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Question:

Give the structures of A, B, and C in the following reactions:
(i) C6H5NO2 Sn + HCl → A
A NaNO2 + HCl → B
B H2O → C
(ii) CH3CN H2O/H+ → A
A NH3 → ΔB
B Br2 + KOH → C

Solution:

(i) The reaction sequence is as follows:

  1. Reduction of Nitrobenzene: Nitrobenzene (C6H5NO2) is reduced by tin (Sn) and hydrochloric acid (HCl) to aniline (C6H5NH2). This is A.
    C6H5NO2 + 6[H] → C6H5NH2 + 2H2O

  2. Diazotization: Aniline (C6H5NH2) reacts with sodium nitrite (NaNO2) and hydrochloric acid (HCl) to form benzenediazonium chloride (C6H5N2Cl). This is B.
    C6H5NH2 + NaNO2 + 2HCl → C6H5N2Cl + NaCl + 2H2O

  3. Hydrolysis: Benzenediazonium chloride (C6H5N2Cl) undergoes hydrolysis in the presence of water (H2O) to yield phenol (C6H5OH). This is C.
    C6H5N2Cl + H2O → C6H5OH + N2 + HCl

(ii) The reaction sequence is as follows:

  1. Hydrolysis of Acetonitrile: Acetonitrile (CH3CN) undergoes acid-catalyzed hydrolysis (H2O/H+) to form acetic acid (CH3COOH). This is A.
    CH3CN + 2H2O → CH3COOH + NH3

  2. Decarboxylation: Acetic acid (CH3COOH) undergoes decarboxylation upon heating with ammonia (NH3) to produce methane (CH4). This is B (although the reaction conditions aren't completely standard for this type of decarboxylation, this is the most likely product given the next step).
    CH3COOH + NH3 → CH3COO-NH4+ → CH4 + CO2 + H2O

  3. Halogenation and elimination: Methane (CH4) reacts with bromine (Br2) in the presence of potassium hydroxide (KOH) in a free radical substitution process followed by elimination. This leads to the formation of bromoform (CHBr3). This is C. Note that under strong basic conditions, CH4 would likely not readily undergo bromination. However, if this is considered a sequence where conditions are adjusted to favor the reaction, a likely intermediate is CH3Br. This then undergoes further bromination, eventually leading to CHBr3 via elimination. There are other possibilities depending on reaction conditions. This answer assumes conditions can be adjusted to obtain this result.
    CH4 + Br2 → CH3Br + HBr
    CH3Br + Br2 → CH2Br2 + HBr
    CH2Br2 + Br2 → CHBr3 + HBr
    (or with excess KOH and heat, could theoretically proceed directly to CHBr3).