
How is propanone converted into (i) \[propan - 2 - ol\] (ii) \[2 - methylpropan - 2 - ol\]
Answer
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Hint:Propanone is an organic compound with a chemical formula ${C_3}{H_6}O$. It is the simplest and smallest member of the ketone family. Its common name is acetone. Acetone is a colorless, highly volatile, and flammable liquid. It has a very pungent odour.
Complete answer:
(i) Propanone to \[propan - 2 - ol\]
Conversion of Propanone to \[propan - 2 - ol\] is done by reduction. Propanone is a carbonyl compound with functional groups.
We have to convert propanone with $CO$ functional group to alcohol with $ - OH$ functional group. When we reduce carbonyl compounds in the presence of a reducing agent, they get converted to alcohol. So we will treat propanone with a reducing agent like lithium aluminum hydride $(LiAl{H_4})$ and Sodium borohydride $(NaB{H_4})$. On treatment with $(LiAl{H_4})$ propanone will be reduced to \[propan - 2 - ol\] . It is also called isopropyl alcohol. The reaction will be:
(ii) Propanone to \[2 - methylpropan - 2 - ol\]
The conversion of Propanone to \[2 - methylpropan - 2 - ol\]is a two-step process. So First we will treat the Ketone with Grignard reagent i.e. $R - MgX$ where $R$ is the methyl group and $X$ is the halogen. After that, we have to do the acidic hydrolysis of the compound. When we react Propanone with $C{H_3} - MgBr$, the reagent will break into two parts $\mathop {C{H_3}}\limits^ - $ and $\mathop {MgBr}\limits^ + $, so oxygen being the electronegative atom will attract the shared electron pairs towards itself creating polarity in the bond. The $\mathop {C{H_3}}\limits^ - $ electrophile will attack there and forms the bond with the carbonyl carbon and $\mathop {MgBr}\limits^ + $ will bond with oxygen forming an intermediate. In the second step, the acidic hydrolysis of the intermediate is done to replace the $O - MgBr$ group with $ - OH$. The reaction of the conversion will be:
Note:
\[propan - 2 - ol\] or isopropyl alcohol is a colorless, flammable organic compound that has a very strong odor. It is used in the manufacturing of antiseptics, disinfectants, and a wide variety of household chemicals.
Complete answer:
(i) Propanone to \[propan - 2 - ol\]
Conversion of Propanone to \[propan - 2 - ol\] is done by reduction. Propanone is a carbonyl compound with functional groups.
We have to convert propanone with $CO$ functional group to alcohol with $ - OH$ functional group. When we reduce carbonyl compounds in the presence of a reducing agent, they get converted to alcohol. So we will treat propanone with a reducing agent like lithium aluminum hydride $(LiAl{H_4})$ and Sodium borohydride $(NaB{H_4})$. On treatment with $(LiAl{H_4})$ propanone will be reduced to \[propan - 2 - ol\] . It is also called isopropyl alcohol. The reaction will be:
(ii) Propanone to \[2 - methylpropan - 2 - ol\]
The conversion of Propanone to \[2 - methylpropan - 2 - ol\]is a two-step process. So First we will treat the Ketone with Grignard reagent i.e. $R - MgX$ where $R$ is the methyl group and $X$ is the halogen. After that, we have to do the acidic hydrolysis of the compound. When we react Propanone with $C{H_3} - MgBr$, the reagent will break into two parts $\mathop {C{H_3}}\limits^ - $ and $\mathop {MgBr}\limits^ + $, so oxygen being the electronegative atom will attract the shared electron pairs towards itself creating polarity in the bond. The $\mathop {C{H_3}}\limits^ - $ electrophile will attack there and forms the bond with the carbonyl carbon and $\mathop {MgBr}\limits^ + $ will bond with oxygen forming an intermediate. In the second step, the acidic hydrolysis of the intermediate is done to replace the $O - MgBr$ group with $ - OH$. The reaction of the conversion will be:
Note:
\[propan - 2 - ol\] or isopropyl alcohol is a colorless, flammable organic compound that has a very strong odor. It is used in the manufacturing of antiseptics, disinfectants, and a wide variety of household chemicals.
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