After the reaction with carbon dioxid, benzoic acid is in the ether layer. But a

Question

After the reaction with carbon dioxid, benzoic acid is in the ether layer. But after adding 3 M NaOH solution benzoic acid will be in the aqueous layer. How is that possible? AND how does this process separates the byproducts biphenyl and benzene and remaining starting material (bromobenzene) from the product benzoic acid? Include a reaction equation as the main part of your answer. (Write water soluble/insoluble under the compounds.) What are the significant signals of benzoic acid in the IR spectra and at what wavenumber do you expect these signals? (Not fingerprint region!) Why do the signals of the aromatic hydrogens of benzoic acid appear downfield in the^1H NMR?

Explanation / Answer

Q5.

A)

The benzoic acid will react with the NaOh in order to form benzoate ions + water from neutralization

note that the benzoate ions, as the name states, are ions in solution, in aqueous solution.

B)

Biphenyl and benzene will be in MOLECULAR state, so they will go to the organic layer while the aquoeus solution remains with benozoate (bencoic acid)

C)

The equation

C6H5COOH(aq) + NaOH(aq) --> NaC6H5COO(aq) + H2O(l)

recall that

Na+(aq) + C6H5COO-(aq) in solution

Q6.

Benzoic acid has --> C=O group in peak 1750

C-OH (alcohol) broad peak in 3200-3400 range

Q7

Because they are shielded, since the ring is closed

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