Question

how to convert acetone to ethanol?

Answer 1

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Answer 2

Steps to convert acetone to ethanol

Step 1:

• Acetone is reduced to propan-2-ol in the presence of a compound called Lithium Alumanuide ($LiAlH_4$).
• The reaction involved is as follows:

$\mathrm{CH}_{3} \mathrm{COCH}_{3} \stackrel{\mathrm{LiAlH}_{4}}{\longrightarrow} \mathrm{CH}_{3} \mathrm{CHOHCH}_{3}$

Step 2:

• Propan-2-ol is heated with concentrated sulfuric acid due to which water molecule from it is removed and yields propene.
• The reaction involved is as follows:

$\mathrm{CH}_{3}-\mathrm{CHOH}-\mathrm{CH}_{3}+\mathrm{H}_{2} \mathrm{SO}_{4} \stackrel{\Delta}{\rightarrow} \mathrm{CH}_{3}-\mathrm{CH}=\mathrm{CH}_{2}$

Step 3:

• Propene is treated with two compounds namely hydrogen bromide (HBr) and hydrogen peroxide ($H_2O_2$).
• It leads to the formation of a compound called 1-Bromopropane as per the following equation:

$C H_{3}-C H=C H_{2} \stackrel{H B r / H_{2} O_{2}}{\longrightarrow} C H_{3}-C H_{2}-C H_{2}-B r$

Step 4:

• 1-Bromopropane is made to react with aqueous KOH due to which the bromine group gets eliminated as KBr and 'OH' group gets attached to it and forms propan-1-ol.
• The equation involved in the reaction is as follows:

$\mathrm{CH}_{3}-\mathrm{CH}_{2}-\mathrm{CH}_{2}-\mathrm{Br} \stackrel{\mathrm{AqKOH}}{\longrightarrow} \mathrm{CH}_{3}-\mathrm{CH}_{2}-\mathrm{CH}_{2}-\mathrm{OH}$

Step 5:

• Oxidation of propan-1-ol in the presence of $KMnO_4$ yields respective acid which is propanoic acid.
• The equation involved is:

$C H_{3}-C H_{2}-C H_{2}-O H \stackrel{[o] / K M n O_{4}}{\longrightarrow} C H_{3}-C H_{2}-C O O H$

Step 6:

• The addition of ammonia to propanoic acid and heating leads to the formation of propanamide.
• When propanamide is applied with Halfmann bromide reaction it leads to the formation of ethyl amine.

Step 7:

• When ethyl amine is treated with sodium nitrite and HCl, it gives ethanol as the product.
• The reaction involved in the reaction is as follows:

$C H_{3}-C H_{2}-N H_{2} \stackrel{\text { NaNO2 } / H C l}{\longrightarrow} C H_{3}-C H_{2}-O H$

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