Question

1,1 дихлорпропан+2KOH

Answer 1

Cl₂HC-CH₂-CH₃ + 2KOH (спиртовой) ⇒ HC≡C-CH₃ + 2KCl + 2H₂O пропин

Cl₂HC-CH₂-CH₃ + 2KOH (водный) ⇒ H(O)C-CH₂-CH₃ + 2KCl + H₂O пропаналь

Answer 2

When 1,1-dichloropropane (C₃H₆Cl₂) reacts with potassium hydroxide (KOH), it undergoes a nucleophilic substitution reaction known as dehydrohalogenation or elimination reaction. The reaction can be represented by the following equation:

C₃H₆Cl₂ + 2KOH → C₃H₄ + 2KCl + 2H₂O

In this reaction, the two chloride atoms (Cl) in 1,1-dichloropropane are replaced by hydroxide ions (OH⁻) from potassium hydroxide. This results in the formation of propene (C₃H₆), potassium chloride (KCl), and water (H₂O).