2-Nitropyrrole is an aromatic compound. In its structure below, highlight the atom with the lone pair in red if the lone pair occupies a p orbital, blue if it occupies an s p orbital, or green if it occupies an s p orbital. 2 3 -NO 2 ༠, H ྾ ཅ
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- For the aromatic compounds below, draw the p-orbitals for the compound.What is the major difference between an antiaromatic and aromatic compound? Antiaromatic compounds have at least one sp³ hybridized atom in the ring. Antiaromatic compounds are not conjugated. Only aromatic compounds have 4n+ 2 pi electrons. The structure must be cyclic for aromatic but not antiaromatic compounds. Aromatic, not Antiaromatic compounds, are planar.Specify whether the two structures are resonance contributors to the same resonance hybrid. Be sure to explain your reasoning. If yes, be sure to specify which is preferred and why. H2C=NH2
- Q3 Draw the structure of the cyclopentadienyl cation (C5H5*). Sketch a Frost diagram of the T-orbital system of the cyclopentadienyl cation. Clearly illustrate bonding, non-bonding and antibonding energy levels, and show electron occupation of orbitals for the ground state. State whether the cation is aromatic or not and explain your answer.Azulene, an isomer of naphthalene, has a remarkably large dipole moment for a hydrocarbon (μ = 1.0 D). Explain, using resonance structures.LYRICA (pregabalin) is an oral pharmaceutical primarily indicated for the management of fibromyalgia and neuropathic pain. At a pH of 11, LYRICA has an overall charge of –1, with most atoms neutral and only one oxygen showing a –1 formal charge. The structure for LYRICA at pH 11 is shown below. Add all the missing lone pairs to the given structure of LYRICA.
- Nicotine present in tobacco has the following structure. Which nitrogen is more basic and explain your choice. CH3 2 ○ N1 is more basic as it is sp2 hybridized and more electronegative because of less S character. O N2 is more basic as it is sp3 hybridized and less electronegative because of less S character. ON1 is more basic as its lone pair has resonance.Chapter 2 [References] H3C HB CH3 На, CH3 CH3 D H3C B НО Cholestanol differs from cholesterol only in the absence of a double bond in ring B. Draw the three-dimensional structure of cholestanol, and then determine the orientation of the following groups: #1: H at the junction of rings C & D with respect to ring C| #2: H at the junction of rings A & B with respect to ring B #3: Methyl at the junction of rings C & D with respect to ring C| O IThe bonding between the carbon and oxygen in a ketone can be described as C(sp2)-0(sp²), o-bond and C(p)-O(p), л-bond. How would you describe the bonding between carbon and oxygen in a ketene? Sketch the orbitals used to make the bonds in the ketene. H a ketene
- Explain how covalent bonds are formed in each of the following compounds in terms of orbital hybridisation and overlap of orbitals (i) Ethene, C2H4 (ii) Ethyne, C2H2In the sketch of the structure of SO2 label all bonds. Drag the appropriate labels to their respective targets. Labels can be used once, more than once, or not at all. : S(p) – O(p) Lone pair in p orbital Lone pair in sp? orbital o : S(p) – 0(sp²) т: S(p) — О(p) T: S(sp²) – O(p) r: S(sp²) – O(p) S(p) – O(sp²)According to Huckel's rule, what is the value of "n" for the aromatic compound shown here? 1 o 2 3 O 4 O 5