For which of the following might you expect aromaticity (geometry permitting)? (a) The annulenes containing up to 20 carbons. (Annulenes are monocyclic compounds of the general formula \([-\mathrm{CH}=\mathrm{CH}-]_{\mathrm{n}}\). (b) The monocyclic polyenes \(\mathrm{C}_{9} \mathrm{H}_{10} \mathrm{C}_{9} \mathrm{H}_{9}^{+}, \mathrm{C}_{9} \mathrm{H}_{9}{ }^{-}\).

Annulenes have a general formula of \([-\mathrm{CH}=\mathrm{CH}-]_{\mathrm{n}}\). We are given that the annulenes contain up to 20 carbon atoms. We will analyze them based on whether they can satisfy Huckel's rule for aromaticity.

To satisfy Huckel's rule, annulenes must have a planar and cyclic structure with a continuous overlapping of p-orbitals, and the total number of delocalized π-electrons must be 4n+2. For each annulene, we can find the total number of π-electrons as every double bond has 2 π-electrons. Therefore: - If n is even, the annulene compound will have n π-electrons. - If n is odd, the annulene compound will have n-1 π-electrons, as one of the carbons will not be double-bonded. Now we will check if the total number of π-electrons in various annulene structures satisfies Huckel's rule (4n+2).

Analyzing the annulenes containing up to 20 carbons in terms of total π-electrons and Huckel's rule: - For n=6: [6]annulene has 6 π-electrons. According to Huckel's rule, the formula becomes 4n + 2 = 6, which yields a solution of n=1 (an integer). Therefore, [6]annulene (benzene) is aromatic. - For higher even values of n: After reaching 6, all other even values of the total number of π-electrons will not satisfy Huckel's rule since these values will no longer be of the form 4n+2. - For odd values of n: In odd annulene cases, there will be n-1 π-electrons, which cannot satisfy Huckel's rule since n-1 is an even number and cannot be written in the form of 4n+2. Thus, the only up to 20 carbons containing annulene species that might exhibit aromaticity is [6]annulene (benzene). All the other annulenes up to 20 carbons will not be aromatic. #Part (b): Monocyclic Polyenes#

The given monocyclic polyenes are \(\mathrm{C}_{9} \mathrm{H}_{10}, \mathrm{C}_{9} \mathrm{H}_{9}^{+}, \mathrm{C}_{9} \mathrm{H}_{9}{ }^{-}\). We will analyze whether these polyenes satisfy Huckel's rule for aromaticity by counting their π-electrons and checking whether they have a cyclic structure with continuous overlapping of p-orbitals.

Analyzing each polyene by counting their π-electrons: 1. \(\mathrm{C}_{9} \mathrm{H}_{10}\): In this molecule, there are 4 double bonds, which give a total of 8 π-electrons. This number is not of the form 4n+2 and thus cannot satisfy Huckel's rule for aromaticity. 2. \(\mathrm{C}_{9} \mathrm{H}_{9}^{+}\): In this cation, there are also 4 double bonds, resulting in a total of 8 π-electrons. Again, this number is not of the form 4n+2 and cannot satisfy Huckel's rule for aromaticity. 3. \(\mathrm{C}_{9} \mathrm{H}_{9}{ }^{-}\): In this anion, there are 5 double bonds with 10 π-electrons, which is not of the form 4n+2. Therefore, this molecule also cannot satisfy Huckel's rule for aromaticity. None of the given monocyclic polyenes can exhibit aromaticity as none of them satisfy the conditions stated in Huckel's rule.