Answer:
2-methyl-2-pentyl-1,3-dioxolane
Explanation:
In this case, we have two reactions:
First reaction:
1-heptyne + mercuric acetate -------> Compound A
Second reaction:
Compound A + HOCH2CH2OH -------> Compound C
<u>First reaction</u>
In the first reaction, we have as a main functional group a triple bond. We have to remember that mercuric acetate in sulfuric acid will produce a ketone. The carbonyl group (C=O) would be placed in the most substituted carbon of the triplet bond (in this case, carbon 2). With this in mind, we will have as a product: heptan-2-one. (See figure 1).
<u>Second reaction</u>
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In this reaction, we have as reagents:
-) Heptan-2-one
-) Ethylene-glycol ![HOCH_2CH_2OH](https://tex.z-dn.net/?f=HOCH_2CH_2OH)
-) Sulfuric acid ![H_2SO_4](https://tex.z-dn.net/?f=H_2SO_4)
When we put ethylene-glycol with a ketone or an aldehyde we will form a cyclic acetal. In this case, this structure would be formed on carbon 2 forming 2-methyl-2-pentyl-1,3-dioxolane. (See figure 2).
I hope it helps!
Answer:
The answer is a gamma particle.
Ionization enthalpy, IE, is also called ionization potential is the ability to remove the electron from the neutral gaseous atom. There is a trend observed in the periodic table for the IE value. As we go from left to right in a period, IE vale increases. While moving from top to bottom in a group, IE value decreases.
- The phenomenon of unexpected drop in IE1 values between Groups 2 and 13, in period 2 and period 4 is due to the introduction of d-orbitals in the case of period 4 elements.
- While moving in the period, there is the constant addition of electrons in the nucleus. The shell sie remains constant while electron pull increases from the nucleus, this leads to a reduction in the size of the atom. As the size decreases, it is difficult to remove the electron from the atom, and thus IE value increases in the case of period 2.
- When we study the case of period 4, there is an introduction of d-electrons. As the inner shell electron increases, there is an increase in the shielding effect. This shielding effect tends to decrease the nuclear attraction between the nucleus and outermost electrons. Ultimately this decreases the IE value in the fourth period. Such a phenomenon is absent in the case of group 2 elements.
- If we speak in terms of orbital energy, the IE value decreases while moving from top to bottom in the period. This is due to the fact that, as we go down in the periodic table, the number of shells increases, and the outermost electron is too far from the nuclear attraction, therefore it can be ejected out easily. This marks a decrease in IE value.
To learn more about ionization refer the link:
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Answer:
See the attachment for the filled in chart.
Let me know if you need any other help:)
Answer:
B
Explanation:
Fluorescence spectroscopy examines the florescence given off by a sample. It is the mechanism whereby a light beam (UV light is most common) is used to excite electrons in the sample resulting in them giving off light, at times this light is visible.