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- For each organic compound in the table below, enter the locant of the highlighted side chain. CH3 CH₂ CH₂ H CH3 · C CH₂- CH CH3 CH₂ CH3- compound CH₂ I CH₂- CH CH | CH₂ CH3- — - CH₂ | CH₂-C. CH3 | — ぎーぎ CH3 locant of highlighted side chain 0 0You measured the carbon-carbon bond lengths of four different compounds in Experiment 8 (Molecular Modeling and IR spectroscopy). Which of the following compounds has the shortest average C-C bond length? Benzene Cyclohexene Cyclohexane O Cyclohexadiene Which of the following statement(s) is(are) true? Select all that apply: When two atoms are bonded together, the distance between atoms is the same all the time. In the IR spectrum, carbonyl groups (C=O) have strong sharp peaks ranging from 1700 cm 1 to 1750 cm-1. When a molecule absorbs an IR radiation, electrons are excited within the molecule and each absorbed electron creates a band in the molecules IR spectrum. IR spectroscopy is useful in confirming the structure of the analyzed compound. Molecule N2 does not absorb IR radiationНо LIGH + Cv (NO3)3 911₂0 ㅋ HH₂0, 120°C32904 24h ount B Used 0.839) - terephthalic and 09 Used 2,0g - Cr (103) 3 9H₂0
- Calculate delta Ho for the rxn: Na2O (s) + SO3 (g) --> Na2SO4 (g) given the ff information: 1. Na (s) + H2O (l) --> NaOH (s) + 1/2 H2 (g) delta Ho = -146 kJ 2. Na2SO4 (s) + H2O --> 2NaOH (s) + SO3 (g) delta Ho = +418 kJ 3. 2Na2O (s) + 2H2 (g) --> 4Na (s) + 2H2O (l) delta Ho = +259 kJ The answer should be -581 kJ. I am very confused. thank you.Molecular Recognition (Supramolecular Chemistry) in organic chemistry deals with the "lock and key " mechanisms that form new molecules. However, little is known of this subject when moleules go beyond 1500 daltons (g/mole) for example macromolecules of polyolefins (polyethylene, polypropylene), thermoplastic polyesters and polyamides (nylon). It is understood that pre-directional H bonding is the mechanism of molecular recognition of macromolecules. Question: What is the immediate and post manifestation of inducing molecular recognition to condensation polymers and ring opening polymerization polymers like thermoplastic polyesters (PET ), and Nylon 66, 6 etc. (polyamides)?Which of the following molecules is represented in this IR spectrum? INFRARED SPECTRUM TRANSMITTANCE 0.8 000 0 ABCD 0.6 0.4 0.2 3000 Wavenumber (cm-1) NIST Chemistry WebBook (http://webbook.nist.gov/chemistry) A B 2000 -NH₂ CH3 OH с 1000 A при OH
- Please explain how to use DEPT spectraHalogenated compounds are particularly easy to identify by their mass spectra because both chlorine and bromine occur naturally as mixtures of two abundant isotopes. Recall that chlorine occurs as 35Cl (75.8%) and 37Cl (24.2%); and bromine occurs as 79Br (50.7%) and 81Br (49.3%). At what masses do the molecular ions occur for the following formulas? What are the relative percentages of each molecular ion? (a) Bromomethane, CH3Br (b) 1-Chlorohexane, C6H13ClWhy do aldehydes, esters, and amides all have a strong absorption in the 1630-1780 cm1 region of their IR spectra? A) The bond between H and the sp³-hybridized C in these functional groups vibrates in this energy range. B) Each of these functional groups has at least two resonance structures, and the different vibrations of the resonance structures give off energy in this region. C) The bond between O and the sp²-hybridized C in these functional groups vibrates at a frequency in this energy range. D) Light at this wavenumber causes the average C to O bond length to increase which causes more of this light to be transmitted. E) An electron in the bond of these functional groups gets excited to the * orbital.
- Which of the following molecules is represented in this IR spectrum? INFRARED SPECTRUM TRANSMITTANCE 0.8 0.6 0.4 0.2 3000 Wavenumber (cm-1) NIST Chemistry WebBook (http://webbook.nist.gov/chemistry) B 2000 -NH₂ CH3 OH с 1000 D OHThe leaves of the Brazilian Tree Senna multijunga contain a number of pryidine alkaloids that inhibit acetylcholinterinase. Two recentyl isolated isomeric compounds have the strcture have the strcture shown below. (NOTE: M=293) Use the mass spectral data provided to determine the precise location of the hydroxyl group in each isomer. Isomer A: EI-MS, m/z(rel. int): 222(20), 150(10), 136(25), 123(100) Isomer B:EI-MS, m/z(re;. int): 236(20), 150(10), 136(25), 123(100)Below are two molecules, cyclohexylamine (A) and aniline (B). When analyzed by Infrared Spectroscopy (IR) two different frequencies are observed for the C-N absorption band. Which molecule will have the lower wavenumber absorption band for the C-N bond. Explain why in 35 words or less. „NH2 NH2 А B
