How to convert acetone to methanol ?
Answers
Step 1:
Acetone is reduced to propan-2-ol in the presence of a compound called Lithium Alumanuide (LiAlH_4LiAlH4 ).
The reaction involved is as follows:
\mathrm{CH}_{3} \mathrm{COCH}_{3} \stackrel{\mathrm{LiAlH}_{4}}{\longrightarrow} \mathrm{CH}_{3} \mathrm{CHOHCH}_{3}CH3COCH3⟶LiAlH4CH3CHOHCH3
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}CH3−CHOH−CH3+H2SO4→ΔCH3−CH=CH2
Step 3:
Propene is treated with two compounds namely hydrogen bromide (HBr) and hydrogen peroxide (H_2O_2H2O2 ).
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 rCH3−CH=CH2⟶HBr/H2O2CH3−CH2−CH2−Br
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}CH3−CH2−CH2−Br⟶AqKOHCH3−CH2−CH2−OH
Step 5:
Oxidation of propan-1-ol in the presence of KMnO_4KMnO4 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 HCH3−CH2−CH2−OH⟶[o]/KMnO4CH3−CH2−COOH
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 HCH3−CH2−NH2⟶ NaNO2 /HClCH3−CH2−OH