In this week's lab, we learned about extraction. This week's "paper" lab started by dissolving p- toluic acid, p-tertbutylphenol and acetanilide in tertbutylmethyl ether. Suppose a student accidentally extracted the mixture first with 0.5- M NaOH (instead of 0.5M NaHCO3). When the get to the end of the experiment (which will discuss more next week), they find that Beaker B does not contain any p-tertbutylphenol, in fac there is no compound dissolved in the solution in Beaker B. Where is the p-tertbutylphenol that should be in Beaker B? Explain what happened. (I don't want a "spilled" it on the countertop explanation. I want an explanation explained b the concepts learned in this lab.) H. N. p-toluic acid pka = 4.2 p-tertbutyl phenol pka = 10.2 acetanilide pka = 24

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Where is the p-tertbuttlphenol that should be in breaker B explain what happened.
In this week's lab, we learned about extraction. This week's "paper" lab started by dissolving p-toluic acid, p-tertbutylphenol, and acetanilide in tertbutylmethyl ether. Suppose a student accidentally extracted the mixture first with 0.5 M NaOH (instead of 0.5 M NaHCO₃). When they get to the end of the experiment (which we will discuss more next week), they find that Beaker B does not contain any p-tertbutylphenol; in fact, there is no compound dissolved in the solution in Beaker B.

Where is the p-tertbutylphenol that should be in Beaker B? Explain what happened. (I don't want a "spilled it on the countertop" explanation. I want an explanation explained by the concepts learned in this lab.)

**Chemical Structures:**

1. **p-toluic acid**
   - Structure: Aromatic ring with a carboxylic acid group and a methyl group.
   - pKₐ = 4.2

2. **p-tertbutyl phenol**
   - Structure: Aromatic ring with a hydroxyl group and a tert-butyl group.
   - pKₐ = 10.2

3. **acetanilide**
   - Structure: Aromatic ring with an amide group (NH and C=O).
   - pKₐ = 24

---

**Explanation:**

The p-tertbutylphenol likely did not dissolve in Beaker B due to the use of 0.5 M NaOH. During the extraction, NaOH deprotonates the more acidic compounds. The p-tertbutylphenol, with a pKₐ of 10.2, was not deprotonated efficiently due to its relatively high pKₐ compared to the solution's pH. Since the solution is strong enough to deprotonate more acidic compounds, the p-tertbutylphenol remained as its neutral form and stayed in the organic layer, not transferring into the aqueous layer in Beaker B.
Transcribed Image Text:In this week's lab, we learned about extraction. This week's "paper" lab started by dissolving p-toluic acid, p-tertbutylphenol, and acetanilide in tertbutylmethyl ether. Suppose a student accidentally extracted the mixture first with 0.5 M NaOH (instead of 0.5 M NaHCO₃). When they get to the end of the experiment (which we will discuss more next week), they find that Beaker B does not contain any p-tertbutylphenol; in fact, there is no compound dissolved in the solution in Beaker B. Where is the p-tertbutylphenol that should be in Beaker B? Explain what happened. (I don't want a "spilled it on the countertop" explanation. I want an explanation explained by the concepts learned in this lab.) **Chemical Structures:** 1. **p-toluic acid** - Structure: Aromatic ring with a carboxylic acid group and a methyl group. - pKₐ = 4.2 2. **p-tertbutyl phenol** - Structure: Aromatic ring with a hydroxyl group and a tert-butyl group. - pKₐ = 10.2 3. **acetanilide** - Structure: Aromatic ring with an amide group (NH and C=O). - pKₐ = 24 --- **Explanation:** The p-tertbutylphenol likely did not dissolve in Beaker B due to the use of 0.5 M NaOH. During the extraction, NaOH deprotonates the more acidic compounds. The p-tertbutylphenol, with a pKₐ of 10.2, was not deprotonated efficiently due to its relatively high pKₐ compared to the solution's pH. Since the solution is strong enough to deprotonate more acidic compounds, the p-tertbutylphenol remained as its neutral form and stayed in the organic layer, not transferring into the aqueous layer in Beaker B.
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