Question

Complete the following reactions and write the names and formulae of A, B, C, D, E, F: (i) \[ \text{Phenol (OH group)} \xrightarrow{\text{conc. HNO}_3} A \] (ii) \[ \text{Phenol} \xrightarrow{CHCl_3 + aq.NaOH} B \] (iii) \[ CH_3-CH_2-CH_2-O-C(CH_3)_2-CH_2-CH_3 \xrightarrow{HI, \Delta} C + D \] (iv) \[ CH_3CH(OH)CH_3 \xrightarrow{CrO_3} E \] (v) \[ CH_3COOH \xrightarrow{(i) LiAlH_4 \, (ii) H_2O} F \]

Hint:

Picric acid forms by nitration, salicylaldehyde by Reimer–Tiemann, ethers split with HI, secondary alcohols oxidise to ketones, and acids reduce to alcohols.

Answer 1

(i) Phenol + conc. HNO\(_3\).
Forms picric acid (2,4,6-trinitrophenol). \[ C_6H_5OH + 3HNO_3 \rightarrow C_6H_2(NO_2)_3OH + 3H_2O \] (ii) Phenol + CHCl\(_3\) + NaOH (Reimer–Tiemann).
Forms salicylaldehyde. \[ C_6H_5OH + CHCl_3 + 3NaOH \rightarrow C_6H_4(OH)(CHO) + 3NaCl + 2H_2O \] (iii) Ether cleavage with HI.
\[ CH_3CH_2CH_2-O-C(CH_3)_2CH_2CH_3 + HI \rightarrow CH_3CH_2CH_2OH + (CH_3)_2CHCH_2CH_2I \] C = Propanol, D = 2-iodo-2-methylbutane.
(iv) Oxidation of isopropanol.
\[ CH_3CH(OH)CH_3 \xrightarrow{CrO_3} CH_3COCH_3 \] E = Acetone.
(v) Reduction of acetic acid.
\[ CH_3COOH \xrightarrow{LiAlH_4} CH_3CH_2OH \] F = Ethanol.
Step 3: Conclusion.
- A = Picric acid.
- B = Salicylaldehyde.
- C = Propanol.
- D = 2-iodo-2-methylbutane.
- E = Acetone.
- F = Ethanol.