Draw and name all possible aromatic compounds with the...

Key Concepts

Structural Determination from Molecular Formulas

Determining viable structures from a given molecular formula involves calculating degrees of unsaturation and applying rules for stable bonding configurations. This approach is key to systematically identifying all possible isomers that satisfy the formula and structural requirements.

Halogenated Aromatic Compounds

Halogenated aromatic compounds are aromatic molecules that contain one or more halogen atoms attached either directly to the aromatic ring or within substituent groups. The presence of halogens like bromine introduces additional isomerism and influences the chemical behavior of the molecules.

Substituted Benzene Derivatives

Substituted benzene derivatives are compounds derived from benzene by replacing one or more hydrogen atoms with other atoms or groups. This concept is essential in understanding how additional groups, such as alkyl or halogen substituents, can be attached to the benzene ring while maintaining aromaticity.

Constitutional Isomerism

Constitutional isomers are molecules that share the same molecular formula but differ in the connectivity of their atoms. In problems requiring the enumeration of possible structures, understanding constitutional isomerism enables one to systematically consider all distinct arrangements that meet the given formula.

Aromaticity

Aromaticity refers to the unique stability exhibited by cyclic, planar molecules with delocalized ?-electrons that obey Huckel's rule. This concept is fundamental when dealing with benzene and its derivatives, as it dictates the structural and electronic properties of the compounds involved.

Positional Isomerism in Aromatic Substitution

Positional isomerism, particularly in aromatic systems, involves the different relative positions (ortho, meta, and para) of substituents attached to a benzene ring. Recognizing and applying these variations is crucial when naming and drawing all possible substituted aromatic compounds.

Transcript

00:01 Solving problem 22 of chapter 15.

00:06 Draw and name all possible aromatic compounds with the formula c8h9 br.

00:12 And it's also telling us that there are 14 structures.

00:18 So let's write down the formula in here.

00:20 We need to satisfy 8 carbons, 9, hydrogen, 1 bromine.

00:27 So we can use benzene that for sure is aromatic.

00:34 We know it has six carbons and six hydrogen, so we need two extra carbon and three extra hydrogen.

00:42 So let's start positioning the bromine in here and let's add two method groups, for example, in this position.

00:53 So now we have eight carbons, nine hydrogens because there are one, two, three, three, three.

00:58 456, 7, 8, 9 and 1 bromine.

01:02 And this structure is going to be called 1, bromo, 2, 3, dimethyl, benzene.

01:20 Ok.

01:21 So now with this structure, we can have many other combination just moving these two methods around.

01:28 So if we number the carbon atoms of the benzene ring as 1, 2, 3, 4, 5, and 6.

01:38 Six we can have for example the one bromo i'm not going to rewrite the name because it's going to be the same i just um will change the numbers so this one is referred to the bromo and actually we always stay as one and these two numbers are referred to the position of the two methyl groups and these are those that will change so we can have for example the one bromo two for the metil benzene the that means we'll have the bromo on carbon number one, the second methyl on carbon number two, and the third methyl will have on carbon number four instead of carbon number three.

02:23 Then we can have one bromu, two, five dimethyl benzene, one bromo, two, six dimethyl benzene, 1 -bromo 3 -4 the methyl -benzin 1 -bromo 3 -5 the methyl -benzin and that's it because if you keep trying moving the metals around other than this combination you will end up with one of these.

02:58 Then another possibility is to actually put the two extra carbons and hydrogen together.

03:07 So for example, we can have, we keep the bromine in here and we put in this position ch2, ch3.

03:19 So the formula is still satisfied because we have eight carbons, nine hydrogen and one bromine.

03:26 And this molecule is going to be called one bromo, one, bromon two.

03:35 This group is going to be called ethyl, so two, ethyl, benzene...

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