The Mannich condensation is a widely used reaction to form highly substituted amines. In the key ste — Organic Chemistry Chemistry Question
Synthesis of a Mannich Base: a Mannich Mystery
The Mannich condensation is a widely used reaction to form highly substituted amines. In the key step in this reaction, an enolate or its equivalent adds to an iminium ion that is often formed in situ from an amine and an aldehyde. In this way, three molecules are condensed to form the final product. In particular, reactions of phenols and formaldehyde in the presence of primary or secondary amines gives rise to benzylic amines, with reaction taking place exclusively in the activated positions ortho or para to the phenol group:
[VISUAL]
In this experiment, you will explore the Mannich reaction between 2,2-dimethyl-1,3-diaminopropane with excess 2,4-di-tert-butylphenol and formaldehyde. Because the starting amine has two primary amino groups, one could envision many different possible Mannich products that could be formed in this reaction. In fact, one product is formed selectively and can be isolated in moderate yield. You will be asked to suggest a structural formula of this product based on its 1H NMR spectra provided below.
Chemicals
* 2,2-Dimethyl-1,3-diaminopropane, NH2CH2C(CH3)2CH2NH2
* 2,4-di-tert-butylphenol, C6H3(C[CH3]3)2OH
* Aqueous formaldehyde, 37% (w/v)
* Ethanol
* Methanol
* Hexane/ethyl acetate mixture for TLC (3:1 v/v)
Equipment and glassware
* Balance (± 0.01 g precision or better)
* Erlenmeyer flask, 125 cm3
* Teflon-coated stirbar
* Hotplate/stirrer
* Graduated cylinder, 10 cm3
* Büchner funnel
* Filter flask and source of vacuum (e.g., water aspirator)
* Silica gel-coated TLC plates and development chamber
* Melting point apparatus
* Ice water bath
* Spatulas
Procedure
1. To the 125 cm3 Erlenmeyer flask add 0.35 g 2,2-dimethyl-1,3-diaminopropane, 2.2 g 2,4-di-tert-butylphenol, 10 cm3 ethanol, and a stirbar. Stir the mixture until it becomes homogeneous, then add 1.0 cm3 37 % aqueous formaldehyde solution.
2. Heat the mixture to a gentle boil, with stirring, on the hotplate/stirrer. Maintain at a gentle boil for 1.5 hr. Alternatively, the heating can be carried out in a round-bottom flask under a reflux condenser, using a heating mantle or oil bath to heat the flask, with the solution maintained at reflux for 1.5 hr.
3. Take the flask off of the hotplate, remove the stir bar from the solution, and allow the reaction mixture to cool to room temperature. If no solid has formed, scratch the inner sides of the flask with a spatula to initiate crystallization. After the solution has reached room temperature, chill the flask in an ice bath for at least 10 minutes.
4. Suction-filter the precipitate on the Büchner funnel. Wash the solid thoroughly with 10 cm3 methanol to remove any unreacted 2,4-di-tert-butylphenol. After the wash, leave the precipitate on the Büchner funnel with the vacuum on (to suck air through the precipitate) for at least 15 min. This serves to dry the solid by evaporating any residual methanol.
5. Scrape the solid into a tared container and measure the yield of product.
6. Characterize the product by its melting point (it is between 200 – 250 °C) and by thin layer chromatography (silica gel, eluting with 3:1 hexane:ethyl acetate (v/v)).
The 1H NMR spectra of the product, recorded in CDCl3 solution at 500 MHz at – 40 °C and at 55 °C, are shown below. For each temperature, the full spectrum from 0 – 2 ppm is shown, then an expansion of the region from 1.5 – 4.5 ppm. Peak positions, where listed, are given in ppm. Some small impurities in the solvent are observable; they are marked with asterisks (*) and should be ignored. Based on these spectra, suggest a structural formula for the observed product.
[VISUAL]
Model Answer
Consider the high-temperature spectrum first. There are two aromatic signals (δ 6.8 and 7.2 ppm) and a downfield peak at ~10 ppm of equal intensity, attributable to the two remaining aromatic hydrogens (after the H ortho to the OH is replaced by a CH2NRR' group) and to the phenolic OH resonance (shifted downfield due to its intramolecular hydrogen bonding to the amine. This corresponds to the fragment shown below:
[VISUAL]
This would also require two tert-butyl singlets (observed at δ 1.3 and 1.4 ppm) and the benzylic CH2 protons (twice the intensity of either aromatic peak, observed at δ 3.6 ppm). To determine how many hydroxybenzyl groups have been added to the amine, consider the C(CH3)2 signal at δ 1.0 ppm; it has an integral of 6H per diamine moiety, and it is about 1.5 times the size of the benzylic resonance at δ 3.6 ppm. That establishes that there are two hydroxybenzyl groups present. The presence of only a single peak for the CH2C(CH3)2CH2 signals (δ 2.2 ppm) shows that the two sides of the molecule are symmetrical, i.e., each nitrogen has undergone one Mannich reaction rather than having one nitrogen react twice. The only peak not accounted for is at δ 3.2 ppm (integral 2H). At first it seems plausible to assign this as NH hydrogens. But consider the low-temperature spectrum: all of the aliphatic resonances (except the tert-butyl peaks) split at –40 ºC to give pairs of resonances. Thus, in the expanded region the peak at δ 3.6 ppm splits into doublets (J = 14 Hz) at δ 3.36 and 3.96; the peak at δ 3.2 ppm splits into doublets (J = 8 Hz) at δ 2.55 and 4.07; and the peak at δ 2.2 ppm splits into doublets (J = 11 Hz) at δ 1.74 and 2.65. (The singlet at δ 1.0 at +55 ºC also splits, into singlets at δ 0.81 and 1.13 ppm.) All of the doublets have coupling constants consistent with geminal 2J values appropriate for HCH' couplings. In contrast, there is no way that NH protons could possibly split into a pair of doublets. Thus there must be a CH2 group (derived from condensation of the diamine with formaldehyde) linking the two nitrogen atoms:
[VISUAL]
Suggest an explanation for the change in appearance of the 1H NMR spectra with temperature.
Model Answer
The six-membered ring in the center of the molecule will adopt a chair conformation. At low temperature, this gives rise to inequivalent axial and equatorial methyl groups and methylene hydrogens (the structure contains a mirror plane, so the two sides are equivalent to each other). Because no symmetry relates the two benzylic hydrogens to each other, they too will be inequivalent (diastereotopic). Since the two hydrogens on each of the methylene groups are inequivalent to each other, they will couple to each other, giving rise to pairs of doublets.
[VISUAL]
As the temperature is raised, the structure begins to undergo chair-to-chair conformer interconversion, which causes all the axial groups to become equatorial and vice versa. (An exception is the nitrogen substituents; since pyramidal inversion at N is fast, the benzyl groups can remain in the equatorial position.) This causes the axial and equatorial hydrogens or methyl groups, and the diastereotopic benzyl protons, to interconvert. When this process is rapid, the pairs of peaks therefore coalesce at a single, time-averaged chemical shift, as is seen at +55 ºC. Since the environments of the aromatic, OH, and tert-butyl protons do not change in this interconversion, those peaks are essentially unaffected by temperature. (The small upfield shift of the OH peak at higher temperature is due to decreasing hydrogen bonding as the temperature is raised.)
Calculate a percent yield of product.
Model Answer
The limiting reagent is the diamine, 0.35 g = 3.4 · 10–3 mol. The theoretical yield of the product (C36H58N2O2, M = 550.87 g mol–1) is 1.89 g. Typical actual yield = 0.46 g, 24%.
Report the melting point and Rf value of the compound.
Model Answer
mp = 231 – 234 ºC. Rf (silica gel, 3:1 hexane:ethyl acetate) = 0.61.