Draw the structure of an alcohol with the molecular formula C5H12O that exhibits the following signals in its 13C NMR spectra: (
a) Broadband decoupled: 73.8 δ, 29.1 δ, and 9.5 δ (b) DEPT-90: 73.8 δ (c) DEPT-135: positive signals at 73.8 δ and 9.5 δ; negative signal at 29.1 δ
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Answer:
The answer to the question is
The rate constant for the reaction is 1.056×10⁻³ M/s
Explanation:
To solve the question, e note that
For a zero order reaction, the rate law is given by
[A] = -k×t + [A]₀
This can be represented by the linear equation y = mx + c
Such that y = [A], m which is the gradient is = -k, and the intercept c = [A]₀
Therefore the rate constant k which is the gradient is given by
Gradient = where [A]₁ = 8.10×10⁻² M and [A]₂ = 1.80×10⁻³ M
= = -0.001056 M/s = -1.056×10⁻³ M/s
Threfore k = 1.056×10⁻³ M/s