\(X\) is the product obtained by hydrolysis of prop-1-yne in the presence of mercuric sulphate under dilute acidic medium. \(Y\) is obtained by the reaction of ethanenitrile with methyl magnesium bromide in dry ether followed by hydrolysis. IUPAC name of product obtained from \(X\) and \(Y\) in presence of barium hydroxide followed by heating is :
Step 1: Understanding the Question:
The problem involves a sequence of reactions to identify two intermediates 'x' and 'y', and then finding the final product formed when they react together under specific conditions. Step 2: Detailed Explanation: 1. Finding 'x':
Hydration of prop-1-yne (\(CH_{3}-C \equiv CH\)) with \(HgSO_{4}/H_{2}SO_{4}\) (Kucherov Reaction):
\[ CH_{3}-C \equiv CH + H_{2}O \xrightarrow{Hg^{2+}/H^{+}} [CH_{3}-C(OH)=CH_{2}] \xrightarrow{\text{tautomerize}} CH_{3}-CO-CH_{3} \]
Product 'x' is Propanone (Acetone).
2. Finding 'y':
Reaction of Ethane nitrile (\(CH_{3}CN\)) with Methyl magnesium bromide (\(CH_{3}MgBr\)):
\[ CH_{3}-C \equiv N + CH_{3}MgBr \rightarrow CH_{3}-C(CH_{3})=NMgBr \xrightarrow{H_{3}O^{+}} CH_{3}-CO-CH_{3} \]
Product 'y' is also Propanone (Acetone).
3. Final Reaction:
'x' and 'y' (both Acetone) are treated with \(Ba(OH)_{2}\) (a base) followed by heating. This is an Aldol Condensation reaction.
Step A: Aldol formation
\[ 2CH_{3}-CO-CH_{3} \xrightarrow{Ba(OH)_{2}} (CH_{3})_{2}C(OH)-CH_{2}-CO-CH_{3} \text{ (4-hydroxy-4-methylpentan-2-one)} \]
Step B: Dehydration (heating)
\[ (CH_{3})_{2}C(OH)-CH_{2}-CO-CH_{3} \xrightarrow{\Delta} (CH_{3})_{2}C=CH-CO-CH_{3} + H_{2}O \]
The structure is \((CH_{3})_{2}C=CH-COCH_{3}\).
IUPAC Name: Numbering starts from the carbonyl side. C1:methyl, C2:keto, C3:double bond, C4:methyl.
Name: 4-Methylpent-3-en-2-one (commonly known as Mesityl oxide). Step 3: Final Answer:
The final product is 4-Methylpent-3-en-2-one.
