Interpretation:
The product of the given reaction is to be determined, and the complete, detailed mechanism of the reaction is to be drawn based on the given
Concept introduction:
In Hofmann elimination reaction, the quaternary ammonium iodide salt is formed when quaternary ammonium is reacted with excess of methyl iodide. This product is then treated with silver oxide, water and heat. The substitution of the iodine by a hydroxyl anion is done by an elimination reaction. This gives the least substituted
In
In addition to chemical shift, a
Complicated splitting patterns can result when a proton is unequally coupled to two or more protons that are different from one another.
The ideal range for
The integration of each signal indicates the number of protons responsible for that signal. The splitting pattern of a signal indicates the number of neighboring protons that are distinct from the protons responsible for that signal. To deduce the structure of an unknown compound, the first step is to find the index of hydrogen deficiency if the molecular formula is given. Based on the data given in the
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EBK ORGANIC CHEMISTRY: PRINCIPLES AND M
- Reagan is doing an atomic absorption experiment that requires a set of zinc standards in the 0.4- 1.6 ppm range. A 1000 ppm Zn solution was prepared by dissolving the necessary amount of solid Zn(NO3)2 in water. The standards can be prepared by diluting the 1000 ppm Zn solution. Table 1 shows one possible set of serial dilutions (stepwise dilution of a solution) that Reagan could perform to make the necessary standards. Solution A was prepared by diluting 5.00 ml of the 1000 ppm Zn standard to 50.00 ml. Solutions C-E are called "calibration standards" because they will be used to calibrate the atomic absorption spectrometer. Table 1: Dilutions of Zinc Solutions Solution Zinc Solution Volume Diluted Solution Concentration used volume (ppm Zn) (mL) (mL) concentration (ppm Zn) Solution concentration A 1000 5.00 50.00 1.00×10² (ppm Zn(NO3)2) 2.90×10² Solution concentration (M Zn(NO3)2 1.53×10-3 B Solution A 5.00 100.00 5.00 C Solution B 5.00 50.00 0.50 7.65×10-6 D Solution B 10.00 50.00…arrow_forwardNonearrow_forwardNonearrow_forward
- Nonearrow_forward(b) Provide the number of peaks in each of the indicated signals ('H NMR) for the compound below. CH3 6 1 H&C. C H₂ H2 3 HA 2 2 4 5 5arrow_forward8. The emission spectrum below for a one-electron (hydrogen-like) species in the gas phase shows all the lines, before they merge together, resulting from transitions to the ground state from higher energy states. Line A has a wavelength of 10.8 nm. BA Increasing wavelength, \ - a) What are the upper and lower principal quantum numbers corresponding to the lines labeled A and B? b) Identify the one-electron species that exhibits the spectrum.arrow_forward
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