Unsaturated molecules such as alkenes really only undergo one reaction - addition reactions. You may recall from your gcse science that small molecules can "add" across the carbon carbon double bond (C=C) to form new saturated molecule. The addition of bromine and an explanation of the mechanism for these addition reactions was discussed on the page covering simple addition reaction. This page covers the addition of hydrogen halides and sulfuric acid across a C=C, we will cover addition to unsymmetrical alkenes.
The hydrogen halides are all gases at room temperature, they include hydrogen chloride (HCl), hydrogen bromide (HBr) and hydrogen iodide (HI). Hydrogen chloride and hydrogen bromide both have a permanent dipole due to large differences in the electronegativity values between hydrogen and these halogens. This aids in an easy reaction in the gas phase with the carbon cabon double bond in an alkene. This reaction produces halogenalkanes, a mechanism for the gas phase reaction is shown below:
In the reaction shown above the partially positively charged hydrogen atom in the hydrogen bromide molecule acts as the electrophile, while the negatively charged bromide ion acts as a nucleophile when it attacks the carbocation. Hydrogen chloride, bromide and iodide are all acidic gases and will readily dissolve in water to form acidic solutions. So if the reaction with alkenes is carried out in aqueous conditions rather than in the gas phase then the hydrogen ion (H+) will act as the electrophile and the halide ion as the nucleophile to produce the halogenalkane.
There are numerous methods of preparing alcohols from alkenes. One method which is often used in industry is called direct hydration. This involves adding a molecule of water across the carbon carbon double bond (C=C) to produce an alcohol. This method requires a phosphoric acid (H3PO4>) catalyst, a high temperature (570K) and a high pressure (65 atmospheres). The mechanism for this reaction is shown below. The mechanism is similar to the other addition reactions to alkenes and proceeds via three steps:
Alcohols can also be produced from the reaction of concentrated sulfuric acid with an alkene. The mechanism for this reaction is shown below, hopefully you will notice it is very similar to previous reactions of alkenes, that we have:
The diols are a family of alcohols
which contain two hydroxyl groups (R-OH) per molecule. Perhaps the most widely used
ethane-1,2-diol (CH2OHCH2OH) which is used as antifreeze and as a coolant in car engines.
Alkenes can be oxidised to form 1,2-diols using cold dilute solutions of potassium permangante as an oxidising agent. The permanganate ion , MnO4- is a powerful oxidising agent due to the fact that the central manganese atom has an oxidation state of +7. The potassium permanganate can be used in neutral, acidic or alkaline conditions to oxidise the alkene to form the diol. The permanganate ion forms a purple solution when dissolved in water. If it is used to oxidise an alkene the colour changes depend on whether the reaction is carried out in acidic, alkaline or neutral solutions.
If an alkene such as ethene (CH2=CH2) is bubbled through an acidified solution of the permanganate ion (MnO4-) then the purple permanagnate ion is reduced to form the pale pink Mn2+ ion, which is so pale as to be almost colourless. While the alkene is oxidised to form the diol.We can show this redox reaction as:
The above two half-equations are clearly not balanced since there are 5 electrons needed to reduce the permanagante ion and only 2 electrons are releasd by the oxidation of ethene. So to balance it simply multiply the top equation by x2 and the bottom equation by x5 to give:
Whereas if ethene gas is bubbled through an alkaline solution of potassium permanganate we have as a final equation for this redox reaction:
The permanganate ion adds to one side of the carbon carbon double bond (C=C) in an alkene, this is called syn addition, we can show this as:
However the complete equations for the oxidation of alkenes by potassium permanganate are NOT required at A-level and we can simply show the equation for the oxidation reaction as: