The Student Room Group

Optical and Geometric Isomeris

Q1: Draw the skeletal formula of a branched-chain aldehyde with molecular formula C5H10O that is optically active.

I know that the molecule has to have a chiral centre to be optically active and it has to have oxygen double-bonded to a carbon on the end to be an aldehyde. What I came up with was:
CH3CH(CH2CH3)CHO
I'm not sure if it's right or not but whatever the right answer is I cannot wrap my head around how to draw the skeletal formula for this.


Q2: Draw the E and Z forms of a structural isomer of C5H10O that shows both optical and geometric isomerism.

I don't even know where to begin with this question, other than the fact that there must be a C=C bond and a chiral centre.

Does anyone know how to do these questions?
Original post by Maci.
Q1: Draw the skeletal formula of a branched-chain aldehyde with molecular formula C5H10O that is optically active.

I know that the molecule has to have a chiral centre to be optically active and it has to have oxygen double-bonded to a carbon on the end to be an aldehyde. What I came up with was:
CH3CH(CH2CH3)CHO
I'm not sure if it's right or not but whatever the right answer is I cannot wrap my head around how to draw the skeletal formula for this.


Q2: Draw the E and Z forms of a structural isomer of C5H10O that shows both optical and geometric isomerism.

I don't even know where to begin with this question, other than the fact that there must be a C=C bond and a chiral centre.

Does anyone know how to do these questions?

The first answer is correct.
For a skeletal formula you just show the carbon -carbon bonds and heteroatoms (not hydrogen)

methylbutanal.jpg
Original post by Maci.

Q2: Draw the E and Z forms of a structural isomer of C5H10O that shows both optical and geometric isomerism.

I don't even know where to begin with this question, other than the fact that there must be a C=C bond and a chiral centre.

Does anyone know how to do these questions?

There does not have to be a C=C bond, only a bond with restricted rotation. This also applies to alicyclic compounds

The IHD remains the same with 1 alicycle so there cannot be a C=O double bond.

optical_and_geometric.jpg

Both of the starred carbon atoms are optically active. Hence, there are 22 = 4 optical isomers.
The hydroxyl group and the methyl group can be either above or below the plane of the ring. This makes two different EZ isomers.
Reply 3
Original post by charco
The first answer is correct.
For a skeletal formula you just show the carbon -carbon bonds and heteroatoms (not hydrogen)

methylbutanal.jpg

Thank you : )

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