Allyl cyanide contains 9sigma and 3pi bonds. | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) The structure of allyl cyanide is $CH_2=CH-CH_2-C \equiv N$. To calculate the total number of bonds: $1$. $CH_2=$ group: $2 \sigma$ ($C$-$H$) + $1

Vedclass · Accepted Answer

$9\sigma, 3\pi$

Vedclass · Answer

(A) The structure of allyl cyanide is $CH_2=CH-CH_2-C \equiv N$. To calculate the total number of bonds: $1$. $CH_2=$ group: $2 \sigma$ ($C$-$H$) + $1 \sigma$ ($C$-$C$) + $1 \pi$ ($C$=$C$) = $3 \sigma, 1 \pi$. $2$. $-CH=$ group: $1 \sigma$ ($C$-$H$) + $1 \sigma$ ($C$-$C$) = $2 \sigma$. $3$. $-CH_2-$ group: $2 \sigma$ ($C$-$H$) + $1 \sigma$ ($C$-$C$) = $3 \sigma$. $4$. $-C \equiv N$ group: $1 \sigma$ ($C$-$C$) + $1 \sigma$ ($C$-$N$) + $2 \pi$ ($C$-$N$) = $2 \sigma, 2 \pi$. Total: $(3+2+3+1) \sigma = 9 \sigma$ bonds and $(1+2) \pi = 3 \pi$ bonds.

Allyl cyanide contains $9\sigma$ and $3\pi$ bonds.

Similar Questions

$CH_3-CH_2-CH=CH_2$ has hybridisation

In which of the following compounds is the underlined carbon atom $sp^3$ hybridized?

The number of $sp^{3}$ hybridised carbons in an acyclic neutral compound with molecular formula $C_{4}H_{5}N$ is.

In the molecule $CH_2=C=CH-CH_3$ (where the atoms are numbered $1, 2, 3,$ and $4$ respectively),the hybridization of carbon $1, 2, 3,$ and $4$ are:

In the reaction,the hybridisation states of carbon atoms $1, 2, 3, 4$ are:
$Br-CH=CH-Br \xrightarrow[Catalyst]{H_2} Br-CH_2-CH_2-Br$

Vedclass Test Series

Exam Paper Generator

Online Exam Module