In the hydrocarbon $CH_{3}-CH=CH-CH_{3}-CH_{2}-C \equiv CH$ The state of hybridization of carbons 1,3 and 5…
In the hydrocarbon
$CH_{3}-CH=CH-CH_{3}-CH_{2}-C \equiv CH$
The state of hybridization of carbons 1,3 and 5 are in the following sequence.
$\mathrm{sp}^3, \mathrm{sp}^2, \mathrm{sp}$
$\mathrm{sp}^2, \mathrm{sp}, \mathrm{sp}^3$
$\mathrm{sp}, \mathrm{sp}^3, \mathrm{sp}_3^2$
$\mathrm{sp}, \mathrm{sp}^2, \mathrm{sp}^3$
Solution
\(\stackrel{sp^3}{\underset{6}{\mathrm{CH}_3}}-\stackrel{sp^2}{\underset{5}{\mathrm{CH}}}= \stackrel{sp^2}{\underset{4}{\mathrm{CH}}}-\stackrel{sp^3}{\underset{3}{\mathrm{CH}_2}}-\stackrel{sp}{\underset{2}{\mathrm{C}}} \equiv \stackrel{sp}{\underset{1}{\mathrm{CH}}}\)
The state of hybridisation of carbon in 1,3 and 5 position are \(s p, s p^3, s p^2\)