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Ethyl cyanide is reduced with SnCl2 and ...

Ethyl cyanide is reduced with `SnCl_2` and HCI, compound (A) is formed, which on hydrolysis gives product (B). The no. of pure orbitals present in the compound(B)

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The correct Answer is:
8
`CH_3 - CH_2 -CN underset(HCI) overset(SaCI_2) to C_2 H_5 -CH_2 -NH_2 overset(H_2 O) to CH_3 -CH_2 -CH_2 - CH_2 -OH +NH_3`
No. Of pure orbitals =8 ( in H- atoms )
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