then the electrons and protons would have a even amount of negetive electric charges
Answer:
2-chloro-4-methylpentanal.
Explanation:
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In this case, according to the chemical compound:
CH3-CH-CH2-CH-CHO
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CH3 Cl
We can see the main functional group is an starting carbonyl, which means this is an aldehyde. Moreover, we can see a Cl-substituent on the second carbon and a methyl substituent on the fourth carbon. Therefore, the IUPAC name turns out: 2-chloro-4-methylpentanal.
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Whenever the fuel is being used up, a star explodes and the energy leakage from a star's core ceases.
Explanation:
The dying star expands in the "Red Giant," before even the inevitable collapse starts, due to nuclear reactions just outside of the core.
It becomes a white dwarf star when the star has almost the same density as the Sun. If it's much larger, a supernova explosion could take place and leave a neutron star away. However, if it is very large–at least three times the Sun's mass–the crumbling core of the star, nothing will ever stop it from crumbling. The star is imploding into a black hole, an endless gravitational loop in space.
Answer:
(2R,3S)-2-ethoxy-3-methylpentane
and
(2S,3S)-2-ethoxy-3-methylpentane
Explanation:
For this case, we will have
as nucleophile. Also, this compound is also in excess. So, we will have as solvent
a protic solvent. Therefore the Sn1 reaction would be favored.
The first step would be the carbocation formation followed by the attack of the nucleophile. In this case both isomers would be produced: R and S (see figure).