Question 26.IE: An acyclic organic compound with the formula C6H12O is found......
An acyclic organic compound with the formula C_{6}H_{12}O is found to be optically active, does not decolorize Br_{2} (\text{in} CCl_{4}), and does not undergo reaction when treated with a mixture of Na_{2}Cr_{2}O_{7} \text{and} H_{2}SO_{4}. However, a reaction does occur when the compound is treated with NaBH_{4}. Identify a compound having these physical and chemical properties. Then draw structures for the two stereoisomers, using dashed and solid wedge symbols. Finally, give acceptable names for the two stereoisomers.
Analyze
There is only one oxygen atom per molecule, so the compound must be an alcohol, ether, aldehyde, or ketone. Also, because there are only 12 hydrogen atoms per molecule, rather than 2 × 6 + 2 = 14, we know that the molecule must contain either a π bond or a ring. We are told the compound is acyclic, and thus the molecule must contain a π bond and not a ring. Alcohols (except tertiary alcohols) and aldehydes are rather easily oxidized, so we suspect the molecule is neither an alcohol (though possibly a tertiary alcohol) nor an aldehyde. The ether linkage is stable in the presence of most oxidizing and reducing agents, and because the compound we need can be reduced, we suspect that that the compound
is not an ether. At this point, we must decide whether the molecule is a ketone or a tertiary alcohol containing a carbon–carbon double bond. We can rule out a carbon–carbon double bond because the compound does not react with Br_{2}, a reaction characteristic of alkenes. The compound we seek is a ketone.
Solve
We have reasoned that the compound is a ketone. The linkage associated with a ketone is
The five remaining carbon atoms are distributed among the two alkyl groups. The carbon of a carbonyl group is bonded to only three groups, and thus it cannot be chiral. The chiral carbon is part of one of the alkyl groups. Both R and R′ must contain at least one carbon atom and so, the maximum number of carbon atoms in R or R′ is four.
Let’s choose R to be the alkyl group that contains the chiral carbon. A chiral carbon atom must be bonded to four different groups and yet R contains no more than four carbon atoms. The only possibility for R is
H_{3}C—\overset{\begin{matrix} H \\ \mid \end{matrix} }{\underset{\begin{matrix} \mid \\ CH_{2}CH_{3} \end{matrix} }{C}}—
We have accounted for five of the six carbon atoms (R contains four carbon atoms and the carbonyl group contains one). Thus, R′ represents a methyl group. A skeletal structure for the molecule is
H_{3}C—\overset{\begin{matrix} H \\ \mid \end{matrix} }{\underset{\begin{matrix} \mid \\ ^{4}CH_{2}^{5}CH_{3} \end{matrix} }{^{3}C}}—\overset{\begin{matrix} O \\ \parallel \end{matrix} }{^{2}C}—^{1}CH_{3}
3-methylpentan-2-one
The longest carbon chain has five carbon atoms with a carbonyl group bonded to C2 and a methyl group bonded to C3. The molecule is 3-methylpentan-2-one. Because C3 is chiral, two enantiomers are possible.
Assess
It may not be obvious that there is only one possibility for R. Try drawing structures of other ketones having the formula C_{6}H_{12}O to convince yourself that there is only one possibility. Had we not been told the compound was acyclic, a unique identification would not have been possible.