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Test: Alpha-carbon Chemistry - MCAT MCQ


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10 Questions MCQ Test - Test: Alpha-carbon Chemistry

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Test: Alpha-carbon Chemistry - Question 1
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Which of the following resonance structures would be the most stable hybrid of the base catalyzed enolate of this molecule?

Detailed Solution for Test: Alpha-carbon Chemistry - Question 1
  • A base catalyzed keto-enol tautomerization would involve the creation of an enolate anion, with resonance structures of a carbanion and an oxyanion.
  • In an enol reaction, the alpha carbon acts as a nucleophile, and would accept the extra pair of electrons to form the carbanion
  • The electronegativity of an element determines if it is better able to accept electrons in its hybridization, and therefore a more electronegative atom will be more stable with a negative formal electric charge and it will contribute more to the reaction product.
  • Oxygen is more electronegative element, and therefore the oxygen with a negative formal charge is the most stable hybrid structure.
Test: Alpha-carbon Chemistry - Question 2
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What is the proper IUPAC name for the following molecule?

Detailed Solution for Test: Alpha-carbon Chemistry - Question 2
  • The aldehyde group would take precedence in numbering the carbons, so we should start numbering from left to right.
  • The hydroxyl group is located on carbon four, but the double bond must be described in naming the molecule.
  • The double bond between carbons changes the name of the molecule from nonanal to nonenal, and it begins at the carbon 2 position so the IUPAC name is 4-hydroxy-2-nonenal.
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Test: Alpha-carbon Chemistry - Question 3
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Which of the following are true statements regarding enolate formation?
I. Overall the more substituted a double bond on an enolate, the less stable the molecule is.
II. Low temperature reactions favor a more rapidly formed enolate
III. Thermodynamic enolates are favored in reactions with strong, sterically hindered bases.

Detailed Solution for Test: Alpha-carbon Chemistry - Question 3
  • Kinetic enolates are more rapidly formed, but are less stable. Thermodynamic enolates are slower to form, but are more stable
  • Saytseff's (or Zaitsev’s) rule states that the more substituted alkene will form in preference to the less substituted molecule. So, generally, the more substituted a double bond is, the more stable the alkene will be.
  • A strong, sterically hindered base will favor the less hindered portion of a molecule, which is more likely to result in a kinetic enolate
  • Lower temperatures favor the formation of the kinetic enolate; therefore statement II is correct.
Test: Alpha-carbon Chemistry - Question 4
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In the presence of a base, which carbon acts as the electrophilic site for this reaction of this molecule?
Detailed Solution for Test: Alpha-carbon Chemistry - Question 4
  • This reaction is the second step of a mixed aldol condensation, after formation of an enolate.
  • The alpha carbon of the ketone will act as a nucleophile, not an electrophile.
  • The oxygen of the aldehyde will be partially negative, and thus not an electrophile.
  • An aldol condensation requires a nucleophilic attack on the carbonyl electrophile at site B.
Test: Alpha-carbon Chemistry - Question 5
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What is the correct product of this reaction?
Detailed Solution for Test: Alpha-carbon Chemistry - Question 5
  • This is a mixed aldol condensation reaction, involving a nucleophilic attack on a carbonyl carbon.
  • The catalytic base will take a hydrogen from the most acidic carbon, which will be the alpha carbon between two oxygens. This will form a carbanion at that alpha carbon.
  • The carbonyl carbon of the cyclopentanone will act as an electrophile, and undergo nucleophilic attack from the alpha carbanion forming a double bond after removal of the water from the intermediate. This gives us a cyclopentane double bonded to the carbon between the two carbonyls groups.
Test: Alpha-carbon Chemistry - Question 6
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Which of the following ring structures will result in an equilibrium that favors the enol form?
Detailed Solution for Test: Alpha-carbon Chemistry - Question 6
  • The enol tautomer of these structures would involve a hydroxyl group at the ketone position, and a double bond formed between the carbonyl and alpha carbon.
  • The keto form is highly favored under normal conditions.
  • The enol form is extra stabilized and therefore favored when the formation of the double bond results in an aromatic ring, such as in phenol.
Test: Alpha-carbon Chemistry - Question 7
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Which of the following are correct products of a retro-aldol reaction on the following molecule?
Detailed Solution for Test: Alpha-carbon Chemistry - Question 7
  • The mechanism of a retro-aldol reaction is the exact reverse of an aldol reaction, therefore we can imagine the bond between the carbonyl carbon and the alpha carbon being broken to form our products.
  • To break our bond, we can imagine 2 hydrogens being added to the alpha carbon. It is important to note, the product shown is the aldol product, and has not undergone condensation.
  • Since we simply stop at the aldol addition product, we can simply move our electrons from the bond between the alpha carbon and hydroxyl carbon and create a second aldehyde group in place of the hydroxl group.
Test: Alpha-carbon Chemistry - Question 8
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Which of the following could undergo an aldol condensation to form the following molecule?
Detailed Solution for Test: Alpha-carbon Chemistry - Question 8
  • The best way to start with this problem is to number the carbons. This molecule has 11 total.
  • The aldol condensation requires two carbonyl carbons and will result in a double bond following the addition of heat. In this case, since there is one double bond in the product, no other double bonds should be present in the molecule.
  • The condensation will result in a double bond between an alpha carbon and a carbonyl carbon to complete the ring, and there will be a 5 carbon chain coming off the ring, therefore there must be a 5 carbon chain attached to the alpha carbon to get the proper product.
Test: Alpha-carbon Chemistry - Question 9
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Given the following reaction, are the products or reactants favored?
Detailed Solution for Test: Alpha-carbon Chemistry - Question 9
  • To find which side of the reaction is favored, we can use the given pKa’s, and plug them into the formulas:
    pKeq = pKa acidL - pKa acidR
    kEq = 10-pKeq
  • 25 - 35 = -10, so our pKeq value is -10. This means that our kEq = 1010
  • Since our KEq >> 1, we can conclude that the reaction will heavily favor the products side.
Test: Alpha-carbon Chemistry - Question 10
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Which of the following would be the proper products of an aldol condensation of this molecule?
Detailed Solution for Test: Alpha-carbon Chemistry - Question 10
  • An aldol condensation would involve the nucleophilic attack of the carbonyl carbon onto the relatively electrophilic alpha carbon.
  • To form the enal product, the oxygen of the carbonyl will become a hydroxyl leaving group, so only one oxygen atom is expected in the product.
  • The final product will have an alpha carbon double bonded to a 4 carbon chain group.
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