Halogenoalkanes (1) Identity and formation

1. Halogenoalkane formation:

How are halogenoalkanes formed?

In this previous post, I described the free radical halogenation of alkanes with specific reference to methane.

So  the overall reaction is:

CH₄    +    Cl₂      =        CH₃Cl      +    HCl  

in the presence of ultraviolet light

Halogenoalkanes can also form from alkenes through the addition of bromine or hydrogen halides.

In two previous posts, I discussed both electrophilic addition of bromine to an alkene (here) and the Markovnikov addition of hydrogen bromide to propene (here).  

Both these reactions create new halogenoalkenes so:

C₂H₄      +    Br₂       =        C₂H₄Br₂

1,2-dibromoethane

CH₃CH=CH₂      +    HBr      =       CH₃CHBrCH₃

2-bromopropane

2. How do haloalkane boiling points compare with alkane boiling points?

Halogenoalkanes are a homologous series of compounds with similar chemical and physical properties like alkanes or alkenes.

They have a gradation of boiling points that are higher than the alkanes.

	methyl-	ethyl-	propyl-	butyl-	pentyl-
	CH3-	CH3CH2-	CH3CH2CH2-	CH3(CH2)3-	CH3(CH2)4-
Alkane	-161.7	-88.6	-42.1	-0.5	36.1
Fluoro	-78.4	-37.7	-2.5	32.5	62.8
Chloro	-24.2	12.3	46.6	78.4	107.8
Bromo	3.6	38.4	71	101.6	129.6
Iodo	42.4	72.3	102.5	130.5	157

You can see three features of the haloalkanes boiling point in this chart.

a) the haloalkanes have boiling points higher than the corresponding alkane.

The reason is not hard to find.

The molar mass of the haloalkane is higher than the corresponding alkane so van der Waals forces are greater because more electrons in the haloalkane.

Also for those haloalkanes where the halogen is more electronegative than hydrogen the greater polar character of the R—X bond means stronger polar forces between haloalkane molecules than between alkane molecules.

b) there is a gradation in boiling point within a haloalkane series. 

The reason here is the lengthening alkyl chain has more electrons so van der Waals forces are greater the longer the chain.

c) for a given alkyl chain: say the butyl- derivatives, the boiling point increases even though the halogen electronegativity decreases from fluorine to iodine!

The increasing number of electrons and increasing van der Waals forces with the longer alkyl chain have a greater effect than bond polarity, overcoming any changes in bond polarity. 

3) Haloalkane and C–X bond properties:

What are the distinctive features of the C—X bond?

[Note: X is often put for a generalized halogen atom.]

    a) Bond length and strength

Bond energies decrease from fluorine to iodine because the C—X bond is lengthening.

The halogen atom increases in radius from fluorine to iodine lengthening the C‑X bond.

average	C—F	C—Cl	C—Br	C—I
bond energy	485	330	275	215

(kJ/mol)

Halogen atom                   64                     99                       114                   133

 radii (nm)

       

     b) Bond polarity

Halogen electronegativity values are higher than that of carbon (except for iodine) so from fluorine to iodine the C—X bond polarity decreases:  

Electronegativity values:    Carbon: 2.5   Fluorine 4.0

                                                                 Chlorine 3.0

                                                                 Bromine 2.8

                                                                 Iodine     2.5

We can visualize this effect like this:

      CH3—F         >          CH3—Cl        >          CH3—Br        >          CH3—I 

      δ+       δ-                    δ+       δ-                    δ+       δ-                   non-polar

      2.5      4.0                  2.5      3.0                  2.5      2.8                  2.5      2.5

Difference:    1.5                             0.5                             0.3                              0.0

Similar to what we said above, for those haloalkanes where the halogen is more electronegative than carbon, the greater polar character of the R—X bond means stronger polar forces between haloalkane molecules and higher boiling points. 

As the halogen electronegativity increases so does its "electron-pulling" power making the molecule more polar increasing the value of its dipole moment.

But in the case of iodomethane, where the electronegativities of both carbon and iodine are the same, then the boiling point is the result of van der Waals forces between molecules only.  

All courses tend to focus on the three middle halogens so you never hear much about the fluoroalkanes until you start studying the effects of refrigerants on the ozone layer in the upper atmosphere.

We will be just looking at the chloro, bromo and iodo compounds (the "frodo" compound is for another time and age!)

The Mole (4) Using the mole to determine equation stoichiometry

Ok so what is a chemical equation?

If a chemical formula reveals the mole ratio of elements in  compound then a chemical equation reveals the ratio of reactants to each other and to the products of a chemical change.

This mole ratio is called the equation stoichiometry.

You always wanted know what that word meant, didn't you!!

Here's an example, a simple one:

2H2O2     =      2H2O   +     O2

You can see here that the mole ratio or stoichiometry is given by the large numbers in front of the chemical formulas.

That's 2 moles hydrogen peroxide decomposing to form 2 moles water and a mole of oxygen gas.

Each large number represents the number of moles (or the amount) of that compound in the balanced symbol equation for the chemical change.

(This blog post describes how to balance combustion equations)

Question is how can anyone measure an equation stoichiometry?

Here is one approach using the mass of a precipitate formed in a chemical reaction.

You can find this example here

But we need a simpler example to illustrate the idea.

Suppose we react a known mass of magnesium with excess hydrochloric acid and measure the volume of hydrogen evolved.

Here's the apparatus you could use.

Then we have a way of comparing the mass of magnesium used with the volume of hydrogen formed.

Here are a typical set of results that could be obtained:

Magnesium (g)       Hydrogen (cm3)

0.24                         240
0.12                         120
0.06                          60
0.10                         100

 If these mass results are converted into amounts in moles then this is what we find:

Amt Magnesium (mol)   Amt Hydrogen (mol)
0.01                                 0.01
0.005                               0.005
0.0025                             0.0025
0.00417                           0.00417

The mole ratio of magnesium to hydrogen in this chemical change is 1:1

We can begin to construct the chemical equation for the reaction between magnesium and hydrogen like this:

Mg    +    x HCl    =      XMgCl2     +     H2

It follows that x is then 2 and X  is 1 and the completed chemical equation for this reaction is

Mg    +     2HCl     =   MgCl2       +     H2

Calculations both at GCSE and Advanced level often require you to calculate masses of reacting substances or products formed.

These questions often rely on you using the molar quantities in the chemical equation for the reaction

Here is a typical example for you:

What mass of carbon dioxide is formed from the complete thermal decomposition of 25g of sodium hydrogen carbonate  NaHCO3?


Write the equation:                            2NaHCO3     =      Na2CO3    +     H2O     +     CO2

Put molar quantities to the symbols:   168g                      106g                 18g               44g                  
(you need to calculate the molar
mass (Mr) of each compound)

From the question. identify the           168g                                                                    44g
compounds involved

Work out what 1g would give               1g                                                                      44/168

Scale up by the quantity in                    25g                                                                  (25*44)/168
the question

                                                                                  Answer 6.55g carbon dioxide
Some example problems:

Zinc chloride can also be prepared in the laboratory by the reaction between zinc and hydrogen chloride gas.
Zn + 2HCl =  ZnCl2 + H2
An impure sample of zinc powder with a mass of 5.68 g was reacted with
hydrogen chloride gas until the reaction was complete. The zinc chloride produced had a mass of 10.7 g.
Calculate the percentage purity of the zinc metal. Give your answer to 3 significant figures.

People who have a zinc deficiency can take hydrated zinc sulfate (ZnSO4.xH2O) as a dietary supplement.
A student heated 4.38 g of hydrated zinc sulfate and obtained 2.46 g of anhydrous zinc sulfate.

Use these data to calculate the value of the integer x in ZnSO4.xH2O Show your working. 

The Mole (3) Using the mole to determine the simplest (empirical) formulae
The Mole (1) Relative Atomic Mass and the Mass Spectrometer
Mole (2) Amount of substance and Molar Mass