ORG-19

Diels–Alder K snapshot

K = [adduct] / ([diene][dienophile]). Cycloaddition equilibrium.

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EquilibriaDiels–Alder

Governing equation

K=[\mathrm{adduct}]/([\mathrm{diene}][\mathrm{dienophile}])

$[adduct] \in \left[1\times 10^{-6}, 2\right]$

where

[adduct]: Adduct (mol/L)
[diene]: Diene (mol/L)
[dienophile]: Dienophile (mol/L)
K: Equilibrium K (L/mol)

The figure, part by part

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①

Governing law — red

K=[\mathrm{adduct}]/([\mathrm{diene}][\mathrm{dienophile}])

$[adduct] \in \left[1\times 10^{-6}, 2\right]$

$[adduct] = 1\times 10^{-6}$ $K = 0.002$

Along the curve

②

Domain of the variable — blue

[adduct]_{\min} = 1\times 10^{-6},\quad [adduct]_{\max} = 2

$[adduct] \in \left[1\times 10^{-6}, 2\right]$

$[adduct] = 1\times 10^{-6}$ $[adduct] = 1\times 10^{-6}$

Along the curve

③

Response curve — green

K=[\mathrm{adduct}]/([\mathrm{diene}][\mathrm{dienophile}])

$[adduct] \in \left[1\times 10^{-6}, 2\right]$

$[adduct] = 1\times 10^{-6}$ $K = 0.002$

Along the curve

④

Rate of change — amber

\frac{\mathrm{d}K}{\mathrm{d}[adduct]}

$[adduct] \in \left[1\times 10^{-6}, 2\right]$

$[adduct] = 0.013$ $K^{\prime} = 2500$

Along the curve

Lecture brief

Historical brief

Hammett (1937) and Taft linear free-energy, E-factor green metrics, Woodward–Fieser UV and Claisen equilibria are how organic chemistry became predictive. The lab is substituent, waste and tautomer. This sheet (ORG-19 — Diels–Alder K snapshot) is the form associated with Diels–Alder. Working symbols: $[adduct]$ , $[diene]$ , $[dienophile]$ $\rightarrow$ $K$ . A pericyclic 6-electron process. Electron-rich dienes and electron-poor dienophiles give large K.

Purpose

Purpose: compute

K

from

[adduct]

[diene]

[dienophile]

in Organic chemistry via

K=[\mathrm{adduct}]/([\mathrm{diene}][\mathrm{dienophile}])

K = [adduct] / ([diene][dienophile]). Cycloaddition equilibrium. Use it when a real organic chemistry question must be answered in SI before a code check.

Live realistic example

In symbols

Live case. Given

[adduct] = 0.080\,\mathrm{mol/L}

[diene] = 0.020\,\mathrm{mol/L}

[dienophile] = 0.020\,\mathrm{mol/L}

, the governing relation

K=[\mathrm{adduct}]/([\mathrm{diene}][\mathrm{dienophile}])

yields

K = 200.000\,\mathrm{L/mol}

. 1:1 cycloaddition, concentrations of the three species. Move a slider: the numbers are this situation, not a canned story.

Calculator

Inputs

Adduct [adduct]0.08 mol/LDiene [diene]0.02 mol/LDienophile [dienophile]0.02 mol/L

Outputs

Equilibrium K K200.000 L/mol

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Narration of this film

1:1 cycloaddition, concentrations of the three species.

A pericyclic 6-electron process. Electron-rich dienes and electron-poor dienophiles give large K.

Reading speed

Watch on YouTube