Question

The number of double bonds present in the isohypsic transformation product of allylic alcohol (Z) is _ _ _. (in integer)

Hint:

Isohypsic isomerization of allylic alcohols usually converts an alkene alcohol into an aldehyde or ketone without changing the total number of double bonds.

Answer 1

Correct Answer: 1

Step 1: Identify the functional group present in compound \(Z\).
The given compound is an allylic alcohol because the hydroxyl group (\(-OH\)) is attached to a carbon adjacent to a carbon-carbon double bond.

Step 2: Understand isohypsic transformation.
Isohypsic transformation is a redox-neutral rearrangement in which the oxidation states of carbon atoms remain unchanged overall.
In allylic alcohols, this transformation generally converts the alcohol into a carbonyl compound through double bond migration.

Step 3: Recall the allylic alcohol isomerization pattern.
A typical allylic alcohol transformation is:
\[ \text{Allylic alcohol} \rightarrow \text{Aldehyde/Ketone} \] The carbon-carbon double bond shifts and forms a carbonyl group.

Step 4: Analyze the change in bonding.
Initially, the molecule contains one
\[ C=C \] double bond.
After isomerization, the alkene double bond disappears and a carbonyl bond
\[ C=O \] is formed.

Step 5: Count the total number of double bonds in product.
The product contains one carbonyl double bond:
\[ C=O \] Thus, the total number of double bonds remains
\[ 1 \]

Step 6: Verify the transformation.
Since one double bond is replaced by another double bond during isomerization, the product contains exactly one double bond.

Step 7: Final conclusion.
Therefore, the number of double bonds present in the isohypsic transformation product is
\[ \boxed{1} \]