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An alkyl halide C5H11Br (A) reacts with ethanolic KOH to give an alkene ‘B’, which reacts with Br2to 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. Complete hydrogenation of ‘D’ yields a straight chain alkane. Identify A and B.
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The correct answer is:
A is CH3—CH2—CH2—CH2—CH2Br and B is CH3—CH2—CH2—CH=CH2
Correct Option is : (A)
A is CH3—CH2—CH2—CH2—CH2Br and B is CH3—CH2—CH2—CH=CH2
A is CH3—CH2—CH2—CH2—CH2Br and B is CH3—CH2—CH2—CH=CH2
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