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    Perkin Reaction

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    produce the E and Z phenylcinnamic acid. The mechanism for this reaction can be seen in figure 1. Table 1 includes all spectroscopic values for both E and Z isomers of phenylcinnamic acid. The IRs for both E and Z can be seen in figure 2 and 3. The NMR can be seen in figure 4 and 5‚ while the mass spec can be seen in figure 6. According to the experimental IR spectroscopy for the E isomer of phenylcinnamic acid the peaks of the wavenumber were at 2952‚ 2518 1672 and 1426 cm-1. The peak at 2952 presents

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    Spectroscopy-1hnmr

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    Electron Spin Nuclear Magnetic Resonance (NMR) Spectroscopy Organic Chemistry BCH3015 • Electron spin is quantized. • The spin quantum number ms has only two allowed values: ms = +½ or –½ • The two directions of spin create oppositely spin create oppositely directed magnetic fields. fields. 61 Proton Spin Organic Chemistry BCH3015 62 Nuclear Spin States • A proton (the nucleus of a 1H atom) also possesses spin. For each nucleus having the spin quantum number

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    3371‚ 3309 C-H 2900‚ 2855 C-H stretch‚ methyl 1451 Ar C=C 1278 Acetal 1H NMR Chemical shift (δ) Multiplicity and Integration Assignment Compound B 1.6401 S‚ 3H C-CHa3 1.8602 S‚ 3H C-CHb3 5.5459 S‚ 2H C(ph)Hc 7.0352 M‚ 10H Ar-Hd Compound A: Meso-hydrobenzoin Compound B: Meso-1‚2-diphenyl-1‚2- ethanediol Discussion and Conclusion One can ascertain from the H NMR the stereoselectivity taking place in the reaction. Since there is not coupling of the Hc’s

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    Lab Report Preparation Guidelines—IR Spectroscopy‚ General Unknowns‚ CH 315. Introduction/Purpose State in simple terms what the purpose was. Since one is issued a vial of an unknown substance‚ the “purpose” as such is not exactly in doubt. The object is simply to state that obvious purpose‚ as well as summarize (a comma-ed list here is fine) the various means‚ tests‚ so forth that are performed in the approximate order that they are performed. (approximately 10 points) Experimental

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    Organic Chem 2 Quiz 1

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    upfield in the NMR?       A. Ha B. Hb C. Hc D. Hd E. He   3. Which of the following compounds does not give a singlet in its NMR spectrum?       A. A B. B C. C   4. Which of the circled protons in the molecules below would absorb furthest downfield in the NMR?       A. A B. B C. C D. D E. E   5. Which of the labeled protons would absorb furthest downfield in an NMR spectrum?       A. A B. B C. C D. D E. E   6. Which of the indicated protons would absorb furthest downfield in an NMR spectrum?    

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    Organic Chemistry II Laboratory (ABCT357/ABCT3757) Lab Manual 2014-2015 Lee Hang Wai‚ Alston (alston.lee@polyu.edu.hk) Dr. Lee Cheng Hao‚ Samuel (chenghao.lee@polyu.edu.hk) Organic Chemistry II Laboratory (ABCT357‚ ABCT3757) (Group one) Contact: Alston Lee (alston.lee@polyu.edu.hk) Tuesday 8:30-11:30 ‚ Laboratory: Y1315 Lab supporting staffs: YK Au / Kan Chan/Arnold Demonstrators: Dr. Samuel Lee‚ Alston Lee‚ Fu Wai Chung‚ Guo Shuai‚ Yuen On Ying‚ Sep 16 Tue Acetylation of α-D-glucopyranose (Expt

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    of hyaluronic acid (HA) with glycidyl methacrylate (GM) is a strategy for producing photocross-linkable macromers to generate biodegradable hydrogels by in-situ injection and light-induced polymerization. Facile synthesis and characterization by 1H NMR spectroscopy of a series of photopolymerizable macromers based on HA-glycidyl methacrylate (HAGM) conjugates with various degrees of methacrylation (DM) ranging from 14% to 90% are described. Aqueous solutions of HAGM were photopolymerized to yield

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    ethyl-3-(4-nitrophenyl)acrylate. The prediction was the closer the substituent was to the aldehyde‚ the greater the ratio of E/Z isomers; benzaldehyde was predicted to result in the most similar ratio of E/Z isomers. The ratios of E/Z isomers were determined by using the NMR spectra obtained. A percent conversion was also obtained. Ethyl-3-(2-nitrophenyl)acrylate had a ratio of 9.03:1‚ ethyl-3-(3-nitrophenyl)acrylate had a ratio of 5.2:1‚ ethyl-3-(4-nitrophenyl)acrylate had a ratio of 1:1 and ethyl cinnamate had a ratio

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    Qualitative Analysis Lab

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    Abstract Chemically active extraction was used to separate two unknown compounds from each other and to determine what compound class each compound belongs. The solid unknown was purified by recrystallization and its melting point range was determined to be 109-111°C. The liquid unknown was purified by simple distillation and its boiling point was determined to be 95°C. An IR spectrum was collected for each of the purified compounds; the solid was determined to be 3-toluic acid and the liquid was

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    fbfdb

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    the product can cause either a kinetic or a thermodynamic product to be produced. What type of reaction conditions or electrophiles will cause one or the other product to be made. What product do you think will be made in this reaction? Describe how NMR can be used to determine the structure of the product the relative concentrations if more than one product is made. Procedure: State exactly what was done during the experiment. Results Provide the following quantities and calculations: Mass

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