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!)