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

### Detailed Solution 1. **Bromination** $$\text{p-}CH_3\text{–}C_6H_4NO_2 \xrightarrow[AcOH]{Br_2} 2\text{-bromo-}4\text{-nitrotoluene}$$ Bromine enters position 2 (or 6, identical) giving $\ce{C_6H_3Br(NO_2)CH_3}$. 2. **Reduction of $NO_2$ to $NH_2$** $$\ce{C_6H_3Br(NO_2)CH_3} \xrightarrow[HCl]{Sn} \ce{C_6H_3Br(NH_2)CH_3}$$ 3. **Diazotisation** $$\ce{C_6H_3Br(NH_2)CH_3} + \ce{NaNO_2} + 2\ce{HCl} \xrightarrow{0\,^\circ C} \ce{C_6H_3Br(N_2^+Cl^-)CH_3} + 2\ce{H_2O}$$ 4. **Reaction with Ethanol** The aryl diazonium oxidises ethanol, itself being converted to phenol: $$\ce{C_6H_3Br(N_2^+Cl^-)CH_3} + \ce{C_2H_5OH} \longrightarrow \ce{C_6H_3Br(OH)CH_3} + \ce{CH_3CHO} + N_2 + HCl$$ 5. **Final Products** $$A = \ce{C_6H_3Br(OH)CH_3} \quad (\text{2-bromo-4-methylphenol})$$ $$B = \ce{CH_3CHO} \quad (\text{acetaldehyde})$$ **Correct option: (4)**

**68.** The products formed in the following reaction s... - Solution & Explanation | iExplain

Detailed Explanation

Step-by-Step Conceptual Walk-Through

Stage	Key Concept	What to Remember	Why it Happens
1. Bromination ( $Br_2$ , AcOH)	Electrophilic Aromatic Substitution (EAS)	$CH_3$ is ortho/para directing and activates the ring. $NO_2$ is meta directing but deactivates.	Position 2 (or 6) is favoured because it is ortho to $CH_3$ (activating) and meta to $NO_2$ (so less deactivation). Ring is symmetric, so only one brominated isomer forms.
2. Reduction (Sn/HCl)	Metal-acid reduction of $NO_2\rightarrow NH_2$	$Sn$ provides electrons; $HCl$ supplies $H^+$ .	$NH_2$ is strongly activating, setting up for diazotisation.
3. Diazotisation ( $NaNO_2$ , $HCl$ , 0 °C)	Formation of Aryl Diazonium Salt	$ArNH_2 + HNO_2 \rightarrow ArN_2^+Cl^- + 2H_2O$	The diazonium group is an excellent leaving group, so it can be replaced by other nucleophiles.
4. Treatment with Ethanol ( $C_2H_5OH$ )	Hydrolysis/Reductive Substitution	In neutral or slightly acidic alcohol, water (from step 3) or alcohol itself acts as nucleophile. The usual outcome is phenol formation; the alcohol is oxidised to $CH_3CHO$ .	$ArN_2^+$ is a mild oxidising agent; it accepts two electrons while ethanol loses two electrons (gets oxidised).

Thus, the aryl diazonium becomes brominated p-cresol (2-bromo-4-methylphenol) and ethanol becomes acetaldehyde.

Simple Explanation (ELI5)

What is happening here?

Put a new sticker (Br) on the ring – the ring already has two stickers (a methyl $CH_3$ and a nitro $NO_2$ ). The bromine likes to sit next to the $CH_3$ sticker.
Turn the angry red sticker ( $NO_2$ ) into a friendly blue one ( $NH_2$ ) – tin ( $Sn$ ) and acid do that.
Change the blue sticker into a magic ticket ( $N_2^+$ ) that can pop off – that is what $NaNO_2$ and $HCl$ at cold temperature do (diazotisation).
Let the ticket pop off inside alcohol (ethanol) – when it pops, an $OH$ group takes its place on the ring, and the alcohol gets a little bit oxidised to apple-smelling acetaldehyde ( $CH_3CHO$ ).

So you get one main ring product (with Br, $CH_3$ , and $OH$ ) and one small side product (acetaldehyde).

Step-by-Step Solution

Detailed Solution

Bromination
$\text{p-}CH_3\text{–}C_6H_4NO_2 \xrightarrow[AcOH]{Br_2} 2\text{-bromo-}4\text{-nitrotoluene}$ Bromine enters position 2 (or 6, identical) giving $\ce{C_6H_3Br(NO_2)CH_3}$ .
Reduction of $NO_2$ to $NH_2$
$\ce{C_6H_3Br(NO_2)CH_3} \xrightarrow[HCl]{Sn} \ce{C_6H_3Br(NH_2)CH_3}$
Diazotisation
$\ce{C_6H_3Br(NH_2)CH_3} + \ce{NaNO_2} + 2\ce{HCl} \xrightarrow{0\,^\circ C} \ce{C_6H_3Br(N_2^+Cl^-)CH_3} + 2\ce{H_2O}$
Reaction with Ethanol
The aryl diazonium oxidises ethanol, itself being converted to phenol: $\ce{C_6H_3Br(N_2^+Cl^-)CH_3} + \ce{C_2H_5OH} \longrightarrow \ce{C_6H_3Br(OH)CH_3} + \ce{CH_3CHO} + N_2 + HCl$
Final Products
$A = \ce{C_6H_3Br(OH)CH_3} \quad (\text{2-bromo-4-methylphenol})$ $B = \ce{CH_3CHO} \quad (\text{acetaldehyde})$

Correct option: (4)

Examples

Example 1

Synthesis of aspirin dyes via diazotisation and coupling reactions

Example 2

Manufacture of paracetamol starting from p-nitrophenol that requires nitro reduction then acylation

Example 3

Use of ethanol as a reducing agent in Tollens test giving silver mirror while ethanol oxidises to acetaldehyde

Visual Representation

References

[1]Clayden, Greeves, Warren & Wothers – Organic Chemistry (Chapter on Aromatic Substitution)
[2]Peter Sykes – A Guidebook to Mechanism in Organic Chemistry
[3]MS Chauhan – Elementary Problems in Organic Chemistry for JEE
[4]NCERT Chemistry Class XII Part II – Unit on Amines and Diazonium Salts