In the following carbon skeleton structure, which carbon atom(s) exhibit ${\mathrm{sp}}^2$ hybridization? (Assume the structure is ${\mathrm{CH}}_2$=${\mathrm{CH}}_1$-${\mathrm{CH}}_2$-${\mathrm{CH}}_3$)

• A. Carbons 1 and 2
• B. Carbons 3 and 4
• C. Carbons 2 and 3
• D. Only carbon 2

Answer: (A)

Verified step by step guidance

1. Identify the carbon atoms in the given structure: CH\_2=CH\_1-CH\_2-CH\_3, where the subscripts indicate the carbon numbering.
2. Recall that sp\^2 hybridization occurs in carbon atoms that form a double bond, involving one sigma bond and one pi bond.
3. Examine the double bond between carbon 1 (CH\_1) and carbon 2 (CH\_2). Both carbons involved in this double bond will be sp\^2 hybridized.
4. Check the other carbons (carbon 3 and carbon 4) which are single bonded only; these carbons are typically sp\^3 hybridized because they form four sigma bonds.
5. Conclude that the carbon atoms exhibiting sp\^2 hybridization are carbons 1 and 2, as they are directly involved in the double bond.