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20 Advanced 3-Step Organic Chemistry Conversions

20 Advanced 3-Step Organic Chemistry Conversions | Chemca

Chemca

Advanced Organic Synthesis

Master 3-Step
Organic Conversions

Elevate your synthetic logic. This curated collection features 20 intricate three-step organic conversions. These sequences combine aliphatic and aromatic reactions from Class 11 and 12 chapters, teaching you how to step-up, step-down, and manipulate functional groups. Click to reveal detailed mechanistic pathways.

1 Convert Ethanol to Propan-2-ol
Step 1: Controlled Oxidation

To add a carbon and branch it, we must synthesize a ketone or aldehyde. Ethanol (a primary alcohol) is subjected to mild oxidation using Pyridinium Chlorochromate ($\ce{PCC}$) in dichloromethane. This stops the oxidation at the aldehyde stage, yielding Ethanal.

Step 2: Grignard Addition (Step-Up)

Ethanal is reacted with Methylmagnesium bromide ($\ce{CH3MgBr}$, a Grignard reagent) in dry ether. The nucleophilic methyl carbanion attacks the electrophilic carbonyl carbon, breaking the pi bond to form a magnesium alkoxide complex.

Step 3: Acid Hydrolysis

The alkoxide intermediate is immediately hydrolyzed with dilute acid ($\ce{H3O+}$). The oxygen atom is protonated, yielding a secondary alcohol with one additional carbon atom: Propan-2-ol.

$$ \text{Step 1: } \ce{CH3CH2OH ->[PCC] CH3CHO} $$ $$ \text{Step 2 & 3: } \ce{CH3CHO + CH3MgBr ->[dry ether] CH3-CH(OMgBr)-CH3 ->[H3O+] CH3-CH(OH)-CH3} $$
2 Convert Benzene to m-Nitroaniline
Step 1: Primary Nitration

Benzene is treated with a standard nitrating mixture (concentrated $\ce{HNO3}$ and concentrated $\ce{H2SO4}$) at 330 K. The electrophilic attack of the nitronium ion ($\ce{NO2+}$) forms Nitrobenzene.

Step 2: Secondary Nitration (Meta-Directing)

The $-\ce{NO2}$ group is strongly electron-withdrawing and meta-directing. A second, more vigorous nitration (fuming $\ce{HNO3}$, conc. $\ce{H2SO4}$, high heat) places a second nitro group at the meta position, yielding m-Dinitrobenzene.

Step 3: Selective Partial Reduction

To convert only one nitro group to an amine while leaving the other intact, we use a selective reducing agent such as Sodium sulfide ($\ce{Na2S}$) or Ammonium hydrogen sulfide ($\ce{NH4HS}$). This selective reduction yields m-Nitroaniline.

$$ \text{Step 1: } \ce{C6H6 ->[HNO3/H2SO4] C6H5NO2} $$ $$ \text{Step 2: } \ce{C6H5NO2 ->[fuming HNO3/H2SO4][\Delta] m-C6H4(NO2)2} $$ $$ \text{Step 3: } \ce{m-C6H4(NO2)2 ->[Na2S or NH4HS] m-NO2-C6H4-NH2} $$
3 Convert Ethanoic acid to Propanenitrile
Step 1: Strong Reduction

To add a carbon (step-up), we must convert the unreactive carboxylic acid to a reactive halide. First, Ethanoic acid is strongly reduced using Lithium Aluminum Hydride ($\ce{LiAlH4}$) followed by hydrolysis, yielding Ethanol.

Step 2: Halogenation

Ethanol is converted into a good substrate for nucleophilic substitution by reacting it with Thionyl chloride ($\ce{SOCl2}$). This replaces the $-\ce{OH}$ group with chlorine, forming Chloroethane (along with gaseous $\ce{SO2}$ and $\ce{HCl}$).

Step 3: Cyanide Substitution (Step-Up)

Chloroethane (2 carbons) is heated with alcoholic Potassium Cyanide ($\ce{KCN(alc)}$). The cyanide ion displaces the chloride ion in an $S_N2$ reaction, extending the chain by one carbon to form Propanenitrile.

$$ \text{Step 1: } \ce{CH3COOH ->[1. LiAlH4][2. H2O] CH3CH2OH} $$ $$ \text{Step 2: } \ce{CH3CH2OH + SOCl2 -> CH3CH2Cl + SO2 ^ + HCl ^} $$ $$ \text{Step 3: } \ce{CH3CH2Cl + KCN (alc) -> CH3CH2CN + KCl} $$
4 Convert Bromomethane to Ethanol
Step 1: Cyanide Substitution (Step-Up)

Bromomethane (1 carbon) must be stepped up to a 2-carbon chain. It is treated with alcoholic Potassium Cyanide ($\ce{KCN}$). The $S_N2$ substitution yields Ethanenitrile ($\ce{CH3CN}$).

Step 2: Complete Hydrolysis

The nitrile is then completely hydrolyzed by boiling with dilute mineral acid ($\ce{H3O+}$). The carbon-nitrogen triple bond is broken completely to form a carboxylic acid group, producing Ethanoic acid.

Step 3: Strong Reduction

Finally, Ethanoic acid is reduced using the powerful reducing agent Lithium Aluminum Hydride ($\ce{LiAlH4}$), followed by aqueous workup, yielding the desired primary alcohol: Ethanol.

$$ \text{Step 1: } \ce{CH3Br + KCN (alc) -> CH3CN + KBr} $$ $$ \text{Step 2: } \ce{CH3CN + 2H2O ->[H+] CH3COOH + NH4+} $$ $$ \text{Step 3: } \ce{CH3COOH ->[1. LiAlH4][2. H3O+] CH3CH2OH} $$
5 Convert Phenol to Aspirin
Step 1: Activation of Phenol

Phenol is reacted with aqueous Sodium Hydroxide ($\ce{NaOH}$) to form Sodium phenoxide. The phenoxide ion is highly activated for electrophilic aromatic substitution.

Step 2: Kolbe's Reaction

Sodium phenoxide undergoes Kolbe's reaction by heating with Carbon dioxide ($\ce{CO2}$) at 400 K under pressure, followed by acidification ($\ce{H+}$). The electrophile ($\ce{CO2}$) attacks the ortho position, forming 2-Hydroxybenzoic acid (Salicylic acid).

Step 3: Acetylation

Salicylic acid is treated with Acetic anhydride ($\ce{(CH3CO)2O}$) in the presence of a few drops of concentrated sulfuric acid as a catalyst. This acetylates the phenolic $-\ce{OH}$ group, forming Acetylsalicylic acid (Aspirin).

$$ \text{Step 1: } \ce{C6H5OH + NaOH -> C6H5ONa + H2O} $$ $$ \text{Step 2: } \ce{C6H5ONa + CO2 ->[400 K][then H+] C6H4(OH)COOH} $$ $$ \text{Step 3: } \ce{C6H4(OH)COOH + (CH3CO)2O ->[H+] C6H4(OCOCH3)COOH + CH3COOH} $$
6 Convert Ethene to Propanoic acid
Step 1: Hydrohalogenation

Ethene (2 carbons) needs an extra carbon. First, it is reacted with Hydrogen Bromide ($\ce{HBr}$). The electrophilic addition across the double bond yields Bromoethane.

Step 2: Cyanide Step-Up

Bromoethane is heated with alcoholic Potassium Cyanide ($\ce{KCN(alc)}$). The $S_N2$ substitution extends the carbon skeleton from two to three atoms, producing Propanenitrile.

Step 3: Complete Hydrolysis

Propanenitrile is boiled with an aqueous mineral acid ($\ce{H3O+}$). The nitrile group undergoes complete hydrolysis, converting the triple bond directly to a carboxylic acid, yielding Propanoic acid.

$$ \text{Step 1: } \ce{CH2=CH2 + HBr -> CH3CH2Br} $$ $$ \text{Step 2: } \ce{CH3CH2Br + KCN (alc) -> CH3CH2CN + KBr} $$ $$ \text{Step 3: } \ce{CH3CH2CN + 2H2O ->[H+] CH3CH2COOH + NH4+} $$
7 Convert Nitrobenzene to Phenol
Step 1: Reduction

Nitrobenzene is reduced to an amine using active metal and acid, specifically Tin and Hydrochloric acid ($\ce{Sn/HCl}$). The $-\ce{NO2}$ group is fully reduced to an $-\ce{NH2}$ group, yielding Aniline.

Step 2: Diazotization

Aniline is treated with a cold, aqueous solution of Sodium Nitrite and Hydrochloric acid ($\ce{NaNO2 + HCl}$) at strictly maintained ice-cold temperatures (0-5°C). This produces the versatile intermediate Benzene diazonium chloride.

Step 3: Hydrolysis

The diazonium salt is highly unstable to heat. Simply warming the aqueous solution of benzene diazonium chloride allows water to act as a nucleophile, replacing the $\ce{-N2+}$ group to yield Phenol, with the evolution of nitrogen gas.

$$ \text{Step 1: } \ce{C6H5NO2 + 6[H] ->[Sn/HCl] C6H5NH2 + 2H2O} $$ $$ \text{Step 2: } \ce{C6H5NH2 + NaNO2 + 2HCl ->[273 K] C6H5N2+Cl- + NaCl + 2H2O} $$ $$ \text{Step 3: } \ce{C6H5N2+Cl- + H2O ->[\Delta] C6H5OH + N2 ^ + HCl} $$
8 Convert Ethanol to Methanamine
Step 1: Vigorous Oxidation

To step down a carbon series, we need an amide. First, Ethanol (2 carbons) is subjected to complete oxidation using acidified Potassium Permanganate ($\ce{KMnO4/H+}$), yielding Ethanoic acid.

Step 2: Amide Formation

Ethanoic acid is treated with Ammonia ($\ce{NH3}$) and heated strongly. This eliminates a water molecule from the intermediate ammonium ethanoate to form Ethanamide (Acetamide).

Step 3: Hoffmann Bromamide Degradation (Step-Down)

Ethanamide is reacted with Bromine and a strong base ($\ce{Br2/KOH}$). This reaction excises the carbonyl carbon entirely, reducing the chain length from two to one, resulting in Methanamine.

$$ \text{Step 1: } \ce{CH3CH2OH + 2[O] ->[KMnO4/H+] CH3COOH + H2O} $$ $$ \text{Step 2: } \ce{CH3COOH + NH3 -> CH3COONH4 ->[\Delta] CH3CONH2 + H2O} $$ $$ \text{Step 3: } \ce{CH3CONH2 + Br2 + 4KOH -> CH3NH2 + K2CO3 + 2KBr + 2H2O} $$
9 Convert Toluene to 2-Phenylethanoic Acid
Step 1: Side-Chain Halogenation

To extend the side chain, we must first activate it. Toluene is reacted with Chlorine gas in the presence of UV light ($\ce{Cl2/h\nu}$). This initiates a free-radical substitution exclusively on the methyl group, producing Benzyl chloride.

Step 2: Cyanide Step-Up

Benzyl chloride is heated with alcoholic Potassium Cyanide ($\ce{KCN}$). The nucleophilic cyanide ion displaces the chloride, extending the side chain by one carbon atom to yield Benzyl cyanide (Phenylethanenitrile).

Step 3: Hydrolysis

The nitrile is completely hydrolyzed by boiling with aqueous mineral acid ($\ce{H3O+}$). The triple bond is converted into a carboxylic acid, producing 2-Phenylethanoic acid.

$$ \text{Step 1: } \ce{C6H5CH3 + Cl2 ->[h\nu] C6H5CH2Cl + HCl} $$ $$ \text{Step 2: } \ce{C6H5CH2Cl + KCN (alc) -> C6H5CH2CN + KCl} $$ $$ \text{Step 3: } \ce{C6H5CH2CN + 2H2O ->[H+] C6H5CH2COOH + NH4+} $$
10 Convert 1-Bromopropane to Propanone
Step 1: Dehydrohalogenation

To move functionalization from the terminal carbon (C-1) to the middle carbon (C-2), we create a double bond. 1-Bromopropane is heated with alcoholic $\ce{KOH}$. The $\beta$-elimination of $\ce{HBr}$ yields Propene.

Step 2: Markovnikov Hydration

Propene is passed through water in the presence of dilute acid ($\ce{H2O/H+}$). Electrophilic addition follows Markovnikov's rule, placing the hydroxyl group on the secondary carbon, yielding Propan-2-ol.

Step 3: Oxidation

The secondary alcohol is oxidized using acidified Potassium Dichromate ($\ce{K2Cr2O7/H+}$) or Chromium trioxide ($\ce{CrO3}$). Secondary alcohols oxidize cleanly into ketones, producing Propanone (Acetone).

$$ \text{Step 1: } \ce{CH3CH2CH2Br + KOH (alc) ->[\Delta] CH3CH=CH2 + KBr + H2O} $$ $$ \text{Step 2: } \ce{CH3CH=CH2 + H2O ->[H+] CH3-CH(OH)-CH3} $$ $$ \text{Step 3: } \ce{CH3-CH(OH)-CH3 ->[K2Cr2O7/H+] CH3-CO-CH3 + H2O} $$
11 Convert Benzene to Styrene
Step 1: Friedel-Crafts Acylation

A 2-carbon chain is introduced to the ring. Benzene reacts with Acetyl chloride ($\ce{CH3COCl}$) and anhydrous Aluminum chloride ($\ce{AlCl3}$). The electrophilic acylium ion attack yields Acetophenone ($\ce{C6H5COCH3}$).

Step 2: Reduction

The ketone group is selectively reduced using Sodium Borohydride ($\ce{NaBH4}$) in ethanol. The hydride ion attacks the carbonyl carbon, producing the secondary alcohol 1-Phenylethanol.

Step 3: Acid-Catalyzed Dehydration

The alcohol is heated with concentrated Sulfuric acid ($\ce{H2SO4}$ at $\Delta$). The elimination of water forms a double bond that is highly conjugated with the aromatic ring, yielding Phenylethene, commonly known as Styrene.

$$ \text{Step 1: } \ce{C6H6 + CH3COCl ->[AlCl3] C6H5COCH3 + HCl} $$ $$ \text{Step 2: } \ce{C6H5COCH3 + 2[H] ->[NaBH4] C6H5CH(OH)CH3} $$ $$ \text{Step 3: } \ce{C6H5CH(OH)CH3 ->[conc. H2SO4][\Delta] C6H5CH=CH2 + H2O} $$
12 Convert Ethanal to Crotonic Acid
Step 1: Aldol Addition

Ethanal ($\ce{CH3CHO}$) is treated with dilute $\ce{NaOH}$. Due to the presence of $\alpha$-hydrogens, two molecules condense. The enolate of one attacks the carbonyl of the other, forming 3-Hydroxybutanal (aldol).

Step 2: Dehydration

Heating the aldol drives the elimination of a water molecule to form a stable, conjugated system. This yields the $\alpha,\beta$-unsaturated aldehyde, But-2-enal (Crotonaldehyde).

Step 3: Selective Oxidation

To convert the aldehyde to a carboxylic acid without cleaving the newly formed carbon-carbon double bond, a mild oxidizing agent is strictly required. Tollens' reagent ($\ce{[Ag(NH3)2]+}$) selectively oxidizes the $-\ce{CHO}$ group, yielding But-2-enoic acid (Crotonic acid).

$$ \text{Step 1: } \ce{2CH3CHO ->[dil. NaOH] CH3CH(OH)CH2CHO} $$ $$ \text{Step 2: } \ce{CH3CH(OH)CH2CHO ->[\Delta] CH3CH=CHCHO + H2O} $$ $$ \text{Step 3: } \ce{CH3CH=CHCHO + 2[Ag(NH3)2]+ + 3OH- -> CH3CH=CHCOO- + 2Ag v + 4NH3 + 2H2O} $$
13 Convert Aniline to p-Bromoaniline
Step 1: Protection (Acetylation)

Direct bromination of aniline yields 2,4,6-tribromoaniline because the $-\ce{NH2}$ group is highly activating. To get the mono-substituted product, we "protect" the amine. Aniline is reacted with Acetic anhydride ($\ce{Ac2O}$) in pyridine, forming Acetanilide. This reduces the activating power via resonance with the carbonyl oxygen.

Step 2: Mono-Bromination

Acetanilide is reacted with Bromine in acetic acid ($\ce{Br2/CH3COOH}$). Due to the bulkiness of the acetyl group, steric hindrance forces the incoming bromine electrophile exclusively to the para position, yielding p-Bromoacetanilide.

Step 3: Deprotection (Hydrolysis)

The acetyl protection is removed by boiling with aqueous acid or base ($\ce{H3O+}$ or $\ce{OH-}$). The amide bond is hydrolyzed, restoring the primary amine group, yielding pure p-Bromoaniline.

$$ \text{Step 1: } \ce{C6H5NH2 + (CH3CO)2O ->[Pyridine] C6H5NHCOCH3 + CH3COOH} $$ $$ \text{Step 2: } \ce{C6H5NHCOCH3 + Br2 ->[CH3COOH] p-Br-C6H4-NHCOCH3 + HBr} $$ $$ \text{Step 3: } \ce{p-Br-C6H4-NHCOCH3 + H2O ->[H+] p-Br-C6H4-NH2 + CH3COOH} $$
14 Convert Aniline to 1,3,5-Tribromobenzene
Step 1: Complete Bromination

Aniline is highly activated. Reacting it directly with aqueous Bromine (Bromine water, $\ce{Br2(aq)}$) results in rapid electrophilic aromatic substitution at all available ortho and para positions simultaneously, precipitating 2,4,6-Tribromoaniline.

Step 2: Diazotization

To remove the amino group, it must be converted to a diazonium salt. The tribromoaniline is treated with cold Sodium Nitrite and Hydrochloric acid ($\ce{NaNO2/HCl}$, 273K), yielding 2,4,6-Tribromobenzene diazonium chloride.

Step 3: Deamination

The diazonium group is completely removed and replaced by a hydrogen atom by reacting the salt with a mild reducing agent, Hypophosphorous acid ($\ce{H3PO2}$) in the presence of water, yielding 1,3,5-Tribromobenzene.

$$ \text{Step 1: } \ce{C6H5NH2 + 3Br2(aq) -> 2,4,6-C6H2Br3(NH2) v + 3HBr} $$ $$ \text{Step 2: } \ce{2,4,6-C6H2Br3(NH2) ->[NaNO2/HCl, 273 K] 2,4,6-C6H2Br3(N2+Cl-)} $$ $$ \text{Step 3: } \ce{2,4,6-C6H2Br3(N2+Cl-) + H3PO2 + H2O -> 1,3,5-C6H3Br3 + N2 ^ + H3PO3 + HCl} $$
15 Convert Benzene to Benzyl Alcohol
Step 1: Friedel-Crafts Alkylation

First, a carbon side-chain must be added to the ring. Benzene reacts with Methyl chloride ($\ce{CH3Cl}$) catalyzed by anhydrous Aluminum chloride ($\ce{AlCl3}$). The electrophilic methyl cation attacks the ring to form Toluene (Methylbenzene).

Step 2: Side-Chain Halogenation

Toluene is reacted with Chlorine gas in the presence of sunlight or UV light ($\ce{Cl2/h\nu}$). This directs the reaction via a free-radical mechanism exclusively to the benzylic position, substituting one hydrogen for a chlorine to yield Benzyl chloride.

Step 3: Nucleophilic Substitution

The benzylic chloride is highly reactive towards $S_N2$ substitution. Boiling benzyl chloride with aqueous Potassium Hydroxide ($\ce{KOH(aq)}$) displaces the chloride ion with a hydroxyl group, yielding Benzyl alcohol.

$$ \text{Step 1: } \ce{C6H6 + CH3Cl ->[AlCl3] C6H5CH3 + HCl} $$ $$ \text{Step 2: } \ce{C6H5CH3 + Cl2 ->[h\nu] C6H5CH2Cl + HCl} $$ $$ \text{Step 3: } \ce{C6H5CH2Cl + KOH(aq) ->[\Delta] C6H5CH2OH + KCl} $$
16 Convert Propan-2-ol to 2-Methylpropan-2-ol
Step 1: Oxidation to Ketone

To create a tertiary alcohol, we must react a Grignard reagent with a ketone. Propan-2-ol (a secondary alcohol) is oxidized using Chromium trioxide in acid ($\ce{CrO3/H+}$) or Jones Reagent, forming Propanone (Acetone).

Step 2: Grignard Addition

Propanone is reacted with Methylmagnesium bromide ($\ce{CH3MgBr}$) in dry ether. The methyl carbanion attacks the central electrophilic carbonyl carbon, forming a tertiary magnesium alkoxide intermediate.

Step 3: Acid Hydrolysis

The alkoxide intermediate is treated with dilute acid ($\ce{H3O+}$). Protonation of the oxygen yields the target tertiary alcohol, 2-Methylpropan-2-ol.

$$ \text{Step 1: } \ce{CH3CH(OH)CH3 ->[CrO3/H+] CH3COCH3} $$ $$ \text{Step 2 & 3: } \ce{CH3COCH3 + CH3MgBr -> CH3-C(OMgBr)(CH3)2 ->[H3O+] CH3-C(OH)(CH3)2} $$
17 Convert Propan-1-ol to Ethanamine
Step 1: Vigorous Oxidation

A step-down reaction is required (3 carbons to 2 carbons). The first milestone is an acid. Propan-1-ol is oxidized using acidified Potassium Permanganate ($\ce{KMnO4/H+}$), driving it directly to Propanoic acid.

Step 2: Amide Synthesis

Propanoic acid is reacted with Ammonia ($\ce{NH3}$) to form an ammonium salt, which is then strongly heated to dehydrate it into Propanamide.

Step 3: Hoffmann Degradation

Propanamide is subjected to Hoffmann bromamide degradation using Bromine and aqueous $\ce{KOH}$. The carbonyl carbon is completely removed as carbonate, yielding the stepped-down primary amine, Ethanamine.

$$ \text{Step 1: } \ce{CH3CH2CH2OH + 2[O] ->[KMnO4/H+] CH3CH2COOH + H2O} $$ $$ \text{Step 2: } \ce{CH3CH2COOH + NH3 -> CH3CH2COONH4 ->[\Delta] CH3CH2CONH2 + H2O} $$ $$ \text{Step 3: } \ce{CH3CH2CONH2 + Br2 + 4KOH -> CH3CH2NH2 + K2CO3 + 2KBr + 2H2O} $$
18 Convert Ethanol to Butane
Step 1: Halogenation

To double the carbon chain symmetrically, we use the Wurtz reaction, which requires an alkyl halide. Ethanol is treated with Thionyl chloride ($\ce{SOCl2}$) in the presence of pyridine, cleanly converting it to Chloroethane.

Step 2: Wurtz Reaction

Chloroethane (2 moles) is reacted with Sodium metal ($\ce{Na}$) in a strictly anhydrous medium (dry ether). The sodium extracts the halogens, allowing the two ethyl radicals to couple together symmetrically, forming Butane.

Step 3 (Alternative, but mathematically 3 if via Grignard):

Note: Standard Wurtz is 2 steps. For 3 steps: 1. $\ce{SOCl2}$ -> Chloroethane, 2. Mg/ether -> Ethyl Mg Chloride, 3. Chloroethane -> Butane (Coupling). Wurtz conceptually combines 2 & 3. Both are highly acceptable in exams.

$$ \text{Step 1: } \ce{CH3CH2OH + SOCl2 -> CH3CH2Cl + SO2 ^ + HCl ^} $$ $$ \text{Step 2: } \ce{2 CH3CH2Cl + 2Na ->[Dry Ether] CH3CH2CH2CH3 + 2NaCl} $$
19 Convert Acetic Acid to Malonic Acid
Step 1: Hell-Volhard-Zelinsky (HVZ) Reaction

Acetic acid ($\ce{CH3COOH}$) possesses $\alpha$-hydrogens. Treatment with Chlorine ($\ce{Cl2}$) in the presence of a small amount of Red Phosphorus ($\ce{P_{red}}$) substitutes one $\alpha$-hydrogen with chlorine, yielding Chloroacetic acid.

Step 2: Cyanide Substitution

The $\alpha$-haloacid is then reacted with aqueous Potassium Cyanide ($\ce{KCN(aq)}$). The cyanide ion nucleophilically substitutes the chlorine atom via an $S_N2$ mechanism, resulting in Cyanoacetic acid.

Step 3: Complete Hydrolysis

The intermediate is boiled with dilute aqueous acid ($\ce{H3O+}$). The nitrile group ($\ce{-CN}$) is completely hydrolyzed into a second carboxylic acid group, producing Propanedioic acid, universally known as Malonic acid.

$$ \text{Step 1: } \ce{CH3COOH + Cl2 ->[P_{red}] Cl-CH2-COOH + HCl} $$ $$ \text{Step 2: } \ce{Cl-CH2-COOH + KCN (aq) -> NC-CH2-COOH + KCl} $$ $$ \text{Step 3: } \ce{NC-CH2-COOH + 2H2O ->[H+] HOOC-CH2-COOH + NH4+} $$
20 Convert Nitrobenzene to Iodobenzene
Step 1: Reduction

Direct iodination of benzene rings is reversible and difficult. Nitrobenzene is first reduced to a primary amine using Tin and Hydrochloric acid ($\ce{Sn/HCl}$), yielding Aniline.

Step 2: Diazotization

Aniline is treated with a cold mixture of Sodium Nitrite and Hydrochloric acid ($\ce{NaNO2 + HCl}$) strictly between 0-5°C (273-278 K) to form Benzene diazonium chloride, which possesses the excellent nitrogen leaving group.

Step 3: Substitution by Iodide

Unlike Sandmeyer reactions, introducing iodine does not require a copper catalyst. The diazonium salt is simply warmed with an aqueous solution of Potassium Iodide ($\ce{KI}$). The iodide ion nucleophilically replaces the diazonium group, liberating nitrogen gas and producing Iodobenzene.

$$ \text{Step 1: } \ce{C6H5NO2 + 6[H] ->[Sn/HCl] C6H5NH2 + 2H2O} $$ $$ \text{Step 2: } \ce{C6H5NH2 + NaNO2 + 2HCl ->[273 K] C6H5N2+Cl- + NaCl + 2H2O} $$ $$ \text{Step 3: } \ce{C6H5N2+Cl- + KI ->[\Delta] C6H5I + N2 ^ + KCl} $$

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