This chemistry question involves key chemical concepts and calculations. The detailed solution below walks through each step, from identifying the reaction type to computing the final answer.
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a) Explain how it is possible for the product to be a mixture of two isomers. The reaction between ethanal (CH_3CHO) and hydrogen cyanide (HCN) forms 2-hydroxypropanenitrile (CH_3CH(OH)CN). In ethanal, the carbonyl carbon is sp^2 hybridized and trigonal planar. When the cyanide ion (CN^-) attacks this carbon, it can approach from either the top or the bottom face of the planar carbonyl group with equal probability. This attack creates a new chiral center at the carbon that was originally part of the carbonyl group, as it becomes bonded to four different groups (CH_3, H, OH, and CN). The formation of a chiral center from a non-chiral starting material in this manner leads to the production of two enantiomers, which are non-superimposable mirror images of each other. b) Draw spatial formulae showing these two isomers. The two isomers are enantiomers of 2-hydroxypropanenitrile. The chiral carbon is the one bonded to the CH_3, H, OH, and CN groups. cc Isomer 1 & Isomer 2 \\ c H \\ | \\ H_3C-OHCNC-H \\ & c H \\ | \\ H_3C-CNOHC-H \\ Using wedge-dash notation to show spatial arrangement: cc Isomer 1 (R-2-hydroxypropanenitrile) & Isomer 2 (S-2-hydroxypropanenitrile) \\ c H \\ | \\ H_3C----OH \\ CNHC \\ & c H \\ | \\ H_3C----CN \\ OHHC \\ c) Why are the isomers present in equal quantities? The carbonyl carbon in ethanal is achiral and planar. The nucleophilic attack by the cyanide ion can occur with equal probability from either face (top or bottom) of this planar carbonyl group. Since both attack pathways are equally likely and energetically equivalent, they lead to the formation of the two enantiomers in a 1:1 ratio. This equimolar mixture of enantiomers is known as a racemic mixture or racemate. Drop the next question.