CHEM 210 Module 2 Exam () |
Biochemistry | Portage Learning (PDF)
,Section 1: Alcohols & Ethers (Q1-15)
1. Which reagent is most effective for converting a primary alcohol to an aldehyde without
overoxidation to carboxylic acid?
A) KMnO₄, H₃O⁺
B) PCC (pyridinium chlorochromate)
C) CrO₃, H₂SO₄ (Jones reagent)
D) Na₂Cr₂O₇, H₂SO₄
Answer-: : B) PCC (pyridinium chlorochromate)
Rationale: PCC is a mild oxidant that stops at the aldehyde stage for primary alcohols. Strong oxidants
like Jones reagent (C) or dichromate (D) continue to carboxylic acids. KMnO₄ (A) is also strong and gives
carboxylic acids.
2. What is the major product when (R)-2-butanol reacts with SOCl₂ in pyridine?
A) (R)-2-chlorobutane
B) (S)-2-chlorobutane
C) 1-chlorobutane
D) Butene mixture
Answer-: : A) (R)-2-chlorobutane
Rationale: SOCl₂ with pyridine proceeds via an SN₂ mechanism with inversion, but here the nucleophile
(Cl⁻) attacks the chlorosulfite intermediate with retention due to neighboring group participation
(formation of cyclic chlorosulfite). Standard result: retention of configuration.
3. Williamson ether synthesis involves:
A) Alkoxide + primary alkyl halide via SN2
B) Alcohol + alkene via acid catalysis
C) Two alcohols with sulfuric acid
D) Ether cleavage with HI
Answer-: : A) Alkoxide + primary alkyl halide via SN2
Rationale: The Williamson synthesis is an SN2 reaction between an alkoxide nucleophile and an alkyl
halide electrophile. Primary halides work best to avoid E2 competition.
4. Which alcohol undergoes dehydration fastest with concentrated H₂SO₄?
A) 1-butanol
B) 2-butanol
C) 2-methyl-2-propanol (t-butanol)
D) 2-methyl-1-propanol
Answer-: : C) 2-methyl-2-propanol (t-butanol)
Rationale: Dehydration rate follows carbocation stability: tertiary > secondary > primary. Tertiary
alcohols form stable 3° carbocations fastest via E1 mechanism under acidic conditions.
,5. The Lucas test differentiates alcohols based on:
A) Oxidation rate
B) Formation of alkyl chlorides (cloudiness)
C) Color change with Cr(VI)
D) Gas evolution
Answer-: : B) Formation of alkyl chlorides (cloudiness)
Rationale: Lucas reagent (ZnCl₂ in concentrated HCl) converts alcohols to alkyl chlorides. Tertiary
alcohols react immediately (cloudiness), secondary within minutes, primary remain clear unless heated.
6. Epoxide opening with NaOCH₃ in CH₃OH proceeds via:
A) SN1 mechanism at more substituted carbon
B) SN2 mechanism at less substituted carbon
C) Acid-catalyzed ring opening
D) Elimination to form alkene
Answer-: : B) SN2 mechanism at less substituted carbon
Rationale: Under basic conditions, strong nucleophile (CH₃O⁻) attacks the less sterically hindered
epoxide carbon via SN2, with inversion at that carbon.
7. Which compound has the highest boiling point?
A) Diethyl ether
B) 1-butanol
C) Pentane
D) Butanal
Answer-: : B) 1-butanol
Rationale: Alcohols have strong intermolecular hydrogen bonding, leading to higher boiling points than
ethers, alkanes, or aldehydes of similar molecular weight.
8. Pinacol rearrangement of 2,3-dimethyl-2,3-butanediol with acid yields:
A) 2,3-dimethyl-1,3-butadiene
B) 3,3-dimethyl-2-butanone (pinacolone)
C) 2,3-dimethyl-2-butene
D) 2,3-dimethylbutanal
Answer-: : B) 3,3-dimethyl-2-butanone (pinacolone)
Rationale: Vicinal diols undergo acid-catalyzed rearrangement: protonation, loss of water, 1,2-methyl
shift, then capture by water to form ketone.
9. What is the product when ethylene oxide reacts with CH₃MgBr followed by H₃O⁺?
A) Ethanol
B) 1-propanol
C) 2-propanol
D) 1-butanol
Answer-: : B) 1-propanol
Rationale: Grignard attacks epoxide at less hindered carbon, giving primary alcohol with two-carbon
extension: CH₃CH₂OMgBr → after protonation → CH₃CH₂OH? Wait, ethylene oxide is a 3-membered ring
(C₂H₄O). Attack gives CH₃CH₂CH₂OMgBr → 1-propanol.
, 10. Crown ethers are useful in organic synthesis because they:
A) Oxidize primary alcohols
B) Complex with metal cations, solubilizing anions
C) Act as strong reducing agents
D) Protect aldehydes from oxidation
Answer-: : B) Complex with metal cations, solubilizing anions
Rationale: Crown ethers bind specific metal ions via ion-dipole interactions, making associated anions
more nucleophilic in organic solvents ("naked anions").
11. Selective cleavage of anisole (methyl phenyl ether) with HI yields:
A) Phenol + CH₃I
B) Iodobenzene + CH₃OH
C) Phenol + CH₄
D) Benzene + CH₃I
Answer-: : A) Phenol + CH₃I
Rationale: Aryl alkyl ethers cleave at the alkyl-oxygen bond (not aryl-oxygen) to give phenol and alkyl
iodide via SN2 on the alkyl group.
12. Which synthesis is most effective for preparing tert-butyl ethyl ether?
A) CH₃CH₂ONa + (CH₃)₃CBr
B) (CH₃)₃CONa + CH₃CH₂Br
C) CH₃CH₂OH + (CH₃)₃COH + H₂SO₄
D) CH₃CH₂OCH₃ + (CH₃)₃CMgBr
Answer-: : B) (CH₃)₃CONa + CH₃CH₂Br
Rationale: Williamson with primary halide (CH₃CH₂Br) and tertiary alkoxide. Option A would give
elimination (E2) from tertiary halide.
13. Hydroboration-oxidation of 1-methylcyclopentene yields:
A) trans-2-methylcyclopentanol
B) cis-2-methylcyclopentanol
C) 1-methylcyclopentanol
D) 3-methylcyclopentanol
Answer-: : C) 1-methylcyclopentanol
Rationale: Hydroboration gives anti-Markovnikov, syn addition. The boron adds to less substituted
carbon, yielding the tertiary alcohol after oxidation.
14. A compound C₅H₁₂O gives a positive iodoform test. Which structure fits?
A) Pentanal
B) 3-pentanone
C) 2-pentanol
D) 2-methyl-2-butanol
Answer-: : D) 2-methyl-2-butanol
Rationale: Iodoform test positive for methyl ketones (CH₃COR) or alcohols oxidizable to methyl ketones
(CH₃CHOHR). 2-methyl-2-butanol oxidizes to 2-methylbutan-2-one? No, it's tertiary and doesn't oxidize.
Let's reconsider: The alcohol must have CH₃CHOH- group. 2-pentanol (C) oxidizes to 2-pentanone
(methyl ketone), so it gives positive iodoform.
Biochemistry | Portage Learning (PDF)
,Section 1: Alcohols & Ethers (Q1-15)
1. Which reagent is most effective for converting a primary alcohol to an aldehyde without
overoxidation to carboxylic acid?
A) KMnO₄, H₃O⁺
B) PCC (pyridinium chlorochromate)
C) CrO₃, H₂SO₄ (Jones reagent)
D) Na₂Cr₂O₇, H₂SO₄
Answer-: : B) PCC (pyridinium chlorochromate)
Rationale: PCC is a mild oxidant that stops at the aldehyde stage for primary alcohols. Strong oxidants
like Jones reagent (C) or dichromate (D) continue to carboxylic acids. KMnO₄ (A) is also strong and gives
carboxylic acids.
2. What is the major product when (R)-2-butanol reacts with SOCl₂ in pyridine?
A) (R)-2-chlorobutane
B) (S)-2-chlorobutane
C) 1-chlorobutane
D) Butene mixture
Answer-: : A) (R)-2-chlorobutane
Rationale: SOCl₂ with pyridine proceeds via an SN₂ mechanism with inversion, but here the nucleophile
(Cl⁻) attacks the chlorosulfite intermediate with retention due to neighboring group participation
(formation of cyclic chlorosulfite). Standard result: retention of configuration.
3. Williamson ether synthesis involves:
A) Alkoxide + primary alkyl halide via SN2
B) Alcohol + alkene via acid catalysis
C) Two alcohols with sulfuric acid
D) Ether cleavage with HI
Answer-: : A) Alkoxide + primary alkyl halide via SN2
Rationale: The Williamson synthesis is an SN2 reaction between an alkoxide nucleophile and an alkyl
halide electrophile. Primary halides work best to avoid E2 competition.
4. Which alcohol undergoes dehydration fastest with concentrated H₂SO₄?
A) 1-butanol
B) 2-butanol
C) 2-methyl-2-propanol (t-butanol)
D) 2-methyl-1-propanol
Answer-: : C) 2-methyl-2-propanol (t-butanol)
Rationale: Dehydration rate follows carbocation stability: tertiary > secondary > primary. Tertiary
alcohols form stable 3° carbocations fastest via E1 mechanism under acidic conditions.
,5. The Lucas test differentiates alcohols based on:
A) Oxidation rate
B) Formation of alkyl chlorides (cloudiness)
C) Color change with Cr(VI)
D) Gas evolution
Answer-: : B) Formation of alkyl chlorides (cloudiness)
Rationale: Lucas reagent (ZnCl₂ in concentrated HCl) converts alcohols to alkyl chlorides. Tertiary
alcohols react immediately (cloudiness), secondary within minutes, primary remain clear unless heated.
6. Epoxide opening with NaOCH₃ in CH₃OH proceeds via:
A) SN1 mechanism at more substituted carbon
B) SN2 mechanism at less substituted carbon
C) Acid-catalyzed ring opening
D) Elimination to form alkene
Answer-: : B) SN2 mechanism at less substituted carbon
Rationale: Under basic conditions, strong nucleophile (CH₃O⁻) attacks the less sterically hindered
epoxide carbon via SN2, with inversion at that carbon.
7. Which compound has the highest boiling point?
A) Diethyl ether
B) 1-butanol
C) Pentane
D) Butanal
Answer-: : B) 1-butanol
Rationale: Alcohols have strong intermolecular hydrogen bonding, leading to higher boiling points than
ethers, alkanes, or aldehydes of similar molecular weight.
8. Pinacol rearrangement of 2,3-dimethyl-2,3-butanediol with acid yields:
A) 2,3-dimethyl-1,3-butadiene
B) 3,3-dimethyl-2-butanone (pinacolone)
C) 2,3-dimethyl-2-butene
D) 2,3-dimethylbutanal
Answer-: : B) 3,3-dimethyl-2-butanone (pinacolone)
Rationale: Vicinal diols undergo acid-catalyzed rearrangement: protonation, loss of water, 1,2-methyl
shift, then capture by water to form ketone.
9. What is the product when ethylene oxide reacts with CH₃MgBr followed by H₃O⁺?
A) Ethanol
B) 1-propanol
C) 2-propanol
D) 1-butanol
Answer-: : B) 1-propanol
Rationale: Grignard attacks epoxide at less hindered carbon, giving primary alcohol with two-carbon
extension: CH₃CH₂OMgBr → after protonation → CH₃CH₂OH? Wait, ethylene oxide is a 3-membered ring
(C₂H₄O). Attack gives CH₃CH₂CH₂OMgBr → 1-propanol.
, 10. Crown ethers are useful in organic synthesis because they:
A) Oxidize primary alcohols
B) Complex with metal cations, solubilizing anions
C) Act as strong reducing agents
D) Protect aldehydes from oxidation
Answer-: : B) Complex with metal cations, solubilizing anions
Rationale: Crown ethers bind specific metal ions via ion-dipole interactions, making associated anions
more nucleophilic in organic solvents ("naked anions").
11. Selective cleavage of anisole (methyl phenyl ether) with HI yields:
A) Phenol + CH₃I
B) Iodobenzene + CH₃OH
C) Phenol + CH₄
D) Benzene + CH₃I
Answer-: : A) Phenol + CH₃I
Rationale: Aryl alkyl ethers cleave at the alkyl-oxygen bond (not aryl-oxygen) to give phenol and alkyl
iodide via SN2 on the alkyl group.
12. Which synthesis is most effective for preparing tert-butyl ethyl ether?
A) CH₃CH₂ONa + (CH₃)₃CBr
B) (CH₃)₃CONa + CH₃CH₂Br
C) CH₃CH₂OH + (CH₃)₃COH + H₂SO₄
D) CH₃CH₂OCH₃ + (CH₃)₃CMgBr
Answer-: : B) (CH₃)₃CONa + CH₃CH₂Br
Rationale: Williamson with primary halide (CH₃CH₂Br) and tertiary alkoxide. Option A would give
elimination (E2) from tertiary halide.
13. Hydroboration-oxidation of 1-methylcyclopentene yields:
A) trans-2-methylcyclopentanol
B) cis-2-methylcyclopentanol
C) 1-methylcyclopentanol
D) 3-methylcyclopentanol
Answer-: : C) 1-methylcyclopentanol
Rationale: Hydroboration gives anti-Markovnikov, syn addition. The boron adds to less substituted
carbon, yielding the tertiary alcohol after oxidation.
14. A compound C₅H₁₂O gives a positive iodoform test. Which structure fits?
A) Pentanal
B) 3-pentanone
C) 2-pentanol
D) 2-methyl-2-butanol
Answer-: : D) 2-methyl-2-butanol
Rationale: Iodoform test positive for methyl ketones (CH₃COR) or alcohols oxidizable to methyl ketones
(CH₃CHOHR). 2-methyl-2-butanol oxidizes to 2-methylbutan-2-one? No, it's tertiary and doesn't oxidize.
Let's reconsider: The alcohol must have CH₃CHOH- group. 2-pentanol (C) oxidizes to 2-pentanone
(methyl ketone), so it gives positive iodoform.
