An unsaturated hydrocarbon A reactswith H₂ in presence of Lindlar's catalyst to give B as a product. B undergoes ozonolysis
to give Ethanal. i) Give the chemical equation for the conversion of compound A to B and B to Ethanal.
ii) Write the IUPAC name of compound A and B. iii) What happen when Breacts with Baeyer's reagent?
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EXPLANATION.
One zero of the polynomial.
(a² + 9)x² + 13x + 6a is reciprocal of the other.
As we know that,
Let, one root = α.
Other root is = 1/α.
Products of the zeroes of the quadratic equation.
⇒ αβ = c/a.
⇒ α x 1/α = 6a/(a² + 9).
⇒ 1 = 6a/(a² + 9).
⇒ a² + 9 = 6a.
⇒ a² - 6a + 9 = 0.
Factorizes the equation into middle term splits, we get.
⇒ a² - 3a - 3a + 9 = 0.
⇒ a(a - 3) - 3(a - 3) = 0.
⇒ (a - 3)(a - 3) = 0.
⇒ (a - 3)² = 0.
⇒ (a - 3) = 0.
⇒ a = 3.
MORE INFORMATION.
Nature of the roots o the quadratic expression.
(1) = Real and unequal, if b² - 4ac > 0.
(2) = Rational and different, if b² - 4ac is a perfect square.
(3) = Real and equal, if b² - 4ac = 0.
(4) = If D < 0 Roots are imaginary and unequal Or complex conjugate.
One zero of the polynomial.
(a² + 9)x² + 13x + 6a is reciprocal of the other.
As we know that,
Let, one root = α.
Other root is = 1/α.
Products of the zeroes of the quadratic equation.
⇒ αβ = c/a.
⇒ α x 1/α = 6a/(a² + 9).
⇒ 1 = 6a/(a² + 9).
⇒ a² + 9 = 6a.
⇒ a² - 6a + 9 = 0.
Factorizes the equation into middle term splits, we get.
⇒ a² - 3a - 3a + 9 = 0.
⇒ a(a - 3) - 3(a - 3) = 0.
⇒ (a - 3)(a - 3) = 0.
⇒ (a - 3)² = 0.
⇒ (a - 3) = 0.
⇒ a = 3.
MORE INFORMATION.
Nature of the roots o the quadratic expression.
(1) = Real and unequal, if b² - 4ac > 0.
(2) = Rational and different, if b² - 4ac is a perfect square.
(3) = Real and equal, if b² - 4ac = 0.
(4) = If D < 0 Roots are imaginary and unequal Or complex conjugate.
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