The equation for the reaction that will take place for each of the halogenoalkanes:
Halogenoalkane + Hydroxide ion Alcohol + Halide ion
C4H9X + OHֿ C4H9OH + Xֿ
As you can see above the number of carbons and hydrogen’s remains the same this is because it has an alkane skeleton of butane and it is only X which is the halogen which is changed. The oxygen atom is highly electronegative compared to a hydrogen atom and as the oxygen has two lone pairs of electrons the bond is polar, where the Oxygen atom has a slightly negative charge and the Hydrogen atom has a slightly positive charge. There is also a difference in electronegativities between a Carbon-Halogen bonds where the halogen is more electronegative. This also makes this bond polar meaning the halogen has a slightly negative charge and the carbon has a slightly positive charge. Due to this polarity the electron deficient carbon atom is attacked by the electron-rich species which in this case is the nucleophile hydroxide ion. As the hydroxide ion and carbon atom are attracted to each other, the carbon-halogen bond breaks so halogen leaves as a halide ion.
As halogenoalkanes are insoluble in water the reaction will have to carried out in the presence of ethanol, which acts as a solvent. This creates a hydrolysis reaction.
There are no colour changes during this reaction as halogenoalkanes are covalently bonded; they give no precipitate of silver halide. Therefore silver nitrate will be used, and gradually the halogen can be detected determined by the colour of the precipitate. If the precipitate is white then this means that chlorine is present, if cream bromine is the halogen present and finally if yellow it is iodine which is present in the halogenoalkane.
Prediction
From my knowledge of the atoms involved in the reaction I am able to make a few predictions as to how the reactions should appear. If you were to take polarity into consideration you would expect the chloro-compound to be the most reactive. Polarity predicts that the C-Cl bond would be the most reactive and the C-I bond the least reactive. This is due to the polarity where the chlorine atom is the most electronegative (the ability of an atom to attract electrons towards itself in a covalent bond). The electronegativity decreases as we go down the group. This is because the shared electrons in the covalent bonds they form get further away from the nucleus. At the same time these electrons feel the same shielded nuclear charge. The shielded nuclear charge is the nuclear charge after allowing for the shielding of the inner electrons, the shielded nuclear charge of the halogens is +7. Below is a table showing the electronegative values of the three halogens.
This means that the polarity of the C-Cl bond is more polar than C-Br and finally the C-I is the least polar bond. From this the reactivity of the halogens involved is as follows; Cl>Br>I.
Bond enthalpy however predicts that the reverse will be true where the iodo-compound will be the most reactive. Below is a table showing the bond enthalpies of the carbon-halogen bond.
Bond enthalpy which is the energy required to break a bond shows us that iodo-compound will have the weakest bond while the bromo-compounds have stronger bonds and then finally chloro-compounds have the strongest bonds. This is due to the atomic radius of the atoms where iodine is the biggest and chlorine has the smallest atomic radius. The outermost electrons feel less of the nuclear charge as the atoms have a larger atomic radius; this means that it easier for other atoms to react with the electrons of the bigger atom. From this the reactivity of the carbon-halogen bond is as follows; C-I>C-Br>C-Cl. Bond enthalpy governs the rate of reactions of the halogenoalkanes.