Alkene compounds are the hydrocarbons which consist of at least one carbon-carbon double bond. Various synthesis methods for the synthesis of alkene are described herein.
Hi Friends, in this article we will discuss about few basic reactions in organic chemistry which are used for the synthesis of alkenes starting from alkyl halides.
Key words: Alkene, Alkyl halide, Dehydrohalogenation, Dehalogenation, Carbocation, Zaitsev product, Hoffmann product.
Alkene compounds are the hydrocarbons which consist of at least one carbon-carbon double bond. They are important intermediates in organic synthesis because they can be converted in to alkyl halides or alcohols by addition reactions and carbonyl compounds by oxidation reactions. There are two important and basic reactions that are useful for synthesis of alkene compounds. They are;
1. Dehydrohalogenation of alkyl halides
2. Dehalogenation of vicinal dihalides
Let’s
see each reaction in detail;
1.
Dehydrohalogenation
of alkyl halides
Dehydrohalogenation
is a elimination reaction. In this reaction, hydrogen and a halogen gets
eliminate to produce carbon-carbon double bond. The reaction takes place by E1
and E2 mechanisms. In organic synthesis for the preparation of alkene compounds
second order elimination (E2) reactions are preferred because E1 reactions are
known to give multiple products due to rearrangements. Nature of the alkyl
halide is also play important role here, usually secondary and tertiary alkyl
halides gives excellent yield of the elimination product.
E1 Mechanism
E1 reaction is unimolecular reaction it means that rate of the reaction is completely depend upon alkyl halide. In this reaction tertiary alkyl halide reacts faster than secondary alkyl halide. E1 reaction consists of two steps;
1. First step is formation of carbocation intermediate.
2. Second
step is deprotonation by weak base
or solvent.
The
reaction follows Zaitsev rule and provides highly substituted alkene product.
Example 1:
1-iodo-1-methylcyclohexane undergoes dehydrohalogenation when heated with methanol. The reaction follows E1 mechanism. In this case after formation of carbocation intermediate, methanol acts as base and causes deprotonation to form alkene product.
Since
the reaction proceeds through carbocation intermediate; it can form multiple
products. Also carbocation can undergo rearrangements to form thermodynamically
stable products.
Example 2:
3-bromo-2-methylpentane undergoes dehydrohalogenation reaction when heated with water.
E2 Mechanism
E2-reaction
is bimolecular elimination reaction that means rate of the reaction is depend
upon concentration of alkyl halide and base. In this reaction strong base is
required. The mechanism consists of single step. In this mechanism H and halide
should be anti periplaner to each
other. The strong base abstract H from the carbon which is next to halide
bonded carbon. In this reaction also Zaitsev product is favored.
Example 3:
Reaction of ter-butyl bromide with sodium methoxide gives alkene product. Here methoxide anion abstract H and elimination of bromide takes place to form alkene compound.
Choice of Base for Elimination Reaction
As we know that for E2 elimination reaction there is need of strong base. Outcome of the reaction is highly substituted alkene. In general alkoxide (RO-) bases are better for dehydrohalogenation reaction. Unhindered bases like ethoxide / methoxide gives highly substituted alkene product. It is known as Zaitsev product. The bulky base like tert-butoxide gives least substituted alkene product. It is called as Hoffmann product.
Dehydrohalogenation of germinal
dihalides can give alkene product. The reaction follows E2 mechanism and strong
base is required for elimination. Bases like potassium hydroxide (KOH) or sodium
amide (NaNH2) are generally used in this reaction.
Example 4:
If there are two H on carbon which
is next to halide bonded carbon; then reaction doesn’t stop at alkene stage and
second dehydrohalogenation gives alkyne product.
Example 5:
2. Dehalogenation of vicinal dihalides
Vicinal
dihalides react with zinc (Zn) in acetic acid to produce alkene compound. Here
the halogen atoms should be “anti” to each other.
Mechanism
The reaction mechanism consists of
two steps;
1. First
step is incersion of metal in to carbon-halogen bond.
2. Second step is elimination of halide to form alkene prouct.
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