to the halogen atom is deprived of its electron, it 
carries a partial positive charge . Thus when electron rich substrates 
called nucleophiles, approach the carbon atom, the halogen atom leaves as a halide 
ion. Hence alkyl halides undergo nucleophilic substitution reaction, also written 

as SN.

     1. Nucleophilic substitution reaction:

          

a. Reaction with aqueous alkali: 

when alkyl halides are refluxed with aqueous alkali, or moist silver oxide, alcohols 

are produced through substitution of the halogen by hydroxide ion.

This reaction is also called “hydrolysis”
     

 Example:     

Note: Tertiary alkyl halides react by   mechanism, i.e. the mechanism where 
the first step is the self ionization forming a carbonium ion (carbocation), an alkyl 
radical that has lost its electron and bear a positive charge on the carbon, which 
immediately adds the nucleophile 

Secondary alkyl halides however react by either  mechanism depending 

on the condition of the reaction while primary alkyl halides react by  (see below).

: Unimolecular Nucleophilic Substitution that takes place in two steps; the 

    reaction rate is determined by the concentration of one molecule.

                   

   Example of  reaction: hydrolysis of tertiary alkyl halides with sodium 

   hydroxide.

                         

  The hydrolysis can also take place when water alone is added to tertiary alkyl halides. 

   In this case water molecules act as nucleophiles.  

                       

     Mechanism: 

Step 1: Self ionization of the alkyl halides to a stable carbonium ion and a halide ion. 

This is the slowest step of the reaction hence it is the rate determining step.

                    

          Step 2: Attack by incoming nucleophile. This is a fast reaction.

                

The potential energy (P.E )against reaction co-ordinate for hydrolysis of tertiary alkyl 
halides is as below (Fig.4.2). Potential energy is the energy stored in chemical bonds 
of a substance, or the energy of an object due to its position.

The diagram below shows that the products formed have lower energy than the 

reactants, this indicates a favorable situation for the reaction to occur spontaneously.

                          

                                                      Figure 4.2:  potential energy diagram

 The potential energy of the system initially increases because the energy is required 
to break C-X bond; but when the stable carbocation is formed, energy is released 
and the potential energy decreases a bit. As the carbocation and the nucleophile 
(OH-) require a minimum energy (activation energy) to collide efficiently, the P.E 
rises again until the transition state is reached, where the carbon-oxygen is being 
formed. When this bond is completely formed, energy is released and the potential 

energy decreases.

: Bimolecular Nucleophilic Substitution that takes place in one step; the 
   reaction rate depends on the concentration of Nu-
  and the concentration of R-X. 
In this mechanism, contrary to mechanism, the intermediate state also called 
“activated complex” comprises both the leaving group and the entering group: in 

the reaction below, the leaving group is X whereas the entering group is Nu. 

       

b. Reaction with sodium alkoxides

Treatment of alkyl halides with sodium alkoxides produces ethers 

(Wiliamson synthesis)

           

c. Reactions with silver salt of carboxylic acid
  When alkyl halides are refluxed with silver salt of carboxylic acid, esters are formed:

(d) Reaction with potassium cyanide
  When alkylhalides are refluxed with KCN, in presence of an alcohol, alkyl nitriles are 

  produced.

        

   Example:

        

Note: This reaction is of practical importance in organic synthesis
             because it is used to increase the length of a carbon chain.

     e. Reaction with silver nitrite

When alkyl halides are refluxed with silver nitrite, a mixture of a nitro alkane and 
alkyl nitrite are obtained as the products. The difference between the two products 
is in the bonds between the nitrite and the alkyl: 
in nitro alkane and C-ONO in alkyl nitrite.

The two products can be separated by fractional distillation.
       
f. Reaction with ammonia and amines

Reaction of alkyl halide with concentrated ammonia produces a mixture of amines
       
The alkyl amine produced can then react with a molecule of alkyl iodide to produce 

a series of substituted amines as shown in the reactions below:

           

Example of reaction: hydrolysis of primary alkyl halides with sodium hydroxide

            
The transition state shows partial C-O bond formation and partial C-I bond cleavage.

    Energy change diagram during the reaction is as below:

                  

The P.E of the system initially increases along AB curve because energy is required 
to break C-I bond; but when C-O is formed, energy is released and this is shown by 
the curve BC. Since energy released by the formation of C-O bond is greater than the 
energy required to break C-I, the products end up with a lower energy compared 
to the reactant and this is favourable for the reaction to occur. At B a maximum 
P.E is reached when C-I bond is partially broken and C-O bond is partially formed. 

This state is called the transition state or activated complex. The energy barrier, 
Ea, which must be overcome in order that the transition state is reached, is called 
the activation energy of the reaction. The P.E of the system then falls along BC 

releasing energy due to the formation of C-O bond. 

Primary alkyl halides prefer  reaction because of the unstable nature of the 
intermediate or the activated complex formed in  mechanism, the primary 

carbonium ion,

                      Table: 4. 1: Summary of alkyl halides reactions.

               

           2) Elimination reactions 

                Activity 4.5.2

1. What is meant by elimination reaction?
2. Can halogenoalkanes undergo elimination reactions? Explain.
3. a. What are the products of an elimination reaction in halogenoalkanes? 
    b. What specific name is given to this reaction?
   c. What are the conditions and reagent required for this type of 

       reaction?

An elimination reaction is where a saturated organic compound loses an atom 
or group of atoms to form an unsaturated organic compound. Elimination is the 
opposite of addition reaction.

Alkyl halides when boiled with alcoholic potassium hydroxide form alkenes by 

elimination reaction. Hence the alkyl halide loses a molecule of the hydrogen halide.

               

Note: Elimination reaction usually occurs in competition with substitution reaction. 
So when chloroethane is treated with a solution of potassium hydroxide two organic 

products are formed depending on the conditions of the reaction.

           

Ethene is formed by elimination reaction while diethyl ether is formed by substitution 

reaction.

Ether is formed by  mechanism in which  is acting as nucleophile.
while Ethene is formed by elimination reaction in which  is acting as base?

         Mechanism:  
              

 Because the two molecules are involved i.e. 
 and  the reaction is bimolecular and since the alkyl halide loses a mole of HCl 
the reaction is called elimination. Hence the reaction is a bimolemolecular elimination (E2).

In competition between and E2 in primary or secondary alkyl halides, the nature 
of the product formed depends on the solvent, temperature, and structure of the 

halide. 

Elimination is favoured by use of high temperature and a strong base e.g alcohol 

instead of water.

For tertiary alkyl halides, elimination occurs by E1 mechanism. In the mechanism, 

the tertiary alkyl halide undergoes ionization first and then later loses a proton.

              

3) Wurtz reaction

Alkyl halides with sodium metal to give alkanes.

These are compounds in which more than one halogen atom is present. There are 
two main types of dihalides.
Gem dihalides: This is where two halogen atoms are attached to the same carbon 

atom.

Example

Vicinal dihalides: Here the two halogen atoms are on adjacent carbon atoms

Example

The reactions of dihalogenoalkanes are similar to those of monohalogenoalkanes 

but require more reagents.

Examples      

Elimination reaction with excess hot alkali produces alkynes

                     

Checking up 4.5

1. Give the structural formula of the main product of each of the following 

      reactions:

             

2. Halogenoalkanes undergo nucleophilic substitution reaction. Discuss 
     this statement.
3. a. What is a nucleophile? Give two examples.
b. Why do nucleophiles attack halogenoalkanes?
c. What two types of reaction are in competition when a 
    halogenoalkane reacts with a nucleophile? Name two products 
    which can be formed from 1-bromopropane by these reactions.

4. 2- Chloro-2-methyl propane reacts with aqueous sodium hydroxide to 
         form 2-methylpropan-2-ol.
a. Draw what should be the energy diagram for the reaction.
b. Write the mechanism for the reaction.
c. (i) Sketch an energy diagram for the reaction of aqueous sodium 
          hydroxide and chloromethane.
    (ii) Outline the mechanism for the reaction.
d. outline the mechanism for the reaction

4.6. Chemical test for the presence of halogenoalkanes

Activity 4.6:

Put 2mL of ethanol into each of 4 test tubes labelled A-D. A is the control tube and 
therefore no alkyl halides are to be added. To B, add 3-4 drops of 1-chlorobutane. 
To C, add 3-4 drops of 1-bromobutane and to D, add 3-4 drops of 1-iodobutane 
using Pasteur pippete. Stopper the tubes and place them in a hot water bath at 
about 
and leave for a few minutes to equilibrate. Working quickly add about 
1mL of silver nitrate solution to each tube. Start the stopwatch and shake the 
tubes to ensure complete mixing.
a. Record your observations 
b. Make a comment about comparison of the reactions of the three 

     halogeno alkane

Halogenoalkanes can be identified due to some tests. The following Table illustrate 

some chemical tests of halogenoalkanes.

Table 4. 2: Chemical test for halogenoalkanes

                     

Checking up 4.6

Given two samples A and B. You carry out the test for haloalkanes and get the 
following results: A form a pale yellow precipitate and B form a white precipitate. 
Which sample represents and which one represents 

Write chemical equations to justify your answer.

4.7. Uses of halogenoalkanes and dangers associated with CFCs

        Activity 4.7

1. Do you know CFCs? If yes what do you know about them?
2. Do CFCs affect directly our health in our daily life? If yes explain how.
3. What are the dangers posed by CFCs?
4. What solutions do you propose or have been proposed to the problem 

     of CFCs

Solvents: 

    

Medicine:

  is used in anesthesia

Agriculture:

• DDT: Dichloro diphenyl trichloroethane is used as insecticide- DDT. Colorless 
chemical pesticide, dichlorodiphenyltrichloroethane, used to eradicate 
disease-carrying and crop-eating insects. It was first isolated in Germany 
in 1874, but not until 1939 did the Swiss Nobel Prize-winning chemist Paul 
Müller recognize it as a potent nerve poison on insects. The product is banned 

in Rwanda. Below is the structure of DDT.

                                                

Home: Refrigeration, perfumes, etc…

Halogenoalkanes which have boiling temperatures just below room temperature 
can easily be liquefied by a slight increase in pressure. Halogenoalkanes containing 

chlorine and fluorine and no hydrogen are Chlorofluorohydrocarbons. Examples are 

They are usually called chlorofluorocarbons or CFCs. In 
addition to having low boiling temperatures, they are non-flammable, odorless, 

stable, non-toxic and solvents.

 • CFCs appeared to be ideal for use as fluids in refrigerators and as solvents 
in aerosol sprays, they were developed in the 1920s as what appeared to 
be ideal replacements for liquid ammonia and liquid Sulphur dioxide, which 
were formerly used as fluids in refrigerators and air-conditioning units. Being 
good solvents, they were also ideal as the solvents in aerosol sprays.
Aerosols were used to dispense insecticides, hairsprays, perfumes and deodorants, 
window-cleaning, polishes, waxes and laundry products. As more and more 
uses were found for these remarkable compounds, CFCs became big business,
with hundreds of thousands of tones being produced yearly. Now they 

are being phased out. These stable, non-toxic compounds are dangerous!

•Their high stability has turned out to be a problem, during all the time that 
the use of CFCs was increasing, no-one thought about what would happen to 
the gases in the atmosphere. Because of their lack of reactivity and insolubility
in water, there is no natural process for removing CFCs. In fact they drift up 
into the stratosphere where ultraviolet light causes photolysis, i.e. a reaction 
cause by light. The chlorine radicals formed in photolysis take part in a chain 

reaction which converts ozone into oxygen.

           

As you can notice, the chain of reaction above results in the decomposition of ozone 
,which does have the capacity to absorb, and stop dangerous UV from reaching the 
Earth into ordinary oxygen. This can be avoided if and only if human activities send 

no CFCs in the atmosphere.

• This reduce the thickness of the ozone layer. Reactions (a) to (c) form a chain. 
This is why one chlorine radical from one CFC molecule can destroy thousands 

of ozone molecules.

And what can be done?

• Replacements for CFCs have been found, because of concern over the 
decrease in the ozone layer, many nations have agreed to cut down the use 
of CFCs. Alternative compounds are already in production. Hydrohalocarbons 
contain at least one hydrogen atom per molecule. The C-H bond can be 
attacked by HO• radicals in the lower atmosphere and the compounds do not 

reach the upper atmosphere. Hydro halocarbons include

• Hydrochlorofluorocarbons,  used in blowing plastics
    foam and used in air-conditioners
• Hydrofluorocarbons, used in air-conditioners and refrigerators. 

HCFs cause no damage to the ozone layer, although they are greenhouse gases.

Checking Up 4.7

1. State four industrial uses of the halogenoalkanes. Why do fluoroalkanes 

     find special uses?

 4.8. End Unit Assessment

1. Which of the following is NOT a halogenoalkane compound:
    a. Tribromo benzene 
    b. 3-iodohexane
    c. 2-chloro-3-methylpentane

    d. 2-bromopentane

2. Choose from a list of words and fill in the missing words in the text 
     below
        Halogenoalkane, iodine, alkyl halide, haloarene, thyroxine
        …………………………..compounds are compounds in which the 
       halogen atoms like chlorine, bromine, ………… or fluorine are 
       attached to a hydrocarbon chain or an aromatic ring. When the halogen 
       atom is attached to a hydrocarbon chain the compound is called an 

        …………………… or ………………………..

3. Answer by True or False
      a. Chloroform is employed as a solvent in paint remover.
     b. Iodoform was used earlier as an antiseptic.
    c. Methyl chloride, methyl bromide, ethyl chloride and some 
         chlorofluoromethanes are gases at room temperature.
    d. The objects which are non-superimposable on their mirror image 
         (like a pair of hands) are said to be chiral and this property is known 
        as chirality. While the objects, which are superimposable on their 

        mirror images are called achiral.

  e. (chloroform): is used as insecticide
  f. DDT: Dichloride diphenyl trichloroethane is used as anesthesia
 g. Halogenoalkanes therefore, although they dissolve more than 
    alkanes, are only slightly soluble in water.
 h. Halogenoalkanes undergo nucleophilic substitution reactions in 
    which the halogen atom is replaced by a nucleophile.
 i. Elimination reaction is where a saturated organic compound loses 
   an atom or group of atoms attached to form unsaturated organic 

   compound.

4. Name the following halides according to IUPAC system and classify 

   them as primary, secondary or tertiary halogenoalkanes

                       

1. Write the structures of the following organic halogen compounds.
     a. 2-chloro-3-methylpentane
    b. 2-chloro-2-methylpropane
   c. 2,3-dichlorobutane
   d. 2-bromo-4-chloropentane

   e. 1,1,2-trichloropropane

2. Why do bromoalkanes react more readily than chloroalkanes?
3. Why does 1-bromopropane react with nucleophiles but propane does 

not?

4. Write the equations for the preparation of 1-iodobutane from
 (a) 1-butanol, (b)1-chlorobutane, (c) but-1-ene

5. Write the structure of the major organic product 

    in each of the following reactions:

             

6. Arrange the compound of each set in order of reactivity towards SN2 
displacement:
a. 2-bromo-2-methylbutane, 1-bromopentane, 2-bromopentane
b. 1-bromo-3-methylbutane, 2-bromo-2-methylbutane, 3-bromo-2-
    methyl butane
c. 1-bromobutane, 1-bromo-2,2-dimethylpropane, 1-bromo-2-

    methyl butane, 1-bromo-3-methylbutane.

7. a) There are four strucural isomers of molecular formula C4

 The formulae of two of these isomers are given.

      

 i. Draw the remaining two structural isomers.

 ii. Give the name of isomer 2

b) All four structural isomers of  undergo similar reactions with ammonia

i. Give the name of the mechanism involved in these reactions.
ii. Draw the structural formula of the product formed by the reaction 
    of isomer 2 with ammonia.
iii. Select the isomer of molecular formula C4
       that would be most 
      reactive with ammonia. State the structural feature of your chosen 
      isomer that makes it the most reactive of the four isomers.
iv. The elimination of HBr from Isomer 1 produces two structural 
      isomers, compounds A and B.
v. Give the reagents and conditions required for this elimination 
    reaction.
c) Ethene, C2H4, reacts with bromine to give 1,2-dibromoethane.
i. Give the name of the mechanism involved.
ii. Show the mechanism for this reaction.
vi. Give the structural formulae of the two isomers, A and B formed by 

      elimination of HBr from isomer 1.