The Student Room Group

how do you know when a stereoisomer isn't present ?

as in when the same group is attached to the one c atom in the double bond
If the two groups single bonded to the same carbon are identical there can't be stereoisomerism at that C=C bond.
Original post by TutorsChemistry
If the two groups single bonded to the same carbon are identical there can't be stereoisomerism at that C=C bond.


So if a was an atom with b and c and two as were bonded to one c in the double bond but b and c were bonded to the other c would it be a stereo isomer and how would you know
Original post by jonjoshelvey21
So if a was an atom with b and c and two as were bonded to one c in the double bond but b and c were bonded to the other c would it be a stereo isomer and how would you know


We need to be a bit more specific
If atom "a" is the whole group, the answer is no.

But if atom "a" has further atoms bonded to it then you need to consider those too.
In short you need to consider the whole group. For example, methyl group -CH3 and ethyl group -C2H5 could lead to stereoisomerism even though they would both have a C as the first atom bonded to the C=C carbon. It's not just about the first aton bonded to the C=C carbon, its about the whole group.
Original post by TutorsChemistry
We need to be a bit more specific
If atom "a" is the whole group, the answer is no.

But if atom "a" has further atoms bonded to it then you need to consider those too.
In short you need to consider the whole group. For example, methyl group -CH3 and ethyl group -C2H5 could lead to stereoisomerism even though they would both have a C as the first atom bonded to the C=C carbon. It's not just about the first aton bonded to the C=C carbon, its about the whole group.


So what I meant was if a was one atom or one group, so either the same group or same atom alone is bonded to a c within the double bond
Original post by jonjoshelvey21
So what I meant was if a was one atom or one group, so either the same group or same atom alone is bonded to a c within the double bond


Then the answer is no, it isn't possible to have stereoisomerism.
For stereoisomerism to be present both carbons must have 2 different groups. (The groups on one carbon may be the same as on the other carbon).
Hi,
Stereoisomers are compounds that have the same structural formula and molecular formula but the spatial orientation of the atoms in each compound varies in space.
So in Yr 12 you learn about a branch of steroisomerism known as geometrical isomerism, which arises as a reults of the resticted rotation of the C=C bond, in halogenoalkanes.
This means that the the atoms/groups attached to each carbon atom in the double bond can be varied, but the carbon the group is attached to in the double bond must remain constant.
When both the top or bottom group on both the left hand side and the right hand side of the C=C take priority due to having the largest overall atomic numbers, we refer to them as being Z- isomers after the german word meaning zussammen- together.
When the highest priority group is on the bottom on one side and on the top on the other side we refer to these isomers as being E-isomers after the german word Eintgargen- opposite.
Sorry i went on a bit of a tangent about the Kahn Ingold prelog rules- whci is still relevant.

In Yr13 you learn about another branch of stereoisomerism known as optical isomerism, this arrises when two molecules both with the same structural formulae, but have no plane of symmetry due to the carbon being asymmetrical, due to having four different groups attached. The carbon atoms which is asymmetric is described as being a chiral centre, and the overall molecule being a chiral molecule. The two isomers formed, are non- superimpsoable mirror images of each other, which means that they cant be put on top of each other and be seen as identical. The two optical isomers- each known as an enantiomer has the ability to roatate plane polairsed light in opposite directions, and as a results we can distinguish between the two enantiomers provided they are not in a racemic mixture.


Feel i strayed slightly off topic. Apologies and hope this helps.

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