Determine the reagents needed and the synthetic intermediate that is formed in the conversion of the secondary alcohol into the
cyanohydrin given below. drag the appropriate labels to their respective targets.
1 answer:
Secondary Alcohol is converted into corresponding Cyanohydrin in two steps;
Step 1:
Oxidation of Sec. Alcohol:
Secondary Alcohols when treated with Sodium Dichromate in acidic medium are converted into corresponding Ketone.
Step 2:
Reduction of Ketone using HCN:
Ketone is reduced to Cyaynohydrin when treated with CN nucleophile. A nucleophillic addition reaction takes place.
Step 1:
Oxidation of Sec. Alcohol:
Secondary Alcohols when treated with Sodium Dichromate in acidic medium are converted into corresponding Ketone.
Step 2:
Reduction of Ketone using HCN:
Ketone is reduced to Cyaynohydrin when treated with CN nucleophile. A nucleophillic addition reaction takes place.
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Answer: 3.0 moles of water are needed to react with a mole of 3.0 of Na
Explanation:
The balanced chemical reaction of sodium with water is as follows:
According to stoichiometry:
2 moles of sodium reacts with = 2 moles of water
Thus 3.0 moles of sodium reacts with =
moles of water
3.0 moles of water are needed to react with a mole of 3.0 of Na
Answer:
76.1%
Explanation:
The reaction that takes place is:
- BaCl₂ + K₂SO₄ → BaSO₄ + 2KCl
First we determine how many moles of each reactant were added:
- BaCl₂ ⇒ 80 mL * 0.812 M = 64.96 mmol BaCl₂
- K₂SO₄ ⇒ 40 mL * 1.52 M = 60.8 mmol K₂SO₄
Thus K₂SO₄ is the limiting reactant.
Using the <em>moles of the limiting reactant</em> we <u>calculate how many moles of BaSO₄ would have been produced if the % yield was 100%</u>:
- 60.8 mmol K₂SO₄ *
= 60.8 mmol BaSO₄
Then we <u>convert that theoretical amount into grams</u>, using the <em>molar mass of BaSO₄</em>:
- 60.8 mmol BaSO₄ * 233.38 mg/mmol = 14189.504 mg BaSO₄
- 14189.504 mg BaSO₄ / 1000 = 14.2 g BaSO₄
Finally we calculate the % yield:
- % yield = 10.8 g / 14.2 g * 100 %
- % yield = 76.1%