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

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Discussion Overview

The discussion revolves around deducing the structures of four organic compounds based on a high school exam question involving various chemical tests. Participants explore the implications of molecular formulas and the reactions involved in specific tests, including the iodoform test, DNPH test, and silver mirror test.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Homework-related

Main Points Raised

  • One participant suggests that FA8 is likely CH3COOH and proposes that FA9 could be either propan-1-ol or propan-2-ol, depending on the outcome of the iodoform test.
  • Another participant questions the rationale behind adding I2 until the solution turns orange/red in test 3, seeking clarification on whether this indicates excess I2.
  • Some participants assert that FA10 and FA11 contain carbonyl functional groups, with one suggesting that FA11 is likely an aldehyde and FA10 a ketone.
  • A participant asks about the mechanism of the iodoform reaction, specifically regarding the presence of hydrogens α to the carbonyl without an α methyl group.
  • One participant notes that the mechanism for the iodoform reaction is not taught, emphasizing that it only works if there is an alpha methyl group present.
  • Another participant explains that the hydrogens are substituted by iodine, leading to a loss of color in I2, but points out that the resulting RCI2- is not a good leaving group, causing the reaction to stop.

Areas of Agreement / Disagreement

Participants express differing levels of understanding regarding the mechanisms of the tests and the structures of the compounds. There is no consensus on the exact identities of FA10 and FA11, and the discussion remains unresolved regarding the specifics of the reactions involved.

Contextual Notes

Participants mention limitations in their knowledge of the mechanisms for certain reactions, indicating that some aspects of the tests may not have been covered in their curriculum.

phantomvommand
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Summary:: Please see picture below

Screenshot 2021-10-17 at 3.11.41 AM.png

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.
 
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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?
 
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mjc123 said:
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?
 
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?
 
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mjc123 said:
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.
 
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?)
 
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