Question #16b44

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Answer 1 · 2016-12-09T20:15:43

The order is $("CH"_3)_3"C-" > "CH"_2"=CH-" > ("CH"_3)_2"CH-" > "CH"_3"CH"_2"-"$

The Cahn-Ingold-Prelog order is based on the atomic numbers of the atoms attached to the chiral centre.

If there is a tie, we go one atom further out and keep going until the tie is broken.

In this problem, the atoms attached to the chiral centre are all carbon atoms, for
a 4-way tie.

To break the tie, we look at the atoms that are directly attached to these carbons.

First trial

$"C1"$ of the vinyl group is attached to another carbon by a $"C=C"$ double bond.

Each carbon in the double bond counts as two carbons.

So, $"C1"$ is attached to $"C, C"$ , and $"H"$ , written as ( $"C,C,H"$ ) with the atoms listed in order of decreasing atomic number.

$"C1"$ of the isopropyl group is attached to $"C, C"$ , and $"H"$ ( $"C,C,H"$ ).

$"C1"$ of the tert-butyl group is attached to ( $"C,C,C"$ ).

$"C1"$ of the ethyl group is attached to ( $"C,H,H"$ ).

We now compare the atoms in parentheses and look at the first point of difference.

We find tert-butyl ( $"C,C,C"$ ) > vinyl ( $"C,C,H"$ ) = isopropyl ( $"C,C,H"$ ) > ethyl ( $"C,H,H"$ ).

So, tert-butyl is Priority 1 and ethyl is Priority 4, but vinyl and isopropyl are tied in the middle.

Second trial

To break the tie, we look at the atoms directly attached to $"C2"$ .

$"C2"$ in the vinyl group is attached to ( $"C,H,H"$ ).

$"C2"$ in the isopropyl group is attached to ( $"H,H,H"$ ).

So the vinyl group wins and gets Priority 2, and the isopropyl group is Priority 3.

The order is $("CH"_3)_3"C-" > "CH"_2"=CH-" > ("CH"_3)_2"CH-" > "CH"_3"CH"_2"-"$