Question

What are X and Z in the following sequence of reactions?

• A. \(\text{X} = \text{H}_3\text{C}-\text{C}\equiv\text{CH}\), \(\text{Z} = \text{PhCHO}\)
• B. \(\text{X} = \text{H}_3\text{C}-\text{C}\equiv\text{C}-\text{CH}_3\), \(\text{Z} = \text{PhCHO}\)
• C. \(\text{X} = \text{Ph}-\text{C}\equiv\text{CH}\), \(\text{Z} = \text{CH}_3\text{CHO}\)
• D. \(\text{X} = \text{Ph}-\text{C}\equiv\text{C}-\text{CH}_3\), \(\text{Z} = \text{CH}_3\text{CHO}\)

Hint:

To find the components of an aldol condensation product, mentally "cleave" the double bond \(\text{C}=\text{C}\):
- Change the \(\text{C}\) on the side without the carbonyl back to a carbonyl (\(\text{C}=\text{O}\)), which gives \(\text{PhCHO}\).
- Add two hydrogens to the other \(\text{C}\) to get the original ketone, \(\text{CH}_3\text{COCH}_3\).

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
The question asks us to identify the alkyne reactant \(\text{X}\) and the carbonyl compound \(\text{Z}\) in a two-step reaction sequence that produces an \(\alpha,\beta\)-unsaturated ketone, specifically 4-phenylbut-3-en-2-one (\(\text{Ph}-\text{CH}=\text{CH}-\text{CO}-\text{CH}_3\)).

Step 2: Key Formula or Approach:
Let's work backward from the final product:
1. The final product is \(\text{Ph}-\text{CH}=\text{CH}-\text{CO}-\text{CH}_3\). This conjugate system is typically formed via a crossed aldol condensation between a non-enolizable aldehyde (like benzaldehyde, \(\text{PhCHO}\)) and a methyl ketone (like acetone, \(\text{CH}_3\text{COCH}_3\)) under basic conditions.
2. The intermediate \(\text{Y}\) must undergo this aldol condensation with reactant \(\text{Z}\).
3. \(\text{Y}\) is synthesized by hydration of an alkyne \(\text{X}\) using Kucherov's reaction conditions (\(\text{HgSO}_4\), \(\text{dil. H}_2\text{SO}_4\), water).

Step 3: Detailed Explanation:
Let's analyze the steps sequentially:
-

Step 1: Analyzing the Kucherov reaction on alkyne X:
- If we choose propyne (\(\text{H}_3\text{C}-\text{C}\equiv\text{CH}\)) as \(\text{X}\):
- Hydration occurs via Markovnikov's addition of water across the triple bond:
\[ \text{H}_3\text{C}-\text{C}\equiv\text{CH} \xrightarrow{\text{Hg}^{2+}/\text{H}^+, \text{H}_2\text{O}} [\text{H}_3\text{C}-\text{C}(\text{OH})=\text{CH}_2] \]
- The enol intermediate tautomerizes rapidly to form the more stable ketone, acetone (\(\text{Y}\)):
\[ [\text{H}_3\text{C}-\text{C}(\text{OH})=\text{CH}_2] \rightleftharpoons \text{H}_3\text{C}-\text{CO}-\text{CH}_3\text{ (Acetone, Y)} \]
-

Step 2: Analyzing the Aldol Condensation with Z:
- Acetone (\(\text{Y}\)) is treated with reactant \(\text{Z}\) in the presence of dilute \(\text{NaOH}\) and heat.
- To obtain \(\text{Ph}-\text{CH}=\text{CH}-\text{CO}-\text{CH}_3\), the active \(\alpha\)-hydrogen of acetone must attack the carbonyl carbon of benzaldehyde (\(\text{Z} = \text{PhCHO}\)):
\[ \text{PhCHO (Z)} + \text{CH}_3-\text{CO}-\text{CH}_3\text{ (Y)} \xrightarrow{\text{dil. NaOH}} \text{Ph}-\text{CH}(\text{OH})-\text{CH}_2-\text{CO}-\text{CH}_3 \]
- Heating causes dehydration of the aldol product to yield the conjugated \(\alpha,\beta\)-unsaturated ketone:
\[ \text{Ph}-\text{CH}(\text{OH})-\text{CH}_2-\text{CO}-\text{CH}_3 \xrightarrow{\Delta} \text{Ph}-\text{CH}=\text{CH}-\text{CO}-\text{CH}_3 + \text{H}_2\text{O} \]
This matches the product shown in the reaction. Thus, \(\text{X}\) is propyne and \(\text{Z}\) is benzaldehyde.

Step 4: Final Answer:
The reactants are \(\text{X} = \text{H}_3\text{C}-\text{C}\equiv\text{CH}\) and \(\text{Z} = \text{PhCHO}\), which matches option (A).