JEE Challenger
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ChemMedAdv 2014 • Paper 2

Matching Aromatic Precursors with Multi Step Synthesis Schemes

Match the four starting substrates (P\textbf{P}, Q\textbf{Q}, R\textbf{R}, S\textbf{S}) given in List I\textbf{List I} with the appropriate multi-step reaction sequences (1\textbf{1}, 2\textbf{2}, 3\textbf{3}, 4\textbf{4}) provided in List II\textbf{List II} and select the correct matching code:

List-IList-II(P) Ethyne (H-C≡C-H)1. Scheme I:?→(i) KMnO4,HO−,heat→(ii) H+,H2O→(iii) SOCl2→(iv) NH3C7H6N2O3(Q) Resorcinol (benzene-1,3-diol)2. Scheme II:?→(i) Sn/HCl→(ii) CH3COCl→(iii) conc. H2SO4→(iv) HNO3→(v) dil. H2SO4,heat→(vi) HO−C6H6N2O2(R) Nitrobenzene3. Scheme III:?→(i) red hot iron, 873 K→(ii) fuming HNO3,H2SO4,heat→(iii) H2S⋅NH3→(iv) NaNO2,H2SO4→(v) hydrolysisC6H5NO3(S) 4-Nitrotoluene4. Scheme IV:?→(i) conc. H2SO4,60∘C→(ii) conc. HNO3,conc. H2SO4→(iii) dil. H2SO4,heatC6H5NO4\begin{array}{ll} \textbf{List-I} & \textbf{List-II} \\[6pt] \text{(P) } \text{Ethyne } (\text{H-C}\equiv\text{C-H}) & \text{1. Scheme I:} \\[4pt] & \quad ? \xrightarrow{\text{(i) } \text{KMnO}_4, \text{HO}^-, \text{heat}} \xrightarrow{\text{(ii) } \text{H}^+, \text{H}_2\text{O}} \xrightarrow{\text{(iii) } \text{SOCl}_2} \xrightarrow{\text{(iv) } \text{NH}_3} \text{C}_7\text{H}_6\text{N}_2\text{O}_3 \\[8pt] \text{(Q) } \text{Resorcinol } (\text{benzene-1,3-diol}) & \text{2. Scheme II:} \\[4pt] & \quad ? \xrightarrow{\text{(i) } \text{Sn/HCl}} \xrightarrow{\text{(ii) } \text{CH}_3\text{COCl}} \xrightarrow{\text{(iii) conc. } \text{H}_2\text{SO}_4} \xrightarrow{\text{(iv) } \text{HNO}_3} \xrightarrow{\text{(v) dil. } \text{H}_2\text{SO}_4, \text{heat}} \xrightarrow{\text{(vi) } \text{HO}^-} \text{C}_6\text{H}_6\text{N}_2\text{O}_2 \\[8pt] \text{(R) } \text{Nitrobenzene} & \text{3. Scheme III:} \\[4pt] & \quad ? \xrightarrow{\text{(i) red hot iron, } 873\text{ K}} \xrightarrow{\text{(ii) fuming } \text{HNO}_3, \text{H}_2\text{SO}_4, \text{heat}} \xrightarrow{\text{(iii) } \text{H}_2\text{S}\cdot\text{NH}_3} \xrightarrow{\text{(iv) } \text{NaNO}_2, \text{H}_2\text{SO}_4} \xrightarrow{\text{(v) hydrolysis}} \text{C}_6\text{H}_5\text{NO}_3 \\[8pt] \text{(S) } 4\text{-Nitrotoluene} & \text{4. Scheme IV:} \\[4pt] & \quad ? \xrightarrow{\text{(i) conc. } \text{H}_2\text{SO}_4, 60^\circ\text{C}} \xrightarrow{\text{(ii) conc. } \text{HNO}_3, \text{conc. } \text{H}_2\text{SO}_4} \xrightarrow{\text{(iii) dil. } \text{H}_2\text{SO}_4, \text{heat}} \text{C}_6\text{H}_5\text{NO}_4 \end{array}
Question Diagram 1

Options

A

P→1;  Q→4;  R→2;  S→3\text{P} \rightarrow 1;\; \text{Q} \rightarrow 4;\; \text{R} \rightarrow 2;\; \text{S} \rightarrow 3

B

P→3;  Q→1;  R→4;  S→2\text{P} \rightarrow 3;\; \text{Q} \rightarrow 1;\; \text{R} \rightarrow 4;\; \text{S} \rightarrow 2

C

P→3;  Q→4;  R→2;  S→1\text{P} \rightarrow 3;\; \text{Q} \rightarrow 4;\; \text{R} \rightarrow 2;\; \text{S} \rightarrow 1

Correct
D

P→4;  Q→1;  R→3;  S→2\text{P} \rightarrow 4;\; \text{Q} \rightarrow 1;\; \text{R} \rightarrow 3;\; \text{S} \rightarrow 2

Topics & Concepts

AminesAminesElectrophilic Aromatic SubstitutionOxidation and Reduction (Organic)

Step-by-Step Solution

To determine the correct match between the starting materials in List-I and the reaction sequences in List-II, let us analyze each scheme:

1. Analysis of Substrate P (Ethyne, HC≡CH\text{HC}\equiv\text{CH}):

  • Scheme III starts with: HC≡CH→red hot Fe, 873 KBenzene (C6H6)\text{HC}\equiv\text{CH} \xrightarrow{\text{red hot Fe, } 873\text{ K}} \text{Benzene } (\text{C}_6\text{H}_6)
  • Benzene undergoes dinitration with fuming HNO3/H2SO4\text{HNO}_3/\text{H}_2\text{SO}_4 under heating to give 1,3-dinitrobenzene: C6H6→fuming HNO3,H2SO4,Δ1,3-dinitrobenzene\text{C}_6\text{H}_6 \xrightarrow{\text{fuming }\text{HNO}_3, \text{H}_2\text{SO}_4, \Delta} \text{1,3-dinitrobenzene}
  • Selective reduction of one nitro group using H2S⋅NH3\text{H}_2\text{S}\cdot\text{NH}_3 (Zinin reduction) yields 3-nitroaniline: 1,3-dinitrobenzene→H2S⋅NH33-nitroaniline\text{1,3-dinitrobenzene} \xrightarrow{\text{H}_2\text{S}\cdot\text{NH}_3} \text{3-nitroaniline}
  • Diazotization of the amino group followed by hydrolysis gives 3-nitrophenol: 3-nitroaniline→H2SO4NaNO2diazonium salt→hydrolysis3-nitrophenol (C6H5NO3)\text{3-nitroaniline} \xrightarrow[\text{H}_2\text{SO}_4]{\text{NaNO}_2} \text{diazonium salt} \xrightarrow{\text{hydrolysis}} \text{3-nitrophenol } (\text{C}_6\text{H}_5\text{NO}_3)
  • The molecular formula matches: C6H5NO3\text{C}_6\text{H}_5\text{NO}_3.
  • Therefore, P→3\text{P} \rightarrow \mathbf{3}.

2. Analysis of Substrate Q (Resorcinol, benzene-1,3-diol):

  • Scheme IV:
    • Resorcinol undergoes disulfonation at 60∘C60^\circ\text{C} to yield 4,6-disulfonic acid derivative of resorcinol: Resorcinol→conc. H2SO4,60∘C4,6-disulforesorcinol\text{Resorcinol} \xrightarrow{\text{conc. }\text{H}_2\text{SO}_4, 60^\circ\text{C}} \text{4,6-disulforesorcinol}
    • The sulfonic acid groups at positions 4 and 6 block these activated positions, directing nitration to position 2: 4,6-disulforesorcinol→conc. HNO3,H2SO42-nitrobenzene-1,3-diol-4,6-disulfonic acid\text{4,6-disulforesorcinol} \xrightarrow{\text{conc. }\text{HNO}_3, \text{H}_2\text{SO}_4} \text{2-nitrobenzene-1,3-diol-4,6-disulfonic acid}
    • Desulfonation using dilute H2SO4\text{H}_2\text{SO}_4 and heat removes the −SO3H-\text{SO}_3\text{H} groups: →dil. H2SO4,Δ2-nitroresorcinol (C6H5NO4)\xrightarrow{\text{dil. }\text{H}_2\text{SO}_4, \Delta} \text{2-nitroresorcinol } (\text{C}_6\text{H}_5\text{NO}_4)
  • The molecular formula matches: C6H5NO4\text{C}_6\text{H}_5\text{NO}_4.
  • Therefore, Q→4\text{Q} \rightarrow \mathbf{4}.

3. Analysis of Substrate R (Nitrobenzene, C6H5NO2\text{C}_6\text{H}_5\text{NO}_2):

  • Scheme II:
    • Reduction with Sn/HCl\text{Sn/HCl} gives aniline: C6H5NO2→Sn/HClC6H5NH2\text{C}_6\text{H}_5\text{NO}_2 \xrightarrow{\text{Sn/HCl}} \text{C}_6\text{H}_5\text{NH}_2
    • Protection of the −NH2-\text{NH}_2 group with CH3COCl\text{CH}_3\text{COCl} gives acetanilide: C6H5NH2→CH3COClC6H5NHCOCH3\text{C}_6\text{H}_5\text{NH}_2 \xrightarrow{\text{CH}_3\text{COCl}} \text{C}_6\text{H}_5\text{NHCOCH}_3
    • Sulfonation with concentrated H2SO4\text{H}_2\text{SO}_4 directs mainly to the para-position: C6H5NHCOCH3→conc. H2SO44-acetamidobenzenesulfonic acid\text{C}_6\text{H}_5\text{NHCOCH}_3 \xrightarrow{\text{conc. }\text{H}_2\text{SO}_4} \text{4-acetamidobenzenesulfonic acid}
    • Nitration with HNO3\text{HNO}_3 occurs ortho to the −NHCOCH3-\text{NHCOCH}_3 group (position 2): →HNO34-acetamido-3-nitrobenzenesulfonic acid\xrightarrow{\text{HNO}_3} \text{4-acetamido-3-nitrobenzenesulfonic acid}
    • Heating with dilute H2SO4\text{H}_2\text{SO}_4 causes desulfonation: →dil. H2SO4,Δ2-nitroacetanilide\xrightarrow{\text{dil. }\text{H}_2\text{SO}_4, \Delta} \text{2-nitroacetanilide}
    • Basic hydrolysis removes the acetyl protecting group to yield 2-nitroaniline: →HO−2-nitroaniline (C6H6N2O2)\xrightarrow{\text{HO}^-} \text{2-nitroaniline } (\text{C}_6\text{H}_6\text{N}_2\text{O}_2)
  • The molecular formula matches: C6H6N2O2\text{C}_6\text{H}_6\text{N}_2\text{O}_2.
  • Therefore, R→2\text{R} \rightarrow \mathbf{2}.

4. Analysis of Substrate S (4-Nitrotoluene, O2N−C6H4−CH3\text{O}_2\text{N}-\text{C}_6\text{H}_4-\text{CH}_3):

  • Scheme I:
    • Oxidation of the benzylic methyl group of 4-nitrotoluene with alkaline KMnO4\text{KMnO}_4 followed by acidification: 4-nitrotoluene→(ii) H+,H2O(i) KMnO4,HO−,Δ4-nitrobenzoic acid\text{4-nitrotoluene} \xrightarrow[\text{(ii) }\text{H}^+, \text{H}_2\text{O}]{\text{(i) }\text{KMnO}_4, \text{HO}^-, \Delta} \text{4-nitrobenzoic acid}
    • Reaction with SOCl2\text{SOCl}_2 converts the carboxylic acid into an acyl chloride: 4-nitrobenzoic acid→SOCl24-nitrobenzoyl chloride\text{4-nitrobenzoic acid} \xrightarrow{\text{SOCl}_2} \text{4-nitrobenzoyl chloride}
    • Reaction with NH3\text{NH}_3 yields 4-nitrobenzamide: 4-nitrobenzoyl chloride→NH34-nitrobenzamide (C7H6N2O3)\text{4-nitrobenzoyl chloride} \xrightarrow{\text{NH}_3} \text{4-nitrobenzamide } (\text{C}_7\text{H}_6\text{N}_2\text{O}_3)
  • The molecular formula matches: C7H6N2O3\text{C}_7\text{H}_6\text{N}_2\text{O}_3.
  • Therefore, S→1\text{S} \rightarrow \mathbf{1}.

Conclusion:

The corresponding matches are: P→3,Q→4,R→2,S→1\text{P} \rightarrow 3, \quad \text{Q} \rightarrow 4, \quad \text{R} \rightarrow 2, \quad \text{S} \rightarrow 1

This corresponds to Option (C).

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