An optically active alkyne A contains 89.52 % C and 10.48 % H. After hydrogenation over a Pd/C catal — Organic Chemistry Chemistry Question
Identification of Organic Compounds
An optically active alkyne A contains 89.52 % C and 10.48 % H. After hydrogenation over a Pd/C catalyst it is converted to 1-methyl-4-propyl cyclohexane. When compound A reacts with CH3MgBr no gas is liberated. Hydrogenation of A over a Lindlar catalyst, followed by ozonolysis and reaction with KMnO4 gives product B whose 13 C-NMR spectrum shows a peak at 207 ppm. Product B reacts with I2 / NaOH and gives a yellow precipitate, which is filtered off. Acidification of the filtrate gives an optically active product C, whose 13C NMR spectrum does not have any peak over 175 ppm.
Give the structures of A, B and C and account for all observations.
Model Answer
### 1. Determination of the Molecular Formula of A
- Elemental analysis:
- C: 89.52% / 12.011 = 7.453 mol
- H: 10.48% / 1.008 = 10.397 mol
- Ratio H/C ≈ 1.395 ≈ 1.4 = 7/5, giving an empirical formula of C5H7.
- Since hydrogenation of A over Pd/C yields 1-methyl-4-propylcyclohexane (which contains 10 carbons), the molecular formula of A must be double the empirical formula: C10H14.
### 2. Unsaturation and Structure of A
- The degree of unsaturation (DBE) for C10H14 is 10 + 1 - (14/2) = 4.
- Pd/C hydrogenation yields 1-methyl-4-propylcyclohexane (which contains 1 ring, DBE = 1), meaning compound A has a cyclohexane ring (1 DBE) and 3 additional degrees of unsaturation.
- Since A is an alkyne, it must contain one triple bond (2 DBE). The remaining 1 DBE is a double bond in the ring.
- Grignard test: Since A does not react with CH3MgBr to liberate methane gas, the triple bond is internal: -C≡C-CH3 (prop-1-yn-1-yl).
- Chirality: Since A is optically active, it must possess a chiral center. C1 (the carbon bearing the prop-1-yn-1-yl group) is a stereocenter because it is attached to -H, -C≡C-CH3, and two different ring paths (-CH2-CH2- and -CH2-CH=).
- Thus, A is 4-methyl-1-(prop-1-yn-1-yl)cyclohexene (or 4-methyl-5-(prop-1-yn-1-yl)cyclohexene, with the double bond at C3=C4 of the ring).
### 3. Reactions and Identification of B and C
- Selective reduction of A over a Lindlar catalyst reduces only the triple bond to a cis-double bond, yielding 4-methyl-5-(cis-prop-1-en-1-yl)cyclohexene.
- Ozonolysis followed by oxidation with KMnO4 cleaves both double bonds (the side chain and the ring):
- The side-chain double bond (-CH=CH-CH3) is cleaved: the carbon attached to the ring becomes -COOH, and acetic acid (CH3COOH) is released as a byproduct.
- The ring double bond at C3=C4 is cleaved: C3 (=CH-) is oxidized to -COOH, while C4 (=C(CH3)-) is oxidized to a methyl ketone (-CO-CH3), opening the ring.
- This yields product B: HOOC-CH2-CH2-CH(COOH-CH2-CH2-CO-CH3 (3-(2-oxopropyl)hexanedioic acid).
- The 13C-NMR peak of B at 207 ppm is characteristic of the ketone carbonyl in the methyl ketone group.
- B undergoes the iodoform reaction (I2/NaOH) to yield a yellow precipitate of iodoform (CHI3).
- Acidification of the filtrate converts the methyl ketone group (-CO-CH3) of B to a carboxylic acid group (-COOH), yielding the tricarboxylic acid product C: HOOC-CH2-CH2-CH(COOH-CH2-CH2-COOH.
- Product C has no peaks above 175 ppm in its 13C-NMR spectrum because it contains only carboxylic acid carbonyls.
- C is optically active because the chiral center C1 is preserved.
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