00:01
Okay, this problem is asking us how many hydrogens are on each of these indicated carbons of morphine.
00:05
So the molecule that i've drawn out in blue is morphine, and then what is drawn in red, those are just arrows.
00:10
I was just too lazy to draw the head of the arrow.
00:14
Okay, but in green i'll be representing the hydrogens.
00:17
Okay, so attached to, i'll start with the bottom left, attached to this carbon.
00:21
That carbon has a bond this way, two bonds that way and a bond that way.
00:26
So again, we want to have all carbons have a full octet, and they can.
00:30
Can achieve that by having four bonds each.
00:32
So because i have a single bond, a single bond, and a double bond, double, single, that corresponds to four bonds.
00:39
None of them are connected to hydrogens, so this one would be zero.
00:43
Okay, and then moving along this way, same thing this way.
00:45
This one is connected to an oxygen, single bonded, carbon that way, and double bond that way.
00:50
None of them are hydrogens, so that one would be zero.
00:53
Okay, what about this one? so that one, let's see.
00:57
We have a carbon bond, a single bond, carbon single bond.
01:00
Carbon single bond, but what about anything else? we don't have any other, we don't have a double bond anywhere, we don't have any single bonds outstretched to another atom, so we know that that must have an implicit hydrogen, right? one that's not shown, so that would be one.
01:16
Okay, what about this one? so that one has three bonds.
01:20
We have one single bond going that way, one going that way, and one going that way, but we don't have a fourth bond.
01:24
And we know that in order to achieve a full octet, they must have four bonds, because i only have three bonds there, that must mean that there is one hydrogen that is not shown.
01:32
So we have one there.
01:33
What about this one? that one, we have a bond that going that way, and then a bond going that way.
01:39
We have two bonds.
01:40
So we need two more.
01:41
That means that we must have two hydrogens there...