Draw the major organic product for the reaction of 1-phenylpropan-1-one with the provided phosphonium ylide in part 1. In part 2
answer the question about the transformation.
2 answers:
Answer:
The major product formed is a benzyl allyl compound, namely 2-methylene propylbenzene
Explanation:
<u>Part 1.</u>
The product formed from this Wittig reaction with phosphonium ylide is a phenylallyl compound, 2-methylene propylbenzene. Structure and reaction synthesis is attached in file.
<u>Part 2</u>
The types of transformation involved are:
- alkylation (as there is a transfer of alkyl group)
- addition (reaction of ylide)
Answer:
2-methylene propylbenzene (major product)
Explanation:
This reaction yields olefins and organophosphorus compound and is also known as wittig olefination. The reactants fits the category of a ketone (1-phenylpropan-1-one) and a Wittig reagent (phosphonium ylide).
Part 1: Answer in the fig. Below.
1-phenylpropan-1-one + phosphonium ylide ➡ phenylallyl compound + 2-methylene propylbenzene (major product)
Part 2.
Classical mechanisms that happened during its transformation are thus;
• nucleophilic addition
• Elimination
• cycloaddition
• [2+2]/retro-[2+2] mechanism.
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We need to measure 20.0 grams of CaCl₂ to prepare 500 mL of 0.360 M solution.
First, we need to determine the required moles of CaCl₂. We have 500 mL (0.500 L) of a 0.360 M solution (0.360 moles of CaCl₂ per liter of solution).
Then, we will convert 0.180 moles to grams using the molar mass of CaCl₂ (110.98 g/mol).
To prepare the solution, we weigh 20.0 g of CaCl₂ and add it to a beaker with enough distilled water to dissolve it. We stir it, heat it if necessary, and when we have a solution, we transfer it to a 500 mL flask and complete it to the mark with distilled water.
We need to measure 20.0 grams of CaCl₂ to prepare 500 mL of 0.360 M solution.
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