Elimination of bromine from 2­-bromobutane results in the formation of

  • Option 1)

    equimolar mixture of 1 and 2-­butene

  • Option 2)

    predominantly 2­-butene

  • Option 3)

    predominantly 1-­butene

  • Option 4)

    predominantly 2­-butyne

 

Answers (1)

As we learnt in

Reaction of alkyl halide with KOH (alc) -

\beta- elimination reaction take place and produces alkenes.

- wherein

CH_3CH_2Br + KOH (alc.) \rightarrow H_2C=CH_2 + KBr + H_2O

 

 The reaction will give us Saytzeff's product predominantly which is 2 Butane.

 


Option 1)

equimolar mixture of 1 and 2-­butene

Incorrect option

Option 2)

predominantly 2­-butene

Correct option

Option 3)

predominantly 1-­butene

Incorrect option

Option 4)

predominantly 2­-butyne

Incorrect option

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