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Organic Chemistry Flashcards

7 cards from real MCAT practice questions. Tap to flip, then mark Knew It or Still Learning — missed cards come back until you master them.

Read the first 7 Organic Chemistry flashcards as text

  1. Which reagent converts a primary alcohol to an aldehyde without over-oxidizing to a carboxylic acid?

    Answer: PCC (pyridinium chlorochromate)

    PCC selectively oxidizes primary alcohols to aldehydes and stops there because it lacks aqueous conditions needed for further oxidation.

  2. In an E2 elimination reaction, what is the required geometric relationship between the leaving group and the beta hydrogen?

    Answer: Anti-periplanar (180°)

    E2 requires an anti-periplanar arrangement (180°) so that the breaking bonds can align with the forming pi bond in a concerted step.

  3. Which type of isomerism is exhibited by cis- and trans-2-butene?

    Answer: Geometric (cis/trans) isomerism

    Cis- and trans-2-butene are geometric isomers because they have the same connectivity but differ in the spatial arrangement of groups around the double bond.

  4. What product forms when an alkene reacts with ozone (O3) followed by a reductive workup (Zn, H2O)?

    Answer: Two carbonyl compounds (aldehydes/ketones)

    Ozonolysis with reductive workup cleaves the double bond to give two carbonyl compounds; aldehydes if the carbon was =CH2, ketones if disubstituted.

  5. Which statement best describes the inductive effect of a fluorine substituent on benzoic acid?

    Answer: It withdraws electrons inductively, lowering pKa

    Fluorine is highly electronegative and withdraws electron density inductively, stabilizing the carboxylate anion and lowering the pKa (increasing acidity).

  6. A compound with the molecular formula C4H8O shows a strong IR absorption near 1715 cm⁻¹ and no broad O–H stretch. What functional group is present?

    Answer: Ketone or aldehyde

    A carbonyl stretch near 1715 cm⁻¹ without an O–H peak indicates a ketone or aldehyde, not an alcohol or carboxylic acid.

  7. Which of the following best explains why sp² hybridized carbons are shorter than sp³ C–C bonds?

    Answer: Greater s-orbital character increases electronegativity and shortens bonds

    Increased s-character in sp² orbitals means electrons are held closer to the nucleus, resulting in shorter, stronger bonds compared to sp³.