Organic Chemistry 2 Quiz: Carbonyl Structure And Reactivity Electrophilicity Resonance

What this quiz covers

This quiz focuses on Carbonyl Structure And Reactivity Electrophilicity Resonance, giving you a quick way to practice the rules, question types, and explanations that matter most for Organic Chemistry 2.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

1. Phosgene > Acetone > Formaldehyde
2. Acetone > Formaldehyde > Phosgene (correct answer)
3. Formaldehyde > Acetone > Phosgene
4. Phosgene > Formaldehyde > Acetone

Question 2

1. Resonance makes the carbonyl carbon electrophilic and stabilizes the developing negative charge on oxygen after attack. (correct answer)
2. Resonance makes nitrogen electrophilic, so acetone donates electrons to $\ce{NH3}$ in the first step.
3. Resonance prevents any tetrahedral intermediate by keeping the carbonyl carbon strictly $sp$-hybridized.
4. Resonance shifts positive charge to oxygen, so $\ce{NH3}$ attacks oxygen to form an O–N bond first.

Question 3

1. Aldehydes are more electrophilic because fewer alkyl groups donate electron density, increasing $\delta^+$ at carbon. (correct answer)
2. Ketones are more electrophilic because alkyl groups withdraw electrons strongly, increasing positive charge on carbon.
3. Aldehydes are less electrophilic because resonance places negative charge on carbon more often than in ketones.
4. Aldehydes and ketones have identical electrophilicity because oxygen's electronegativity is unchanged by substituents.

Question 4

1. Aldehydes are more reactive because they have less steric hindrance and a more electrophilic carbonyl carbon. (correct answer)
2. Ketones are more reactive because two alkyl groups stabilize the carbonyl carbon's positive charge by resonance.
3. Aldehydes are less reactive because resonance makes their carbonyl oxygen less electronegative than in ketones.
4. Aldehydes and ketones are equally reactive because nucleophilic addition depends only on solvent, not structure.

Question 5

1. I > II > III (correct answer)
2. I > III > II
3. III > II > I
4. II > III > I

Question 6

1. A structure with a double bond between C and O, a single bond between C and N, and no formal charges.
2. A zwitterionic structure with a single bond between C and O (with a negative charge), and a double bond between C and N (with a positive charge).
3. A zwitterionic structure with a double bond between C and O (with a positive charge), a single bond between C and N, and a negative charge on the oxygen.
4. A zwitterionic structure with a C-O single bond, a carbocation at the carbonyl carbon, and a negative charge on the nitrogen. (correct answer)

Question 7

1. It decreases reactivity by placing a positive charge on the oxygen, which inductively pulls electron density away from the nucleophile.
2. It has no net effect on reactivity because the protonation is a rapid, reversible equilibrium that precedes the rate-determining step.
3. It significantly increases reactivity by converting the carbonyl into a much stronger electrophile with substantial positive charge on the carbon. (correct answer)
4. It decreases reactivity by sterically blocking the face of the carbonyl group, preventing the approach of the methanol nucleophile.

Question 8

1. Less electrophilic, because the carbonyl carbon receives resonance donation from two oxygen atoms, delocalizing its positive character more effectively. (correct answer)
2. More electrophilic, because the inductive effect of the second electronegative oxygen atom makes the carbon more electron-poor.
3. Essentially the same, because resonance from one oxygen atom at a time is the dominant stabilizing factor in both molecules.
4. More electrophilic, because the repulsion between the two oxygen lone pairs prevents either from effectively donating into the carbonyl π-system.

Question 9

1. Acetone
2. 3-Pentanone
3. 3,3-Dimethyl-2-butanone (Pinacolone)
4. 2,4-Dimethyl-3-pentanone (Diisopropyl ketone) (correct answer)

Question 10

1. The nitrogen atom in the amide is more electronegative than the carbon atom in acetone, which weakens the C=O bond.
2. The two methyl groups on the nitrogen sterically hinder the carbonyl, causing the bond to lengthen and vibrate more slowly.
3. Resonance donation of the nitrogen's lone pair into the carbonyl π-system gives the C=O bond significant single-bond character. (correct answer)
4. The C-N bond is shorter and stronger than a C-C bond, which inductively strengthens the adjacent C=O bond.

Question 11

1. The pKa data is irrelevant to the electrophilicity of the neutral molecules, as it only measures the stability of the protonated forms.
2. Acetamide is a stronger base, which indicates that its carbonyl carbon is less electron-deficient and therefore less electrophilic. (correct answer)
3. Acetone is a stronger base, which indicates that its carbonyl carbon is more electron-deficient and therefore more electrophilic.
4. Acetamide is a stronger base, which indicates that its carbonyl carbon is more electron-deficient and therefore more electrophilic.

Question 12

1. More electrophilic than both, as the inductive effects of two heteroatoms are additive and dominate over resonance.
2. Less electrophilic than both, because the powerful resonance donation from nitrogen is supplemented by donation from oxygen.
3. Less electrophilic than an ester, but more electrophilic than an amide. (correct answer)
4. More electrophilic than an ester, but less electrophilic than an amide.

Question 13

1. The carbonyl carbon in the lactone is more electropositive due to poor resonance stabilization in the constrained ring.
2. The acyclic ester is better solvated by water, which deactivates it toward attack by hydroxide.
3. The hydrolysis reaction relieves torsional strain that is inherent in the planar five-membered ring structure. (correct answer)
4. The ethoxy group of the acyclic ester is a stronger electron donor than the ring-bound oxygen of the lactone.

Question 14

1. Resonance makes carbonyl oxygen electrophilic, so water attacks oxygen to form a stable oxonium without C–O bond formation.
2. Resonance and polarization keep carbon electrophilic, enabling water attack and stabilizing the developing positive charge after protonation. (correct answer)
3. Resonance prevents addition because the $\pi$ bond cannot be disrupted in any step of the mechanism.
4. Resonance forces water to act as an electrophile, so the carbonyl attacks water to form a carbocation.

Question 15

1. Hydride attacks oxygen because oxygen is the electrophile in the dominant resonance contributor.
2. Hydride attacks carbon because carbon bears $\delta^+$ from C=O polarization, giving an alkoxide intermediate. (correct answer)
3. Hydride cannot add because resonance makes the carbonyl carbon fully neutral and nonreactive.
4. Hydride adds to the methyl group in acetone because it is less hindered than the carbonyl carbon.

Question 16

1. Aldehydes are more reactive because they are less hindered and their carbonyl carbon is more electron-poor. (correct answer)
2. Ketones are more reactive because steric hindrance increases nucleophile attraction to the carbonyl carbon.
3. Aldehydes are less reactive because resonance makes the C=O bond nonpolar compared with ketones.
4. Both are equally reactive because sterics and electronics cancel exactly in all aldehyde/ketone pairs.

Question 17

1. Aldehydes are more reactive because they are less hindered and have fewer electron-donating groups reducing carbonyl electrophilicity. (correct answer)
2. Ketones are more reactive because two substituents increase resonance, forcing a larger positive charge onto oxygen.
3. Aldehydes are less reactive because their resonance contributors place negative charge on carbon more than ketones do.
4. Aldehydes and ketones are equally reactive because nucleophilic addition depends only on nucleophile strength.

Question 18

1. Nucleophilic addition is favored because the carbonyl carbon is electrophilic from C=O polarization and resonance contributors. (correct answer)
2. Nucleophilic addition is disfavored because the carbonyl carbon is nucleophilic due to oxygen-to-carbon resonance donation.
3. Nucleophilic addition occurs at oxygen because oxygen is the electrophilic atom in the carbonyl resonance hybrid.
4. Reactivity is determined solely by temperature; resonance and polarization do not affect nucleophilic addition rates.

Question 19

1. Nucleophiles attack oxygen because oxygen bears $\delta^+$ from resonance, making it the electrophilic center.
2. Nucleophiles attack carbon because the polarized C=O bond gives carbon $\delta^+$, forming a tetrahedral alkoxide. (correct answer)
3. Nucleophiles attack carbon only if the carbonyl lacks resonance; resonance otherwise blocks addition completely.
4. Nucleophiles attack the alkyl substituent because it is more electron-poor than the carbonyl carbon.

Question 20

1. Resonance removes polarization, making the carbonyl carbon non-electrophilic and preventing nucleophilic attack.
2. Resonance places positive charge on oxygen, so oxygen is the main electrophilic site for nucleophiles.
3. Resonance contributes to $\delta^+$ at carbon, maintaining electrophilicity while stabilizing intermediates like alkoxides. (correct answer)
4. Resonance makes carbonyls behave like alkenes, favoring electrophilic addition instead of nucleophilic addition.