\( \mathrm{C_2H_5OH}\xrightarrow{\ \ X\ \ }\mathrm{C_2H_5Cl}\xrightarrow{\ \ Y\ \ }\mathrm{C_2H_5CN} \)
X and Y used in the above reaction are :
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → SOCl2 and KCN
The conversion needs two steps: the −OH group of ethanol is first replaced by Cl, and the Cl is then replaced by −CN.
First, ethanol is treated with thionyl chloride. The −OH is replaced by Cl, and the by-products \( \ce{SO2} \) and HCl are gases that escape, so pure ethyl chloride is left:
\( \displaystyle \ce{C2H5OH + SOCl2 -> C2H5Cl + SO2 ^ + HCl ^} \)
Then ethyl chloride is heated with alcoholic KCN. KCN is ionic and supplies free \( \ce{CN^{-}} \) ions; the cyanide ion attacks through carbon in an SN2 substitution and gives the alkyl cyanide (nitrile):
\( \displaystyle \ce{C2H5Cl + KCN} \xrightarrow{\text{ethanol, heat}} \ce{C2H5CN + KCl} \)
So X = SOCl2 and Y = KCN.
Why the other options are incorrect
Option (1) HgCl2 and AgCN — Incorrect. \( \ce{HgCl2} \) does not convert an alcohol into an alkyl chloride, and AgCN is mainly covalent (Ag–C bond), so it attacks through nitrogen and gives the isocyanide \( \ce{C2H5NC} \), not \( \ce{C2H5CN} \).
Option (3) Cl2 and KCN — Incorrect. KCN is right for the second step, but chlorine does not replace the −OH group of an alcohol; \( \ce{Cl2} \) substitutes H atoms of alkanes in light (free-radical chlorination).
Option (4) ZnCl2/HCl and AgCN — Incorrect. HCl with anhydrous \( \ce{ZnCl2} \) can make \( \ce{C2H5Cl} \) from ethanol, but AgCN in the second step gives ethyl isocyanide, \( \ce{C2H5NC} \).
NCERT (2026-27 edition):
From Alcohols – “The hydroxyl group of an alcohol is replaced by halogen on reaction with concentrated halogen acids, phosphorus halides or thionyl chloride. Thionyl chloride is preferred because in this reaction alkyl halide is formed along with gases SO2 and HCl. The two gaseous products are escapable, hence, the reaction gives pure alkyl halides.”