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.
Its magnifying power is: 4X 5X 9X 20X. A 4-inch, f/5 telescope has a 1-inch eyepiece focal. Its magnifying power is 9x. This answer has been confirmed as correct and helpful.
Answer:
Follows are the solution to this question:
Explanation:
Please find the image file of the chemical reaction in the attachment:
In a water medium, the CH3- type CH 3Li is a heavy nucleophile that attacks the carbonyl carbon atom to form the alkoxide ion, which will then be protonated to form alcohol.
<u>Answer:</u> The mole ratio of H : O in ammonium nitrate is 4 : 3.
<u>Explanation:</u>
We are given a compound named ammonium nitrate having formula 
There are 3 elements in this compound which are nitrogen, hydrogen and oxygen.
To calculate the mole ratio, we write the ratio of their subscripts. For this compound, it is:
The mole ratio of H and O for this compound is 4 : 3.
Answer:
S = 1.1 × 10⁻⁹ M
Explanation:
NaCl is a strong electrolyte that dissociates according to the following expression.
NaCl(aq) → Na⁺(aq) + Cl⁻(aq)
Given the concentration of NaCl is 0.15 M, the concentration of Cl⁻ will be 0.15 M.
We can find the molar solubility (S) of AgCl using an ICE chart.
AgCl(s) ⇄ Ag⁺(aq) + Cl⁻(aq)
I 0 0.15
C +S +S
E S 0.15+S
The solubility product (Ksp) is:
Ksp = 1.6 × 10⁻¹⁰ = [Ag⁺].[Cl⁻] = S (0.15 + S)
If we solve the quadratic equation, the positive result is S = 1.1 × 10⁻⁹ M
