Answer:
A = 2-iodo-2,3-dimethylbutane
B = Ethanol
C = Iodoethane (also called ethyl-iodide)
Explanation:
2-Ethoxy-2,3-dimethylbutane reacts with conc. HI to cleave the oxy-functional group.
On one end, ethanol is formed and on the other hand, 2-iodo-2,3-dimethylbutane is formed.
But ethanol reacts further with conc HI to give iodoethane.
Therefore,
A = 2-iodo-2,3-dimethylbutane
B = Ethanol
C = Iodoethane (also called ethyl-iodide)
This is all shown in the attached image.
Hope this Helps!!!
Electrical--> chemical --> thermal --> mechanical
Answer:
9-10 ppm.
0.2-0.4 ppm.
Explanation:
The proton on the aldehyde group will appear at approximately 9-10 ppm whereas the methylene peak on the alcohol is the only peak 0.2-0.4 ppm for either compound. Aldehydes and aromatics are quite distinctive in the Nuclear magnetic resonance (NMR). Aldehydes show up from 9-10 ppm, usually as a small singlet; aromatic protons show up from 6.5-8.5 ppm. NMR spectroscopy is the use of NMR to study the physical, chemical, and biological properties of matter.
Answer:
1 may is the answer for the question
I think 14 are produced because if you go up by that you get it
