The carbon-halogen bond and physical properties

The bond that decides everything

The carbon-halogen bond is the part of a haloalkane that reacts. Every halogen is more electronegative than carbon. Pauling's scale gives carbon 2.55, fluorine 3.98, chlorine 3.16, bromine 2.96 and iodine 2.66. So the halogen pulls the shared electrons towards itself. Carbon gets a small positive charge, and the halogen gets a small negative charge.

We write this as $\delta ^{+}\text{C}\text{-}\text{X}\delta ^{-}$. A nucleophile (a species with a pair of electrons to give) is attracted to the carbon with the positive charge. That is the beginning of every reaction in the next course.

Down the group, the bond gets longer and weaker

The atoms get bigger from fluorine to iodine. The carbon-halogen bond gets longer, and a longer bond is easier to break.

• $\text{CH}_{3}\text{-}\text{F}$: length $139\ \text{pm}$, bond enthalpy $452\ \text{kJ/mol}$, dipole moment $1.85\ \text{D}$.
• $\text{CH}_{3}\text{-}\text{Cl}$: length $178\ \text{pm}$, bond enthalpy $351\ \text{kJ/mol}$, dipole moment $1.87\ \text{D}$.
• $\text{CH}_{3}\text{-}\text{Br}$: length $193\ \text{pm}$, bond enthalpy $293\ \text{kJ/mol}$, dipole moment $1.81\ \text{D}$.
• $\text{CH}_{3}\text{-}\text{I}$: length $214\ \text{pm}$, bond enthalpy $234\ \text{kJ/mol}$, dipole moment $1.62\ \text{D}$.

NCERT prints slightly different dipole moments, such as $1.860\ \text{D}$ for chloromethane. Reference tables disagree by a few hundredths of a debye and by a few kilojoules per mole in the enthalpies. The trends are the same in all of them, and the trends are what the exams test.

Look at the dipole moments. Fluorine is the most electronegative, yet chloromethane has the largest dipole moment. A dipole moment is charge times distance. The C-F bond has a bigger charge separation but is much shorter, so the two effects nearly cancel. The practice below works out the charge separation from the dipole moment and the bond length.

Physical properties

Boiling points. A haloalkane boils higher than the alkane it comes from. Two attractions hold the molecules together: dipole-dipole forces from the polar bond, and London dispersion forces, which grow with the number of electrons and the size of the molecule. The boiling point rises as we go from $\text{R}\text{-}\text{F}$ to $\text{R}\text{-}\text{I}$ and as the chain gets longer.

Set the chain to 3 carbons and slide the halogen from fluorine to iodine. The boiling point climbs steeply, because the molar mass and the number of electrons rise. Then fix the halogen and change the chain. Each extra carbon adds about 30 K.

Branching lowers the boiling point. A branched molecule is more compact, so it has less surface to touch its neighbours. The four isomers of $\text{C}_{4}\text{H}_{9}\text{Br}$ show this.

• 1-Bromobutane: 375 K.
• 2-Bromobutane: 364 K.
• 1-Bromo-2-methylpropane: 364 K.
• 2-Bromo-2-methylpropane: 346 K.

Compared with an alcohol, a haloalkane of similar molar mass boils lower. Bromoethane boils at 312 K and ethanol at 351 K. Alcohols form hydrogen bonds. Haloalkanes cannot.

Density. Chloroalkanes with one halogen are lighter than water. Bromo and iodo compounds are heavier. The densities of the three 1-halobutanes at $20\ ^\circ\text{C}$ are $0.886\ \text{g/mL}$ for the chloride, $1.276\ \text{g/mL}$ for the bromide and $1.617\ \text{g/mL}$ for the iodide. Density also rises as more halogen atoms are added: $\text{CH}_{2}\text{Cl}_{2}$ is $1.33\ \text{g/mL}$, $\text{CHCl}_{3}$ is $1.49\ \text{g/mL}$ and $\text{CCl}_{4}$ is $1.59\ \text{g/mL}$.

Solubility. Haloalkanes dissolve only slightly in water. To dissolve, a molecule would have to break the hydrogen bonds between water molecules. The new attractions between haloalkane and water are weaker, so little is gained. They dissolve well in organic solvents, such as ether, benzene and alcohol, where the attractions are of the same strength.

Practise with new numbers

Try it in Python

The first part of the program works out the charge separation for the four methyl halides, as in the practice. The second part uses the boiling points of three chloroalkanes to predict the next one.

Common mistakes

• Saying fluorine gives the biggest dipole moment because it is most electronegative. Check the bond length too. Chlorine gives the largest value.
• Thinking the C-I bond is strongest because iodine is the largest atom. It is the weakest of the four.
• Predicting that a haloalkane dissolves in water because it is polar. It cannot form hydrogen bonds, and it dissolves very little.
• Forgetting the effect of branching. Among isomers, the more branched halide boils lower.

Remember these

• The C-X bond is polar. Carbon is $\delta^+$ and the halogen is $\delta^-$.
• From C-F to C-I the bond gets longer and weaker. The enthalpy falls from about 452 to 234 kJ/mol.
• Boiling points rise with the molar mass and with the number of electrons, and fall with branching.
• Haloalkanes are not soluble in water, but dissolve in organic solvents. Bromo and iodo compounds are denser than water.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. Which has the higher boiling point, 1-chloropropane or 1-iodopropane? Why?

2. Why is 2-bromo-2-methylpropane lower boiling than 1-bromobutane?

3. Why is bromoethane much less soluble in water than ethanol?

4. Order the bond enthalpies of $\text{CH}_{3}\text{-}\text{Cl}$, $\text{CH}_{3}\text{-}\text{Br}$ and $\text{CH}_{3}\text{-}\text{I}$ from highest to lowest.

References

• NCERT, Chemistry Part II for Class XII, the chapter Haloalkanes and Haloarenes, on the nature of the C-X bond and physical properties.
• Morrison and Boyd, Organic Chemistry, the chapter on alkyl halides, for the data tables.
• OpenStax, Organic Chemistry, Chapter 10 (Organohalides).