00:01
For problem 14, we are on iupac naming.
00:04
Okay.
00:05
So we all have this cycloalkans in here.
00:11
And let's just get going.
00:13
So part part a right here we got six, a six -membered ring, right? one, two, three, four, five, six.
00:25
Okay.
00:26
So i just started counting because it's pretty symmetrical.
00:30
So regardless of which branch point you start counting on the other one's going to end up on carbon 4 so we got a methyl group branching off of carbon 1 and another one branching off of carbon 4 okay so we're going to write that as 1 4 right one on carbon 1 and 1 and you have two methyl groups so that's to give us dimethyl.
01:04
Dymethyl and we got six carbons on our ring.
01:08
So cyclo -hexane, okay.
01:15
Hex -4 -6 and cyclo -for being a ring.
01:19
Okay.
01:20
So that's part a.
01:21
Part b, we got a five -membered ring, and we got an ethyl group...