Q.Match the reactions given in Column I with the types of reactions given in Column II.
Column I:
(i)

The key idea is Markovnikov Addition for reaction (ii) — the hydrogen of HBr adds to the carbon with more hydrogens, giving the more stable carbocation and the observed 2-bromopropane.

Reasoning steps:

1. (i) $\mathrm{C_6H_5Cl}$ with $\mathrm{Fe/Cl_2}$ is an electrophilic aromatic substitution (chlorination of benzene ring).
2. (ii) $\mathrm{CH_3-CH=CH_2 + HBr}$ follows Markovnikov’s rule — H adds to the terminal carbon, Br to the internal carbon — this is electrophilic addition.
3. (iii) $\mathrm{C_6H_5-CH(I)-CH_3}$ with $\mathrm{OH^-}$ forms a benzylic carbocation (stabilised by resonance), then $\mathrm{OH^-}$ attacks — this is nucleophilic substitution ($\mathrm{S_N1}$). …

Let’s go through each reaction one by one, understanding why the mechanism is what it is.

1. (i) $\mathrm{C_6H_5Cl \xrightarrow{Fe/Cl_2}}$ 1,2-dichlorobenzene + 1,2,4-trichlorobenzene

Chlorobenzene is an aromatic ring. The reagent is chlorine gas with iron (which generates $\mathrm{FeCl_3}$, a Lewis acid). This is a classic electrophilic aromatic substitution — the $\mathrm{Cl^+}$ electrophile attacks the ring, and the iron catalyst activates the chlorine molecule. The products are ortho-dichlorobenzene and 1,2,4-trichlorobenzene (the chlorine already present is ortho/para-directing). So this matches (b).

2. (ii) $\mathrm{CH_3-CH{=}CH_2 + HBr \rightarrow CH_3-\underset{\underset{\displaystyle Br}{|}}{CH}-CH_3}$

Propene reacts with HBr. The double bond is electron-rich and attacks the electrophilic proton of HBr, forming a carbocation. The more stable carbocation (secondary, in this case) forms, and then bromide ion attacks that carbocation. This is electrophilic addition — the alkene is the nucleophile, HBr is the electrophile. The product follows Markovnikov’s rule (Br goes to the more substituted carbon). So this matches (d).

3. (iii) $\mathrm{C_6H_5-CH(I)-CH_3 \xrightarrow{OH^-} C_6H_5-CH(OH)-CH_3}$

This is 1-iodo-1-phenylethane reacting with hydroxide ion to give 1-phenylethanol. The iodine is on a benzylic carbon — that carbon can form a relatively stable benzylic carbocation. The reaction proceeds via an S$_N$1 mechanism: first the C–I bond breaks (slow step) to give a benzylic carbocation, then OH$^-$ attacks (fast step). The product shows inversion is not required; racemization would occur. So this matches (e).

4. (iv) 1-chloro-4-nitrobenzene $+\ \mathrm{NaOH} \rightarrow$ 4-nitrophenol …

Concept: Reaction Mechanisms in Organic Chemistry

This question tests your ability to identify the reaction type from the given reactants, conditions, and products. The method below will help you match each reaction correctly.

Method: Mechanism Signature Analysis

Step 1 – Identify the substrate and reagent type

• Is the substrate aromatic or aliphatic?
• Is the reagent electrophilic, nucleophilic, or basic?

Step 2 – Look for key conditions

• Presence of a catalyst (e.g., Fe/Cl₂, alc. KOH)
• Solvent or temperature hints (e.g., OH⁻, heat)

Step 3 – Predict the bond-making/breaking pattern

• Does the reaction involve addition, substitution, or elimination?
• Is the mechanism polar (ionic) or radical?

Step 4 – Match with the correct column II label

Applying the method:

(i) $\mathrm{C_6H_5Cl \xrightarrow{Fe/Cl_2}}$ dichloro + trichlorobenzene

• Aromatic ring + Fe/Cl₂ → Electrophilic aromatic substitution (Cl⁺ attacks ring)
• Match: (b)

(ii) $\mathrm{CH_3CH=CH_2 + HBr \rightarrow CH_3CHBrCH_3}$

• Alkene + HX → Markovnikov addition
• Electrophilic addition
• Match: (d)

(iii) $\mathrm{C_6H_5CH(I)CH_3 \xrightarrow{OH^-} C_6H_5CH(OH)CH_3}$

• Secondary alkyl iodide + OH⁻ → substitution via carbocation (benzylic stabilisation)
• Nucleophilic substitution (S$_N$1)
• Match: (e)

(iv) 1-chloro-4-nitrobenzene + NaOH → 4-nitrophenol …

Here are the common mistakes students make when matching these reaction types, along with how to avoid each.

Mistake 1: Confusing the mechanism in (i) — Halogenation of Chlorobenzene

• The Mistake: Students often see $\mathrm{Fe/Cl_2}$ and immediately think "Electrophilic substitution" is correct, but they fail to check the product. They might also confuse this with nucleophilic substitution because a halogen is present.
• Why it happens: The reagent $\mathrm{Fe/Cl_2}$ generates $\mathrm{Cl}^+$ (an electrophile). However, the product is a mixture of di- and tri- chlorobenzenes. This is a classic case of electrophilic aromatic substitution where the existing chlorine atom is an ortho/para-directing group.
• How to avoid: Always check the reagent and the product.
  • $\mathrm{Fe/Cl_2}$ or $\mathrm{FeBr_3/Br_2}$ always means electrophilic aromatic substitution (EAS) on an aromatic ring.
  • If the product has more halogen atoms on the ring, it's EAS. If the product replaces a halogen with another group (like -OH), it's nucleophilic aromatic substitution (NAS).
  • Correct Match: (i) → ** (b) Electrophilic aromatic substitution**.

Mistake 2: Misidentifying the addition in (ii) — Markovnikov vs. Anti-Markovnikov

• The Mistake: Students see an alkene ($\mathrm{CH_3-CH=CH_2}$) reacting with HBr and assume it's a simple addition. They often forget to check which carbon the bromine attaches to.
• Why it happens: The product shows bromine on the secondary carbon (the middle carbon). This follows Markovnikov's rule (the hydrogen adds to the carbon with more hydrogens). This is a classic electrophilic addition.
• How to avoid: When you see an alkene + HX (or $\mathrm{H_2O/H^+}$), always check the product's structure.
  • If the halogen (or -OH) goes to the more substituted carbon, it's electrophilic addition (Markovnikov).
  • If it goes to the less substituted carbon, it's a free radical addition (anti-Markovnikov, requiring peroxide).
  • Correct Match: (ii) → ** (d) Electrophilic addition**.

Mistake 3: Forgetting the "Aromatic" condition in (iii) — $\mathrm{S_N1}$ vs. $\mathrm{S_N1}$-like

• The Mistake: Students see a halogen leaving and $\mathrm{OH^-}$ attacking, and they immediately tick "Nucleophilic substitution ($\mathrm{S_N1}$)". They are correct, but they often miss the key detail that makes this a special case.
• Why it happens: The carbon bearing the iodine is a benzylic carbon (directly attached to a benzene ring). The carbocation formed ($\mathrm{C_6H_5-CH^+-CH_3}$) is highly stabilized by resonance with the benzene ring. This makes it an extremely fast $\mathrm{S_N1}$ reaction.
• How to avoid: Recognize the benzylic and allylic positions. They are the only cases where $\mathrm{S_N1}$ is very fast and common. For a simple alkyl halide, $\mathrm{OH^-}$ (a strong base) would usually cause elimination, not substitution.
  • Correct Match: (iii) → (f) Nucleophilic substitution ($\mathrm{S_N1}$).

Mistake 4: Confusing (iv) with Electrophilic Substitution

• The Mistake: Students see a benzene ring with a chlorine and think "Electrophilic substitution" because they associate halogens with that reaction. They ignore the strong nucleophile ($\mathrm{NaOH}$) and the product (phenol).
• Why it happens: This is a classic nucleophilic aromatic substitution (NAS). The nitro group (-$\mathrm{NO_2}$) at the para position is a strong electron-withdrawing group. It activates the ring for attack by a nucleophile (like $\mathrm{OH^-}$) by stabilizing the negative charge in the intermediate (Meisenheimer complex).
• How to avoid: Look for the activating group. …