Match List I with List II
|
LIST I |
LIST II |
| A. Phenols | I. Lucas Test |
| B. Ethers | II. Reimer Tiemann reaction |
| C. Alcohols | III. Williamson's Synthesis |
| D. 2, 4, 6 − trinitrophenols | IV. Picric acid |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → A-II, B-III, C-I, D-IV
A. Phenols — This matches II. Reimer Tiemann reaction. Phenol with chloroform and NaOH gets a –CHO group at the ortho position:
\( \displaystyle \ce{C6H5OH + CHCl3 + 4NaOH -> NaOC6H4CHO + 3NaCl + 3H2O} \xrightarrow{\ce{H+}} o\text{-}\ce{HOC6H4CHO} \)
B. Ethers — This matches III. Williamson’s Synthesis. An alkyl halide reacts with a sodium alkoxide:
\( \displaystyle \ce{R-X + R'-ONa -> R-O-R' + NaX} \)
C. Alcohols — This matches I. Lucas Test. With conc. HCl + \( \ce{ZnCl2} \), tertiary alcohols give turbidity at once, secondary in a few minutes, and primary not at room temperature:
\( \displaystyle \ce{ROH + HCl} \xrightarrow{\ce{ZnCl2}} \ce{RCl + H2O} \)
D. 2, 4, 6 − trinitrophenols — This matches IV. Picric acid. Phenol nitrated with concentrated nitric acid gives 2,4,6-trinitrophenol:
\( \displaystyle \ce{C6H5OH + 3HNO3} \xrightarrow{\text{conc. }\ce{H2SO4}} \ce{C6H2(NO2)3OH + 3H2O} \)