How to convert $C l - {({C H}_{2})}_{4} - C l$ into hexan-1, 6-diamine?

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Solution

When $1, 4 - d i c h l o r o b u t a n e$ reacting with $e t h o n o l i c N A C N$ gives $N C - (C H_{2})_{4} - C N$ .
Again this is made to react with $H 2$ in presence of Nickel for reduction Which gives the final product $h e x a n - 1, 6 - d i a m i n e$

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Similar questions

C l - (C H_{2})_{4} - C l

How will you convert?

(i) Benzene into aniline

(ii) Benzene into N, N-dimethylaniline

(iii) Cl−(CH₂)₄−Cl into hexan-1, 6-diamine?

C o

A l

[C o (C l) (e n)_{2}] C l

K_{3} [A l (C_{2} O_{4})_{3}]

{C l}^{-}

{C l}_{2}

C l

C l_{(g)}^{-}

C l_{(g)}

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