Q.64. 'x' is the product which is obtained by the hydrolysis of prop-1-yne in the presence of mercur

Q: What is the correct answer for this exam Chemistry Q64?

The correct answer is B. Sol. For 'x': Prop-1-yne ($H_3C-C\equiv CH$) undergoes hydration (Kucherov reaction) in the presence of $HgSO_4/dil. H_2SO_4$ to give acetone ($H_3C-CO-CH_3$). So, $x = H_3C-CO-CH_3$.For 'y': Ethanenitrile ($H_3C-C\equiv N$) reacts with methyl magnesium bromide ($CH_3MgBr$) followed by hydrolysis to give acetone ($H_3C-CO-CH_3$). So, $y = H_3C-CO-CH_3$.The question asks for the product obtained fr

Q: Q.64. 'x' is the product which is obtained by the hydrolysis of prop-1-yne in the presence of mercuric sulphate under dilute acidic medium at 333 K. 'y' is the product which is obtained by the…

The correct answer is B. Sol. For 'x': Prop-1-yne ($H_3C-C\equiv CH$) undergoes hydration (Kucherov reaction) in the presence of $HgSO_4/dil. H_2SO_4$ to give acetone ($H_3C-CO-CH_3$). So, $x = H_3C-CO-CH_3$.For 'y': Ethanenitrile ($H_3C-C\equiv N$) reacts with methyl magnesium bromide ($CH_3MgBr$) followed by hydrolysis to give acetone ($H_3C-CO-CH_3$). So, $y = H_3C-CO-CH_3$.The question asks for the product obtained fr

Q: What are the options for this this exam Chemistry question?

A. 2-Methylpent-4-en-3-one · B. 4-Methylpent-3-en-2-one · C. 4-Methylpent-1-ene · D. 2-Methylpent-3-one

Question

Q.64. 'x' is the product which is obtained by the hydrolysis of prop-1-yne in the presence of mercuric sulphate under dilute acidic medium at 333 K. 'y' is the product which is obtained by the reaction of ethane nitrile with methyl magnesium bromide in dry ether followed by hydrolysis. IUPAC name of product obtained from 'x' and 'y' in the presence of barium hydroxide followed by heating is:

FirstInTest · Accepted Answer

Sol. For 'x': Prop-1-yne ($H_3C-C\equiv CH$) undergoes hydration (Kucherov reaction) in the presence of $HgSO_4/dil. H_2SO_4$ to give acetone ($H_3C-CO-CH_3$). So, $x = H_3C-CO-CH_3$.For 'y': Ethanenitrile ($H_3C-C\equiv N$) reacts with methyl magnesium bromide ($CH_3MgBr$) followed by hydrolysis to give acetone ($H_3C-CO-CH_3$). So, $y = H_3C-CO-CH_3$.The question asks for the product obtained from 'x' and 'y' in the presence of barium hydroxide followed by heating. This is an Aldol condensation reaction between two molecules of acetone. $2 H_3C-CO-CH_3 \xrightarrow{Ba(OH)_2, \Delta} H_3C-CO-CH_2-C(CH_3)_2-OH \xrightarrow{-	ext{H}_2O} H_3C-CO-CH=C(CH_3)_2$.The product is 4-methylpent-3-en-2-one.

Answer

2-Methylpent-4-en-3-one

Answer

4-Methylpent-3-en-2-one

Answer

4-Methylpent-1-ene

Answer

2-Methylpent-3-one

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