Question
Consider the following structure:
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ {\text{O}}\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \|\\\ ^1\text{C}_3-\ ^2\text{CH}_2-\ ^3\text{C}-\ ^4\text{C}_2-\ ^5\text{C}\equiv\ ^6\text{CH}$
  1. How many σ and π-bonds are present in this compound?
  2. Arrange carbon no. 2, 3, 5 in decreasing order of s-character.
  3. Which atoms have same hybrid state?

Answer

$\ \ \ \ \ \ \ \ {\text{H}\ \ \ \ \text{H}\ \ \ \ \ \text{O}\ \ \ \ \ \text{H}}\\ \ \ \ \ \ \ \ \ \ |\ \ \ \ \ \ |\ \ \ \ \ \ \|\ \ \ \ \ \ \ | \\\text{H}-\text{C}-\text{C}-\text{C}-\text{C}-\text{C}\equiv\text{CH}\\\ \ \ \ \ \ \ \ \ |\ \ \ \ \ \ \ |\ \ \ \ \ \ \ \ \ \ \ \ \ \ | \\\ \ \ \ \ \ \ \ \text{H}\ \ \ \ \ \text{H}\ \ \ \ \ \ \ \ \ \ \ \ \text{H}$
  1. $14\sigma$ and $3\pi$ bonds.
  2. ii. $5>3>2$ is decreasing order of $s$-character of carbon atoms, i.e. $s p>s p^2>s p^3 s p$ has $50 \% s$-character, $s p$ has $33 \%$ s-character, sp has $25 \%$ s-character.
  3. Carbon number 1, 2, 4 are $\mathrm{sp}^3$ hybridised. Carbon number 5 and 6 are sp hybridised.

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