The reducing agent can approach the carbonyl face of camphor by forming a one carbon bridge (known as an exo attack) or a two carbon bridge (termed endo).
The two resultant stereoisomers are known as isoborneol and borneol (from exo attack) (from endo attack). Gas chromatography (GC) analysis may be used to calculate the ratio of each isomeric alcohol in the mixture. Unfortunately, IR analysis does not permit this.
The stereochemistry of the reaction is regulated in stiff cyclic compounds like camphor and norcamphor by protecting one side of the carbonyl group from the reagent's assault. The hydrogen atom is added to the endo side, creating the exo alcohol isoborneol, while the methyl groups on the one-carbon bridge of camphor screen the approach of the hydride from the "top" or exo side of the two-carbon bridge. You will be asked to guess the main isomeric alcohol created by the norcamphor hydride reduction later in the lab report.
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An ice cube when you heat it it can to water and water is liquid of course
Answer:
Helium is created from hydrogen in the sun's core.
Four hydrogen-1 nuclei fuse to produce
- one helium-4 nucleus, two neutrons,
- two positrons, and
- two electron neutrinos.
Explanation:
Step One:
.
Two hydrogen-1 nuclei fuse. One proton will convert to a neutron. The products will be
- one hydrogen-2 nucleus,
- one positron, and
- one electron neutrino.
Step Two:
.
There are plenty of hydrogen-1 nuclei available in the core of the sun. The hydrogen-2 nucleus from step one will fuse with a hydrogen-1 nucleus. The product is
Step Three
.
Two helium-3 nuclei from step two react with each other. The products are:
- one helium-4 nucleus, and
- two hydrogen-1 nuclei.
The overall reaction will be:
.
In other words, hydrogen nuclei in the core of the sun fuse together to form helium.
It easier to remove electrons from a large element(bottom of the periodic table) because there further away from the nucleus.
The answer is fossils. they find and study them
