Having trouble deducing 4 organic structures (High school exam qn)

[phantomvommand]

Summary:: Please see picture below

The dotted line at Test 3 means to continue using the above solution, and add NaOH with the teat pipette. The dotted line separating tests 3 and 4 should be a hard line, which means to use a fresh sample of FA10 and perform the test. It is also known that FA10 is different from FA11.

I do not have the experimental results, which you are supposed to obtain and fill in the blank boxes. However, one can still deduce the most likely structures given the highly restrictive molecular formula; not that many structures even fit the molecular formula.

I have deduced FA8 to be CH3COOH. FA9 can be propan-1-ol or propan-2-ol, depending on whether the iodoform test (test3) gives the yellow ppt of CHI3.

For test 3, I am not sure why they want us to add I2 until the solution turns orange/red. Does this mean adding I2 in excess?

What might FA10 and FA11 be? Could someone please explain all reactions happening in the tests, especially for FA11?

Thanks so much.

 

[mjc123]

These are all standard tests you should have learned about before being given a question like this. If you know about the iodoform test, you ought to know about the DNPH and silver mirror tests. Do you know what functional groups they are testing for?

 

[phantomvommand]

These are all standard tests you should have learned about before being given a question like this. If you know about the iodoform test, you ought to know about the DNPH and silver mirror tests. Do you know what functional groups they are testing for?

Yes, FA11 and FA10 have carbonyl functional groups. It is likely FA11 is an aldehyde (hence the silver mirror), and FA10 should be a ketone (it is not FA11). However, what is happening in test 3 for FA11, which should be propanal?

 

[mjc123]

Do you know the mechanism for the iodoform reaction? What would happen if you have hydrogens α to the carbonyl, but not an α methyl group?

 

[phantomvommand]

Do you know the mechanism for the iodoform reaction? What would happen if you have hydrogens α to the carbonyl, but not an α methyl group?

No, the mechanism is not taught. We are only taught that the only way it works is if it has an alpha methyl group.

 

[mjc123]

The hydrogens are substituted by I (hence the loss of the colour of I₂). But unlike CI₃^-, RCI₂^- is not a good enough leaving group, so the reaction stops there.
(Note: I there is capital I, not small L as I'm sure you realize. Can't we have a font that distinguishes them?)