00:02
This is the answer to chapter 21, problem number 90, from the smith organic chemistry textbook.
00:10
And this problem asks us several different questions.
00:16
So it talks about maltose.
00:20
And so first for a, we're asked to label the acetal and hemacetal carbons.
00:27
So the acetal carbon is this one right here.
00:34
Acetal and in green down here this, oops, it's not pointing to a carbon.
00:43
This is the hemacetyl carbon.
00:49
Okay, yeah, so that's a.
00:51
Just remember acetal is connected to two ether oxygens.
00:57
Hemacetyl is connected to one ether and one alcohol.
01:00
So then for b, we are asked to perform a series of reactions.
01:08
So for the first one, for b1, we're asked treatment with h3o plus.
01:18
So that's going to hydrolyze the linkage between these two rings.
01:24
And so what we will end up with is this.
01:56
And this, the squiggle, means that it means either or.
02:08
So this alcohol could be equatorial or axial.
02:14
And so it'll be a mixture, really, of equatorial and axial there.
02:20
So then for part two of b, we are asked to use methanol and hcl.
02:33
Methanol and hdl.
02:37
And so just bear with me while i draw this.
02:44
So this is not going to affect this top ring at all.
02:51
Oh, h, oh, h, oh, h, oh, h.
03:05
Right, so the top ring really isn't affected at all.
03:11
Really, the bottom ring is not affected very much.
03:15
The only thing that's going to change is that that hemicatal is actually going to become a full acetyl.
03:26
It's going to replace the alcohol with a methoxide.
03:33
That doesn't look good at all.
03:37
Right.
03:47
So there will be a methoxide here.
03:50
And again, it can be actually.
03:51
Or equatorial och -c -h -3.
03:57
Right...