Elimination of bromine from $2 - b r o m o b u t a n e$ results in the formation of:

Equimolar mixture of

1

and

2 - b u t e n e

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P r e d o m i n a n t l y 2 - b u t e n e

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P r e d o m i n a n t l y 1 - b u t e n e

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P r e d o m i n a n t l y 2 - b u t y n e

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Solution

The correct option is A $P r e d o m i n a n t l y 2 - b u t e n e$

In this elimination reaction of alkyl halide major product is produced according to Saytzeff’s rule. This states that when two alkenes may be formed, then alkene which is most substituted one predominates. Therefore, predominantly (80%) 2-butene will be produced.

The correct option is B.

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Similar questions

2

1

1

2

1 - b u t e n e

c i s - 2 - b u t e n e

t r a n s - 2 - b u t e n e

Butyne-1 and Butene-1 can be distinguished by

1 - B u t e n e + H B r H_{2} O_{2} - -- \to h ν 1 - B r o m o B u t a n e

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