4.1.3 (c,d) Stereoisomerism in Alkenes

Syllabus

(c) (i) explanation of the terms:

• stereoisomers (compounds with the same structural formula but with a different arrangement in space)

• E/Z isomerism (an example of stereoisomerism, in terms of restricted rotation about a double bond and the requirement for two different groups to be attached to each Carbon atom of the C=C group)

• cis–trans isomerism (a special case of E/Z isomerism in which two of the substituent groups attached to each Carbon atom of the C=C group are the same)

{Use of E as equivalent to trans and Z as equivalent to cis is only consistently correct when there is an H on each Carbon atom of the C=C bond.}

(ii) use of Cahn–Ingold–Prelog (CIP) priority rules to identify the E and Z stereoisomers

{Assigning CIP priorities to double or triple bonds within R groups is not required}

(d) determination of E/Z or cis–trans stereoisomers of an organic molecule, given its structural formula

What does this mean?

Because of the lack of rotation of C=C double bonds these two molecules are different.

But they have the same formula so they are isomers.

And the atoms are arranged in the same order with the only difference being the spatial arrangement of the bonds - hence Stereo-isomers

So they'll need different names despite both being But-2-ene.

Trans means across (Trans-Atlantic etc) , Cis means together.

So we have Cis But-2-ene when the Hydrogen atoms are together on the same side of the double bond.

And Trans But-2-ene when the Hydrogen atoms are on the opposite sides of the double bond.

Unfortunately, this naming system now only applies when the groups on both sides of the C=C bond are identical.

The E-Z system could be applied to the above, where Z stands for Zusammen - same side.

And E stands for Entgegen -opposite in German.

The examiner won't expect you to know the German but they will expect you to use and understand E-Z when naming compounds

There are rules for how to work out which isomer is E or Z when you don't have identical groups on both sides of the C=C double bond.

These are the Cahn, Ingold & Prelog (CIP) rules.

In the diagram above Br has a higher priority than F because its Atomic number is larger.

And Cl has a higher priority than H because its Atomic number is larger.

So the left molecule is Z 1-Bromo-2-Chloro-1-Fluoroethene because the two high pripority atoms are on the same side.

And the right molecule is E 1-Bromo-2-Chloro-1-Fluoroethene because the two high pripority atoms are on opposite sides.

The above molecule is 3-methyl pent-2-ene but is it E or Z.

On the right the C atom in CH₃ takes priority over the H atom.

On the left both atoms are C, but the ethyl group takes priority over the methyl group.

So the high priority groups are on the same side making it a Z isomer.

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Exam-style Questions

1. (i) Name the alkene CH3CH2CH=CH2

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(ii) Explain why CH3CH2CH=CH2 does not show geometrical isomerism.

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(iii) Draw an isomer of CH₃CH₂CH=CH₂ which does show geometrical isomerism.

(iv) Draw another isomer of CH₃CH₂CH=CH₂ which does not show geometrical isomerism.

(Total 4 marks)

2. The Alkene CH₃CH=C(CH₃)CH₂CH₃ reacts with Hydrogen Bromide to form 3-bromo-3-methylpentane, CH₃CH₂CBr(CH₃)CH₂CH₃, as the major product.

(a) Give the name of this alkene and state the type of stereoisomerism shown by this compound.

Name of Alkene .....................................................................................................................................................

Type of stereoisomerism .....................................................................................................................................(2)

Answers

1. (i) But-1-ene (1)

(ii) two H on one Carbon of double bond (1)

(iii) CH₃CH=CHCH₃ (1)

(iv)

(1)

[4]

2. (a) Name of alkene 3-methylpent-2-ene (1)

Type of stereoisomerism geometrical or E-Z (1)

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