Question

This question related to Diels-Alder Reaction of Anthracene and Maleic Anyhydride.

1. Show the product of the reaction of water with the anhydride product

2. The procedure warns against adding a lot of xylene during recystalization. Why?

3. What is the differences in the carbonyl region of the IR between the starting material and the product.

4. What specific band in the IR is unique to the product and confirms that you have made the product and do not have a mixture of the starting material? Explain.

5. Predict the product of the self-dimerization(via Diels-Alder reaction) of cyclopentadiene.;

Answer 1

Answer 1 and 5Answer: 2

The use of xylene in the Diels-Alder Reaction is warmed because it acts as a solvent for reaction. Xylene is much reactive solvent and hence reactants are more soluble in it then the product, therefore, the product is crystallise from the solvent and not from the reactant.

Also, it has higher boiling point than most of the other organic solvent. this high boiling point allows the reaction to reflux at a high temperature to yield product.

hence it is warned to use excess xylene during the Diels-Alder Reaction.

Answer: 3

The Diels Alder reaction of maleic anhydride and anthracene forms something which closely resembles succinic anhydride. Anhydrides generally have two carbonyl resonances in the 1700-1875 wavenumber region. These stretches are coupled - one is a symmetric stretch and one is an asymmetric stretch. Carbonyl stretches of 5-member cyclic anhydrides usually vibrate at higher frequencies due to strain in the ring. Succinic anhydride (plain) has absorption at 1865 and 1782 wavenumbers. Compare this to acetic anhydride, which has streches at around 1818 and 1750 wavenumbers. The symmetric stretch is the one at lower frequency.

Conjugation to a carbonyl reduces the double bond character and causes the carbonyl to shift to lower frequencies.

Also, C=C stretches are usually fairly weak, but the double bond is cis-orientation and the C=C shows IR spectra at 1600 to 1650 wavenumbers.

Answer: 4

Conjugation to a carbonyl reduces the double bond character and causes the carbonyl to shift to lower frequencies. In acylic anhydrides this effect is pretty substantial (30 to 40 wavenumbers), but it seems the addition of a double bond does not impact much the position of the stretches in 5-membered cyclic anhydrides. It is possible this is because the ring strain is by far the dominant effect. Even so the 1 wavenumber shift to higher frequency MAY be indicative of a loss of the double bond in maleic anhydride when formation of the adduct occurs. However this is so small I wouldn't use this alone.

The other area you may want to look is the double bond itself. C=C stretches are usually fairly weak, but the double bond here is cis-orientation which will help. Typically the C=C stretch shows up in the vicinity of 1600 to 1650 wavenumbers. In a strained system, the shift is again shifted to lower frequencies (to a point). From the NIST/Adrich spectra it appears that the reasonably sharp, intense peak at approximately 1595 wavenumbers may be due to this C=C stretch - but you should be able to provide more specific value from your data. In the adduct, this peak should disappear because there's no longer a double bond to stretch.