68. The products formed in the following reaction sequence are: Starting compound: C6H4NO2CH3 (p-nitrotoluene). Reagents: 1. Br2, AcOH 2. Sn, HCl 3. NaNO2, HCl, 273K 4. C2H5OH. C6H4NO2CH3 --(i-iv)--> A + B. Option 1: A = C6H3Br(OH)CH3, B = C6H3Br(OEt)CH3. Option 2: A = C6H3Br(OEt)CH3, B = CH3COOH. Option 3: A = C6H4BrCH3, B = CH3CHO. Option 4: A = C6H3Br(OH)CH3, B = CH3CHO.

Key ideas you must know

1. Directive effects in EAS (Electrophilic Aromatic Substitution)

   • $\text{CH}_3$ is ortho/para-directing & activating.
   • $\text{NO}_2$ is meta-directing & strongly deactivating.
   • When both are present, the position that satisfies both directives (here, the 2-position) is favoured.

2. Reduction of nitro to amine

   • Sn/HCl converts $\text{NO}_2 \rightarrow \text{NH}_2$.

3. Diazotisation

   • Aromatic $\text{NH}_2$ + $\text{NaNO}_2/\text{HCl}$ at 0 – 5 °C gives an aryl diazonium chloride $\text{Ar–N}_2^+\,\text{Cl}^-$.
   • This ion is an excellent leaving group and can be replaced by many nucleophiles or even by hydrogen.

4. De-diazotisation by ethanol

   • Ethanol is a mild reducing agent.
   • It converts $\text{Ar–N}_2^+$ into $\text{Ar–H}$, itself being oxidised to $\text{CH}_3CHO$.

Logical chain of thought for the problem

1. Where will Br add?
   Evaluate directing effects → position-2 (ortho to CH₃, meta to NO₂) wins.
2. What does Sn/HCl do?
   Just change $\text{NO}_2$ to $\text{NH}_2$.
3. Diazotisation step
   Replace $\text{NH}_2$ with $\text{N}_2^+$, no change to Br or CH₃.
4. Ethanol treatment
   Think of reduction: $\text{N}_2$ escapes, plain H appears; ethanol → acetaldehyde.
5. Compare to options
   Product ring: $\text{CH}_3$ at C-1, $\text{Br}$ at C-2 → $\text{o-bromotoluene}$ ($\text{C}_6\text{H}_4\text{BrCH}_3$) + $\text{CH}_3\text{CHO}$.
   This matches Option 3.

What is happening in the reaction chain?

Imagine you have a toy ring (the benzene ring) with two stickers already on it:

• a yellow "NO₂" sticker (which doesn't like new friends)
• a red "CH₃" sticker (which loves new friends and tells them to sit next to it).

Now four different workers visit the ring one after another:

1. Bromine painter (Br₂ in AcOH) – He wants to add a brown Br sticker. Because the red CH₃ sticker is very friendly (activating) and the yellow NO₂ sticker is rather unfriendly (deactivating), the Br sticker ends up next to the CH₃ sticker.
2. Tin-acid repairman (Sn/HCl) – He secretly transforms the unfriendly yellow NO₂ sticker into a blue NH₂ sticker (much friendlier).
3. Nitrite magicians (NaNO₂/HCl, 0 °C) – They turn that blue NH₂ sticker into a magic balloon N₂⁺ that is ready to fly away.
4. Ethanol cleaner (C₂H₅OH) – The balloon pops, taking the sticker’s place with just a plain spot (H), and ethanol gets slightly used up, changing into acetaldehyde (CH₃CHO).

So, after all four workers finish, the ring now only has:

• the original red CH₃ sticker,
• the new brown Br sticker next to it,
• nothing where the yellow sticker used to be.

Side reward: you also get a little bottle of acetaldehyde from the ethanol cleaner’s work.

Step-by-step calculation of products

1. Bromination
   Reactant: $p$-nitrotoluene ($1$-methyl-$4$-nitrobenzene).
   Directive analysis: $\text{CH}_3 \ (o,p) \quad \text{and} \quad \text{NO}_2 \ (m)$ Common free position = 2 (or 6, identical).
   Product: 2-bromo-4-nitrotoluene.

2. Reduction
   $\text{Sn/HCl}:\; \text{NO}_2 \rightarrow \text{NH}_2$
   Product: 2-bromo-4-aminotoluene.

3. Diazotisation
   $\text{NaNO}_2/\text{HCl},\; 0 ^\circ \text{C}:\; \text{NH}_2 \rightarrow \text{N}_2^+\text{Cl}^-$
   Product: 2-bromo-4-diazonium toluene chloride.

4. Treatment with ethanol
   $\text{Ar–N}_2^+ + \text{CH}_3CH_2OH \;\longrightarrow\; \text{Ar–H} + \text{CH}_3CHO + N_2 + HCl$
   Therefore, diazonium is replaced by H.

5. Final organic products
   (i) Aromatic product A: 2-bromotoluene (\big(\text{C}_6\text{H}_4\text{BrCH}_3\big)).
   (ii) Side product B: acetaldehyde ((\text{CH}_3CHO)).

Hence, the correct option is Option 3.