Compound \(\mathrm{A}\left(\mathrm{C}_{9} \mathrm{H}_{12}\right)\) shows prominent peaks in its mass spectrum at \(\mathrm{m} / \mathrm{z} 120\) and 105. Compound \(\mathrm{B}\) (also \(\mathrm{C}_{9} \mathrm{H}_{12}\) ) shows prominent peaks at \(m / z 120\) and 91 . On vigorous oxidation with chromic acid, both compounds give benzoic acid. From this information, deduce the structural formulas of compounds A and B.

Short Answer

Expert verified
Answer: Compound A has the structural formula of para-Methyl Toluene or para-Xylene (\(\displaystyle \mathrm{C}_{6} \mathrm{H}_{4}( \mathrm{H}\mathrm{C})\left(\mathrm{CH}_{3}\right)_{2}\)), and compound B has the structural formula of ortho-Ethyl Toluene (\(\displaystyle \mathrm{C}_{6} \mathrm{H}_{4}\left(\mathrm{CH}_{3}\right)\left(\mathrm{CH}_{2} \mathrm{CH}_{3}\right)\)).

Step by step solution

01

Deduce possible formulas based on mass spectral data

Prominent peak at m/z 120 is due to molecular ion or parent peak, so the molecular weight of both compounds A and B is 120. Given that both compounds have molecular formula C9H12, we can proceed further. Compound A displays a prominent peak at m/z 105, which is a difference of 15 units from the parent peak. Most likely, this difference corresponds to the loss of a methyl (CH3) group. Compound B shows a prominent peak at m/z 91, that is a difference of 29 units from the parent peak. This difference likely corresponds to the loss of an ethyl (C2H5) group.
02

Consider oxidation product - benzoic acid

We know that both compounds A and B, on vigorous oxidation with chromic acid, give benzoic acid. Benzoic acid has the molecular formula C7H6O2. This tells us that our original compounds A and B should have two additional carbons when compared to benzoic acid, with one of them likely derived from the CH3 or C2H5 group losses observed in the mass spectrum.
03

Deduce the structure of compound A

Combining the information from Steps 1 and 2, we can deduce that compound A is a derivative of benzoic acid with an extra methyl group and one extra carbon. Adding this extra carbon to the benzoic acid structure as a side chain and methyl group at the end gives the structure for toluene which is C7H8. With a methyl group at the para-position (opposite to the methyl group in toluene), the structure will be C9H12. Thus, the structure of compound A is para-Methyl Toluene or para-Xylene: para-Xylene (A): \(\displaystyle \mathrm{C}_{6} \mathrm{H}_{4}( \mathrm{H}\mathrm{C})\left(\mathrm{CH}_{3}\right)_{2}\)
04

Deduce the structure of compound B

Following a similar approach as compound A, compound B is a derivative of benzoic acid, but with an extra ethyl group instead of a methyl group. Also, we will add an extra carbon to the benzene ring. One possibility is to add ethyl to a methylbenzene (Toluene) at the ortho position (next to the methyl group). This gives us the structure: ortho-Ethyl Toluene (B): \(\displaystyle \mathrm{C}_{6} \mathrm{H}_{4}\left(\mathrm{CH}_{3}\right)\left(\mathrm{CH}_{2} \mathrm{CH}_{3}\right)\) Both para-Xylene (A) and ortho-Ethyl Toluene (B) have a molecular formula of C9H12 and exhibit the correct loss of CH3 and C2H5 groups in their mass spectra. They also produce benzoic acid after vigorous oxidation with chromic acid, as required by the given information.

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