Question

How do I know whether the diol was meso or racemic using the NMR spectrum? Do we need to see the NMR spectrum of the acetonide product? How do they different between the compounds from meso and racemic?

The Synthesis of 2,2-Dimethyl-1,5- Dioxolane; The Acetonide Derivative of a Vicinal Diol PRELAB EXERCISE: Draw the most stable structures of meso- and racemic-1,2-stilbenediol using Newman projections. Draw the same two isomers with the hydroxyl groups eclipsed. In Chapter 55, yellow benzil was reduced with sodium borohydride to the color- less diol. The melting point of the product (137°C) indicates that (10,25)-(meso)- hydrobenzoin has been produced. But how do we know that this is the isomer produced? The object of this experiment is to demonstrate, using logic based on nuclear magnetic resonance (NMR) spectral data, that the product is indeed the meso isomer and not the racemic isomer. The diol is reacted with acetone to form a crystalline acetonide (a ketal) that will have either structure 1 or structure 2. An examination of the NMR spectrum of this derivative will reveal whether or not the two methyl groups of the acetonide are equivalent, and thus whether the diol is meso or racemic. Rather than simply explaining each step in this process and the reasoning behind it, you will discover them for yourself by answering the questions at the end of this chapter. By consid- ering each question in turn, you will see how it is possible to determine the stereo- chemistry of the starting diol. OH OH TI C-C + CH; -C-CH3 HH FeCl3 H3C) НАС, CH3 CH and/or + H2O CoHS CH CoHS CH H 2

Answer 1

meso-1,2-diphenyl-1,2-ethanediol (2,2-dimethyl-4,5-diphenyl-1,3-dioxolane)

1H-NMR: cis-2,2-Dimethyl-4,5-diphenyl-1,3-dioxolane(meso)

300 MHz, CDCl3

delta [ppm] mult. atoms assignment

1.62 s 3 H CH3

1.84 s 3 H CH3

5.53 s 2 H CH-O (product)

6.8-7.2 m 10 H CH (arom., product)

trans-2,2-Dimethyl-4,5-diphenyl-1,3-dioxolane (Racemic)

nmr data

delta nb atoms multipliciy

7.456 4 H m(CH aromatic)

7.346 2 H m CH aromatic

7.69975 4 H m CH aromatic

4.741 2 H d CH-O

1.399 6 H s CH3

AS We can se in the readings of NMR of meso and Racemic compound, the chemical shift of CH-O

is different in both due to arrangement of the phenyl group.In meso compound it shows signal at 5.53 ppm but in Racemic compound it shows signal at 4.741 ppm. because H of CH-O in meso compound is less shielded than H of CH-O of Racemic compound due to this it (meso) shows high chemical shift than CH-O of Racemic compound. And CH3 in trans (Racemic)complex show only one signal whereas CH3 in cis (meso)shows 2 signal. so we can distinguished meso and Racemic compound with the help of NMR.