Passage Chloramphenicol (Compound 1) Is an Antibiotic That Acts Against Gram-Positive

Passage
Chloramphenicol (Compound 1) is an antibiotic that acts against gram-positive and gram-negative bacteria by blocking peptidyl transfer, which inhibits protein synthesis by bacterial ribosomes.  However, it tastes bitter, so most patients cannot tolerate it.  As a result, synthesis of hydroxyl-substituted chloramphenicol analogues has become of interest because ester chloramphenicol analogues taste better than chloramphenicol.  Substituents on the primary hydroxyl group of chloramphenicol analogues are typically hydrolyzed quickly in vivo to regenerate Compound 1.  Compound 3, a chloramphenicol analogue, was synthesized by the transesterification of Compound 1 and vinyl propionate (Compound 2) , catalyzed by the enzyme Lip_BA lipase (Figure 1) .
Figure 1  Transesterification of chloramphenicol catalyzed by Lip_BAAlthough enzyme catalysis allows for efficient regiospecific and stereospecific bond formation, the reaction conditions for the conversion efficiency of Compound 1 to Compound 3, as well as the purity of Compound 3, must also be optimized.  The reaction solvent, temperature, and time were studied to find the conditions that gave the highest conversion rate and purity.  Each parameter was altered individually while all others were kept constant, and the enzyme concentration was 4.5 g/L in each trial.  The results of this study are shown in Table 1.  The purity of Compound 3 was found to be over 90% when 1,4-dioxane was used and over 70% when dichloromethane was used.Table 1  Transesterification Conditions Optimization
The transesterification reaction was monitored by thin-layer chromatography (Figure 2) , and the product was confirmed by ¹H NMR.  The ¹H NMR spectra of Compounds 1 and 3 can be distinguished by the shift of H_e and H_f due to the addition of the electronegative ester functional group.
Figure 2  Thin-layer chromatography of reaction mixture
Adapted from F. Dong et al., "Transesterification Synthesis of Chloramphenicol." Molecules. ©2017 MDPI.
-Based on the structure of Compound 1 provided in Figure 1, which stretch gives a narrow absorption of weak intensity at approximately 3100 cm⁻¹ in the infrared spectrum?

A) The C=C stretch
B) The C=O stretch
C) The C-H_a stretch
D) The C-H_e stretch

Correct Answer:

Verified

Unlock this answer now
Get Access to more Verified Answers free of charge

Access For Free