\( \text{Cyclopropanol}\xrightarrow[\ce{H2SO4}]{\text{Conc.}}\mathrm{A}\xrightarrow{\ce{Cl2/H2O}}\mathrm{B}\xrightarrow{\ce{KCN}}\mathrm{C}\xrightarrow{\ce{H2O/H^+}}\mathrm{D} \)
Identify D
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 2-hydroxycyclopropane-1-carboxylic acid
Each reagent carries out one known step on the cyclopropane ring:
First, concentrated H2SO4 dehydrates cyclopropanol (acid-catalysed elimination of water) to cyclopropene, A:
\( \displaystyle \ce{C3H5OH} \xrightarrow{\text{conc. }\ce{H2SO4}} \ce{C3H4 + H2O} \)
Then Cl2/H2O (hypochlorous acid, HOCl) adds across the C=C double bond, giving the chlorohydrin 2-chlorocyclopropan-1-ol, B.
\( \displaystyle \ce{C3H4 + HOCl -> C3H4(OH)Cl} \)
Next, KCN replaces Cl by CN (nucleophilic substitution), giving 2-hydroxycyclopropane-1-carbonitrile, C.
\( \displaystyle \ce{C3H4(OH)Cl + KCN -> C3H4(OH)CN + KCl} \)
Finally, acid hydrolysis converts the nitrile group into a carboxylic acid group, giving D, which keeps the −OH group and carries −COOH on the next carbon:
\( \displaystyle \ce{C3H4(OH)CN + 2H2O + H^+ -> C3H4(OH)COOH + NH4^+} \)
Why the other options are incorrect
Option (1) cyclopropanol — Incorrect. This is the starting compound; the sequence adds a carbon (from CN) and so cannot return to it.
Option (2) 2-hydroxycyclopropane-1-carbonitrile — Incorrect. This is C, the nitrile before hydrolysis; \( \ce{H2O} \)/\( \ce{H^{+}} \) converts −CN to −COOH.
Option (4) cyclopropanecarboxylic acid — Incorrect. It has lost the −OH group introduced by HOCl; nothing in the sequence removes that hydroxyl group.