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  • Consider the following reaction,  <img alt="\text{ethanol}\xrightarrow{{\text{PBr}_3 }}\text{X}\xrightarrow{{\text{alc}.\,\,\text{KOH}}}\text{Y}\xrightarrow[{(\text{ii})\text{H}_2 \t

Consider the following reaction, 

\text{ethanol}\xrightarrow{{\text{PBr}_3 }}\text{X}\xrightarrow{{\text{alc}.\,\,\text{KOH}}}\text{Y}\xrightarrow[{(\text{ii})\text{H}_2 \text{O},\,\text{heat}}]{{(\text{i})\text{H}_2 \text{SO}_4 \text{room}\,\text{temperature}}}\text{Z};

the product Z is:

Option: 1

CH3CH2 - O - CH2 - CH3


Option: 2

CH3 - CH2 - O - SO3H


Option: 3

CH3CH2OH


Option: 4

CH2 = CH2


Answers (1)

best_answer

PBr3 act as a halogenating agent that converts -OH group into -Br. Then  Alc. KOH is a dehydrohalogenation agent. H2S04 and H20 converts an olefin into alcohol.

                     

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