Chemistry Determination of the molecular structure of two organic compounds

[Sommerfeld_01]

Homework Statement

(It's a translation from spanish). A compound 'A' has the molecular formula C7H8O. This compound is insoluble in water, diluted HCl, and an aqueous solution of NaHCO3. However, it is soluble in an aqueous solution of NaOH. When compound 'A' is treated with bromine water at room temperature, compound 'B' is rapidly formed with the molecular formula C7H5OBr3. What are the structures of compounds 'A' and 'B'?

Relevant Equations

C7H8O + Br2 + H2O ---> C7H5OBr3

I've figured out that both compounds have 4 unsaturations. The fact that they are insoluble in water makes me think they might have a benzene ring. When bromine water is added (I found out online that it's Br2 in water), perhaps some hydrogens from the benzene ring might be replaced. However, I'm not entirely sure how to use the rest of the information to determine the position of the oxygen and the extra carbon, or if the reaction I'm suggesting is correct. So, I would really appreciate it if someone could point me to some sources to learn how to do this.

 

[jim mcnamara]

There are very good resources - google C7H8O then follow the link to Pubchem - this gets you further along.
Formulas written like C7H8O can represent more than one molecule or isomers sometimes.

Otherwise it is a matter of trial and error and not necessarily efficient.

 

[TeethWhitener]

The solubility information tells you about what functional groups might be present in the molecule. For instance, solubility in acids generally denotes basic functional groups, while solubility in base denotes acidic functional groups.

Also, do you know what kinds of organic compounds react with bromine water? Specifically, what kinds of benzene derivatives? (you're on the right track with the benzene ring)

 

[Sommerfeld_01]

Your explanation led me to consider C6H5CH2OH (benzyl alcohol) as compound 'A,' which should have a slight acidic nature. Moreover, I recently learned that benzene derivatives capable of reacting with bromine water are those containing activating groups and it should generate 1,3,5-Tribromophenylmethanol. I truly appreciate your guidance.

 

[TeethWhitener]

I’m not sure this is right. The -CH2OH group is not particularly activating to the phenyl ring. There is a far more strongly activating group that you should consider.

 

[Sommerfeld_01]

Sorry for not answering sooner. I've been a little busy, so I've been taking a deeper look at activating and deactivating groups. I believe the answer to the problem might be 3-methylphenol for compound 'A', and 2,4,6-tribromo-5-methylphenol for compound 'B'. Again, I'm thankful for the feedback

 

[TeethWhitener]

Why would the methyl group have changed position upon reaction with bromine water?

Edit: to be clear, you’re likely not going to get any stereochemical information out of these experiments. But you should still think about the question above as it relates to your OP.

 

[Sommerfeld_01]

Sorry, I made a mistake in naming it. Although it really don't change the position of the methyl in the name I gave, since it's a cyclic structure, in both cases the methyl it's just a carbon away from the OH, but you're right I should have started counting towards the methyl group, so the name of compound 'B' would be 2,4,6-tribromo-3-methylphenol