Q.Draw the structures of major monohalo products in each of the following reactions:

1. Cyclohexanol + SOCl₂ → SOCl₂ converts alcohols to alkyl chlorides. With pyridine the reaction follows an SN₂-type path (inversion of configuration); without pyridine it proceeds through the SNi mechanism (internal nucleophilic substitution, retention of configuration). The major product is chlorocyclohexane.

   

2. 1-Ethyl-4-nitrobenzene + Br₂ → heat/UV Heat/UV light promotes free-radical benzylic bromination. The benzylic C–H bond (CH₂ group) is selectively brominated. Major product: 1-(1-bromoethyl)-4-nitrobenzene ($p$-$O_2NC_6H_4CHBrCH_3$).
3. 4-(Hydroxymethyl)phenol + HCl → heat The benzylic –OH is more reactive than the phenolic –OH. Under acidic conditions, the benzylic alcohol undergoes SN₁ to give the benzylic chloride. Major product: 4-(chloromethyl)phenol ($p$-$HOC_6H_4CH_2Cl$).
4. 1-Methylcyclohex-1-ene + HI → Concept: Markovnikov Addition — the proton adds to the less substituted carbon of the double bond, placing the iodide on the more substituted carbon. The tertiary carbocation intermediate is more stable. Major product: 1-iodo-1-methylcyclohexane.
5. CH₃CH₂Br + NaI → NaI in acetone drives an SN₂ reaction (Finkelstein reaction). Iodide is a better nucleophile and more soluble in acetone. Major product: CH₃CH₂I (ethyl iodide).
6. Cyclohexene + Br₂ → heat/UV light …

(i) Cyclohexanol $+ \ SOCl_2 \rightarrow$

Thionyl chloride converts an alcohol's $-OH$ into a chloride. With no pyridine present (as here), the reaction proceeds through the $S_Ni$ (internal nucleophilic substitution) pathway: the alcohol first forms a chlorosulfite ester, and the chloride ion that is released stays associated with the same face of the carbon it left from, delivering the chlorine back to that same face — giving retention of configuration. (With pyridine added, the freed chloride ion instead attacks from the opposite face for a clean $S_N2$-style inversion.) Since cyclohexanol's C1 is not a stereocentre, retention or inversion makes no visible difference to the product here — either way the product is chlorocyclohexane.

(ii) 1-Ethyl-4-nitrobenzene ($p$-$O_2NC_6H_4CH_2CH_3$) $+ \ Br_2 \xrightarrow{\text{heat or UV light}}$

Heat/UV light homolyses $Br_2$ into bromine radicals, favouring free-radical substitution at the most stabilised C–H bond. The benzylic C–H (on the ethyl group's carbon directly attached to the ring) gives a radical stabilised by resonance into the ring, so bromination occurs there rather than anywhere else on the chain. The nitro group's directing effect governs electrophilic aromatic substitution, not this radical pathway, so it plays no role in where the Br ends up. Major product: 1-(1-bromoethyl)-4-nitrobenzene.

(iii) 4-(Hydroxymethyl)phenol ($p$-$HOC_6H_4CH_2OH$) $+ \ HCl \xrightarrow{\text{heat}}$

Two $-OH$ groups are present: a phenolic $-OH$ directly on the ring, and a benzylic $-CH_2OH$ on the side chain. The phenolic C–O bond has partial double-bond character from resonance with the ring and resists substitution under these mild conditions. The benzylic alcohol, by contrast, readily protonates and loses water to form a benzylic carbocation (stabilised by the ring), which chloride then attacks. Major product: 4-(chloromethyl)phenol — the phenolic $-OH$ is untouched.

(iv) 1-Methylcyclohex-1-ene $+ \ HI \rightarrow$ …

Here is the solution method for each reaction, following the concept-first approach.

Method: Functional Group Transformation & Regioselectivity Analysis

This method involves identifying the functional group, predicting the reaction mechanism (SN1, SN2, E2, electrophilic addition, free radical substitution), and applying the relevant rule (Markovnikov, anti-Markovnikov, Zaitsev, etc.) to determine the major product.

(i) Cyclohexanol $+ \ SOCl_2 \rightarrow$

• Step 1: Identify the functional group. Alcohol ($-OH$).
• Step 2: Identify the reagent. $SOCl_2$ (thionyl chloride) is a classic reagent for converting alcohols to alkyl chlorides.
• Step 3: Determine the mechanism. This proceeds via an SN2 mechanism (with inversion of configuration if the carbon is chiral). For cyclohexanol, the OH is on a secondary carbon.
• Step 4: Draw the product. The OH is replaced by Cl.
• Major product: Chlorocyclohexane

(ii) 1-Ethyl-4-nitrobenzene $+ \ Br_2 \xrightarrow{\text{heat or UV light}}$

• Step 1: Identify the functional group. Alkyl side chain (ethyl group) attached to an aromatic ring with a nitro ($-NO_2$) group.
• Step 2: Identify the reagent and conditions. $Br_2$ with heat or UV light indicates free radical substitution (not electrophilic aromatic substitution).
• Step 3: Determine the site of reaction. The reaction occurs on the benzylic carbon (the carbon directly attached to the benzene ring) because the benzylic radical is highly stabilized by resonance with the ring.
• Step 4: Draw the product. One hydrogen on the benzylic carbon is replaced by bromine.
• Major product: 1-Bromo-1-(4-nitrophenyl)ethane ($p$-$O_2NC_6H_4CHBrCH_3$)

(iii) 4-(Hydroxymethyl)phenol $+ \ HCl \xrightarrow{\text{heat}}$

• Step 1: Identify the functional groups. Two OH groups: one is phenolic (on the ring), one is benzylic (on the side chain).
• Step 2: Identify the reagent and conditions. $HCl$ with heat.
• Step 3: Determine reactivity. The benzylic alcohol is much more reactive than the phenolic OH. The benzylic carbocation is highly stabilized by resonance with the ring, making it an excellent candidate for SN1 reaction.
• Step 4: Draw the product. The benzylic OH is replaced by Cl. The phenolic OH remains unchanged.
• Major product: 4-(Chloromethyl)phenol ($p$-$HOC_6H_4CH_2Cl$)

(iv) 1-Methylcyclohex-1-ene $+ \ HI \rightarrow$

• Step 1: Identify the functional group. Alkene ($C=C$).
• Step 2: Identify the reagent. $HI$ (hydrogen halide).
• Step 3: Apply Markovnikov's rule. In the addition of HX to an unsymmetrical alkene, the hydrogen atom adds to the carbon with the greater number of hydrogen atoms, and the halogen adds to the carbon with the fewer hydrogen atoms.
• Step 4: Determine the product. The double bond is between C1 (with a methyl group, no H) and C2 (with one H). H adds to C2, I adds to C1.
• Major product: 1-Iodo-1-methylcyclohexane

(v) $CH_3CH_2Br + NaI \rightarrow$

• Step 1: Identify the functional group. Alkyl halide (bromide).
• Step 2: Identify the reagent. $NaI$ (sodium iodide) in acetone.
• Step 3: Determine the mechanism. This is a classic Finkelstein reaction, an SN2 process. Iodide is a good nucleophile and a good leaving group. Acetone is a polar aprotic solvent that favors SN2.
• Step 4: Draw the product. Bromine is replaced by iodine.
• Major product: Iodoethane ($CH_3CH_2I$)

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This set of reactions tests your ability to distinguish between substitution, addition, and free-radical mechanisms. The most common mistakes come from misidentifying the reaction type or misapplying Markovnikov’s rule.

Here’s a breakdown of each reaction, the typical errors, and how to avoid them.

(i) Cyclohexanol $+ \ SOCl_2 \rightarrow$

Common mistake:

Students treat this as an elimination or oxidation. They draw cyclohexene or cyclohexanone.

Why it happens:

$SOCl_2$ is often associated with dehydration or chlorination, but the exact mechanism matters.

Correct approach:

$SOCl_2$ converts alcohols to alkyl chlorides. Without pyridine, the reaction proceeds through the internal $S_Ni$ pathway (the alcohol forms a chlorosulfite ester, and the released chloride delivers the chlorine back to the SAME face it left from — retention of configuration). With pyridine present, the freed chloride instead attacks from the opposite face, giving a clean $S_N2$-style inversion.

• Product: Chlorocyclohexane
• Since C1 of cyclohexanol is not a stereocentre, retention vs. inversion makes no visible difference to the product here — either pathway gives the same chlorocyclohexane.

How to avoid the mistake:

• Memorize: $SOCl_2$ + alcohol → alkyl chloride, via $S_Ni$ (retention, no pyridine) or an $S_N2$-style inversion (with pyridine).
• Do not assume elimination unless a strong base is present.

(ii) 1-Ethyl-4-nitrobenzene $+ \ Br_2 \xrightarrow{\text{heat or UV light}}$

Common mistake:

Students draw electrophilic aromatic substitution (bromination on the ring).

Why it happens:

$Br_2$ with a catalyst ($FeBr_3$) gives ring bromination. But here, heat/UV light changes the mechanism.

Correct approach:

• Heat/UV light initiates free-radical substitution at the benzylic position (the carbon next to the ring).
• The nitro group is meta-directing and deactivates the ring, but that is irrelevant here — the reaction is not on the ring.
• Product: 1-(1-Bromoethyl)-4-nitrobenzene ($p$-$O_2NC_6H_4CHBrCH_3$)

How to avoid the mistake:

• Always check the reaction conditions:
  • $Br_2$ + catalyst → aromatic substitution
  • $Br_2$ + heat/UV → free-radical at benzylic or allylic position
• The benzylic radical is highly stabilized, so it is the preferred site.

(iii) 4-(Hydroxymethyl)phenol $+ \ HCl \xrightarrow{\text{heat}}$

Common mistake:

Students replace the phenolic $-OH$ with $Cl$, or replace both $-OH$ groups.

Why it happens:

Phenol $-OH$ does not undergo substitution easily because of resonance stabilization.

Correct approach:

• The benzylic alcohol ($-CH_2OH$) reacts via $S_N1$ (benzylic carbocation is stable).
• The phenolic $-OH$ remains unchanged.
• Product: 4-(Chloromethyl)phenol ($p$-$HOC_6H_4CH_2Cl$)

How to avoid the mistake:

• Remember: Phenolic $-OH$ is not a good leaving group under acidic conditions.
• Only the side-chain $-OH$ (alcoholic) reacts.

(iv) 1-Methylcyclohex-1-ene $+ \ HI \rightarrow$

Common mistake:

Students add $H$ and $I$ randomly, or place $I$ on the less substituted carbon.

Why it happens:

Misapplication of Markovnikov’s rule — they think “$H$ goes to the carbon with more $H$’s” but forget that the more substituted carbon gets the positive charge.

Correct approach:

• Markovnikov addition: $H^+$ adds to the less substituted alkene carbon (to form the more stable tertiary carbocation).
• $I^-$ then attacks the carbocation.
• Product: 1-Iodo-1-methylcyclohexane

How to avoid the mistake:

• Always draw the carbocation intermediate.
• The more stable carbocation (tertiary > secondary > primary) determines where the nucleophile ($I^-$) goes.

(v) $CH_3CH_2Br + NaI \rightarrow$

Common mistake:

Students think no reaction occurs, or draw elimination.

Why it happens:

$NaI$ is a weak base, so elimination is unlikely. But students sometimes forget the Finkelstein reaction.

Correct approach:

• This is an $S_N2$ reaction: $I^-$ displaces $Br^-$ (iodide is a better nucleophile).
• Product: $CH_3CH_2I$ (ethyl iodide) + $NaBr$ …