Draw the curved arrow mechanism for the formation of an acetal from acidic methanol and 4-methylpentan-2-one in the fewest steps
. When given the choice, draw the arrows that lead to the resonance structures with full octets around each atom other than hydrogen. Do not show any inorganic byproducts or counterions. Reagents needed for each step are provided in the boxes.
1 answer:
Answer:
See explanation below
Explanation:
This reaction is known as Ketone hydrolisis in acid medium. This involves the formation of an hemi cetal, and then, the acetal. This is often used to convert ketones or aldehydes in ethers.
The first step involves the reaction with the acid. The carbonile reacts with the acid and forms an alcohol there. The next step is the reaction of the alcohol, in this case, the methanol to form the hemi cetal. Then in the third step, we repeat the first step, using acid to turn the OH group into a great leaving group such water. Then the water leaves the molecule, leaving the space wide open in the next step for methanol, and the acetal is formed.
See picture for the curved arrow mechanism
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Answer:
1) After adding 15.0 mL of the HCl solution, the mixture is before the equivalence point on the titration curve.
2) The pH of the solution after adding HCl is 12.6
Explanation:
10.0 mL of 0.25 M NaOH(aq) react with 15.0 mL of 0.10 M HCl(aq). Let's calculate the moles of each reactant.
There is an excess of NaOH so the mixture is before the equivalence point. When HCl completely reacts, we can calculate the moles in excess of NaOH.
NaOH + HCl ⇒ NaCl + H₂O
Initial 2.5 × 10⁻³ 1.5 × 10⁻³ 0 0
Reaction -1.5 × 10⁻³ -1.5 × 10⁻³ 1.5 × 10⁻³ 1.5 × 10⁻³
Final 1.0 × 10⁻³ 0 1.5 × 10⁻³ 1.5 × 10⁻³
The concentration of NaOH is:
NaOH is a strong base so [OH⁻] = [NaOH].
Finally, we can calculate pOH and pH.
pOH = -log [OH⁻] = -log 0.040 = 1.4
pH = 14 - pOH = 14 - 1.4 = 12.6