Free practice final
Organic Chemistry 1 practice final (free, with answers)
An Organic Chemistry 1 final usually covers bonding and acid-base reasoning, conformations and stereochemistry, substitution and elimination mechanisms, and the reactions of alkenes and alkynes. Work each question on paper before you open the answer, and when you miss one, redraw the structure or mechanism until the explanation makes sense.
Topics
- Structure and Bonding
- Acids and Bases
- Conformations
- Stereochemistry
- Substitution Reactions
- Elimination Reactions
- Alkene Reactions
- Alkyne Reactions
15 questions
Try each one before you open its answer.
1. In acetonitrile, CH₃–C≡N, what is the hybridization of the CH₃ carbon and of the nitrile carbon, in that order?
- A sp² and sp
- B sp³ and sp
- C sp³ and sp²
- D sp and sp
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Answer: B. sp³ and sp
The CH₃ carbon makes four σ bonds, so it is sp³. The nitrile carbon has only two electron domains (a single bond to carbon and a triple bond to nitrogen), so it is sp and linear.
2. How many degrees of unsaturation (rings plus π bonds) does a compound with the molecular formula C₈H₉NO have?
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Answer: 5
Degrees of unsaturation = (2C + 2 + N − H − X)/2 = (16 + 2 + 1 − 9)/2 = 5. A benzene ring (one ring plus three π bonds) and one C=O would account for all five.
3. Which compound is the strongest acid?
- A CH₃COOH
- B ClCH₂COOH
- C Cl₂CHCOOH
- D CH₃CH₂OH
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Answer: C. Cl₂CHCOOH
Electronegative chlorine atoms pull electron density away from the carboxylate and stabilize the conjugate base by induction, and two chlorines do this more than one. Dichloroacetic acid has a pKa near 1.3, versus about 2.9 for chloroacetic acid, 4.8 for acetic acid and 16 for ethanol.
4. Acetic acid (pKa 4.76) is mixed with ammonia. The pKa of the ammonium ion, NH₄⁺, is 9.25. What is the equilibrium constant for CH₃COOH + NH₃ ⇌ CH₃COO⁻ + NH₄⁺? Round to two significant figures.
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Answer: Keq ≈ 3.1 × 10⁴
For an acid-base reaction, Keq = 10^(pKa of the product acid − pKa of the reactant acid) = 10^(9.25 − 4.76) = 10^4.49 ≈ 3.1 × 10⁴. The large value means the equilibrium lies far to the side of the weaker acid, NH₄⁺.
5. In the more stable chair conformation of cis-1-tert-butyl-4-methylcyclohexane, where are the two substituents?
- A tert-Butyl equatorial, methyl axial
- B tert-Butyl axial, methyl equatorial
- C Both equatorial
- D Both axial
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Answer: A. tert-Butyl equatorial, methyl axial
In a cis-1,4-disubstituted cyclohexane, one group must be axial and the other equatorial in either chair, so "both equatorial" is impossible. The bulky tert-butyl group suffers far worse 1,3-diaxial strain than methyl, so it takes the equatorial position.
6. What is the relationship between (2R,3S)-2,3-dichloropentane and (2S,3R)-2,3-dichloropentane?
- A Diastereomers
- B Identical (a meso compound)
- C Constitutional isomers
- D Enantiomers
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Answer: D. Enantiomers
Every stereocenter is inverted, so the two are nonsuperimposable mirror images. This compound cannot be meso because C2 and C3 carry different groups (CH₃ on one side, CH₂CH₃ on the other), so the molecule has no internal mirror plane.
7. A pure enantiomer has a specific rotation of +40.0°. A sample of the same compound, measured under the same conditions, has a specific rotation of +30.0°. What percentage of the sample is the (+) enantiomer?
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Answer: 87.5%
Enantiomeric excess = 30.0/40.0 = 75%. The sample is 75% excess (+) plus 25% racemic mixture (12.5% of each enantiomer), so the (+) enantiomer is 75% + 12.5% = 87.5%.
8. (R)-2-Bromobutane is treated with sodium azide (NaN₃) in DMSO. What is the major product?
- A (R)-2-Azidobutane
- B (S)-2-Azidobutane
- C A racemic mixture of (R)- and (S)-2-azidobutane
- D But-2-ene, formed by E2
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Answer: B. (S)-2-Azidobutane
Azide is a strong nucleophile but a weak base, and DMSO is polar aprotic, so a secondary bromide reacts by SN2 with backside attack and inversion. The azide group takes the same priority rank that bromine had, so inverting the geometry changes the label from R to S.
9. A tertiary alkyl bromide reacts with sodium azide in a polar protic solvent by an SN1 mechanism. If the alkyl bromide concentration is doubled and the azide concentration is tripled, how does the initial rate change?
- A It does not change
- B It triples
- C It doubles
- D It increases sixfold
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Answer: C. It doubles
The SN1 rate law is rate = k[alkyl halide], because the slow step is ionization to the carbocation and the nucleophile only joins afterward. Doubling the substrate doubles the rate, and the azide concentration has no effect.
10. 2-Bromo-2-methylbutane is heated with potassium tert-butoxide in tert-butanol. What is the major product?
- A 2-Methylbut-1-ene
- B 2-Methylbut-2-ene
- C 3-Methylbut-1-ene
- D 2-tert-Butoxy-2-methylbutane
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Answer: A. 2-Methylbut-1-ene
The bulky tert-butoxide base removes the more accessible hydrogen on the terminal CH₃ group, giving the less substituted (Hofmann) alkene. A small base such as ethoxide would instead favor the more substituted Zaitsev product, 2-methylbut-2-ene, and substitution is minor because the substrate is tertiary and the base is hindered.
11. trans-1-Chloro-2-methylcyclohexane undergoes E2 elimination with sodium ethoxide. What is the major alkene product?
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Answer: 3-Methylcyclohexene
E2 on a cyclohexane needs the leaving group and a β-hydrogen to be anti-periplanar, meaning both axial. When Cl is axial in the trans isomer, the methyl group on C2 is also axial, so the hydrogen on C2 is equatorial and cannot be removed; only C6 has an axial hydrogen, which gives 3-methylcyclohexene rather than the more substituted 1-methylcyclohexene.
12. What is the product of adding Br₂ to cis-but-2-ene?
- A meso-2,3-Dibromobutane
- B (2R,3R)-2,3-Dibromobutane only
- C 1,2-Dibromobutane
- D A racemic mixture of (2R,3R)- and (2S,3S)-2,3-dibromobutane
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Answer: D. A racemic mixture of (2R,3R)- and (2S,3S)-2,3-dibromobutane
Bromine adds anti through a cyclic bromonium ion, and the bromide can open either carbon with equal likelihood. Anti addition to a cis alkene gives the (2R,3R) and (2S,3S) enantiomers in equal amounts; the same anti addition to trans-but-2-ene would give the meso compound.
13. Which reagents convert 3,3-dimethylbut-1-ene into 3,3-dimethylbutan-2-ol as the major product?
- A H₂SO₄ in water
- B BH₃·THF, then H₂O₂ and NaOH
- C Hg(OAc)₂ in water, then NaBH₄
- D OsO₄, then NaHSO₃
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Answer: C. Hg(OAc)₂ in water, then NaBH₄
Oxymercuration-demercuration adds water with Markovnikov orientation through a mercurinium ion, so no carbocation forms and no rearrangement occurs. Acid-catalyzed hydration goes through a secondary carbocation that undergoes a methyl shift to give 2,3-dimethylbutan-2-ol, hydroboration gives the anti-Markovnikov 3,3-dimethylbutan-1-ol, and OsO₄ gives a diol.
14. Propyne has a pKa of about 25. Which base can deprotonate it essentially completely? (pKa of each conjugate acid: H₂O 15.7, NH₃ 38, CH₃OH 15.5, H₂CO₃ 6.4.)
- A NaOH
- B NaNH₂
- C NaOCH₃
- D NaHCO₃
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Answer: B. NaNH₂
An acid-base equilibrium favors the side with the weaker acid. Only ammonia (pKa 38) is a weaker acid than propyne (pKa 25), so only amide ion pulls the equilibrium toward the acetylide; the other bases have conjugate acids stronger than the alkyne.
15. What is the major organic product when hex-1-yne is treated with H₂SO₄, H₂O and HgSO₄?
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Answer: Hexan-2-one (2-hexanone)
Water adds with Markovnikov orientation, putting the OH on the internal carbon and giving an enol. The enol tautomerizes to the more stable ketone, so the carbonyl ends up on C2.