Question

Which of the following reactions with produce 1-butanol as one of the products? (Assume excess lithium aluminum hydride and water.) A 1) LiAlH4 +LiAIH, HO 2) HO 1) LiAlH4 +LiAlH, +HO 2) HO 1) LiAlH4 + LiAlH +HO 2) HO OH =0 1) LiAlH4 +LiAIH4 +HO 2) HO = 1) LiAlH4 +LiAIH4 +HO N(CH3)2 2) HO

Answer 1

Lithium aluminium hydride is a strong reducing agent and it reduces many functional groups like acid, ester, carbonyl groups, etc. A. Lithium aluminium hydride reduces a ketone to form a secondary alcohol. OH + LiAlH4 + H2O LAH converts aldehyde into primary alcohols. + LiAlH4 + H2O

LAH reduces the carboxylic acid and covert it into primary alcohol. + LiAlH4 + H2O - OH LAH converts the given acid chloride into a primary alcohol. + + Lia LiAlH4 + H2O 'c - Lithium aluminium hydride reduces the given amide into tertiary amine.

+ LiAlH4 + H2O N(CH3)2 N(CH3)2 Therefore, reactions B, C and D produces 1-Butanol.

Answer 2

A) will be reduced to secondary alcohol (2-propanol)

B) will be reduced to 1-butanol.

C) will be reduced to 1-butanol as excess of LiAlH4 and H2O is used, First acid will be reduced to aldehyde and then aldehyde will be reduced to alcohol.

D) will be reduced to 1-butanol as excess of LiAlH4 and H2O is used. Eirst acid chloride will be reduced to aldehyde, then aldehyde will be reduced to alcohol.

E) will be reduced to amine.

1) LiAIH4 ne bonne 2) H20 1) LIAIH4 OH 2) H20 1) LIAIH LIAIH4 Mou OH 2) H2O OH LiAIH4 1) LIAIH4 weer nder OH H2O ci 2) H20 1) LIAIH4 - 2) H20