Consider the reaction below to answer the following question: This reaction is an example of: A. A Michael reaction B. A Robinso
1 answer:
The reaction is missing, so i have attached it.
Answer:
C. An intramolecular aldol condensation
Explanation:
From the attached image showing the reaction, we can see that the left hand side of the reaction has 2 carbonyl groups which are the double bonds attached to the oxygen atoms.
Now, on the right hand side, we can see that a six member ring has been formed.
This 6 member ring is produced because one of the carbonyls on the left hand side was deprotonated at the alpha position thereby serving as a nucleophile, which then attacks the carbon in the other carbonyl.
This process is possible because it underwent the processes of deprotonation, intramolecular aldol addition, proton transfer and elimination to yield the right hand side product which is ɑ,β-unsaturated carbonyl compound.
Thus, the correct answer is option C
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Answer:
Butan-2-one
Explanation:
1. 1700 cm⁻¹
A strong peak near 1700 cm⁻¹ is almost certainly a carbonyl (C=O) group.
2. Triplet-quartet
A triplet-quartet pattern indicates an ethyl group.
The 2H quartet is a CH₂ adjacent to a CH₃. The peak normally occurs at δ 1.3, but it is shifted 1.2 ppm downfield to δ 2.47 by an adjacent C=O group.
The 3H triplet at δ 1.05 is the methyl group. It, too, is shifted downfield from its normal position at δ 0.9. The effect is smaller, because the methyl group is further from the carbonyl.
3. 3H(s) at δ 2.13
This indicates a CH₃ group with no adjacent hydrogen atoms.
It is shifted 0.8 ppm downfield to δ 2.13 by the adjacent C=O group.
4. Identification
The identified pieces are CH₃CH₂-, -(CO)-, and -CH₃. There is only one way to put them together: CH₃CH₂-(C=O)-CH₃.
The compound is butan-2-one.
