An alkyl halide C5H11Br gives only 2- Pentene when treated with alcoholic KOH. Write the condensed
formula and equation for the reaction.
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Answer:
C5H11Br (koh alcohlic) → C5H10
An alkyl halide C5H11Br (A) reacts with ethanolic KOH to give an alkene 'B', which reacts with Br2 to give a compound 'C', which on dehydrobromination gives an alkyne 'D'. On treatment with sodium metal in liquid ammonia one mole of 'D' gives one mole of the sodium salt of 'D' and half a mole of hydrogen gas.
Explanation:
condenced formula of 2 pentene
CH2≈CHCHCH2CH3
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