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Considering the state of hybridization of carbon atoms, find out the molecule among the following which is linear?
A
$CH_3-CH_2-CH_2-CH_3$
B
$CH_3-CH=CH-CH_3$
C
$CH_3-C\equiv C-CH_3$
D
$CH_2=CH-CH_2-C\equiv CH$
Detailed Solution
An sp hybridised carbon forms two σ bonds at $180^\circ$ to each other.
In but-2-yne, $CH_3-C\equiv C-CH_3$, both triple-bonded carbons are sp hybridised, so the four carbon atoms C–C≡C–C lie in one straight line.
In butane all carbons are $sp^3$ (bond angle $109.5^\circ$), giving a zig-zag chain.
In but-2-ene the double-bonded carbons are $sp^2$ (bond angle $120^\circ$), so the chain is bent.
Pent-1-en-4-yne contains $sp^2$ and $sp^3$ carbons as well, so the molecule is not linear.
Hence the linear molecule is $CH_3-C\equiv C-CH_3$.
In but-2-yne, $CH_3-C\equiv C-CH_3$, both triple-bonded carbons are sp hybridised, so the four carbon atoms C–C≡C–C lie in one straight line.
In butane all carbons are $sp^3$ (bond angle $109.5^\circ$), giving a zig-zag chain.
In but-2-ene the double-bonded carbons are $sp^2$ (bond angle $120^\circ$), so the chain is bent.
Pent-1-en-4-yne contains $sp^2$ and $sp^3$ carbons as well, so the molecule is not linear.
Hence the linear molecule is $CH_3-C\equiv C-CH_3$.