Propose structures for aromatic hydrocarbons that meet the...

Key Concepts

Electrophilic Aromatic Substitution (EAS)

Electrophilic aromatic substitution is a key reaction mechanism in which an electrophile replaces a hydrogen atom on an aromatic ring. This substitution process is governed by the stability of the intermediate sigma complex and the electronic nature of substituents already present on the ring.

Aromatic Compounds

Aromatic compounds are cyclic molecules with delocalized ?-electrons that follow Huckel's rule, which gives them enhanced stability and unique chemical reactivity. Their planar structures and conjugated systems form the basis for understanding their behavior in many organic reactions.

Substituent Effects and Directing Groups

The presence of substituents on an aromatic ring influences both the reactivity and the position at which new substituents are introduced during EAS. Activating groups typically direct incoming electrophiles to ortho and para positions, while deactivating groups tend to direct them to the meta position, impacting the number and type of isomers formed.

Molecular Symmetry and Regioselectivity

Symmetry in aromatic molecules plays a crucial role in determining the number of unique substitution products. A symmetric substitution pattern may lead to a single product if all positions are equivalent, whereas asymmetric substitution results in multiple isomers. This concept helps predict and rationalize the outcomes of substitution reactions in substituted aromatic rings.

Structure Determination from Molecular Formula

Analyzing a molecular formula for aromatic compounds involves calculating degrees of unsaturation to infer possible rings and double bonds, and proposing structures consistent with observed reactivity. This process often integrates elements of symmetry, substituent effects, and known reaction mechanisms to determine plausible structural isomers.

Transcript

00:01 Solving problem 23 of chapter 15.

00:06 Proposed structures for aromatic hydrocarbons that meet the following descriptions.

00:12 So let's start from the first one.

00:19 C9h12, which gives only 1 c9h11 br.

00:31 C9h11 br.

00:34 On substitution of an hydrogen on the aromatic ring with bromine.

00:39 So the problem is already telling us we have an aromatic ring.

00:46 And these are six carbons, so we need extra three carbons.

00:52 And this compound is obtained by substitution of one of the hydrogen connected, of the benzene ring, so not the hydrogen connected to other substitutes.

01:04 So if we put three metal groups that will account for a total of nine carbons, for example, one in here, one in here, and one in here.

01:21 Now we have nine carbons, as the formula is asking for.

01:26 We have 12 hydrogen, one, two, three, four, five, six, seven, eight, nine, ten, eleven, and twelve.

01:36 So this structure is fine.

01:38 Now, if we think about substituting one of these hydrogens, so either this one, this one, or this one, we obtain, for example, ch3, ch3, for example, let's say we are substituting this one, so we have a br in here.

02:08 Now, it doesn't matter if we substitute the bromine with the hydrogen in this position, this position or this position, because we would obtain three structures that are exactly the same, so they are equivalent.

02:22 And this is because if you image, for example, dividing the molecule in two in this direction or in this direction or in this direction, the molecule is symmetric.

02:35 So whatever substitution on this carbon, this carbon or this carbon will give the same structure.

02:44 Okay, this one was the first one.

02:46 Then the second one will be c10h14, c10h14 that gives only one c10 h13 cl.

03:08 So to give c -10 h -13 cl.

03:17 Always by substituting one of the hydrogen on the benzene ring with the chlorine.

03:23 So for the second one, let's draw the benzene ring with the double bonds.

03:30 Now we have six carbon, we need to add extra four carbons.

03:35 So let's image to add four methyl groups, for example.

03:40 Ch3, okay.

03:44 So each of this line, it means a ch3.

03:46 I'm not going to rewrite all of them.

03:50 Okay, so in this case, if we substitute one of the hydrogen on the benzene ring, for example, this one or this one with the chlorine, we will obtain either the same structure with the chlorine in here or the chlorine in here.

04:06 They are going to be exactly the same because if we cut the molecule, in this direction or in this direction, they are symmetrical or better.

04:24 If we cut the molecule in this direction is going to be symmetric, either you put the chlorine in here or the chlorine in here.

04:36 So this structure is going to give only one structure with the chlorine.

04:44 Then in this case, we can also have, for example, this one.

04:55 So where this one is a ch2, ch3, ch2, ch3 for a total of 6, 7, 8, 9, 10, 10 carbons and 14 hydrogens.

05:08 So now we can substitute the hydrogen with the chlorine in these positions.

05:15 And again, either of these four possible structure are going to be equivalent, because the molecule again is symmetric.

05:26 Then we can also have this one.

05:34 So 6, 7, 8, 9 and 10 carbons and 14 hydrogen.

05:40 So just each of this line is a methyl, a ch3...

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