Structural Elucidation 1

Structural Elucidation questions are usually quite challenging as it requires students to be familiar with all the organic reactions in syllabus and identify the unknown compounds.

We need a very systematic approach to deduce the observations and identify the compounds.

Let's look at this example:

We have a few unknown compounds (from F to J) to identify, so let's do this part by part.

Deduce Functional Groups in F (C₈H₁₂O)

F undergoes acid base reaction with sodium to form H₂ gas, so either alcohol or phenol is present (acid is eliminated since F has only 1 oxygen).

F cannot be oxidised by dichromate so it'll have either a tertiary alcohol or phenol functional group as both cannot be oxidised.

F has a close C to H ratio, which suggests a high degree of unsaturation in F.

This is also confirmed in the reaction of F by KMnO₄ to form G, H and CO₂.

Hence alkene functional groups are present in F.

Also, phenol is rejected since it does not undergo oxidative cleavage and the oxygen in F is due to a tertiary alcohol.

Deduce Functional Groups in G (C₄H₆O₅) and H (C₂H₂O₄)

F undergoes oxidative cleavage with KMnO₄ to form G, H and 2 CO₂.

Products of oxidation have to be stable to oxidation and products of oxidative cleavage are usually ketone (1 oxygen), acid (2 oxygen) and CO₂.

This means we can deduce the number and type of functional groups in G and H.

G has 5 oxygens so should have 2 acid groups and 1 ketone.

However recall there is a tertiary alcohol in F which is resistant to oxidation, hence it will not be oxidised by KMnO₄ and that odd oxygen should be due to tertiary alcohol instead of ketone.

Hence G has 2 acid groups and 1 tertiary alcohol group.

H has 4 oxygens so should have 2 acid groups.

Deduce H (C₂H₂O₄)

H can be further oxidised to CO₂ and this means H must be ethanedioic acid (COOH)₂ which is one of the 2 acids in A Level Chemistry syllabus that can be oxidised to CO₂ and H₂O.

The other acid that can be oxidised to CO₂ and H₂O is methanoic acid, HCOOH.

Deduce J (C₃H₂O₅), I (C₄H₄O₄) and G (C₄H₆O₅)

G reacts with conc H₂SO₄ to form I (C₄H₄O₄).

Comparing molecular formula we can deduce this is an elimination of H₂O from tertiary alcohol in G to form alkene in I.

I undergoes oxidative cleavage with KMnO₄ to form J (C₃H₂O₅) and CO₂.

The additional oxygen in J should be due to a ketone hence we can draw out the structure for J (HOOCCOCOOH).

The CO₂ evolved is due to a terminal alkene carbon (CH₂=)

So we can work backwards to figure out the structure of I and G.

Remember G has a tertial alcohol so we have to put the -OH group on the tertiary carbon instead of methyl carbon.

Deduce F (C₈H₁₂O)

Now we can list down the products of oxidative cleavage of F and deduce the different parts of alkene that will form F.

A simple way to figure this out is to remove the oxygen from the products of oxidative cleavage.

Now we have 4 fragments that we need to piece together.

Thankfully fragments 1 and 2 are symmetrical, which means that putting everything together:

3-1-2-4 and 3-2-1-4 are the same compound F.

It's always a good idea to verify our answers by checking the molecular formula given in the question.

Topic: Organic Chemistry, A Level Chemistry, Singapore

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