Question
Allyl isocyanide has :
A.
$$9\sigma \,{\text{and}}\,4\pi \,{\text{bonds}}$$
B.
$$8\sigma \,{\text{and}}\,5\pi \,{\text{bonds}}$$
C.
$$9\sigma ,3\pi \,{\text{and}}\,2\,{\text{non - }}\,{\text{bonds}}\,{\text{electrons}}$$
D.
$$8\sigma ,3\pi \,{\text{and}}\,4\,{\text{non - }}\,{\text{bonds}}\,{\text{electrons}}$$
Answer :
$$9\sigma ,3\pi \,{\text{and}}\,2\,{\text{non - }}\,{\text{bonds}}\,{\text{electrons}}$$
Solution :
$$C{H_2} = CH - C{H_2} - N\mathop = \limits^ \to C:$$
The above structure of allyl isocyanide clearly shows $$5C - H\left( \sigma \right),2C - C\left( \sigma \right),1C - N\left( \sigma \right),1N - C\left( \sigma \right),1C - C\left( \pi \right),2N - C\left( \pi \right)$$ bonds, i.e., $$9\sigma $$ and $$3\pi $$ bonds in all. There are 2 non-bonded electrons on the C-atom (co-ordinate bond between $$N$$ and $$C,$$ the electron pair of $$N$$ is shifted towards $$C$$ ) .