8) Explain the observation that biphenyl uv spectrum show overlapping orbitals of the two rings, but that of the dimesityl derivative showed absorption that correspond to two isolated 1,3,5-trimethylbenzene rings
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- -Ocimene is a pleasant-smelling hydrocarbon found in the leaves of certain herbs. It has the molecular formula C10H16 and a UV absorption maximum at 232 nm. On hydrogenation with a palladium catalyst, 2,6-dimethyloctane is obtained. Ozonolysis of -ocimene, followed by treatment with zinc and acetic acid, produces the following four fragments: (a) How many double bonds does -ocimene have? (b) Is -ocimene conjugated or nonconjugated? (c) Propose a structure for -ocimene. (d) Write the reactions, showing starting material and products.Treatment of compound E (molecular formula C4H8O2) with excess CH3CH2MgBr yields compound F (molecular formula C6H14O) after protonation with H2O. E shows a strong absorption in its IR spectrum at 1743 cm-1. F shows a strong IR absorption at 3600–3200 cm-1. The 1H NMR spectral data of E and F are given. What are the structures of E and F?Compound E signals at 1.2 (triplet, 3 H), 2.0 (singlet, 3 H), and 4.1 (quartet, 2 H) ppmCompound F signals at 0.9 (triplet, 6 H), 1.1 (singlet, 3 H), 1.5 (quartet, 4 H), and 1.55 (singlet, 1 H) ppmFriedel–Crafts alkylation of benzene with (R)-2-chlorobutane and AlCl3 affords sec-butylbenzene. Would you expect the product to exhibit optical activity? Explain, with reference to the mechanism.
- why do (R)-3-chloro-1-methylcycloprop-1-ene and 5-Bromo-1,3-Cyclohexadiene react differently with cyanide (why doesn't cyanide pass through its usual mechanism for these two products and what did you obtain these specific products)The following compound, 2-hydroxycycloheptatrienone, does not give all the usual carbonyl group reactions. (i) Explain this apparent anomaly. Furthermore, explain the significant difference between the relative stability of this compound and its 3- and 4-hydroxy isomers. (ii) What would you expect to be the influence of adding (a) a nitro group as a substituent on the 7-membered ring or (b) an alkyl group, on the degree of aromaticity of the above compound. (iii) Based on your understanding of the hydrogen bonding concept, which isomer would you expect to have a higher m.p, (assuming they are both solids) between 2- and 4- hydroxycycloheptatrienone? Explain your choice.Reaction of (CH3)3CCHO with (C6H5)3P=C(CH3)OCH3, followed by treatment with aqueous acid, affords R (C7H14O). R has a strong absorption in its IR spectrum at 1717 cm−1 and three singlets in its 1H NMR spectrum at 1.02 (9 H), 2.13 (3 H), and 2.33 (2 H) ppm. What is the structure of R? We will learn about this reaction in Chapter 18.
- Although the carbonyl absorption of cyclic ketones generally shifts to higher wavenumber with decreasing ring size, the C = O of cyclopropenone absorbs at lower wavenumber in its IR spectrum than the C = O of cyclohex-2-enone. Explain this observation by using the principles of aromaticity.Wolff-Kishner reduction of compound W gave compound A. Treatment of A with m-chloroperbenzoic acid gave B which on reduction with LiAH4 gave C. Oxidation of compound C with chromic acid gave D (C9H14O). Suggest the structures for A, B, C, and D.(a) Which diastereomer of oct-4-ene yields a mixture of two enantiomers, (4R,5R)- and (4S, 5S)-4,5-dibromooctane on reaction with Br2? (b) Which diastereomer of oct-4-ene yields a single meso compound, (4R, 5S)-4,5-dibromooctane?
- Propose a structure consistent with each set of data.a. Compound A:Molecular formula: C8H10OIR absorption at 3150–2850 cm–11H NMR data: 1.4 (triplet, 3 H), 3.95 (quartet, 2 H), and 6.8–7.3 (multiplet, 5 H) ppm b. Compound B:Molecular formula: C9H10O2IR absorption at 1669 cm–11H NMR data: 2.5 (singlet, 3 H), 3.8 (singlet, 3 H), 6.9 (doublet, 2 H), and 7.9 (doublet, 2 H) ppmSN1 substitution and E1 elimination frequently compete in the same reaction.(a) Propose a mechanism and predict the products for the solvolysis of2-bromo-2,3,3-trimethylbutane in methanolCompounds X has the formula C7H15Cl; Y is C7H15Br.X undergoes base-promoted E2 elimination to give a single alkene product Z. Y likewise reacts under similar conditions to give a single alkene product that is isomeric with ZCatalytic hydrogenation of Z affords 3-ethylpentane.X readily reacts in SN2 fashion with sodium iodide in acetone. Y does not undergo a similar SN2 reaction. Propose structures for X and Y.