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How could you convert butanoic acid into the following compounds?
Write each step showing the reagents needed.
(a) 1-Butanol (b) 1-Bromobutane (c) Pentanoic acid
(d) 1-Butene (e) Octane

Short Answer

Expert verified

The conversion of butanoic acid into the following compounds can be explained.

Step by step solution

01

Conversion of butanoic acid into 1-Butanol

The hydrolysis with acids can happen when treated with LiAlH4, there is a reduction of butanoic acid to 1-Butanol. The steps involved are,

CH3CH2CH2CO2H2.H3O+1LiAlH4CH3CH2CH2CH2OH

02

Conversion of butanoic acid into 1-Bromobutane

The hydrolysis with acids can happen when treated with LiAlH4, there is a reduction of butanoic acid to 1-Butanol. Treat it with PBr3, there is a conversion of alcohol into 1-bromobutane. The steps involved are,

CH3CH2CH2CO2H2.H3O+1LiAlH4CH3CH2CH2CH2OHPBr2CH3CH2CH2CH2Br
03

Conversion of butanoic acid into Pentanoic acid

The hydrolysis with acids can happen when treated with LiAlH4, there is a reduction of butanoic acid to 1-Butanol. Treat it with PBr3, there is a conversion of alcohol into 1-bromobutane. The butanenitrile hydrolysis with aqueous acids results in pentanoic acid formation.The steps involved are,

CH3CH2CH2CO2H2.H3O+1LiAlH4CH3CH2CH2CH2OHPBr2CH3CH2CH2CH2Br2.H3O+1.NaCNCH3CH2

04

Conversion of butanoic acid into 1-Butene

The hydrolysis with acids can happen when treated with LiAlH4, there is a reduction of butanoic acid to 1-Butanol. Treat it with PBr3, there is a conversion of alcohol into 1-bromobutane. Treatment of 1-bromobutane with KOH, eliminates HBr to get 1-butene.The steps involved are,

CH3CH2CH2CO2H2.H3O+1LiAlH4CH3CH2CH2CH2OHPBr2CH3CH2CH2CH2BralcoholicKOHCH3CH2CH

05

Conversion of butanoic acid into Octane

The hydrolysis with acids can happen when treated with LiAlH4, there is a reduction of butanoic acid to 1-Butanol. Treat it with PBr3, there is a conversion of alcohol into 1-bromobutane. The conversion of 1-bromobutane into lithiumdibutylcopper in addition to octane. The steps involved are,

CH3CH2CH2CO2H2.H3O+1LiAlH4CH3CH2CH2CH2OHPBr2CH3CH2CH2CH2Br2Ll2CH3CH2CH2CH2-Li+

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