Question

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The following two drawings are resonance structures of one compound. a-o But the following two drawings are not resonance structures Not resonance structures They are, in fact, two different compounds. Choose the correct explanation(s). Select all that apply. Benzene does not have three C-C single bond and three C-C double bonds. In fact, all six C-C bonds of the ring have the same bond order and same length. Benzene have three C- single bond and three C-C double bonds. All the six C-C bonds of the ring does not have the same bond order and same length Cyclooctatetraene has four isolated bonds and the molecule adopts tub shape to avoid aromaticity Cyclooctatetraene has four isolated x bonds and the molecule adopts tub shape to avoid anti-aromaticity In cyclooctatetraene, some of the C-C bonds are double bonds and some of the C-c bonds are single bonds Cyclooctatetraene does not have four C- single bond and four C-C double bonds. In fact, all the eight bonds of the ring have the same bond order and same length Therefore, the two methyl groups can be seperated by a C-C single bond or a C C double bond and these two possibilities represent different compounds. Therefore, the two methyl groups can be seperated by a C-C single bond or a C C double bond and these two possibilities represent resonance structures

Answer 1

1.

They are resonance structures because they are interconvertible when moving electrons So: Benzene have three C-C single bond and three C = C double bonds. All six C-C bonds of the ring does not have the same bond order and same length.

2.

They are not resonance structures but isomers 21 E Z 3E Z 3 (12,32,52,7Z)-1,2-dimethylcycloocta-1,3,5,7-tetraene (1E,37,52,52, 7E)-1,2-dimethylcycloocta-1,3,5,7-tetraene So: The two methyl group can be separate by a C-C single bond or C=C double and these two possibilities represent different compounds.

3

The resonance structure is

4

-H 2 electrons a cyclopropenyl cation Polygon Method for cyclopropenyl cation: 3-member ring Antibonding Energy Bonding Ground state

5.

H-H Antibonding Molecular Orbital LUMO "Lowest Unoccupied Molecular Orbital" 1s 1s Energy H-H Bonding Molecular Orbital HOMO "Highest Occupied Molecular Orbital" H₂ H I So: The Highest Occupied Molecular Orbital is: The bonding Molecular Orbital

6.

Na, NH3(0) CH, OH meta-xylene So: meta-xylene

7.

: 0:1 LOR Each atom of carbon, oxygen and nitrogen adopts sp hybridization. The relocation of an unshared electron pair of oxygen completes the electron sextet required for aromaticity according to Hückel rule LOR So: It's aromatic

8.

- H + Conjugate base Highly stabilized 6 en 4n+2 Aromatic -H* On Conjugate base Highly unstable 8 et 4n Antiaromatic

9.

Aromatic compounds are organic compounds composed of carbon and hydrogen atoms arranged in ring structures with delocalized pi electrons (4n+2). The antiaromatic compounds are cyclic, flat and completely conjugated molecules, but consist of 4n a electrons Non-aromatic compounds are molecules that lack one or more of the requirements to be aromatic: flat and cyclic structure, fully conjugated system C-sp It is not cyclic, therefore it is not aromatic Cyclic, conjugated but with a sp carbon, therefore it is non-aromatic Cyclic, conjugated but with 4n a electrons, therefore it antiaromatic Cyclic, conjugated with 4n+2 i electrons, therefore it aromatic Cyclic, conjugated with 4n+2 i electrons, therefore it aromatic

9

—N EO Act as a base —ОН Act as a base The lone pair on the nitrogen atom in compound B is deslocalized which in estabilishing aromaticity