For the given reaction, OH 1 H2C = CH-CH-CH3 [X] will be [XH C = CH
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Br
H
2
C=CH−CH
2
−OH
HBr
excess
CH
3
−
CH
∣
−CH
2
−Br
In the above reaction, the alkene reacts with an excess of hydrogen bromide to give the final product. In this double bond is removed & Br is added to its place. Also, −OH is removed for −Br group. It follows Markownikov's Rule for the formation of the product.
Explanation:
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