Question

when trans-4-iodo cyclohexanol is treated with koh, the product loses the broad ir stretch centered at 3300 cm-1 and does not exhibit any optical activity. however, when trans-3- iodo cyclohexanol is treated with koh, the product is optically active and retains the broad stretch at 3300 cm-1 . draw the two starting materials and propose a mechanism for their reactions with koh that is consistent with these observations

Answer 1

When trans-4-iodo cyclohexanol is treated with KOH, the alcohol group is eliminated, leading to an alkene with no optical activity. However, trans-3-iodo cyclohexanol undergoes a substitution reaction, preserving the chiral center and maintaining optical activity and a broad IR absorption at 3300 cm-1.

Step-by-step explanation:

The observation that trans-4-iodo cyclohexanol treated with KOH loses the broad IR stretch at 3300 cm-1 suggests the conversion of an alcohol to an alkene, where the hydroxyl group is eliminated and a double bond is formed. This is consistent with the disappearance of the O-H stretch in the IR spectrum and the lack of optical activity indicating the product is not chiral. On the other hand, when trans-3-iodo cyclohexanol is treated with KOH and the product retains the broad stretch at 3300 cm-1 and is optically active, it suggests that the alcohol has undergone substitution rather than elimination, preserving the chiral center in the cyclohexanol molecule.