What are the rules to EZ configuration? (Entgegen-Zusammen)

I know that the substituent having elements with the highest atomic number are those that are "involved" I also know the rule with how uncommon isotopes get highest priority before anything else. However, what do I do here?
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Here the #2^(nd)# carbon (counting from left to right) has methyl groups both on the top and bottom. How should I determine which takes highest priority? I am having trouble with this, so it would be very much appreciated if someone could mention some other rules besides the ones I said. Thank you!

2 Answers
Jun 28, 2017

Answer:

(4E, 6E)-2,6-dimethyloct-2, 4, 6-triene

Explanation:

There are 3 C=C bonds to choose from. The left one cannot be used as the left hand side both have the same combined atomic mass on each bond, one methyl group.

The middle C=C bond can be used, as the left hand side only has methylethene group, while the right hand side has the exact same thing. They are also on opposite sides, so it will be of E notation.

The right C=C bond only has a methyl group on the right hand side, the left hand side has a methyl group and a 2-methylpent-2, 4-ene group. The right hand side methyl group and 2-methylpent-2, 4-ene group are also on opposite sides, giving an E notation.

The whole formula is 2,6-dimethyloct-2, 4, 6-triene (without EZ notation). 2,6-dimethyl coming from the 2 methyl groups being on the 2nd and 6th carbon atoms in the main carbon chain. Oct comes from there being 8 carbon atoms in the main carbon chain. 2, 4, 6-triene comes from the C=C bonds being on the 2nd, 4th and 6th carbon atoms in the main carbon chain.

As there are E notations at the C=C bonds on the 4th and 6th carbon atoms in the main carbon chain, the actual name is (4E, 6E)-2,6-dimethyloct-2, 4, 6-triene. The (4E, 6E) shows the E notation on the 4th and 6th carbon atoms in the main carbon chain.

Jun 29, 2017

Answer:

Warning! Long Answer. Here's what I get.

Explanation:

Ignoring configuration, the name of the compound is 2,6-dimethylocta-2,4,6-triene.

Name

Rules for assigning #E,Z# configurations

  1. Assign each substituent on the double bond a priority.
  2. If the two groups of higher priority are on opposite sides of the double bond, the configuration is #E#.
  3. If the two groups of higher priority are on the same side of the double bond, the configuration is #Z#.

Now, let's apply these rules to the three double bonds in the molecule.

The 2,3-double bond

This double bond has no #E,Z# priority because the two groups attached to #"C2"# are methyl groups.

Neither group has priority over the other.

The 4,5-double bond

4,5

The two atoms directly attached to #"C4"# are #"H"# and #"C3"#. #"C3"# has higher priority.

The two atoms directly attached to #"C5"# are #"H"# and #"C6"#. #"C6"# has higher priority.

Since #"C3"# and #"C6"# are on opposite sides of the double bond, the configuration is #E#.

The 6,7-double bond

6,7

The two atoms directly attached to #"C6"# are #"C5"# and the #"C"# of the methyl group (a tie!).

To break the tie, we must go to the atoms next further out.

The methyl carbon is attached to three #"H"# atoms. We list them as #"(H,H,H")#.

#"C5"# is attached to #"C4"# (the double bond counts as two carbons) and an #"H"#, so we list it as #"(C,C,H")# in order of decreasing atomic number.

Then we look for the first point of difference. The first #"C"# of #"C5"# is higher priority than the first #"H"# of the methyl group, so #"C5"# takes precedence.

The two atoms directly attached to #"C7"# are #"H"# and #"C8"#. #"C8"# has higher priority.

Since #"C5"# and #"C8"# are on opposite sides of the double bond, the configuration is #E#.

The complete name of the compound is (#4E,6E#)-2,6-dimethylocta-2,4,6-triene.