A $H_{3}C-CN\rightarrow 140°CNaCNH_{3}C-CH_{2}-NH_{2}$
B $H_{3}C-C∥O-NH_{2}\rightarrow Br_{2}/NaOHH_{3}C-NH_{2}$
C $H_{3}C-CN\rightarrow 90°CH_{2}NiH_{3}C-CH_{2}-NH_{2}$
D $H_{3}C-CN\rightarrow 140°CH_{2}/NiH_{3}C-CH_{2}-NH_{2}$
Correct Answer: Option C
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