Formation of Alkenes from Alcohols by Dehydration - ELIMINATION
Pass the vapour of the alcohol over heated aluminium oxide, or mix the alcohol with concentrated sulphuric acid
or concentrated phosphoric acid and heat to 180oC. The alcohol is dehydrated. In effect, an H atom and an –
OH group are removed from adjacent carbon atoms, leaving a double bond. This process of removing atom or
atoms from adjacent carbon atoms leaving a double bond is the exact opposite of addition and is called
elimination.
The first step is the protonation of the OH group followed by the loss of a water molecule leaving a
carbocation. This carbocation then loses a proton to produce the alkene. So a proton is regenerated and the
reaction is therefore acid catalysed.
The mechanism of this reaction IS required.
Cyclohexanol forms cyclohexene and ethanol forms ethene.
+ H+
CH3 CH2 OH
CH2 = CH2
+
H2O
Some alcohols can form more than one alkene. Butan2ol forms but1ene and but2ene because, during
the second stage where a proton leaves the carbocation to form a double bond, the proton can leave from one of
two carbon atoms and consequently form one of two isomers of butene.
Forms but–1–ene
forms but–2–ene
Of the three alcohols with the formula C4 H9 OH, the tertiary alcohol, 2–methylpropan–2–ol has the fastest rate
of dehydration, with butan–2–ol next and, finally butan–1–ol last. This is because the tertiary alcohol produces
a tertiary carbocation intermediate that is more stable than the secondary which is more stable than the primary
carbocation intermediate formed by butan–1–ol.
1