Allyl isocyanide has

Question

Allyl isocyanide has (A) 9 σ and 4 π bonds (B) 8 σ and 5 π bonds (C) 9 σ, 3 π and 2 non-bonded electrons. (D) 8 σ, 3 π and 4 non-bonded electrons.

Tardigrade Admin · Accepted Answer

CH 2= CH - CH 2- N ≡ C In it we find, 5 C - H (σ bonds ), 1 C - C (σ bond), 1 C - N (σ bond ), 1 C = C (1 σ and 1 π bonds ), i.e., a total of 9 σ and 3 π bonds and two non-bonded electrons are present on C - atom of N ≡ C :

Q. Allyl isocyanide has

$9 σ$ and $4 π$ bonds

$8 σ$ and $5 π$ bonds

$9 σ, 3 π$ and 2 non-bonded electrons.

$8 σ, 3 π$ and 4 non-bonded electrons.

$C H_{2} = C H - C H_{2} - N \equiv C$

In it we find, $5 C - H (σ$ bonds $), 1 C - C$ (\sigma bond), $1 C - N (σ$ bond $), 1 C = C (1 σ$ and $1 π$ bonds $)$ , i.e., a total of $9 σ$ and $3 π$ bonds and two non-bonded electrons are present on $C$ - atom of $N \equiv C$ :

Q. Allyl isocyanide has

9σ and 4π bonds

8σ and 5π bonds

9σ,3π and 2 non-bonded electrons.

8σ,3π and 4 non-bonded electrons.

CH2​=CH−CH2​−N≡C In it we find, 5C−H(σ bonds ),1C−C (\sigma bond), 1C−N(σ bond ),1C=C(1σ and 1π bonds ), i.e., a total of 9σ and 3π bonds and two non-bonded electrons are present on C - atom of N≡C :

$9 σ$ and $4 π$ bonds

$8 σ$ and $5 π$ bonds

$9 σ, 3 π$ and 2 non-bonded electrons.

$8 σ, 3 π$ and 4 non-bonded electrons.

$C H_{2} = C H - C H_{2} - N \equiv C$

In it we find, $5 C - H (σ$ bonds $), 1 C - C$ (\sigma bond), $1 C - N (σ$ bond $), 1 C = C (1 σ$ and $1 π$ bonds $)$ , i.e., a total of $9 σ$ and $3 π$ bonds and two non-bonded electrons are present on $C$ - atom of $N \equiv C$ :