The first dissociation for H2X:
H2X +H2O ↔ HX + H3O
initial 0.15 0 0
change -X +X +X
at equlibrium 0.15-X X X
because Ka1 is small we can assume neglect x in H2X concentration
Ka1 = [HX][H3O]/[H2X]
4.5x10^-6 =( X )(X) / (0.15)
X = √(4.5x10^-6*0.15)
∴X = 8.2 x 10-4 m
∴[HX] & [H3O] = 8.2x10^-4
the second dissociation of H2X
HX + H2O↔ X^2 + H3O
8.2x10^-4 Y 8.2x10^-4
Ka2 for Hx = 1.2x10^-11
Ka2 = [X2][H3O]/[HX]
1.2x10^-11= y (8.2x10^-4)*(8.2x10^-4)
∴y = 1.78x10^-5
∴[X^2] = 1.78x10^-5 m
The major alkene product that results when n,n-dimethylhexan-2-amine undergoes cope elimination is hexene or hex-1-ene.
The reaction in which an amine is oxidize to an intermediate called an N-oxide which , when heated , acts as base in an intramolecular elimination reaction. The oxidation of tertiary amine into N-oxide is called cope reaction.
This elimination gives the less substituted alkene along with more substituted alkene which is Zaitsev product.
Example: Cope elimination of n,n-dimethylhexan-2-amine form hexene.
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Explanation:
