00:01
Okay, so for problem 17, okay, we have de -gloose and de -galactose, and we're reducing both of these.
00:08
Okay, so we're reducing using nabh4 and water, okay, and we're doing that for both.
00:29
Okay, so once you reduce those, basically what we're doing is taking that aldehyde and converting it to a ch2 -o -h.
00:41
Okay so we're doing that for both de -glucose and the galactose so let's go ahead and draw that so we have ch2 oh for both sides right so the only thing you're changing here really is carbon one everything else stays the same okay so we'll do the same thing for the galactose so we'll have a ch2 oh for carbon one and ch2 oh for carbon and then we'll have those four carbons in between.
01:37
Okay, almost done here.
01:41
All right, so the question stem is telling us that after this reduction, the resulting molecule from the glucose is optically active.
01:59
So it is optically active...