Answer:
The Dehydrohaogenation of 1-bromo propane with alcoholic KOH gives propene which on again hydrohalogenation with HBr gives 2-bromo propane due to Markonikove's rule for addition.
Explanation:
Answer:
Ka = ( [H₃O⁺] . [F⁻] ) / [HF]
Explanation:
HF is a weak acid which in water, keeps this equilibrium
HF (aq) + H₂O (l) ⇄ H₃O⁺ (aq) + F⁻ (aq) Ka
2H₂O (l) ⇄ H₃O⁺ (l) + OH⁻ (aq) Kw
HF is the weak acid
F⁻ is the conjugate stron base
Let's make the expression for K
K = ( [H₃O⁺] . [F⁻] ) / [HF] . [H₂O]
K . [H₂O] = ( [H₃O⁺] . [F⁻] ) / [HF]
K . [H₂O] = Ka
Ka, the acid dissociation constant, includes Kwater.
