First, recognize that this is an elimination reaction in which hydroxide must leave and a double bond must form in its place. It is likely an E2 reaction. Here is an efficient mechanism:
1) Pre-reaction: Protonate the -OH to make it a good leaving group, water. H2SO4 or any strong H+ donor works. The water is positively charged but still connected to the compound.
2) E2: Use a sterically hindered base, such as tert-butoxide (tButO-) to abstract the hydrogen from the secondary carbon. [You want a sterically hindered base because a strong, non-sterically hindered base could also abstract a hydrogen from one of the two methyl groups on the tertiary carbon, and that leads to unwanted products, which is not efficient]. As the proton of hydrogen is abstracted, water leaves at the same time, creating an intermediate tertiary carbocation, and the 2 electrons in the C-H bond immediately are used to make a double bond towards the partial positive charge.
In the products we see the major product and water, as expected. Even though you have an intermediate, remember that an E2 mechanism technically happens in one step after -OH protonation.
Mitosis is where the cell divides its previously-copied DNA and cytoplasm to make two new, identical daughter cells.
Answer:
1. Option A. Beta decay minus 0 -1e
2. Option B. Alpha 4 2He
3. Option A. Beta decay minus 0 -1e
Explanation:
1. 66 29Cu is undergoing beta decay minus since it produces a daughter nuclei having the same mass number and the atomic number increased by 1 i.e 66 30Zn
2. 238 92U is undergoing alpha decay since the daughter nuclei produced has a decrease of 4 in the mass number and a decrease of 2 in the atomic number ie 234 90Th
3. 14 6C is undergoing beta decay minus since the daughter nuclei produced has the same mass number and the atomic number increased by 1 i.e 14 7N
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Answer:
C) Conjugate Acid
Explanation:
H3O is the conjugate acid of H2O and X is the conjugate base of HX. The reaction going forward is the reaction of an acid and a base.