Draw the Fischer projection of the product of the oxidation of d-galactose at C1.

Video transcript

Welcome back everyone to another video which of the following illustrates the fissure projection of the product formed when the manos is oxidized at the carbon one position. And we're given four answer choices. So let's begin this problem by simply drawing the structure of the nanos. And we want to draw a fissure projection. Now, carbon number one is the position where we simply have an aldehyde functional group cho followed by carbon number two. So we're going to introduce the horizontal lines showing the bonding pattern for every other carbon starting with carbon. Number two, we want to recall that on the left side, we have hydroxyl on the right side, we have hydrogen followed by carbon number three. Once again, we have hydroxyl on the left hygen, on the right. And then there would be two hydroxyl groups or the next two carbons that are on the right side with hygen being on the left. And of course, the final carbon atom simply corresponds to C H2O H. Now, according to this problem, we are performing an oxidation reaction at carbon number one. And since we have an aldehyde, we have to recall that whenever we oxidize an aldehyde, we are producing a carboxylic acid, simply speaking, we're adding an additional oxygen atom. So cho becomes coo H or simply speaking co2 H and that would be our product for this problem. Now, if we look at the answer choices, we first of all want to make sure that we have the correct arrangement of hydroxy groups or the answer choices. So A has correct arrangement of hydroxy groups. Now, b well, it has the correct arrangement of hydroxy groups and we can immediately exclude C and D because the hydroxy groups are not consistent with the structure of the mammals. And between A and B, we first will notice that A has an aldehyde functional group. So we haven't performed any oxidation process in B sciences. We have actually oxidized the carbon at the very bottom from a primary alcohol into a carboxylic acid. This leaves us with the option B as the correct answer. We noticed that we have formed the carboxylic acid functional group coo H and now the final carbon remains the same chtoh. So it's basically unchanged. And this means that B is the correct answer to this problem. Thank you for watching.