Q17-67E Dehydration of trans-2-methylcyc... [FREE SOLUTION]

Chapter 17: Q17-67E (page 567)

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!)

Short Answer

Expert verified

The 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.

Step by step solution

Structure of trans-2-methylcyclopentanol

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 groupand the hydrogen at the fifth carbon is an anti-hydroxyl group.

Step2:Trans-2-methyl undergoes dehydration reaction with POCl3

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.