How do you name this compound using IUPAC system (including steps)?

Question

And including stereochemistry (cis trans or R S )

I know you start numbering from the double bond because there are no functional groups like -OH, you take the longest carbon chain which is six carbons in this question, so now it should be 3-bromo-3-methyl-1-hexene. My quarrel is with the stereochemistry and how do you name the compound regarding the widge and the dash (cis and trans or R and S)

I also know CIP priorities, it's just the widge and the dash being at the middle of the compound that confuses me.

trans-2-bromo-2-vinyl-pentane ?
R-2-bromo-2-vinyl-pentane ?

Answer 1

1. Identify the parent chain. In this case you have a 6-carbon chain starting at the alkene, wrapping through the chiral center and ending off to the right. 6 makes the prefix 'hex'; the alkene starts at the '1' position; thus, hex-1-ene.

2. Substituents. In this case they're both at the 3-position so they'll be listed in alphabetical order: 3-bromo-3-methyl

3. Chiral Center at the 3 position. Numbering the legs of the center according to IUPAC rules gives (1) to the bromine atom, (2) to the carbon atom double bonded to the next carbon atom, (3) the carbon atom single-bound to the next carbon atom, and (4) the methyl group. Rotate this tetrahedral image in our head to where you're looking at the molecule with numbers 1-3 towards you. The numbers rotate clockwise, thus the chiral center is (3​R)

4. String it all together.

(3​R)-3-bromo-3-methylhex-1-ene.

Answer 2

When I complete my reading, I guess, I understand your problem of imagine stereochemistry. When we draw a structure, three different lines we use to show stereochemistry of the bonds. They represent as follows:

• Solid lines are on plane of your paper;
• Wedges are above the paper; and
• Dashed lines are below the paper.

That means, $\ce{-Br}$ (highest priority group) is coming out of the paper towards you. Ethylene group (second priotity) is going out inside of the paper away from you. Methyl (least priority) and propyl (third priority) groups are on the paper. Thus, if you look towards methyl group from chiral carbon, you see rotation go from $\ce{-Br}$ to ethylene group to propyl group, making it clockwise rotation. Thus stereoconfiguration is (R)-. The rest is according to the answer by Michael Green and comments by orthocresol. Hope this help you understand the stereochemistry.