ORGANIC CHEMISTRY ACS 2025 MULTICHOICE ANSWERED
EXAM QUESTIONS WITH DETAILED RATIONALES
When naming a carboxylic acid, what is the first step?
A. Count substituents on a ring.
B. Name the alkyl substituent on the α-carbon.
C. Find the longest continuous carbon chain containing the carboxylic acid group.
D. Locate the most substituted double bond.
Rationale: The parent chain must include the carboxyl carbon to name the acid correctly.
Which suffix (prefix for the anion) is used for carboxylate derivatives (esters/anion)?
A. -ane
B. -ene
C. -oate
D. -ol
Rationale: Carboxylate salts and esters use “-oate” (e.g., ethyl acetate).
How is an ester named?
A. Name the acid part first, then the alcohol part.
B. Use only the acid root and add -ol.
C. Name the alkyl group from the alcohol first, then the acid part as a carboxylate.
D. Number the ring substituents, then add -ester.
Rationale: Esters are named R′ O–C(=O)–R: alkyl (R′) + carboxylate (R-oate).
The terms ortho, meta, para apply when a benzene derivative has:
A. Three identical substituents.
B. Exactly two substituents.
C. A fused aromatic ring.
D. A carbonyl group.
Rationale: Ortho/meta/para describe relative positions of two substituents on benzene.
In general, which resonance contributor is more stable?
A. The one with more separated charges.
B. The one with fewer covalent bonds.
C. The one with more covalent (bonding) character and fewer separated charges.
D. The one with maximum positive charge.
Rationale: Resonance contributors with more bonding and less charge separation are lower in
energy.
,ESTUDYR
Why is charge separation in resonance contributors unfavorable?
A. It increases covalent bonding.
B. It always forms radicals.
C. It costs energy and reduces contributor stability.
D. It produces aromaticity.
Rationale: Separate charges increase electrostatic energy, lowering resonance contributor
weight.
Which statement about molecular orbitals is true?
A. Antibonding π is lower in energy than bonding π.
B. Bonding π orbital is lower in energy than antibonding π orbital.
C. σ orbitals are always higher than π orbitals.
D. Antibonding orbitals are always populated in ground state.
Rationale: Bonding MOs stabilize (lower energy) versus antibonding MOs (higher energy).
Which is lower in energy: σ orbital or π orbital (for same atoms)?
A. π orbital
B. σ orbital
C. They are always equal.
D. Depends only on hybridization, not type.
Rationale: σ bonds (head-on overlap) are generally stronger and lower in energy than π (side-
on).
Which reagent is commonly used to convert an alkene into an epoxide (maintaining
stereochemistry)?
A. HBr
B. KMnO4 (hot)
C. mCPBA (a peracid)
D. LiAlH4
Rationale: Peracids (mCPBA) epoxidize alkenes stereospecifically (syn retention).
Reagents for hydroboration of alkenes typically include:
A. BH3 in ether (BH3·THF or BH3·Et2O).
B. HBr / peroxide.
C. Both A and later H2O2/NaOH for oxidation.
D. Ozone then Zn.
Rationale: Hydroboration uses BH3 then H2O2/NaOH to give anti-Markovnikov alcohols.
Anti-Markovnikov addition refers to:
A. H adding to most substituted carbon.
, ESTUDYR
B. H adding to the carbon with the fewest hydrogens.
C. Addition only to aromatic rings.
D. Radical polymerization only.
Rationale: Anti-Markovnikov places H on more substituted carbon’s opposite; borane additions
do that.
In hydroboration, H and B add to the same face of the alkene — this is called:
A. Anti addition.
B. Syn addition.
C. Radical addition.
D. Electrophilic addition.
Rationale: Hydroboration is concerted syn addition (same face).
Markovnikov’s rule states that in HX addition:
A. H adds to less substituted carbon always.
B. H adds to carbon already bearing more H’s; halide goes to more substituted carbon.
C. Halide adds to least substituted carbon.
D. Reaction proceeds only via radicals.
Rationale: Carbocation stability dictates adding proton to yield the more stable carbocation.
Reagent for aromatic sulfonation is typically:
A. HCl gas.
B. SO3 (fuming H2SO4) or concentrated H2SO4.
C. NaOH.
D. NBS.
Rationale: Sulfonation uses SO3/H2SO4 to install –SO3H.
Which substituents are typically activating (EDG) and direct ortho/para?
A. –NO2, –CF3.
B. –OH, –OR, –NH2, –NR2, alkyl (R).
C. –COOH, –CHO.
D. –CN, –SO3H.
Rationale: Electron-donating groups push electron density into ring and direct ortho/para.
Which groups are deactivating (EWG) and typically meta-directing?
A. –OH and –OR.
B. –NH2 only.
C. –NO2, –CF3, –COOR, –CHO, –COOH, –CN.
D. Alkyl groups.
Rationale: EWGs withdraw electron density via resonance/induction, lowering reactivity.