ORGANIC CHEMISTRY 2 ACS FINAL 2025 MULTICHOICE ANSWERED EXAM QUESTIONS WITH DETAILED RATIONALES

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ORGANIC CHEMISTRY 2 ACS FINAL 2025 MULTICHOICE ANSWERED
EXAM QUESTIONS WITH DETAILED RATIONALES
How can an aldehyde be protected during reactions?
A. Convert to a hemiacetal with base
B. Oxidize to carboxylic acid
C. Form an acetal by reacting with a diol under acid (H⁺)
D. Reduce to an alcohol with NaBH₄
Rationale: Acetals (from diol + acid) protect carbonyls; reversible by acid hydrolysis.

What does permanganate (KMnO₄, NaOH) commonly do to aldehydes and 1° alcohols?
A. Reduce them to alkanes
B. Convert to esters
C. Oxidize to carboxylic acids
D. Convert to imines
Rationale: KMnO₄ is a strong oxidant that oxidizes aldehydes/primary alcohols to acids.

How is an acetal (protecting group) hydrolyzed back to the carbonyl?
A. NaBH₄, EtOH
B. H₂O, H⁺ and heat (acidic hydrolysis)
C. PCC, CH₂Cl₂
D. NaOH, heat
Rationale: Acetals are acid-labile and revert to carbonyl + diol under aqueous acid.

At about what IR frequency do O–H or N–H stretches appear?
A. 1700 cm⁻¹
B. 3000 cm⁻¹
C. ~3400 cm⁻¹
D. 2100 cm⁻¹
Rationale: Broad O–H/N–H stretches center near 3300–3500 cm⁻¹ (≈3400 cm⁻¹).

What IR frequency range corresponds to sp³ C–H stretches?
A. 3100–3000 cm⁻¹
B. 3000–2850 cm⁻¹
C. 3300 cm⁻¹
D. 1700 cm⁻¹
Rationale: Alkane C–H stretches fall ~2850–3000 cm⁻¹.

What IR region corresponds to sp² C–H stretches?
A. 2850–3000 cm⁻¹

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B. 3300 cm⁻¹
C. 3100–3000 cm⁻¹
D. 2100 cm⁻¹
Rationale: Alkenic/aromatic C–H appear slightly above 3000 cm⁻¹.

At what IR frequency do sp C–H stretches (alkynes) appear?
A. 3000 cm⁻¹
B. 2850 cm⁻¹
C. ~3300 cm⁻¹
D. 2100 cm⁻¹
Rationale: Terminal alkyne C–H stretches are sharp near 3300 cm⁻¹.

Where does a C≡C or C≡N triple bond show up in IR?
A. 1700 cm⁻¹
B. 1500 cm⁻¹
C. ~2100 cm⁻¹
D. 3300 cm⁻¹
Rationale: Triple bonds give medium bands ~2100–2250 cm⁻¹.

At what IR frequency is a C=O carbonyl stretch typically observed?
A. 1500 cm⁻¹
B. 2100 cm⁻¹
C. ~1700 cm⁻¹
D. 3300 cm⁻¹
Rationale: Carbonyl stretches center near 1700 cm⁻¹ (varies with functional group).

At what approximate IR frequency do C=C or aromatic C–C stretches appear?
A. 2100 cm⁻¹
B. ~1600 cm⁻¹
C. 3300 cm⁻¹
D. 1700 cm⁻¹
Rationale: C=C and aromatic skeletal vibrations are near 1500–1600 cm⁻¹.

Where do aromatic C–C stretches often appear in IR?
A. 3300 cm⁻¹
B. 2100 cm⁻¹
C. ~1500 cm⁻¹
D. 1700 cm⁻¹
Rationale: Aromatic C–C in-ring stretches around 1450–1600 cm⁻¹ with bands near 1500.

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What IR pattern indicates a carboxylic acid?
A. Sharp O–H at 3400 only
B. C=O ~1720 cm⁻¹ plus broad strong O–H band 2500–3300 cm⁻¹
C. C≡O band at 2100 cm⁻¹
D. No distinctive bands
Rationale: Acids show strong carbonyl and very broad hydrogen-bonded O–H absorption.

What IR pattern is characteristic of an amide?
A. Only a 1700 cm⁻¹ band
B. Carbonyl (below 1700) plus two N–H bands around 3400 cm⁻¹
C. Sharp triple bond at 2100 cm⁻¹
D. Broad O–H at 3000 cm⁻¹
Rationale: Amide N–H stretches (~3300–3500) plus amide C=O ~1650 cm⁻¹.

Where is an sp³ proton (alkyl H) typically found in ¹H NMR (ppm)?
A. 7–8 ppm
B. 3–4 ppm
C. ~1 ppm
D. 10 ppm
Rationale: Alkyl protons appear near 0.5–2 ppm, often ≈1 ppm average.

A proton on a carbon attached to a π system (allylic) shows around:
A. 1 ppm
B. ~2 ppm
C. 7–8 ppm
D. 10 ppm
Rationale: Allylic/α to π systems shift protons downfield to ~1.5–3 ppm (≈2 ppm).

Protons on carbons attached to oxygen (R–CH₂–O) appear around:
A. 7–8 ppm
B. 0–1 ppm
C. 3–4 ppm
D. 10–12 ppm
Rationale: Electronegative O deshields adjacent protons, giving δ ~3–4 ppm.

Where do sp² alkene protons resonate in ¹H NMR?
A. 0–1 ppm
B. ~5–6 ppm
C. 7–8 ppm