To determine which organic compound has the same hybridization as its combustion product \((\text{CO}_2)\), let's examine the hybridization states of the carbon atoms in each compound and compare them with that in carbon dioxide.
The carbon atom in \(\text{CO}_2\) is surrounded by two double bonds (O=C=O). The hybridization of carbon in \(\text{CO}_2\) is sp because it forms two sigma bonds and has a linear geometry.
In ethane, each carbon atom forms three sigma bonds with hydrogen and one sigma bond with the other carbon atom. The hybridization is sp^3, which involves four sigma bonds.
In ethene, each carbon atom forms two sigma bonds with hydrogen and one sigma bond with the other carbon atom, resulting in a pi bond as well. The hybridization is sp^2, which involves three sigma bonds.
In ethyne, each carbon atom forms one sigma bond with hydrogen and one with the other carbon atom, along with a triple bond (two pi bonds) between carbon atoms. The hybridization is sp, involving two sigma bonds.
In ethanol, the hybridization of the carbon atoms primarily engaged in bonding with hydrogen and other carbon or oxygen atoms is sp^3 for the carbon atoms because they form four sigma bonds.
The only compound with the same hybridization as \(\text{CO}_2\) is Ethyne, where carbon is sp hybridized.
Major products A and B formed in the following reaction sequence , are