1.2 When 2-chlorobutane is strongly heated in the presence of concentrated sodium hydroxide, the major product formed is ... A but-1-ene. B but-2-ene. C butan-1-ol. D butan-2-ol.

To determine the major product formed when 2-chlorobutane is strongly heated in the presence of concentrated sodium hydroxide, we need to consider the reaction mechanism involved. 1. **Identify the Reaction Type**: The reaction of 2-chlorobutane with concentrated sodium hydroxide is typically an elimination reaction (E2 mechanism) rather than a substitution reaction (S_N2 or S_N1). In this case, the hydroxide ion acts as a base, abstracting a proton from the β-carbon (the carbon adjacent to the one bearing the leaving group, Cl). 2. **Identify the Possible Products**: - When 2-chlorobutane undergoes elimination, it can form alkenes. The possible alkenes that can be formed are: - But-1-ene (from the elimination of HCl from the 1st carbon) - But-2-ene (from the elimination of HCl from the 2nd carbon) 3. **Determine the Major Product**: - In elimination reactions, the more substituted alkene is generally favored due to greater stability (Zaitsev's rule). - But-2-ene is more substituted than but-1-ene, as it has a double bond between two carbon atoms that each have two substituents (methyl groups), while but-1-ene has a terminal double bond. 4. **Conclusion**: Therefore, the major product formed when 2-chlorobutane is strongly heated in the presence of concentrated sodium hydroxide is but-2-ene. The correct answer is **B but-2-ene**.