Question: 67: What is the hybridization shown by \mathrm{C}_{1} and \mathrm{C}_{2} carbons, respectively in the given compound?
\mathrm{OHC}-\mathrm{CH}=\mathrm{CH}-\mathrm{CH}_{2} \mathrm{COOCH}_{3}(1) \mathrm{sp}^{2} and \mathrm{sp}^{3}
(2) \mathrm{sp}^{2} and \mathrm{sp}^{2}
(3) s p^{3} and s p^{2}
(4) \mathrm{sp}^{3} and \mathrm{sp}^{3}
Answer: Option (1)
Explanation:
In the given compound, \mathrm{C}_{1} is the carbon of the aldehyde group \mathrm{OHC}.
This carbon is involved in a carbon–oxygen double bond and has three regions of electron density.
Therefore, the hybridization of \mathrm{C}_{1} is \mathrm{sp}^{2}.
\mathrm{C}_{2} corresponds to the \mathrm{CH}_{2}
carbon present after the double bond.
This carbon forms four single bonds and has four regions of electron density.
Hence, the hybridization of \mathrm{C}_{2} is \mathrm{sp}^{3}.
Therefore, the correct hybridizations are \mathrm{sp}^{2}
for \mathrm{C}_{1} and \mathrm{sp}^{3} for \mathrm{C}_{2}.
Hence, option (1) is correct.