Question

An optically active alkyne A (C10H14) can be catalytically hydrogenated to butylcyclohexane. Treatment of A with C2H5MgBr liberates no gas. Catalytic hydrogenation of A over Pd/C in the presence of quinoline poison and treatment of the product B with O3 and then H2O2 gives an optically active tricarboxylic acid C8H12O6. (A tricarboxylic acid is a compound with three –CO2H groups.) Give the structure of A (without stereochemistry).

Answer 1

The reaction of A with aqueous H2SO4 in the presence of Hg(ii) to give two ketones is a characteristic hydration of an internal alkyne, and is a reaction you need to know. The alkyne is not symmetric because it gives two different products depending on where the H adds and where the OH adds on the two C#C carbons where # represents a triple bond. This experiment also tells you that two of the three pi bonds in A are part of a triple bond. Since there is only one pi bond not accounted for, it must be a in an alkene which only requires one pi bond.

The KMnO4 reaction cleaves triple bonds to give carboxylic acids from each of the two carbons in the C#C. It also cleaves C=C bonds to give ketones from any alkene carbons that have two other C's attached to them and carboxylic acids from any alkene carbons that have an H on them. Since there is one ketone and 3 carboxylic acids, there appears to be an alkene like the following present:

C........C
..\....../
...C=C
../......\
C........H

Note that the products shown account for all 10 carbons, but there is no ring. This means the ring was cleaved by the KMNO4, so the C=C seems to be in the ring. Since five member rings are common, the structure I drew probably is part of a cyclopentene that, when cleaved gave the ketone on the next to last C to right in your second structure, and the carboxylic acid on the left end of that structure. Note that there are 5 carbons between these including tthe two carbonyl carbons. The CO2H on the right end of the butanoic acid and the CO2H on the right end of the keto diacid must have been joined by the C#C in the original structure, A.

A seems to be

/.\\_C#C-CH2CH2CH3
\_/

Compounds C and D would have a C=O on the C next to the ring or the C attached to the CH2CH2CH3, but it is not possible to say which is C and which is D from the information given. The other end of the C#C will have become a CH2 in C and D.

B formed in the hydrogenation will be pentylcyclopentane

Answer 2

Treatment of A with C2H5MgBr liberates no gas beacuse A does not have any carbonyl group or teriminal alkyne group.

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An optically active alkyne A (C10H14) can be catalytically hydrogenated to butylcyclohexane. Here, both the double bond and triple bond are reduced.

Catalytic hydrogenation of A over Pd/C in the presence of quinoline poison to form compound B. Here triple bond can only reduced this reagent not affect on the double bond. Treatment of the product B with O3 and then H2O2 gives an optically active tricarboxylic acid, C8H12O6.

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Therefore, the expected compound A is shown below:

Answer 3