Animated Solution for Chemistry - Carbonyl Compounds: Which one of the following carbonyl compounds cannot be prepared by addition of water on an alkyne in the presence of ${\text{HgSO}}_4$ and ${\text{H}}_2{\text{SO}}_4$?

Imagine you have an alkyne and you want to transform it into a carbonyl compound. The classic way to do this is by adding water across the triple bond. But alkynes are stubborn; they don't react with water easily. This is where our chemical catalysts, mercuric sulfate (${\text{HgSO}}_4$) and sulfuric acid (${\text{H}}_2{\text{SO}}_4$), come to the rescue. This specific hydration is famously known as the Kucherov Reaction.

When water adds to the alkyne, it strictly obeys Markovnikov's rule. The hydroxyl ($\text{-OH}$) group attaches to the more substituted carbon atom (the one with fewer hydrogens). This initial addition forms an enol (an alkene with an alcohol group). However, enols are notoriously unstable. They rapidly undergo a process called tautomerization, shifting a proton to form a much more stable carbon-oxygen double bond, resulting in a ketone.

Let's break down the options to see which one defies this rule.

Option (a): Acetaldehyde (${\text{CH}}_3\text{CHO}$) If we start with the simplest alkyne, ethyne ($\text{H}-\text{C}\equiv \text{C}-\text{H}$), it is perfectly symmetric. Adding water gives vinyl alcohol, which tautomerizes directly into acetaldehyde. So, ethyne is the only alkyne that yields an aldehyde upon hydration.

Options (b) and (d): Methyl Ketones Cyclohexyl methyl ketone and butan-2-one are both methyl ketones. Any terminal alkyne (like cyclohexyl ethyne or but-1-yne) will undergo Markovnikov addition. The $\text{-OH}$ group goes to the internal carbon, and after tautomerization, you are guaranteed to get a methyl ketone. These are perfectly valid products.

Propanal

Option (c): Propanal (${\text{CH}}_3{\text{CH}}_2\text{CHO}$) Propanal is a three-carbon aldehyde. To synthesize it from an alkyne, you would have to start with propyne (${\text{CH}}_3-\text{C}\equiv \text{C}-\text{H}$). However, if you subject propyne to the Kucherov reaction, Markovnikov's rule dictates that the $\text{-OH}$ group attaches to the middle carbon.

This yields acetone, not propanal! To get propanal, you would need an anti-Markovnikov addition of water, which requires a completely different set of reagents, specifically hydroboration-oxidation (using ${\text{B}}_2{\text{H}}_6$ followed by ${\text{H}}_2{\text{O}}_2/{\text{OH}}^{-}$).

Therefore, propanal cannot be prepared using ${\text{HgSO}}_4$ and ${\text{H}}_2{\text{SO}}_4$.