a) To begin this structure you'll start with a pentene ring, hence the prefix "cyclo-". We then place the double bond between carbons 1 and 2. Lastly, for the "dimethyl", we place two methyl groups on carbon 3.
b) For this structure you begin with a carbon chain with 6 carbons. A bromine atom can be added to the end or beginning of the chain depending on which way you number your carbons. The double bond can then be placed between the 2nd and 3rd carbon because of the "2-hexene". Finally, two methyl groups will be placed on carbons 2 and 3 to fulfill the "dimethyl".
c) The first thing to recognize in this structure's name is "ether". This insinuates that there will be an oxygen between two carbons. Vinyl means that the oxygen will be attached to the carbon on the triple bond. Finally, an ethyl group would also be attached to the other side of the oxygen.
d) Alcohol in this compound's name automatically means there will be an -OH in the structure. However, the allyl part means the -OH will be attached to the carbon next to the carbon with the double bond.
Before reading you explanation, I tried drawing these structures for practice. I found that I need more practice than I thought I did. I had forgotten what Vinyl and Allyl meant and this was the kind of thing I was unlikely to focus studying because I thought I knew everything. So thank you for the reminder!
ReplyDeleteThis comment has been removed by the author.
ReplyDeleteWhenever I do these types of problems I always like to break it up into parts. For some reason I would get into drawing cyclo molecules when it was not even mentioned during the first few days of class. I guess I just thought benzene rings and cycloalkanes looked cool! One thing that I learned was to underline each part. I’ll show you what I mean:
ReplyDeletea) 3,3-dimethyl cyclo pentene
b) 6-bromo 2,3-dimethyl 2-hexene
c) ethy vinyl ether
d) allyl alcohol
One thing that can lead to a mistake is if you don’t read carefully. Knowing your groups is essential but attention to detail is key. A and B I think is the only one that would have been trickily if you did not see the –ene and you were on the Final where your mind was not just focusing on alkenes.
I would have to agree with Daniel, I always start these types of problems by breaking the name down into the smallest groups possible. I also had a really bad habit of drawing cyclo - whatever, even when it wasn't in the name. I then like to look at the end of the name and work backwards, so for example, 3,3-dimethyl cyclo pentene... Pent - 5, ene - a double bond (at the beginning because there isn't a number for it), cyclo - I close the pentene, and finish by adding two methyls on Carbon #3. If you just take your time, read the problem carefully, and apply all of the little things that give the compound its name, you can't go wrong.
ReplyDelete