The MCAT (Medical College Admission Test) includes a section on organic chemistry. This section evaluates the candidate's understanding of fundamental chemical principles, reaction mechanisms, and analytical techniques.This section is essential for assessing problem-solving skills and the ability to apply organic chemistry concepts in biological and medical contexts.l.toLowerCase().replace(/\s+/g,"-")" id="3e6549e5-ad14-42d7-a83b-604be5cb109c" data-toc-id="3e6549e5-ad14-42d7-a83b-604be5cb109c">MCAT Organic Chemistry Practice QuestionsIt is crucial to practise test questions to excel in the MCAT exam. Many previous test-takers agree that solving practice problems before exam day helps improve confidence and familiarity with question patterns. By attempting MCAT-style questions, students become accustomed to different question formats and answer choices, enhancing their problem-solving abilities.Regular practice also helps identify weak areas and improve performance through targeted study. Keep a notepad and pen ready, and challenge yourself with MCAT organic chemistry practice questions to strengthen your preparation!l.toLowerCase().replace(/\s+/g,"-")" id="cc4181e9-221b-4532-a6e2-266be7efbcf7" data-toc-id="cc4181e9-221b-4532-a6e2-266be7efbcf7">1. Which functional group exhibits the strongest intermolecular forces and highest boiling point?A) AldehydeB) KetoneC) Carboxylic acidD) EtherAnswer: C) Carboxylic acidExplanation: Carboxylic acids have strong hydrogen bonding, leading to higher boiling points compared to aldehydes, ketones, and ethers. Hydrogen bonding significantly increases the energy required for vaporization.l.toLowerCase().replace(/\s+/g,"-")" id="caea0e4a-78ff-4689-970b-4a1bc2a1e8d7" data-toc-id="caea0e4a-78ff-4689-970b-4a1bc2a1e8d7">2. What reagent selectively oxidizes a primary alcohol to an aldehyde without further oxidation?A) PCC (Pyridinium chlorochromate)B) KMnO₄C) CrO₃D) H₂CrO₄Answer: A) PCCExplanation: PCC stops oxidation at the aldehyde stage. Other strong oxidizers like KMnO₄ and CrO₃ further oxidize the aldehyde to a carboxylic acid.l.toLowerCase().replace(/\s+/g,"-")" id="e690f9c0-f1e9-4a8f-b6e1-1c56a528d65e" data-toc-id="e690f9c0-f1e9-4a8f-b6e1-1c56a528d65e">3. What is the hybridization of the carbonyl carbon in an aldehyde?A) spB) sp²C) sp³D) none of the aboveAnswer: B) sp²Explanation: The carbonyl carbon in an aldehyde forms three sigma bonds and one π bond, giving it sp² hybridization and trigonal planar geometry.l.toLowerCase().replace(/\s+/g,"-")" id="5b1c6db6-ff07-48f0-82ec-0de5bc116208" data-toc-id="5b1c6db6-ff07-48f0-82ec-0de5bc116208">4. Which type of isomers differ in connectivity of atoms?A) Conformational isomersB) Configurational isomersC) Constitutional isomersD) EnantiomersAnswer: C) Constitutional isomersExplanation: Constitutional isomers have the same molecular formula but different connectivity (e.g., ethanol vs. dimethyl ether).l.toLowerCase().replace(/\s+/g,"-")" id="bd006969-367c-4fa9-9dad-0f26adf55de5" data-toc-id="bd006969-367c-4fa9-9dad-0f26adf55de5">5. In thin-layer chromatography (TLC), which type of compound has the highest Rf value?A) Polar compoundB) Nonpolar compoundC) Ionic compoundD) Amphoteric compoundAnswer: B) Nonpolar compoundExplanation: Nonpolar compounds travel farther on the TLC plate because the stationary phase (silica gel) is polar, and polar compounds interact more with it, leading to lower Rf values.l.toLowerCase().replace(/\s+/g,"-")" id="aa74492b-6152-4d74-acf7-131b6b472220" data-toc-id="aa74492b-6152-4d74-acf7-131b6b472220">6. Which molecule is the best nucleophile in an SN2 reaction in a protic solvent?A) F⁻B) Cl⁻C) Br⁻D) I⁻Answer: D) I⁻Explanation: In protic solvents, larger ions like I⁻ are less solvated and more nucleophilic, while smaller ions (F⁻) are heavily solvated and weaker nucleophiles.l.toLowerCase().replace(/\s+/g,"-")" id="6abf2c03-4d76-405b-b831-dc7c0e9efcc1" data-toc-id="6abf2c03-4d76-405b-b831-dc7c0e9efcc1">7. Which spectroscopy technique is best for identifying functional groups in a compound?A) Mass spectrometryB) Infrared (IR) spectroscopyC) UV-Vis spectroscopyD) NMR spectroscopyAnswer: B) Infrared (IR) spectroscopyExplanation: IR spectroscopy detects specific bond vibrations, allowing the identification of functional groups (e.g., O-H stretch around 3200–3600 cm⁻¹).l.toLowerCase().replace(/\s+/g,"-")" id="9038ef93-6aee-4ac6-a311-72fb4147ad49" data-toc-id="9038ef93-6aee-4ac6-a311-72fb4147ad49">8. Which reaction mechanism involves the formation of a carbocation intermediate?A) SN1B) SN2C) E2D) Diels-AlderAnswer: A) SN1Explanation: The SN1 reaction is a two-step mechanism: (1) Leaving group leaves, forming a carbocation, and (2) Nucleophile attacks.l.toLowerCase().replace(/\s+/g,"-")" id="0b886b58-8534-4a2d-9cb7-1ba2d0dbe5fd" data-toc-id="0b886b58-8534-4a2d-9cb7-1ba2d0dbe5fd">9. Which compound is the most acidic?A) CH₄B) CH₃OHC) CH₃COOHD) NH₃Answer: C) CH₃COOHExplanation: Acetic acid (CH₃COOH) is the strongest acid due to the resonance stabilization of its conjugate base.l.toLowerCase().replace(/\s+/g,"-")" id="e85e383a-466c-4492-b6db-353e47415fc7" data-toc-id="e85e383a-466c-4492-b6db-353e47415fc7">10. Which condition favors an E2 reaction over an SN2 reaction?A) Strong base, primary substrateB) Weak base, primary substrateC) Strong base, tertiary substrateD) Weak nucleophile, secondary substrateAnswer: C) Strong base, tertiary substrateExplanation: E2 dominates with strong bases and hindered substrates (e.g., tertiary halides), while SN2 prefers primary substrates.l.toLowerCase().replace(/\s+/g,"-")" id="a4d65337-7e71-45d4-bad0-ad8f36074f0f" data-toc-id="a4d65337-7e71-45d4-bad0-ad8f36074f0f">11. Which compound has the lowest pKa?A) EthanolB) PhenolC) Acetic acidD) AmmoniaAnswer: C) Acetic acidExplanation: Acetic acid is more acidic than ethanol, phenol, and ammonia, with a pKa around 4.8 due to resonance stabilization of its conjugate base.l.toLowerCase().replace(/\s+/g,"-")" id="77aa911a-4834-4bc0-959a-99bb772aedab" data-toc-id="77aa911a-4834-4bc0-959a-99bb772aedab">12. Which reaction mechanism leads to inversion of configuration?A) SN1B) SN2C) E1D) E2Answer: B) SN2Explanation: SN2 is a concerted reaction where the nucleophile attacks from the opposite side, causing inversion of configuration.l.toLowerCase().replace(/\s+/g,"-")" id="f4ddf957-7698-4629-a46f-c854dc1bcc40" data-toc-id="f4ddf957-7698-4629-a46f-c854dc1bcc40">13. Which reagent converts an alkene into an epoxide?A) PCCB) OsO₄C) mCPBAD) H₂/PdAnswer: C) mCPBAExplanation: mCPBA (meta-chloroperoxybenzoic acid) oxidizes alkenes to epoxides.l.toLowerCase().replace(/\s+/g,"-")" id="92fbfd3c-bc1c-499f-8fc8-077b921ef88c" data-toc-id="92fbfd3c-bc1c-499f-8fc8-077b921ef88c">14. Which compound undergoes nucleophilic acyl substitution most readily?A) AmideB) EsterC) Acid chlorideD) Carboxylate ionAnswer: C) Acid chlorideExplanation: Acid chlorides have the best leaving group (Cl⁻), making them the most reactive toward nucleophilic attack.l.toLowerCase().replace(/\s+/g,"-")" id="4a4c8043-3fc6-406a-b5c5-a6c55756a397" data-toc-id="4a4c8043-3fc6-406a-b5c5-a6c55756a397">15. Which solvent is best for an SN2 reaction?A) WaterB) AcetoneC) MethanolD) Acetic acidAnswer: B) AcetoneExplanation: SN2 favours polar aprotic solvents like acetone, which do not solvate nucleophiles strongly.l.toLowerCase().replace(/\s+/g,"-")" id="c763d224-c21c-4963-8311-c26db5c7ded7" data-toc-id="c763d224-c21c-4963-8311-c26db5c7ded7">16. Which reagent is used for the Markovnikov addition of water to an alkene?A) BH₃/THF, H₂O₂B) H₃O⁺C) NaNH₂D) HBr, peroxideAnswer: B) H₃O⁺Explanation: Acid-catalyzed hydration (H₃O⁺) follows Markovnikov’s rule, adding OH to the more substituted carbon.l.toLowerCase().replace(/\s+/g,"-")" id="89e95408-1250-4cc0-8303-1f90ce0e120c" data-toc-id="89e95408-1250-4cc0-8303-1f90ce0e120c">17. What happens when benzene reacts with Br₂/FeBr₃?A) Bromine is added to all carbonsB) Bromine adds to the para positionC) Bromine substitutes a hydrogenD) Benzene is reducedAnswer: C) Bromine substitutes for hydrogenExplanation: Electrophilic aromatic substitution (EAS) replaces hydrogen with bromine in the presence of FeBr₃.l.toLowerCase().replace(/\s+/g,"-")" id="bc873d14-59de-42b6-82e9-7e6eed740fb1" data-toc-id="bc873d14-59de-42b6-82e9-7e6eed740fb1">18. Which reaction converts a ketone to an alcohol?A) PCCB) LiAlH₄C) H₂CrO₄D) Br₂Answer: B) LiAlH₄Explanation: LiAlH₄ is a strong reducing agent that converts ketones to secondary alcohols.l.toLowerCase().replace(/\s+/g,"-")" id="ecff1588-f402-40c2-bf3b-b3ea739dc2ea" data-toc-id="ecff1588-f402-40c2-bf3b-b3ea739dc2ea">19. Which molecule is more soluble in water?A) HexaneB) EthanolC) BenzeneD) CyclohexaneAnswer: B) EthanolExplanation: Ethanol has a hydroxyl (-OH) group, allowing hydrogen bonding with water.l.toLowerCase().replace(/\s+/g,"-")" id="59e16240-b13d-44cb-9d59-eac8e6cd633c" data-toc-id="59e16240-b13d-44cb-9d59-eac8e6cd633c">20. Which of the following Newman projections represents the most stable conformation of butane?A) Eclipsed conformation with methyl groups at 0°B) Eclipsed conformation with methyl groups at 120°C) Gauche conformation with methyl groups at 60°D) Anti conformation with methyl groups at 180°Answer: D) Anti conformation with methyl groups at 180°Explanation: The anti-conformation (staggered with methyl groups at opposite sides, 180° apart) is the most stable due to minimal steric hindrance. Eclipsed conformations are less stable due to torsional strain.l.toLowerCase().replace(/\s+/g,"-")" id="c17f1e85-87d2-4863-a6ba-51240e9d3352" data-toc-id="c17f1e85-87d2-4863-a6ba-51240e9d3352">21. How many stereo centres does the compound shown below have?l.toLowerCase().replace(/\s+/g,"-")" id="2c2fcd58-f40a-4523-979a-4a9ec4bfd720" data-toc-id="2c2fcd58-f40a-4523-979a-4a9ec4bfd720">(CH₃)₂CHCH (OH) CH₂BrA) 0B) 1C) 2D) 3Answer: B) 1Explanation: A stereo centre (chiral centre) is a carbon with four different groups. Here, the CH (OH) CH₂Br carbon is the only chiral centre.l.toLowerCase().replace(/\s+/g,"-")" id="8cd7ef4d-2cc3-423b-a13b-a0369c2b5098" data-toc-id="8cd7ef4d-2cc3-423b-a13b-a0369c2b5098">22. What is the correct IUPAC name for the following compound?l.toLowerCase().replace(/\s+/g,"-")" id="31df0da1-ccd7-41f6-a252-1ad3140ee08c" data-toc-id="31df0da1-ccd7-41f6-a252-1ad3140ee08c">CH₃CH₂CH (Br) CH₃A) 2-BromobutaneB) 1-BromobutaneC) 3-BromobutaneD) 1-Bromo-2-methylpropaneAnswer: A) 2-BromobutaneExplanation: Numbering the longest chain from the end closest to the substituent, bromine is on carbon 2, making it 2-bromobutane.l.toLowerCase().replace(/\s+/g,"-")" id="7ea89a89-b508-4844-a16b-884fd14134f6" data-toc-id="7ea89a89-b508-4844-a16b-884fd14134f6">24. What is the major product of the following reaction?l.toLowerCase().replace(/\s+/g,"-")" id="0ac5bf18-c4dd-4a17-85ce-aad085920baf" data-toc-id="0ac5bf18-c4dd-4a17-85ce-aad085920baf">Cyclohexene + Br₂ in CCl₄A) 1, 2-DibromocyclohexaneB) 1-BromocyclohexaneC) 1, 4-DibromocyclohexaneD) CyclohexanolAnswer: A) 1, 2-DibromocyclohexaneExplanation: Br₂ in CCl₄ leads to the anti-addition of bromine across the double bond, forming a trans-1, 2-dibromo cyclohexane.l.toLowerCase().replace(/\s+/g,"-")" id="bf7965f4-34da-4da7-a8ed-aabbe086859e" data-toc-id="bf7965f4-34da-4da7-a8ed-aabbe086859e">25. Which diene undergoes the fastest Diels-Alder reaction?A) 1,3-ButadieneB) 1,3-CyclopentadieneC) 1,3-HexadieneD) 1,4-PentadieneAnswer: B) 1, 3-CyclopentadieneExplanation: Cyclic dienes react faster in Diels-Alder reactions due to locked s-cis conformation, which is essential for the reaction.l.toLowerCase().replace(/\s+/g,"-")" id="34007554-2d4f-4fbf-b1eb-124432e9041e" data-toc-id="34007554-2d4f-4fbf-b1eb-124432e9041e">26. Identify the major product of the following reaction:l.toLowerCase().replace(/\s+/g,"-")" id="c79fe3de-3cc3-4487-a9f3-7633f57d8e79" data-toc-id="c79fe3de-3cc3-4487-a9f3-7633f57d8e79">CH₃CH₂C≡CH + HBrA) CH₃CH₂C (Br)=CH₂B) CH₃CH₂CH=CHBrC) CH₃CH₂C(Br)₂CH₃D) CH₃CH₂C(Br)=CHBrAnswer: A) CH₃CH₂C(Br)=CH₂Explanation: HBr adds to the alkyne via Markovnikov's rule, placing Br on the more substituted carbon, forming a bromoalkene.l.toLowerCase().replace(/\s+/g,"-")" id="ad12aec9-ca66-4e03-82ff-69f741db526d" data-toc-id="ad12aec9-ca66-4e03-82ff-69f741db526d">27. Which structure represents meso-2, 3-dibromobutane?A) A molecule with two chiral centres and a plane of symmetryB) A molecule with two chiral centres and no symmetryC) A molecule with no chiral centresD) A molecule with two chiral centres and a double bondAnswer: A) A molecule with two chiral centres and a plane of symmetryExplanation: Meso compounds have chiral centres but are achiral due to symmetry.l.toLowerCase().replace(/\s+/g,"-")" id="e89b5683-b9c1-48f1-9e2b-f0562317caf3" data-toc-id="e89b5683-b9c1-48f1-9e2b-f0562317caf3">28. Which compound exhibits the highest dipole moment?A) Carbon tetrachloride (CCl₄)B) Acetone (CH₃COCH₃)C) Benzene (C₆H₆)D) Ethane (C₂H₆)Answer: B) AcetoneExplanation: Acetone has a polar C=O bond, making it the most polar among these molecules.l.toLowerCase().replace(/\s+/g,"-")" id="2161c34e-1977-40d5-8585-55b9c0ae0961" data-toc-id="2161c34e-1977-40d5-8585-55b9c0ae0961">29. Which compound undergoes the fastest SN2 reaction?A) CH₃BrB) CH₃CH₂BrC) CH₃CH₂CH₂BrD) CH₃CH₂CH₂CH₂BrAnswer: A) CH₃BrExplanation: SN2 prefers primary halides, and methyl halides react the fastest due to no steric hindrance.l.toLowerCase().replace(/\s+/g,"-")" id="fb28e8a2-1c1d-4151-85bc-4de1c060b85e" data-toc-id="fb28e8a2-1c1d-4151-85bc-4de1c060b85e">30. Which of the following compounds would exhibit the highest rate of racemization in an acidic solution?A) 2-ChlorobutaneB) 2-Bromo-2-methylpropaneC) 2-Chloro-2-methylpropaneD) 2-IodobutaneAnswer: C) 2-Chloro-2-methylpropaneExplanation: Racemization occurs via carbocation formation. 2-Chloro-2-methylpropane forms a highly stable tertiary carbocation, leading to rapid racemization.l.toLowerCase().replace(/\s+/g,"-")" id="3f73edda-a1e9-4739-84f5-da2dbc9f69fe" data-toc-id="3f73edda-a1e9-4739-84f5-da2dbc9f69fe">31. Given the following reaction, what is the expected major product?l.toLowerCase().replace(/\s+/g,"-")" id="b8763b1c-d459-4e55-8f20-2ee11dac9917" data-toc-id="b8763b1c-d459-4e55-8f20-2ee11dac9917">(CH₃)₃CBr + NaOCH₃ →?A) (CH₃)₃COCH₃B) (CH₃)₃CCH₃C) (CH₃)₂C=CH₂D) CH₃CH₂CBr (CH₃)₂Answer: C) (CH₃)₂C=CH₂Explanation: Since the substrate is tertiary, E2 elimination dominates over SN2 when using a strong base (NaOCH₃).l.toLowerCase().replace(/\s+/g,"-")" id="72ebcba7-e35e-441a-85ab-ab582bc931b1" data-toc-id="72ebcba7-e35e-441a-85ab-ab582bc931b1">32. The major product of the following reaction is:l.toLowerCase().replace(/\s+/g,"-")" id="a81227e0-2fa6-414b-ab57-250b006a188f" data-toc-id="a81227e0-2fa6-414b-ab57-250b006a188f">PhCH₂Br + Mg (ether) →?A) PhCH₂MgBrB) PhCH₃C) PhCH₂OHD) PhMgBrAnswer: A) PhCH₂MgBrExplanation: The reaction forms a Grignard reagent, phenylmethylmagnesium bromide (PhCH₂MgBr).l.toLowerCase().replace(/\s+/g,"-")" id="f066c4a2-8657-428f-a3f4-8a851707bee8" data-toc-id="f066c4a2-8657-428f-a3f4-8a851707bee8">33. In which of the following reactions does a free radical intermediate play a key role?A) CH₄ + Cl₂ → CH₃Cl + HClB) CH₃CH₂Br + OH⁻ → CH₃CH₂OHC) CH₃COOH + SOCl₂ → CH₃COClD) CH₃CH=CH₂ + HBr → CH₃CHBrCH₃Answer: A) CH₄ + Cl₂ → CH₃Cl + HClExplanation: This is free radical halogenation, initiated by UV light, where Cl• radicals are formed.l.toLowerCase().replace(/\s+/g,"-")" id="07c66b35-9f1e-4a2e-9c5d-2d0c659c1255" data-toc-id="07c66b35-9f1e-4a2e-9c5d-2d0c659c1255">34. How many unique signals would be expected in the ¹H NMR spectrum of 2, 2-dimethylbutane?A) 2B) 3C) 4D) 5Answer: B) 3Explanation: 2, 2-Dimethylbutane has three distinct proton environments due to molecular symmetry.l.toLowerCase().replace(/\s+/g,"-")" id="239655d5-49da-4253-84af-ef50a7c2ee5d" data-toc-id="239655d5-49da-4253-84af-ef50a7c2ee5d">35. What is the hybridization of the carbon in the isocyanide functional group (-NC)?A) spB) sp²C) sp³D) sp³dAnswer: A) spExplanation: The carbon in -NC forms a triple bond with nitrogen, requiring sp hybridization.l.toLowerCase().replace(/\s+/g,"-")" id="0acbe41f-9a40-4357-950b-0e80412ef325" data-toc-id="0acbe41f-9a40-4357-950b-0e80412ef325">36. What is the expected fragmentation pattern of butanone (CH₃COCH₂CH₃) in mass spectrometry?A) Peaks at m/z = 43 and 58B) Peaks at m/z = 29 and 44C) Peaks at m/z = 15 and 30D) Peaks at m/z = 50 and 75Answer: A) Peaks at m/z = 43 and 58Explanation: Butanone undergoes McLafferty rearrangement, producing m/z 43 (CH₃CO⁺) and 58 (CH₃COCH₂⁺).l.toLowerCase().replace(/\s+/g,"-")" id="4fecf4ac-a891-481f-98b3-a98a438c4249" data-toc-id="4fecf4ac-a891-481f-98b3-a98a438c4249">37. Which of the following molecules is most reactive in an electrophilic aromatic substitution reaction?A) NitrobenzeneB) BenzeneC) TolueneD) Benzoic acidAnswer: C) TolueneExplanation: The methyl group is electron-donating, activating the benzene ring towards EAS.l.toLowerCase().replace(/\s+/g,"-")" id="497b06c3-51ed-430f-949e-8c0824fbbb04" data-toc-id="497b06c3-51ed-430f-949e-8c0824fbbb04">38. What is the correct order of increasing acidity for the following compounds?I. CH₃COOHII. CF₃COOHIII. CH₃CH₂OHA) III < I < IIB) I < III < IIC) III < II < ID) II < I < IIIAnswer: A) III < I < IIExplanation: CF₃COOH is the strongest acid due to the electron-withdrawing effect of fluorine.l.toLowerCase().replace(/\s+/g,"-")" id="a0065365-3c98-4c15-976b-a9758a811e6f" data-toc-id="a0065365-3c98-4c15-976b-a9758a811e6f">39. Which of the following statements about infrared spectroscopy is correct?A) A strong absorption around 3300 cm⁻¹ indicates a C≡C bondB) The C=O stretch appears around 1650 cm⁻¹C) The O-H stretch of alcohol appears broad and strong near 3200–3600 cm⁻¹D) The C-H stretch of an alkane appears below 2000 cm⁻¹Answer: C)Explanation: The O-H stretch of alcohol appears broad and strong near 3200–3600 cm⁻¹l.toLowerCase().replace(/\s+/g,"-")" id="7f6dddbc-81a5-434f-b993-72819309c333" data-toc-id="7f6dddbc-81a5-434f-b993-72819309c333">40. What is the major product of the following reaction?l.toLowerCase().replace(/\s+/g,"-")" id="898cc8ad-114d-4de8-92e1-ebad9411e364" data-toc-id="898cc8ad-114d-4de8-92e1-ebad9411e364">Ph-C≡C-H + NaNH₂ →?A) Ph-CH=CH₂B) Ph-C≡C⁻ Na⁺C) Ph-C≡C-CH₃D) Ph-CH₂-C≡CHAnswer: B) Ph-C≡C⁻ Na⁺Explanation: NaNH₂ deprotonates terminal alkynes, forming an acetylide anion.l.toLowerCase().replace(/\s+/g,"-")" id="07a88511-8feb-4976-8b4a-941e66373e22" data-toc-id="07a88511-8feb-4976-8b4a-941e66373e22">41. Which of the following describes the product of an aldol condensation reaction followed by dehydration?A) An alcoholB) A β-hydroxy aldehydeC) An α, β-unsaturated carbonyl compoundD) A carboxylic acidAnswer: C) An α, β-unsaturated carbonyl compoundExplanation: Aldol condensation forms a β-hydroxy ketone/aldehyde, which dehydrates to form an α, β-unsaturated carbonyl.l.toLowerCase().replace(/\s+/g,"-")" id="c7a1bcec-3802-45bc-9e73-73a4cd8e2dec" data-toc-id="c7a1bcec-3802-45bc-9e73-73a4cd8e2dec">42. The major product of the reaction between Benzaldehyde and acetophenone under basic conditions is:A) BenzoinB) α, β-Unsaturated ketoneC) Benzoic acidD) PhenylacetoneAnswer: B) α, β-Unsaturated ketoneExplanation: This is a Claisen-Schmidt condensation, yielding a β-unsaturated ketone.l.toLowerCase().replace(/\s+/g,"-")" id="a3cef8f3-d480-4305-ae18-d06f0a0e5774" data-toc-id="a3cef8f3-d480-4305-ae18-d06f0a0e5774">43. What is the order of increasing reactivity for the following carboxylic acid derivatives?I. Acyl chlorideII. EsterIII. AmideA) III < II < IB) I < III < IIC) II < I < IIID) III < I < IIAnswer: A) III < II < IExplanation: Acyl chlorides are the most reactive, followed by esters, and amides are the least reactive due to resonance stabilization.l.toLowerCase().replace(/\s+/g,"-")" id="fa97d4b7-2b50-4bb5-b2b6-7137071e39b6" data-toc-id="fa97d4b7-2b50-4bb5-b2b6-7137071e39b6">44. Which of the following statements about SN1 and SN2 reactions is FALSE?A) SN1 reactions occur faster in polar protic solvents.B) SN2 reactions result in inversion of configuration at the reaction centre.C) SN1 reactions involve a carbocation intermediate.D) SN2 reactions prefer highly branched substrates to avoid steric hindrance.Answer: D) SN2 reactions prefer highly branched substrates to avoid steric hindrance.Explanation: SN2 reactions occur via a single-step, backside attack mechanism, which bulky groups hinder. They are the fastest with primary substrates.l.toLowerCase().replace(/\s+/g,"-")" id="cb9e98a0-e62a-4830-948a-678b7e6ca513" data-toc-id="cb9e98a0-e62a-4830-948a-678b7e6ca513">45. In a Diels-Alder reaction, which conformation must the diene be in to react?A) S-transB) S-cisC) ChairD) PlanarAnswer: B) S-cisExplanation: The diene in a Diels-Alder reaction must be in the s-cis conformation to allow orbital overlap with the dienophile.l.toLowerCase().replace(/\s+/g,"-")" id="ce11cfc3-e306-4e47-b8ac-b1e72274374c" data-toc-id="ce11cfc3-e306-4e47-b8ac-b1e72274374c">46. Which of the following correctly explains why benzene is more stable than 1, 3, and 5-hexatriene?A) Benzene has localized π electrons.B) Benzene has a higher heat of hydrogenation.C) Benzene exhibits complete delocalization of π electrons.D) Benzene undergoes addition reactions readily.Answer: C) Benzene exhibits complete delocalization of π electrons.Explanation: Aromatic stabilization arises from continuous π-electron delocalization over the entire ring, making benzene much more stable.l.toLowerCase().replace(/\s+/g,"-")" id="b738d154-5101-495d-b2d6-a099a9704678" data-toc-id="b738d154-5101-495d-b2d6-a099a9704678">47. A compound exhibits IR absorption at 1700 cm⁻¹ and a broad peak at 3200–3500 cm⁻¹. What is the most likely functional group present?A) AlcoholB) Carboxylic acidC) AldehydeD) KetoneAnswer: B) Carboxylic acidExplanation: Carboxylic acids show a strong C=O stretch (~1700 cm⁻¹) and a broad O-H stretch (~3200–3500 cm⁻¹) due to hydrogen bonding.l.toLowerCase().replace(/\s+/g,"-")" id="d5d1ea35-c7eb-437d-ab98-7f825d90b2fa" data-toc-id="d5d1ea35-c7eb-437d-ab98-7f825d90b2fa">48. A reaction follows first-order kinetics and produces an optically inactive product from an optically active reactant. What is the most likely mechanism?A) SN1B) SN2C) E2D) Radical substitutionAnswer: A) SN1Explanation: SN1 reactions involve carbocation formation, allowing racemization, which leads to optical inactivity in the product.l.toLowerCase().replace(/\s+/g,"-")" id="7f382cf8-ad5b-4b3d-95d0-7053d43d021e" data-toc-id="7f382cf8-ad5b-4b3d-95d0-7053d43d021e">49. Which of the following best explains why tertiary alcohols react more rapidly with HCl than primary alcohols?A) The tertiary alcohol is more acidic.B) The tertiary alcohol forms a more stable carbocation.C) The tertiary alcohol undergoes a concerted mechanism.D) The tertiary alcohol has greater steric hindrance.Answer: B) The tertiary alcohol forms a more stable carbocation.Explanation: In acidic conditions, tertiary alcohols react via an SN1 mechanism, where the tertiary carbocation is highly stable, leading to faster reaction rates.l.toLowerCase().replace(/\s+/g,"-")" id="ce856bdd-e625-4e3f-b50a-5f89c4ca493d" data-toc-id="ce856bdd-e625-4e3f-b50a-5f89c4ca493d">50. Which of the following best describes the electronic structure of a carbonyl carbon in aldehydes and ketones?A) spB) sp²C) sp³D) sp² with resonance stabilizationAnswer: B) sp²Explanation: The carbonyl carbon in aldehydes and ketones is sp² hybridized with a planar geometry and π-electron delocalization.l.toLowerCase().replace(/\s+/g,"-")" id="a37e483d-c178-4de3-acec-beae7a24691e" data-toc-id="a37e483d-c178-4de3-acec-beae7a24691e">51. Which of the following molecules will exhibit optical activity?A) 2-ButanolB) 2-ButanoneC) ButanalD) 2-Methylpropanoic acidAnswer: A) 2-ButanolExplanation: Optical activity requires a chiral centre (a carbon with four different groups). 2-Butanol has a chiral center at C-2.l.toLowerCase().replace(/\s+/g,"-")" id="abe3129a-f1ae-4cd5-9b4d-65f66a0709c3" data-toc-id="abe3129a-f1ae-4cd5-9b4d-65f66a0709c3">52. Which of the following best explains why amines act as nucleophiles?A) They have an empty p-orbital.B) They have a lone pair of electrons.C) They have a resonance-stabilized structure.D) They contain a highly electronegative nitrogen.Answer: B) They have a lone pair of electrons.Explanation: Amines have a lone pair on nitrogen, making them excellent nucleophiles in organic reactions.l.toLowerCase().replace(/\s+/g,"-")" id="62e32381-59d2-4984-a339-2f4eee98d098" data-toc-id="62e32381-59d2-4984-a339-2f4eee98d098">53. Which of the following describes the relationship between (R)-2-butanol and (S)-2-butanol?A) DiastereomersB) EnantiomersC) Meso compoundsD) Conformational isomersAnswer: B) EnantiomersExplanation: (R) - and (S)-2-butanol are non-superimposable mirror images, making them enantiomers.l.toLowerCase().replace(/\s+/g,"-")" id="26f54bbb-132a-4918-be89-4e9dcd068d25" data-toc-id="26f54bbb-132a-4918-be89-4e9dcd068d25">54. A researcher performs an organic reaction where a benzene with an alcohol substituent becomes a benzene with an aldehyde substituent. Which IR spectroscopy peak would appear in the product?A) Broad, 3500 cm⁻¹B) Sharp, 3500 cm⁻¹C) Broad, 1700 cm⁻¹D) Sharp, 1700 cm⁻¹Answer: D Explanation: The aldehyde product contains a C=O bond, which appears as a sharp peak at ~1700 cm⁻¹ in IR spectroscopy.l.toLowerCase().replace(/\s+/g,"-")" id="32ec25f1-c458-4220-a23c-3c46fed9527c" data-toc-id="32ec25f1-c458-4220-a23c-3c46fed9527c">55. Why do protic solvents decrease nucleophilicity?A) They are sterically bulky.B) They alter the electrophile's properties.C) They hydrogen bond with the nucleophile.D) They introduce a net positive charge.Answer: CExplanation: Protic solvents (e.g., water and alcohol) form hydrogen bonds with nucleophiles, reducing their ability to attack electrophiles. l.toLowerCase().replace(/\s+/g,"-")" id="ba9eb718-7142-4933-b7a4-37c95d1ad440" data-toc-id="ba9eb718-7142-4933-b7a4-37c95d1ad440">56. A researcher studies a molecule with three chiral centres. After purification, a racemic mixture is obtained. Which stereochemical designations could describe the mixture?A) S, S, S and R, S, RB) R, S, S and S, R, RC) R, S, R and S, R, RD) S, S, R and S, S, SAnswer: BExplanation: A racemic mixture contains equal amounts of two enantiomers, which have opposite stereochemistry at every chiral centre.l.toLowerCase().replace(/\s+/g,"-")" id="2ccc98f6-0d44-45d8-bfa3-9e16afaaff25" data-toc-id="2ccc98f6-0d44-45d8-bfa3-9e16afaaff25">57. Which of the following elements is MOST likely to be a strong nucleophile?A) Hydroxide ion (OH⁻)B) Water (H₂O)C) Ethanol (C₂H₅OH)D) Tert-butanol (C₄H₉OH)Answer: AExplanation: The hydroxide ion (OH⁻) is a strong nucleophile due to its negative charge and lack of steric hindrance.l.toLowerCase().replace(/\s+/g,"-")" id="21842405-9f96-41d9-903f-aaa8f33c58c4" data-toc-id="21842405-9f96-41d9-903f-aaa8f33c58c4">58. Which functional group is most likely to undergo nucleophilic acyl substitution?A) Primary alcoholB) AldehydeC) KetoneD) Carboxylic acidAnswer: DExplanation: Carboxylic acids undergo nucleophilic acyl substitution because their carbonyl carbon is electrophilic, and the hydroxyl (-OH) group can be replaced by a nucleophile.