Draw the major condensation product obtained by treatment of ethyl 3-methylbutanoate with sodium ethoxide in ethanol.
1 answer:
Answer:
ethyl 3-ethoxy-3-hydroxy-2-isopropyl-5-methyl hexanoate
Explanation:
In this case, we have a <u>very strong base</u> (sodium ethoxide). Therefore, this compound will remove a hydrogen from ethyl 3-methyl butanoate generating a <u>carbanion</u>.
This carbanion, can attack <u>another ethyl 3-methyl butanoate molecule </u>on the carbonyl group generating a new C-C bond and producing a negative charge in the oxygen.
Then the ethanol can <u>protonate</u> the molecule generating an "OH" group and the ethoxide.
See figure 1
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<h3>Answer:</h3>
0.0463 mol KCl
<h3>General Formulas and Concepts:</h3><u>Math</u>
<u>Pre-Algebra</u>
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
- Left to Right
<u>Chemistry</u>
<u>Atomic Structure</u>
- Reading a Periodic Table
- Using Dimensional Analysis
<u>Step 1: Define</u>
3.45 g KCl
<u>Step 2: Identify Conversions</u>
Molar Mass of K - 39.10 g/mol
Molar Mass of Cl - 35.45 g/mol
Molar Mass of KCl - 39.10 + 35.45 = 74.55 g/mol
<u>Step 3: Convert</u>
- Set up:
- Multiply/Divide:
<u>Step 4: Check</u>
<em>Follow sig fig rules and round. We are given 3 sig figs.</em>
0.046278 mol KCl ≈ 0.0463 mol KCl