Question

The correct reagents to carry out the following conversion $HC \equiv CCH_2CH_2CH_3 \rightarrow CH_3-\overset{O}{\underset{||}C}-CH_2CH_2CH_3$

Answer 1

HgSO_4, H_2SO_4

Answer 2

The conversion is from 1-pentyne ($HC \equiv CCH_2CH_2CH_3$) to 2-pentanone ($CH_3-\overset{O}{\underset{||}C}-CH_2CH_2CH_3$). This is a hydration reaction of a terminal alkyne.

There are two main methods for the hydration of alkynes:

1. Acid-catalyzed hydration with Mercury(II) sulfate ($\text{HgSO}_4$) and Sulfuric acid ($\text{H}_2\text{SO}_4$):

   This reaction follows Markovnikov's rule, meaning the hydroxyl group ($\text{OH}$) adds to the more substituted carbon of the alkyne and the hydrogen adds to the less substituted carbon. For terminal alkynes, this typically leads to the formation of a ketone (except for acetylene, which gives acetaldehyde).

   For 1-pentyne:

   $HC \equiv CCH_2CH_2CH_3 \xrightarrow{\text{HgSO}_4, \text{H}_2\text{SO}_4, \text{H}_2O}$

   The $\text{OH}$ adds to C-2 and $\text{H}$ adds to C-1, forming an enol:

   $CH_2=C(OH)CH_2CH_2CH_3$ (enol form)

   This enol rapidly tautomerizes to the more stable keto form:

   $CH_2=C(OH)CH_2CH_2CH_3 \rightleftharpoons CH_3-\overset{O}{\underset{||}C}-CH_2CH_2CH_3$ (2-pentanone)

   This matches the desired product.

2. Hydroboration-oxidation ($\text{BH}_3 \cdot \text{THF}$ followed by $\text{H}_2\text{O}_2/\text{OH}^-$):

   This reaction follows an anti-Markovnikov addition, meaning the hydroxyl group ($\text{OH}$) adds to the less substituted carbon of the alkyne. For terminal alkynes, this leads to the formation of an aldehyde.

   For 1-pentyne:

   $HC \equiv CCH_2CH_2CH_3 \xrightarrow{1. \text{BH}_3 \cdot \text{THF}} \xrightarrow{2. \text{H}_2\text{O}_2/\text{OH}^-}$

   This would lead to an enol with the $\text{OH}$ on C-1:

   $HO-CH=CHCH_2CH_2CH_3$ (enol form)

   This enol would tautomerize to an aldehyde:

   $O=CH-CH_2CH_2CH_2CH_3$ (pentanal)

   This is not the desired product (2-pentanone).

Therefore, the correct reagents to carry out the conversion of 1-pentyne to 2-pentanone are $\text{HgSO}_4$ and $\text{H}_2\text{SO}_4$ (in the presence of water).

The reaction can be represented as:

$$\text{HC} \equiv \text{CCH}_2\text{CH}_2\text{CH}_3 \xrightarrow{\text{HgSO}_4, \text{H}_2\text{SO}_4, \text{H}_2\text{O}} \text{CH}_3-\overset{\text{O}}{\underset{||}{\text{C}}}-\text{CH}_2\text{CH}_2\text{CH}_3$$

Explanation:

The conversion of 1-pentyne to 2-pentanone requires the hydration of a terminal alkyne following Markovnikov's rule. Acid-catalyzed hydration using $\text{HgSO}_4/\text{H}_2\text{SO}_4$ achieves this, forming an enol intermediate that tautomerizes to the desired ketone. Hydroboration-oxidation would yield an aldehyde (pentanal) via anti-Markovnikov addition.