Courses
Courses for Kids
Free study material
Offline Centres
More
Store

# In the following reaction,${C_2}{H_2}\xrightarrow[{HgS{O_4}/{H_2}S{O_4}}]{{{H_2}O}}X \rightleftharpoons C{H_3}CHO$What is $X$ ?(a) $C{H_3}C{H_2}OH$ (b) ${H_3} - O - C{H_3}$(c) $C{H_3}C{H_2}CHO$(d) $C{H_2} = CHOH$

Last updated date: 13th Jun 2024
Total views: 373.5k
Views today: 9.73k
Verified
373.5k+ views
Hint: The reactant alkyne reacts with ${H_2}O$ in presence of $HgS{O_4}/{H_2}S{O_4}$ and forms a product. So, we have to think about the reaction in which alkyne reacts with ${H_2}O$ in presence of $HgS{O_4}/{H_2}S{O_4}$ and then tautomerism because the compound ( $\rightleftharpoons$ this sign shows that this compound is tautomerising itself) is changing itself.

So, we have to start with the reaction in which alkyne reacts with ${H_2}O$ in presence of $HgS{O_4}/{H_2}S{O_4}$. So, that reaction is Alkyne Hydration.
Alkyne hydration:
So, we try to learn these reaction with an example,
So, let’s take an example,
$C{H_3} - C \equiv C - H + {H_2}O \to C{H_3}COC{H_3}$
Mechanism:

So, in case of unsymmetrical alkynes, the addition of water in the alkyne in according to the Markovnikov’s Rule, which states that the $O -$ atom will attach with the carbon having less number of $H -$ atoms and the two $H -$ atoms will attach with the carbon having greater number of $H -$ atoms.
But In these cases, the alkyne is symmetric.
So, we don’t have to follow any Markovnikov’s Rule,
And the reaction will be,
$H - C \equiv C - H\xrightarrow[{HgS{O_4}/{H_2}S{O_4}}]{{{H_2}O}}C{H_3}CHO$
Now, the $C{H_3}CHO$ will tautomerise itself,
Tautomerism: When an aldehyde having one hydrogen atom, adjacent to the carbonyl group (i.e, $\alpha$ carbon), this hydrogen can move to the oxygen atom of the carbonyl group and the double bond between the carbonyl group is shifted to the adjacent to the carbonyl group (i.e, $\alpha$ carbon). This type of movement of bonds is called tautomerism and the compounds are called tautomers.
So, $C{H_3}CHO$ will tautomerise itself as
${C_2}{H_2}\xrightarrow[{HgS{O_4}/{H_2}S{O_4}}]{{{H_2}O}}C{H_2} = CHOH \rightleftharpoons C{H_3}CHO$
Hence, the correct option is (D) $LiF$.
Note: While alkyne is reacting with ${H_2}O$ in presence of $HgS{O_4}/{H_2}S{O_4}$ , we have to use the Markovnikov’s Rule when the alkyne is un-symmetric and when $\alpha$ carbon is present, we must have to think once about the tautomerism.