Organic Chemistry - Standalone book
Organic Chemistry - Standalone book
10th Edition
ISBN: 9780073511214
Author: Francis A Carey Dr., Robert M. Giuliano
Publisher: McGraw-Hill Education
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Chapter 18, Problem 43P
Interpretation Introduction

Interpretation:

The key intermediate, compound B formed in the reaction is to be identified and the other compound formed in the reaction is to be shown.

Concept introduction:

Ketal is an organic compound with general formula R2C(OR')2. In the formula, 2

alkoxy groups (OR') are bonded with same carbon atom in between the chain. The R' and R groups may be same or different.

In intramolecular aldol condensation, the self condensation occurs to form cyclic enone.

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Can you please help mne with this problem. Im a visual person, so can you redraw it, potentislly color code and then as well explain it. I know im given CO2 use that to explain to me, as well as maybe give me a second example just to clarify even more with drawings (visuals) and explanations.
Part 1. Aqueous 0.010M AgNO 3 is slowly added to a 50-ml solution containing both carbonate [co32-] = 0.105 M and sulfate [soy] = 0.164 M anions. Given the ksp of Ag2CO3 and Ag₂ soy below. Answer the ff: Ag₂ CO3 = 2 Ag+ caq) + co} (aq) ksp = 8.10 × 10-12 Ag₂SO4 = 2Ag+(aq) + soy² (aq) ksp = 1.20 × 10-5 a) which salt will precipitate first? (b) What % of the first anion precipitated will remain in the solution. by the time the second anion starts to precipitate? (c) What is the effect of low pH (more acidic) condition on the separate of the carbonate and sulfate anions via silver precipitation? What is the effect of high pH (more basic)? Provide appropriate explanation per answer
Part 4. Butanoic acid (ka= 1.52× 10-5) has a partition coefficient of 3.0 (favors benzene) when distributed bet. water and benzene. What is the formal concentration of butanoic acid in each phase when 0.10M aqueous butanoic acid is extracted w❘ 25 mL of benzene 100 mL of a) at pit 5.00 b) at pH 9.00

Chapter 18 Solutions

Organic Chemistry - Standalone book

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