Chemistry · General chemistry II · Worked example

Assign E or Z when cis and trans disagree

2-Bromobut-2-ene, ${CH}_{3} C (Br) = {CHCH}_{3}$ , is drawn with its two ${CH}_{3}$ groups on opposite sides of the double bond. Is it E or Z, and can it be called trans?

List the groups on each carbon

Carbon 2 carries $Br$ and ${CH}_{3}$ . Carbon 3 carries ${CH}_{3}$ and $H$ . Three groups besides hydrogen sit on the double bond.

Rank the groups on carbon 2

Bromine has atomic number $35$ and carbon $6$ , so $Br$ outranks ${CH}_{3}$ . The size of the groups plays no part.

Rank the groups on carbon 3

Carbon, $6$ , beats hydrogen, $1$ , so ${CH}_{3}$ outranks H.

Compare the sides

The ${CH}_{3}$ groups are on opposite sides, so the $Br$ on carbon 2 is on the same side as the ${CH}_{3}$ on carbon 3. Those are the two higher-ranked groups, and same side makes it Z.

(2Z)-2-bromobut-2-ene

Decide about trans

The name trans-2-bromobut-2-ene would describe the ${CH}_{3}$ groups, but with three groups on the double bond there is no agreed reference for cis or trans. The systematic name uses Z, and calling this compound trans invites the reader to picture the wrong letter.

Result

Z. The compound is (2Z)-2-bromobut-2-ene, and cis or trans does not apply.

Your turn

1. One like the example

2-Chlorobut-2-ene, ${CH}_{3} C (Cl) = {CHCH}_{3}$ , is drawn with its two ${CH}_{3}$ groups on the same side of the double bond. Is it E or Z?

Show the answer and explanation

E. The compound is (2E)-2-chlorobut-2-ene.

On carbon 2, chlorine, $17$ , outranks ${CH}_{3}$ . On carbon 3, ${CH}_{3}$ outranks $H$ . With the ${CH}_{3}$ groups together, the $Cl$ sits opposite the ${CH}_{3}$ on carbon 3, so the higher-ranked groups are on opposite sides.

2. A variation

Maleic acid, $HOOC - CH = CH - COOH$ , has its two $- COOH$ groups on the same side of the double bond. Give its systematic name with E or Z.

Show the answer and explanation

(2Z)-But-2-enedioic acid.

Each carbon carries one $- COOH$ and one H, and the $- COOH$ outranks the H. Both higher-ranked groups are on the same side, so it is Z. Here cis and Z agree, because each carbon carries one group and one hydrogen.

3. Somewhere new

In ClFC=CBrI, one carbon carries $Cl$ and $F$ and the other carries $Br$ and $I$ . The $Cl$ and the $Br$ are drawn on the same side. Is the alkene E or Z?

Show the answer and explanation

E. The compound is (E)-1-bromo-2-chloro-2-fluoro-1-iodoethene, which Chalk writes (1E).

On the first carbon, $Cl$ , $17$ , outranks $F$ , $9$ . On the second, $I$ , $53$ , outranks $Br$ , $35$ . The $Cl$ is on the same side as the $Br$ , so it is on the other side from the $I$ , and the higher-ranked pair is apart.

Keep exploring

Draw it, then choose Analyze → Stereochemistry, and Chalk puts a Z beside the double bond. Under Analyze → Target enter trans-2-bromobut-2-ene and turn Check on. Chalk marks the name as not systematic, because a double bond with three groups takes E or Z. The Practice buttons import each practice structure without a name, for your own name to be checked.

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Practice the key terms as flashcards, then open the example in its tool, change a value and keep a useful result on your board.

Study the 5 key terms Draw (2Z)-2-bromobut-2-ene Practice 1: import the structure and check your name Practice 2: import maleic acid and check your name Practice 3: import the structure and check your name Open worked example on a board Stereochemistry rules in Chemistry Reference

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