Given below are two statements : Consider the following reaction

The reaction shown is the classical “cyanohydrin formation” in which hydrogen cyanide adds to the carbonyl group of an aldehyde or a ketone:

$$\mathrm{R{-}C(=O){-}R'} + HCN \;\longrightarrow\; \mathrm{R{-}C(OH)(CN){-}R'}$$

Here the substrate is $$CH_3CHO$$ (ethanal). Hence the product is $$CH_3CH(OH)CN$$.

Statement I: “The product obtained in the above reaction possesses a chiral centre.”
  • In $$CH_3CH(OH)CN$$ the carbon marked with an asterisk below is attached to four different groups: $$CH_3{-},\; {-}OH,\; {-}CN,$$ and $$H{-}$$.
  • Because the four substituents are different, that carbon is a stereogenic (asymmetric) centre, making the molecule chiral.
Therefore Statement I is true.

Statement II: “HCN adds to the carbonyl group by a nucleophilic-addition mechanism in which $$CN^-$$ first attacks the electrophilic carbonyl carbon and the resulting alkoxide is then protonated.”
  • Mechanism steps:
    1. Generation of the nucleophile: $$HCN \rightleftharpoons H^+ + CN^-$$ (in the presence of a base such as $$NaCN$$).
    2. Nucleophilic attack: $$CN^-$$ attacks the partially positive carbon of the $$C=O$$ bond, giving an alkoxide ion.
    3. Protonation: the alkoxide abstracts a proton (from $$HCN$$ or another source) to yield the cyanohydrin.
  • These are the standard steps of nucleophilic addition to a carbonyl, so Statement II is also true.

Since both statements correctly describe the reaction and its product, the correct choice is:

Option D which is: Both Statement I and Statement II are true