Identifying Stationary Points: Phenol has two easily identifiable and distinguishable stationary points. Identify both of these stationary points and show that they are either minima or transition state points. Please provide a Ball-And-Stick image with RHF energy (in Hartree) for these two structures at the HF/3-21G level.
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- Please answer this NEATLY, COMPLETELY, and CORRECTLY for an UPVOTE. Long and detailed explanations are not needed. Keep it short, brief, and direct because I only need the answers as soon as possible. Calculate the % ee of a mixture containing 80% of one enantiomer and 20% of the other? A. 80B. 120C. 60D. 90Please give clear handwritten answer of all subparts of question no. 5 i will give you upvote.Comparing Minima: Catechol has three primary "in the ring plane" distinguishable stationary point conformations with different relative hydroxyl orientations while hydroquinone has only two. a) Identify and image the structures of all these stationary points. Provide RHF energy (in Hartree) for all these structures at the HF/3-21G level. b) Compare the global minimum energies for catechol and hydroquinone. Explain the potential energy difference between these conformers of catechol and hydroquinone.
- help with the following ochem question part a, b, and c regarding the NMR for the compound.... (5S, 7R)-7-bromo-,3,9-diethyl-6,6-dimethylundec-3,8-dien-5-ol(1) 4-Phenyl Phenol (2) KOH in ethanol (1M) (3) Benzyl chloride (4) 4-benzyloxy biphenyl (1) (2) Name (3) (4) Williamson Ether Synthesis Physical Properties of Regents and Products Percent Yield: Structural Formula Structure and Formula KOM C₁₂ H₂O C₂ B₂ CI Volume (mL) Solution: 10mL Molar Mass (g/mol) 0.6mL 170. al 56.1056 126.58 260.34 Mass (8) 0.8566 Density (g/ml) C₂H₂O Quantities of Reagents Used and Products Formed 2.162 1.11 1.1 1.1 Melting Pt. (°C) 163 Solute: -39 136-137 Moles 0.01 0.005033 0.00521 Boiling Pt. ("C) 306.4 179 N/A Limiting Reagent Check ✓ Actual Theoretical Melting Poind: 126.3°C-131.0%(a) Below are 'H NMR, 19C NMR and IR spectra of ONE of four compounds: toluene, methyl benzoate, stilbene and hex-5-en-3-one. Which of the following compounds is represented by these spectra. OCH, CH3 toluene methyl benzoate stilbene hex-5-en-3-one 1H NMR spectrum (in CDCI3) No. (ppm) Value [8.02 B.14] 7.54 . 7.64] 741.7.51] (3.913.99 0.222 0.117 0.232 0.368 0.15: 0.16 0.10 0.105 0.05 0.00 6.00 0.12 0.23 0.22 7.80 7.30 745 7.40 8.10 8.00 0.23 0.37
- Cineole is the chief component of eucalyptus oil; it has the molecular formula C₁0H180 and contains no double or triple bonds. It reacts with hydrochloric acid to give the dichloride shown: Cineole C10H180 edit structure... HCI CI XX Deduce the structure of cineole. Be sure to include stereochemistry.The pka values of OH group in 4-cyanophenol and 4-nitrophenol are identical BUT in the di-metbylsbstituted compounds C and D differ significantly. The cyano derivative C is more acidic than compound D. Please briefly explain why this is the case (you can use words and drawings). Hint: delocalization... A B ОН OH OH ОН CN NO2 CN NO2Please answer this NEATLY, COMPLETELY, and CORRECTLY for an UPVOTE. Long and detailed explanations are not needed. Keep it short, brief, and direct because I only need the answers. What is the percentage of each of the two enantiomers present in a mixture having a 40% enantiomeric excess (ee)?
- The pk, of phenol (C,H5OH) is 10.0. When a methyl (CH3) group is attached to the ring, the pK, increases, as shown here for the ortho, meta, and para isomers. OH ОН ОН ОН CH3 (a) Explain why the pk, values of all three isomers are higher than the pk, of phenol itself. (b) Explain why the meta isomer has the lowest pk, of CH3 Phenol CH3 the three isomers. pKa = 10.0 10.3 10.1 10.2A problem in dyeing fabrics is the degree of fastness of the dye to the fabric. Many of the early dyes were surface dyes; that is, they did not bond to the fabric, with the result that they tended to wash off after repeated laundering. Indigo, for example, which gives the blue color to blue jeans, is a surface dye. Color fastness can be obtained by bonding a dye to the fabric. The first such dyes were the so-called reactive dyes, developed in the 1930s for covalently bonding dyes containing-NH, groups to cotton, wool, and silk fabrics. In the first stage of the first-developed method for reactive dyeing, the dye is treated with cyanuric chloride, which links to the fabric through the amino group of the dye. The remaining chlorines are then displaced by the-OH groups of cotton (cellulose) or the-NH, groups of wool or silk (both proteins). CI -Cotton Dye-NH, Но-Сotton N' CI CI Dye-NH CI Dye-NH 'N. 0-Cotton Cyanuric chloride A reactive dye Dye covalently bonded to cotton Propose a…Consider all of the following data and answer parts (A) through (C): When ketone A is treated with an equivalent of CH3MgI in Et₂0 followed by acid (H30+) workup, two constitutional isomers B and C (C5H100) are isolated. The minor isomer B is a ketone. Isomer C decolorizes solutions of Br₂ and permagnate (i.e. it's an alkene - Chem210). When treated with HCl(aq), C gives two constitutional isomers D and E (C5H9Cl) - Chem210. Chloride D, the major product gives 2-methyl-1,3-butadiene when treated with potassium t-butoxide - Chem210. Compound B also is identical to the product obtained from the oxymercuration of 1-pentyne. (A) Identify compounds A through E by structure. (B) Classify both the AB and A → C. (C) Write mechanisms for the AC and C D + E reactions.