Predict the major products of the following reactions. 1. \(\frac{(1) L D A}{(2) C H_{2}=C H C H_{2} B r}\) 2. \(\frac{(1) L D A}{(2) C H_{3} C H_{2} I}\) 3. \(\frac{(1) L D A}{(2) C H_{3} I}\) Equation transcription: Text transcription: frac{(1) L D A}{(2) C H{2}=C H C H_{2} B r} frac{(1) L D A}{(2) C H{3} C H{2} I} frac{(1) L D A}{(2) C H{3} I}
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Textbook Solutions for Organic Chemistry
Question
Show reaction sequences (not detailed mechanisms) that explain these transformations:
\(\mathrm{CH}_{2} \mathrm{O}+2\)
\(\frac{(1) \mathrm{NaOEt}}{(2) \mathrm{H}^{+}}\)
\(\mathrm{COOH}\)
\(\mathrm{CH}_{2}(\mathrm{COOEt})_{2})\(\frac{\mathrm{NaOEt}}{\mathrm{H}_{3} \mathrm{O}^{+}}\)
Solution
Solution:
Here we have to explain the following transformation
Step 1
(formaldehyde) (ethyl methyl malonate) (2-methyl-2-en-4-oxobenzoic acid)
In this reaction formaldehyde on addition with one molecule of ethyl methyl malonate undergoes aldol condensation reaction in presence of sodium ethoxide and form 3-ethoxy-2-methylpropan-2,3,4-trione. This on addition with another molecule of ethyl methyl malonate undergoes michael addition reaction in presence of sodium ethoxide and forms an adduct which undergoes aldol condensation reaction followed by decarboxylation and form 2-methyl-2-en-4-oxobenzoic acid.
(2-methyl-2-en-4-oxobenzoic acid)
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