Q17-67E
Question
Dehydration of trans-2-methylcyclopentanol with POCl3 in pyridine yields predominantly 3-methylcyclopentene. Is the stereochemistry of this dehydration syn or anti? Can you suggest a reason for formation of the observed product? (Make molecular models!)
Step-by-Step Solution
VerifiedThe reaction is as follows:
If the molecule undergoes syn elimination then the proton at the second carbon is removed and forms a 1-methyl-1-cyclopentene.
But in the reaction, 3-methylcyclopentene is formed. Therefore, the compound undergoes anti-elimination.
The activation energy of a syn elimination is too high when compared to an anti-elimination and thus the anti-elimination reaction is favored in spite of the less stable product.
The structure of the compound trans-2-methylcyclopentanol is as follows:
In the above compound, the hydrogen at the second carbon is syn to a hydroxyl group and the hydrogen at the fifth carbon is an anti-hydroxyl group.
Trans-2-methyl undergoes a dehydration reaction with POCl3 in pyridine to form a 3-methylcyclopentene.
The reaction is as follows:
If the molecule undergoes syn elimination then the proton at second carbon is removed and forms a 1-methyl-1-cyclopentene.
But in the reaction 3-methylcyclopentene if formed. Therefore, the compound undergoes anti-elimination.
The activation energy of a syn elimination is too high when compared to an anti-elimination and thus the anti-elimination reaction is favored in spite of the less stable product.