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Propose a structure for an aromatic hydrocarbon, C9H12, that can form two C9H11Cl products on substitution of a hydrogen on the aromatic ring with chlorine. You do not have to consider stereochemistry. In cases where there is more than one answer, just give one. Ignore the ortho, para-directing effects of the alkyl groups in answering this question. Consider only the number of nonequivalent hydrogens on the aromatic ring.

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7 votes

Final answer:

A suitable structure that fits the criteria C9H12 and can form two distinct C9H11Cl products is 1-ethyl-3-methylbenzene. This arrangement provides two nonequivalent hydrogen positions for substitution by chlorine.

Step-by-step explanation:

To determine a possible structure for an aromatic hydrocarbon with formula C9H12 that can form two different C9H11Cl products upon chlorination, we must design a molecule with two types of nonequivalent hydrogens. Given that we are ignoring ortho, para-directing effects, we're solely focusing on the number of distinct hydrogen atoms that can be substituted.

An example of such a structure could be an ethylbenzene derivative. Ethylbenzene has a seven carbon aromatic (benzene) and a two-carbon ethyl group (-CH2-CH3) attached to it. If we place the ethyl group such that there is only one hydrogen on the benzene ring that is equivalent to another hydrogen, we can satisfy the conditions of the problem. For the structure C9H12, this can be achieved if the ethyl group is attached to the ring in a position adjacent to some of the remaining hydrogens, specifically the meta position to prevent equivalence.

Thus, one suitable structure for C9H12 is 1-ethyl-3-methylbenzene. In this case, there would be two nonequivalent hydrogens on the benzene ring that could be substituted by chlorine to yield two distinct C9H11Cl products.

User Amera
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3 votes

Answer:

Step-by-step explanation:

From the given study,

the molecular formula of the given aromatic compound = C9H12

the first process we suppose to carry out is to calculate the double bond equivalence.

So, the double bond equivalence
DBE = {N_C+1-(((N_H+N_(Cl))-N_H)/(2)) }


DBE = {9+1-(((12+0)-0)/(2)) }

DBE = 4

Hence, the aromatic compound possesses 4 double bonds, this signifies that the might be a presence of benzene ring.

The substitutional products of the aromatic hydrocarbon C9H12 can be seen in the attached file below.

Propose a structure for an aromatic hydrocarbon, C9H12, that can form two C9H11Cl-example-1
User Brian D
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