00:01
Okay, problem seven, we're drawing this structure of arabino's, okay, and we're given the systematic name for it.
00:09
Okay, so starting with the pentanol part, okay, so that means we have five carbons, and our carbon one is, of course, going to be an aldehyde.
00:24
Okay, so that signifies our aldehyde for carbon one.
00:29
So this would be your aldehyde for carbon 1 and we're going to have five carbons total.
00:36
So as usual, we're going to end with a ch2 -o -h and we'll have three carbons in between.
00:47
Okay.
00:48
So the way i want to do this is to figure out beforehand which would be priority number two.
00:57
Because as usual in sugars, that oh group is always priority number one and the hydrogen is priority.
01:03
Number four so it's all about which one is priority priority number two or three okay so since this aldehyde is double bonded to oh this is going to be always priority number two for the structure okay and this structure below it would be priority number three okay so let's start with carbon number two we have the configuration to r.
01:37
So we have r for carbon 2, which means we would want technically a clockwise direction, but because this is a fisher projection, if we have a clockwise direction, that is actually going to be s...