Consider the following sequence of reactions-H 2 C - CH - CH - CH 2 Br 21 mole -60- C A i- KCN excess - ii- H 2 - NiThe end product B is-A- H 2 N - CH 22- CH - CH - CH 22- NH 2B- H 2 N - CH 26- NH 2C- H 2 C - CH - C H - CH 22- NH 2H 2 C - CH - C NHD- NC - CH 2- CH - CH - CH 2- CN

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Reduction of Amides,Oximes and Azides with LAH

Consider the ...

Question

Consider the following sequence of reactions:
$H_{2} C = C H - C H = C H_{2} B r_{2} (1 mole) - ------ \to 60^{o} C (A) i . K C N (excess) - -------- \to i i . H_{2} / N i$

The end product B is:

H_{2} N - (C H_{2})_{2} - C H = C H - (C H_{2})_{2} - N H_{2}

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H_{2} N - (C H_{2})_{6} - N H_{2}

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N C - C H_{2} - C H = C H - C H_{2} - C N

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H_{2} C = C H - C | N H_{2} H - (C H_{2})_{2} - N H_{2}

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C H_{3} (C H_{2})_{4} C H_{3} A n h y . A l C l_{3} / H C l - --------- \to C H_{3} - C H - (C H_{2})_{2} - C H_{3} + C H_{3} C H_{2} - C H - C H_{2} - C H_{3}

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C H_{3} 2 - m e t h y l p e n t a n e

C H_{3} 3 - m e t h y l p e n t a n e

C H_{2} = C H - C H = C H - + C H_{2}

C H_{2} = C H - + C H - C H = C H_{2}

C H_{2} = C H - + C H - - C H - + C H_{2}

+ C H_{2} - C H = C H - - C H - + C H_{2}

I U P A C

C H_{3} - C H = C H - C H_{2} - C H = C H - C H - C H_{2} - C H = C H_{2}

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C_{2} H_{5}

C H_{3} - C H = C H - C H_{2} O H P C C - -- \to Product

I . ⊖ C H_{2} - \oplus C H - C H = C H - O - C H_{3}

I I . ⊖ C H_{2} - C H = C H - \oplus C H - O - C H_{3}

I I I . C H_{2} = C H - ⊖ C H - \oplus C H - O - C H_{3}

I V . C H_{2} = C H - ⊖ C H - C H = \oplus O - C H_{3}

C H_{2} = C H - C H = C H - O - C H_{3}

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