JEE MAIN - Chemistry (2019 - 9th April Morning Slot - No. 2)
The major product of the following reaction is :
$${\rm{C}}{{\rm{H}}_3}{\rm{CH = CHC}}{{\rm{O}}_2}{\rm{CH_3 }}\buildrel {LiAl{H_4}} \over \longrightarrow $$
$${\rm{C}}{{\rm{H}}_3}{\rm{CH = CHC}}{{\rm{O}}_2}{\rm{CH_3 }}\buildrel {LiAl{H_4}} \over \longrightarrow $$
CH3CH2CH2CHO
CH3CH2CH2CO2CH3
CH3CH = CHCH2OH
CH3CH2CH2CH2OH
Explanation
LiAlH4 reduces esters to alcohols but does not
reduce C = C.
$${\rm{C}}{{\rm{H}}_3}{\rm{CH = CHC}}{{\rm{O}}_2}{\rm{CH_3 }}\buildrel {LiAl{H_4}} \over \longrightarrow $$ CH3CH = CHCH2OH + CH3OH
$${\rm{C}}{{\rm{H}}_3}{\rm{CH = CHC}}{{\rm{O}}_2}{\rm{CH_3 }}\buildrel {LiAl{H_4}} \over \longrightarrow $$ CH3CH = CHCH2OH + CH3OH
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