An organic compound (A) with molecular formula C8H8O forms an orange-red precipitate with 2,4-DNP reagent and gives yellow precipitate on heating with iodine in the presence of sodium hydroxide. It neither reduces tollens' or fehlings' reagent, nor does it decolourise bromine water or Baeyer's reagent. On drastic oxidation with chromic acid, it gives a carboxylic acids (B) having molecular formula C7H6O2. Identify the compounds (A) and (B) and explain the reactions involved.
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● Answers -
Compound A => Butan-2-one
Compound B => Butan-2-ol
● Explanation -
Molecular formula C4H8O indicates X being either aldehyde or ketone.
As A doesn't reduce Tollen's reagent, it must be a ketone.
Postive iodoform test indicates methyl group at one end of ketone A.
Then compound A must be CH3-CO-C2H5 i.e. butan-2-one.
Butan-2-one (A) on reduction with LiAlH4 gives butan-2-ol (B).
CH3-CO-C2H5 ---> CH3-CH(OH)-C2H5
Butan-2-ol (B) on hydrolysis with conc.H2SO4 gives but-2-ene.
CH3-CH(OH)-C2H5 ---> CH3-CH=CH-CH3
Hope this helps you...
● Answers -
Compound A => Butan-2-one
Compound B => Butan-2-ol
● Explanation -
Molecular formula C4H8O indicates X being either aldehyde or ketone.
As A doesn't reduce Tollen's reagent, it must be a ketone.
Postive iodoform test indicates methyl group at one end of ketone A.
Then compound A must be CH3-CO-C2H5 i.e. butan-2-one.
Butan-2-one (A) on reduction with LiAlH4 gives butan-2-ol (B).
CH3-CO-C2H5 ---> CH3-CH(OH)-C2H5
Butan-2-ol (B) on hydrolysis with conc.H2SO4 gives but-2-ene.
CH3-CH(OH)-C2H5 ---> CH3-CH=CH-CH3
Hope this helps you...
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