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20 Master-Level 4-Step Organic Conversions

20 Master-Level 4-Step Organic Conversions | Chemca

Chemca

Synthesis Mastery Level

Master 4-Step
Organic Conversions

The pinnacle of synthetic logic. This collection features 20 highly complex four-step organic conversions, combining deep mechanistic knowledge across Class 11 and 12 chemistry. Learn how to sequence step-ups, step-downs, protections, and position shifts. Click to unveil the complete pathways.

1 Convert Methane to Propanone
Step 1: Free Radical Halogenation

Methane is highly unreactive. We first activate it by reacting it with Chlorine gas in the presence of diffuse sunlight ($h\nu$). This free-radical substitution yields Chloromethane.

Step 2: Cyanide Step-Up

Chloromethane is heated with alcoholic Potassium Cyanide ($\ce{KCN}$). The $S_N2$ nucleophilic substitution replaces the chloride ion and extends the carbon chain to two atoms, yielding Ethanenitrile (Acetonitrile).

Step 3: Grignard Nucleophilic Addition

Ethanenitrile is reacted with Methylmagnesium bromide ($\ce{CH3MgBr}$) in dry ether. The nucleophilic methyl carbanion attacks the electrophilic carbon of the nitrile group, forming an unstable Imine magnesium complex.

Step 4: Acid Hydrolysis

The imine complex is hydrolyzed by boiling with dilute mineral acid ($\ce{H3O+}$). The carbon-nitrogen double bond cleaves, substituting the nitrogen with oxygen to yield the 3-carbon ketone, Propanone (Acetone), along with ammonia.

$$ \text{Step 1: } \ce{CH4 + Cl2 ->[h\nu] CH3Cl + HCl} $$ $$ \text{Step 2: } \ce{CH3Cl + KCN(alc) -> CH3CN + KCl} $$ $$ \text{Step 3: } \ce{CH3C\equiv N + CH3MgBr -> CH3-C(CH3)=NMgBr} $$ $$ \text{Step 4: } \ce{CH3-C(CH3)=NMgBr + 2H2O ->[H+] CH3-CO-CH3 + NH3 + Mg(OH)Br} $$
2 Convert Benzene to m-Bromophenol
Step 1: Electrophilic Nitration

Benzene is treated with a nitrating mixture ($\ce{HNO3 / H2SO4}$) at 330K to place a strongly meta-directing group on the ring, yielding Nitrobenzene.

Step 2: Meta-Bromination

Nitrobenzene is reacted with Bromine ($\ce{Br2}$) and a Lewis acid catalyst ($\ce{FeBr3}$). Because the $-\ce{NO2}$ group deactivates ortho/para positions, the incoming bromine electrophile is directed to the meta position, yielding m-Bromonitrobenzene.

Step 3: Reduction

The nitro group is reduced to a primary amine using active metal and acid ($\ce{Sn / HCl}$). This yields m-Bromoaniline.

Step 4: Diazotization and Hydrolysis

m-Bromoaniline is diazotized using cold $\ce{NaNO2/HCl}$ (0-5°C) to form the diazonium salt. This salt is subsequently warmed with water ($\ce{H2O, \Delta}$), replacing the diazonium group with a hydroxyl group to yield m-Bromophenol.

$$ \text{Step 1: } \ce{C6H6 ->[HNO3/H2SO4] C6H5NO2} $$ $$ \text{Step 2: } \ce{C6H5NO2 + Br2 ->[FeBr3] m-Br-C6H4-NO2 + HBr} $$ $$ \text{Step 3: } \ce{m-Br-C6H4-NO2 + 6[H] ->[Sn/HCl] m-Br-C6H4-NH2 + 2H2O} $$ $$ \text{Step 4: } \ce{m-Br-C6H4-NH2 ->[1. NaNO2/HCl, 273 K][2. H2O, \Delta] m-Br-C6H4-OH + N2 ^} $$
3 Convert Ethanol to Propan-1-ol
Step 1: Halogenation

To perform a step-up reaction on a primary alcohol, we first convert it to a halide. Ethanol reacts with Thionyl chloride ($\ce{SOCl2}$) in pyridine to yield Chloroethane.

Step 2: Cyanide Step-Up

Chloroethane is heated with alcoholic Potassium Cyanide ($\ce{KCN}$). The nucleophilic substitution increases the carbon chain from 2 to 3 atoms, forming Propanenitrile.

Step 3: Reduction to Amine

Propanenitrile is strongly reduced using Lithium Aluminum Hydride ($\ce{LiAlH4}$) in ether. The carbon-nitrogen triple bond is fully hydrogenated, yielding the primary amine, Propan-1-amine.

Step 4: Nitrous Acid Reaction

Aliphatic primary amines react with Nitrous acid ($\ce{HNO2}$, generated from $\ce{NaNO2 + HCl}$) to form highly unstable aliphatic diazonium salts, which immediately decompose in water to yield primary alcohols. This yields Propan-1-ol and nitrogen gas.

$$ \text{Step 1: } \ce{CH3CH2OH + SOCl2 -> CH3CH2Cl + SO2 ^ + HCl ^} $$ $$ \text{Step 2: } \ce{CH3CH2Cl + KCN (alc) -> CH3CH2CN + KCl} $$ $$ \text{Step 3: } \ce{CH3CH2CN + 4[H] ->[LiAlH4] CH3CH2CH2NH2} $$ $$ \text{Step 4: } \ce{CH3CH2CH2NH2 + HNO2 -> CH3CH2CH2OH + N2 ^ + H2O} $$
4 Convert Calcium Carbide to Butan-2-one
Step 1: Hydrolysis to Alkyne

Calcium carbide ($\ce{CaC2}$) reacts vigorously with water at room temperature. The hydrolysis yields the simplest alkyne, Ethyne (Acetylene), and calcium hydroxide.

Step 2: Formation of Acetylide

The terminal hydrogen of ethyne is weakly acidic. Passing ethyne gas through a solution of strong base Sodium amide ($\ce{NaNH2}$) in liquid ammonia extracts a proton, forming the nucleophilic Sodium acetylide ($\ce{HC\equiv C^-Na^+}$).

Step 3: Alkylation (Chain Extension)

Sodium acetylide is reacted with Bromoethane ($\ce{CH3CH2Br}$). The acetylide ion nucleophilically attacks the alkyl halide, extending the chain by two carbons to form the terminal alkyne, But-1-yne.

Step 4: Kucherov Hydration

But-1-yne undergoes hydration in the presence of Mercuric sulfate and dilute Sulfuric acid ($\ce{HgSO4 / H2SO4}$). The Markovnikov addition of water forms an enol intermediate that immediately tautomerizes into the ketone, Butan-2-one.

$$ \text{Step 1: } \ce{CaC2 + 2H2O -> HC\equiv CH + Ca(OH)2} $$ $$ \text{Step 2: } \ce{HC\equiv CH + NaNH2 -> HC\equiv CNa + NH3} $$ $$ \text{Step 3: } \ce{HC\equiv CNa + CH3CH2Br -> HC\equiv C-CH2CH3 + NaBr} $$ $$ \text{Step 4: } \ce{HC\equiv C-CH2CH3 + H2O ->[Hg^2+ / H+] CH3-CO-CH2CH3} $$
5 Convert Benzene to p-Fluorotoluene
Step 1: Friedel-Crafts Alkylation

Benzene is reacted with Methyl chloride ($\ce{CH3Cl}$) catalyzed by anhydrous $\ce{AlCl3}$. This places a methyl group on the ring, yielding Toluene. The methyl group is an activating, ortho/para directing group.

Step 2: Selective Nitration

Toluene is nitrated ($\ce{HNO3 / H2SO4}$). Due to steric hindrance at the ortho position, the para-isomer predominates. We isolate the major product, p-Nitrotoluene.

Step 3: Reduction

The nitro group is selectively reduced using Tin and Hydrochloric acid ($\ce{Sn / HCl}$). This converts the $-\ce{NO2}$ to an $-\ce{NH2}$, producing p-Toluidine (p-Methylaniline).

Step 4: Balz-Schiemann Reaction

p-Toluidine is diazotized ($\ce{NaNO2 / HCl}$ at 273K), and then reacted with Fluoroboric acid ($\ce{HBF4}$) to form a diazonium fluoroborate precipitate. Heating this dry salt ($\Delta$) decomposes it to yield p-Fluorotoluene, expelling $\ce{N2}$ and $\ce{BF3}$.

$$ \text{Step 1: } \ce{C6H6 + CH3Cl ->[AlCl3] C6H5CH3 + HCl} $$ $$ \text{Step 2: } \ce{C6H5CH3 ->[HNO3/H2SO4] p-CH3-C6H4-NO2} \text{ (major)} $$ $$ \text{Step 3: } \ce{p-CH3-C6H4-NO2 + 6[H] ->[Sn/HCl] p-CH3-C6H4-NH2 + 2H2O} $$ $$ \text{Step 4: } \ce{p-CH3-C6H4-NH2 ->[1. NaNO2/HCl, 273K][2. HBF4, \Delta] p-CH3-C6H4-F + N2 ^ + BF3} $$
6 Convert Ethene to Butan-1-ol
Step 1: Hydrohalogenation

Ethene (2 carbons) reacts with Hydrogen Bromide ($\ce{HBr}$) to yield Bromoethane. This sets up the molecule for Grignard reagent formation.

Step 2: Grignard Formation

Bromoethane is reacted with Magnesium metal in anhydrous ether. This produces the nucleophilic Grignard reagent, Ethylmagnesium bromide ($\ce{CH3CH2MgBr}$).

Step 3: Oxirane Ring Opening (2-Carbon Extension)

The Grignard reagent is reacted with Ethylene oxide (Oxirane, $\ce{C2H4O}$). The highly strained 3-membered ring is attacked by the nucleophilic ethyl group, opening the ring to form a 4-carbon magnesium alkoxide complex.

Step 4: Acid Hydrolysis

The alkoxide intermediate is treated with dilute acid ($\ce{H3O+}$). Protonation of the terminal oxygen yields the 4-carbon primary alcohol, Butan-1-ol.

$$ \text{Step 1: } \ce{CH2=CH2 + HBr -> CH3CH2Br} $$ $$ \text{Step 2: } \ce{CH3CH2Br + Mg ->[Dry Ether] CH3CH2MgBr} $$ $$ \text{Step 3: } \ce{CH3CH2MgBr + \underset{Oxirane}{\ce{CH2-CH2-O}} -> CH3CH2CH2CH2OMgBr} $$ $$ \text{Step 4: } \ce{CH3CH2CH2CH2OMgBr + H3O+ -> CH3CH2CH2CH2OH + Mg(OH)Br} $$
7 Convert Propan-2-ol to Propanoic Acid
Step 1: Dehydration

A secondary alcohol cannot be directly oxidized to a carboxylic acid with the same carbon count. We must move the functional group. Propan-2-ol is heated with concentrated $\ce{H2SO4}$ at 443 K, undergoing $\beta$-elimination to form Propene.

Step 2: Anti-Markovnikov Halogenation

To place the functional group on the terminal carbon, propene is reacted with $\ce{HBr}$ in the presence of an organic peroxide (Kharasch effect). This free-radical addition strictly yields the primary halide, 1-Bromopropane.

Step 3: Nucleophilic Substitution

1-Bromopropane is boiled with aqueous Potassium Hydroxide ($\ce{KOH(aq)}$). The $S_N2$ substitution replaces the bromide with a hydroxyl group, yielding the primary alcohol, Propan-1-ol.

Step 4: Strong Oxidation

The primary alcohol is subjected to vigorous oxidation using acidified Potassium Permanganate ($\ce{KMnO4/H+}$), converting the terminal $-\ce{CH2OH}$ group completely to a carboxyl group, yielding Propanoic acid.

$$ \text{Step 1: } \ce{CH3-CH(OH)-CH3 ->[conc. H2SO4][443 K] CH3-CH=CH2 + H2O} $$ $$ \text{Step 2: } \ce{CH3-CH=CH2 + HBr ->[Peroxide] CH3-CH2-CH2Br} $$ $$ \text{Step 3: } \ce{CH3-CH2-CH2Br + KOH(aq) ->[\Delta] CH3-CH2-CH2OH + KBr} $$ $$ \text{Step 4: } \ce{CH3-CH2-CH2OH + 2[O] ->[KMnO4/H+] CH3-CH2-COOH + H2O} $$
8 Convert Benzene to 2-Phenylethanoic Acid
Step 1: Alkylation

Benzene is reacted with Methyl chloride and anhydrous $\ce{AlCl3}$ via Friedel-Crafts alkylation. This introduces a 1-carbon side chain to the ring, forming Toluene.

Step 2: Free-Radical Halogenation

Toluene is treated with Chlorine gas under UV light ($\ce{Cl2/h\nu}$). Substitution occurs exclusively at the benzylic position, yielding Benzyl chloride ($\ce{C6H5CH2Cl}$).

Step 3: Cyanide Extension

Benzyl chloride is heated with alcoholic $\ce{KCN}$. The cyanide ion performs an $S_N2$ attack, extending the side chain by one carbon to yield Benzyl cyanide ($\ce{C6H5CH2CN}$).

Step 4: Hydrolysis

The nitrile is completely hydrolyzed by boiling with dilute mineral acid ($\ce{H3O+}$). The $-\ce{CN}$ group converts to a $-\ce{COOH}$ group, yielding 2-Phenylethanoic acid.

$$ \text{Step 1: } \ce{C6H6 + CH3Cl ->[AlCl3] C6H5CH3 + HCl} $$ $$ \text{Step 2: } \ce{C6H5CH3 + Cl2 ->[h\nu] C6H5CH2Cl + HCl} $$ $$ \text{Step 3: } \ce{C6H5CH2Cl + KCN(alc) -> C6H5CH2CN + KCl} $$ $$ \text{Step 4: } \ce{C6H5CH2CN + 2H2O ->[H+] C6H5CH2COOH + NH4+} $$
9 Convert Nitrobenzene to p-Nitroaniline
Step 1: Reduction

Because direct nitration of nitrobenzene yields the meta isomer, we must change the directing group. Nitrobenzene is fully reduced using Tin and HCl ($\ce{Sn / HCl}$) to form Aniline, which is strongly ortho/para directing.

Step 2: Amine Protection

Direct nitration of aniline causes oxidation and tar formation. The amine group is "protected" by acetylation. Aniline reacts with Acetic anhydride in pyridine to form Acetanilide, moderating its reactivity.

Step 3: Nitration

Acetanilide is safely nitrated using a cold mixture of $\ce{HNO3 / H2SO4}$. Due to the steric bulk of the acetyl group, the para isomer dominates, yielding p-Nitroacetanilide.

Step 4: Deprotection (Hydrolysis)

The protecting acetyl group is removed by acid or base hydrolysis ($\ce{H3O+}$ or $\ce{OH-}$, $\Delta$). This restores the original amino group, yielding pure p-Nitroaniline.

$$ \text{Step 1: } \ce{C6H5NO2 + 6[H] ->[Sn/HCl] C6H5NH2 + 2H2O} $$ $$ \text{Step 2: } \ce{C6H5NH2 + (CH3CO)2O ->[Pyridine] C6H5NHCOCH3 + CH3COOH} $$ $$ \text{Step 3: } \ce{C6H5NHCOCH3 + HNO3 ->[H2SO4] p-NO2-C6H4-NHCOCH3 + H2O} $$ $$ \text{Step 4: } \ce{p-NO2-C6H4-NHCOCH3 + H2O ->[H+] p-NO2-C6H4-NH2 + CH3COOH} $$
10 Convert Ethene to 2-Hydroxypropanoic Acid (Lactic Acid)
Step 1: Hydration

Ethene is hydrated by passing it through water with dilute sulfuric acid catalyst ($\ce{H2O / H+}$). This electrophilic addition forms Ethanol.

Step 2: Mild Oxidation

Ethanol is subjected to controlled, mild oxidation using Pyridinium Chlorochromate ($\ce{PCC}$) in dichloromethane. This oxidizes the primary alcohol specifically to an aldehyde, forming Ethanal.

Step 3: Cyanohydrin Formation (Step-Up)

Ethanal undergoes nucleophilic addition with Hydrogen Cyanide ($\ce{HCN}$). The cyanide ion attacks the carbonyl carbon, creating a new C-C bond and forming Ethanal cyanohydrin (2-Hydroxypropanenitrile).

Step 4: Hydrolysis

The nitrile group of the cyanohydrin is fully hydrolyzed by boiling with dilute mineral acid ($\ce{H3O+}$). The $-\ce{CN}$ converts to a $-\ce{COOH}$, leaving the adjacent $-\ce{OH}$ intact. The product is 2-Hydroxypropanoic acid (Lactic acid).

$$ \text{Step 1: } \ce{CH2=CH2 + H2O ->[H+] CH3CH2OH} $$ $$ \text{Step 2: } \ce{CH3CH2OH ->[PCC] CH3CHO} $$ $$ \text{Step 3: } \ce{CH3CHO + HCN -> CH3-CH(OH)-CN} $$ $$ \text{Step 4: } \ce{CH3-CH(OH)-CN + 2H2O ->[H+] CH3-CH(OH)-COOH + NH4+} $$
11 Convert Ethanol to 2-Methylpropan-2-ol
Step 1: Mild Oxidation

Ethanol (2 carbons) is oxidized to an aldehyde using $\ce{PCC}$, yielding Ethanal.

Step 2: 1st Grignard Addition (to Secondary Alcohol)

Ethanal is reacted with Methylmagnesium bromide ($\ce{CH3MgBr}$) followed by acid hydrolysis ($\ce{H3O+}$). The nucleophilic addition steps up the chain by 1 carbon, forming the secondary alcohol, Propan-2-ol.

Step 3: Secondary Oxidation

Propan-2-ol is oxidized using acidified Potassium Dichromate ($\ce{K2Cr2O7/H+}$) or Chromium trioxide. This cleanly forms a ketone, Propanone (Acetone).

Step 4: 2nd Grignard Addition (to Tertiary Alcohol)

Propanone undergoes a second Grignard addition with another mole of Methylmagnesium bromide ($\ce{CH3MgBr}$) followed by hydrolysis ($\ce{H3O+}$). This branches the chain further, yielding the tertiary alcohol, 2-Methylpropan-2-ol.

$$ \text{Step 1: } \ce{CH3CH2OH ->[PCC] CH3CHO} $$ $$ \text{Step 2: } \ce{CH3CHO ->[1. CH3MgBr][2. H3O+] CH3-CH(OH)-CH3} $$ $$ \text{Step 3: } \ce{CH3-CH(OH)-CH3 ->[CrO3/H+] CH3-CO-CH3} $$ $$ \text{Step 4: } \ce{CH3-CO-CH3 ->[1. CH3MgBr][2. H3O+] (CH3)3C-OH} $$
12 Convert Phenol to p-Hydroxyacetophenone
Step 1: Acid-Base Salt Formation

Direct Friedel-Crafts acylation of phenol gives poor yields due to complexation with the $\ce{AlCl3}$ catalyst. We must protect the $\ce{-OH}$. Phenol reacts with $\ce{NaOH}$ to form Sodium phenoxide.

Step 2: Williamson Ether Synthesis

Sodium phenoxide is reacted with Methyl iodide ($\ce{CH3I}$). The phenoxide ion nucleophilically attacks the methyl carbon, forming Methoxybenzene (Anisole). The ether oxygen still activates the ring but won't ruin the catalyst.

Step 3: Friedel-Crafts Acylation

Anisole is reacted with Acetyl chloride and anhydrous $\ce{AlCl3}$. The methoxy group is ortho/para directing. The less sterically hindered para isomer, p-Methoxyacetophenone, is isolated as the major product.

Step 4: Ether Cleavage

The methyl ether is cleaved to restore the phenol group. Boiling with concentrated Hydroiodic acid ($\ce{HI}$) cleaves the aliphatic $\ce{O-CH3}$ bond (the aromatic $\ce{C-O}$ bond is too strong), yielding methyl iodide and p-Hydroxyacetophenone.

$$ \text{Step 1: } \ce{C6H5OH + NaOH -> C6H5ONa + H2O} $$ $$ \text{Step 2: } \ce{C6H5ONa + CH3I -> C6H5OCH3 + NaI} $$ $$ \text{Step 3: } \ce{C6H5OCH3 + CH3COCl ->[AlCl3] p-CH3O-C6H4-COCH3 + HCl} $$ $$ \text{Step 4: } \ce{p-CH3O-C6H4-COCH3 + HI ->[\Delta] p-HO-C6H4-COCH3 + CH3I} $$
13 Convert Propan-1-ol to 2,3-Dimethylbutane
Step 1: Halogenation

Propan-1-ol reacts with Thionyl chloride ($\ce{SOCl2}$) to replace the hydroxyl group with chlorine, forming the primary halide, 1-Chloropropane.

Step 2: Dehydrohalogenation

To shift the functional group to the middle carbon (necessary for the branched target), 1-chloropropane is heated with alcoholic $\ce{KOH}$. The $\beta$-elimination forms Propene.

Step 3: Markovnikov Addition

Propene reacts with Hydrogen Bromide ($\ce{HBr}$). The electrophilic addition follows Markovnikov's rule, placing the halogen on the secondary carbon to yield 2-Bromopropane.

Step 4: Wurtz Coupling

Two moles of 2-Bromopropane are reacted with Sodium metal in dry ether (Wurtz reaction). The two isopropyl radicals couple together symmetrically at the secondary carbons, yielding the highly branched 2,3-Dimethylbutane.

$$ \text{Step 1: } \ce{CH3CH2CH2OH + SOCl2 -> CH3CH2CH2Cl + SO2 ^ + HCl ^} $$ $$ \text{Step 2: } \ce{CH3CH2CH2Cl + KOH(alc) ->[\Delta] CH3CH=CH2 + KCl + H2O} $$ $$ \text{Step 3: } \ce{CH3CH=CH2 + HBr -> CH3-CH(Br)-CH3} $$ $$ \text{Step 4: } \ce{2 CH3-CH(Br)-CH3 + 2Na ->[Dry Ether] CH3-CH(CH3)-CH(CH3)-CH3 + 2NaBr} $$
14 Convert Benzene to m-Chlorophenol
Step 1: Nitration (Meta-Directing Setup)

Benzene is nitrated using $\ce{HNO3 / H2SO4}$. This installs a strongly deactivating, meta-directing nitro group, yielding Nitrobenzene.

Step 2: Meta-Chlorination

Nitrobenzene is reacted with Chlorine gas and a Lewis acid catalyst ($\ce{FeCl3}$ or anhydrous $\ce{AlCl3}$). The electrophile ($\ce{Cl+}$) is directed to the meta position, yielding m-Chloronitrobenzene.

Step 3: Reduction

The nitro group must now be converted into a hydroxyl group. First, it is reduced to an amine using active metal/acid ($\ce{Sn / HCl}$), forming m-Chloroaniline.

Step 4: Diazotization and Hydrolysis

m-Chloroaniline is diazotized ($\ce{NaNO2 / HCl}$ at 273K) to form the diazonium salt. This solution is then gently warmed with water ($\ce{H2O, \Delta}$). The diazonium group is expelled as $\ce{N2}$ gas and replaced by an $-\ce{OH}$ group, yielding m-Chlorophenol.

$$ \text{Step 1: } \ce{C6H6 ->[HNO3/H2SO4] C6H5NO2} $$ $$ \text{Step 2: } \ce{C6H5NO2 + Cl2 ->[FeCl3] m-Cl-C6H4-NO2 + HCl} $$ $$ \text{Step 3: } \ce{m-Cl-C6H4-NO2 + 6[H] ->[Sn/HCl] m-Cl-C6H4-NH2 + 2H2O} $$ $$ \text{Step 4: } \ce{m-Cl-C6H4-NH2 ->[1. NaNO2/HCl, 273K][2. H2O, \Delta] m-Cl-C6H4-OH + N2 ^} $$
15 Convert Methane to Propanenitrile
Step 1: First Halogenation

Methane (1 carbon) must eventually become a 3-carbon nitrile. It is first activated via free-radical chlorination ($\ce{Cl2 / h\nu}$) to yield Chloromethane.

Step 2: Wurtz Coupling

Chloromethane is reacted with Sodium metal in dry ether. The Wurtz reaction symmetrically couples two methyl radicals, doubling the carbon chain to yield Ethane (2 carbons).

Step 3: Second Halogenation

Ethane is reactivated via another free-radical chlorination ($\ce{Cl2 / h\nu}$). This replaces one hydrogen with chlorine to form the primary halide, Chloroethane.

Step 4: Cyanide Step-Up

Chloroethane is heated with alcoholic $\ce{KCN}$. The $S_N2$ substitution extends the chain by adding the third carbon atom (from the cyanide group), yielding Propanenitrile.

$$ \text{Step 1: } \ce{CH4 + Cl2 ->[h\nu] CH3Cl + HCl} $$ $$ \text{Step 2: } \ce{2 CH3Cl + 2Na ->[Dry Ether] CH3-CH3 + 2NaCl} $$ $$ \text{Step 3: } \ce{CH3-CH3 + Cl2 ->[h\nu] CH3CH2Cl + HCl} $$ $$ \text{Step 4: } \ce{CH3CH2Cl + KCN(alc) -> CH3CH2CN + KCl} $$
16 Convert Nitrobenzene to 1,3,5-Tribromobenzene
Step 1: Reduction (Activating the Ring)

The target has three meta-arranged bromines relative to each other, which implies they must have been directed by a powerfully activating ortho/para director that was later removed. Nitrobenzene is reduced using $\ce{Sn / HCl}$ to yield Aniline.

Step 2: Exhaustive Bromination

Aniline is treated with aqueous Bromine (Bromine water). The highly activated ring undergoes rapid substitution at all free ortho and para positions, precipitating 2,4,6-Tribromoaniline.

Step 3: Diazotization

The amino group has served its directing purpose and must be removed. The tribromoaniline is diazotized using cold $\ce{NaNO2 / HCl}$ at 273K, yielding 2,4,6-Tribromobenzene diazonium chloride.

Step 4: Deamination

The diazonium group is entirely removed and replaced by a hydrogen atom. This is achieved by reacting the salt with a mild reducing agent, Hypophosphorous acid ($\ce{H3PO2}$) and water. This yields the final product, 1,3,5-Tribromobenzene.

$$ \text{Step 1: } \ce{C6H5NO2 + 6[H] ->[Sn/HCl] C6H5NH2 + 2H2O} $$ $$ \text{Step 2: } \ce{C6H5NH2 + 3Br2(aq) -> 2,4,6-C6H2Br3(NH2) v + 3HBr} $$ $$ \text{Step 3: } \ce{2,4,6-C6H2Br3(NH2) ->[NaNO2/HCl][273 K] 2,4,6-C6H2Br3(N2+Cl-)} $$ $$ \text{Step 4: } \ce{2,4,6-C6H2Br3(N2+Cl-) + H3PO2 + H2O -> 1,3,5-C6H3Br3 + H3PO3 + HCl + N2 ^} $$
17 Convert Ethene to Crotonic Acid (But-2-enoic acid)
Step 1: Hydration

Ethene (2 carbons) must be dimerized via Aldol condensation to reach 4 carbons. First, it is hydrated ($\ce{H2O / H+}$) to form Ethanol.

Step 2: Mild Oxidation

Ethanol is mildly oxidized using Pyridinium Chlorochromate ($\ce{PCC}$) to form the requisite aldehyde containing $\alpha$-hydrogens: Ethanal.

Step 3: Aldol Condensation & Dehydration

Ethanal is treated with dilute $\ce{NaOH}$ to undergo Aldol addition, forming 3-hydroxybutanal. Upon heating ($\Delta$), it dehydrates to form the $\alpha,\beta$-unsaturated aldehyde, But-2-enal (Crotonaldehyde).

Step 4: Selective Oxidation

The aldehyde group must be oxidized to a carboxylic acid without disturbing the carbon-carbon double bond. A mild oxidizing agent like Tollens' reagent ($\ce{[Ag(NH3)2]+}$) is used, successfully yielding But-2-enoic acid (Crotonic acid).

$$ \text{Step 1: } \ce{CH2=CH2 + H2O ->[H+] CH3CH2OH} $$ $$ \text{Step 2: } \ce{CH3CH2OH ->[PCC] CH3CHO} $$ $$ \text{Step 3: } \ce{2 CH3CHO ->[1. dil. NaOH][2. \Delta] CH3-CH=CH-CHO + H2O} $$ $$ \text{Step 4: } \ce{CH3-CH=CH-CHO ->[Tollens' Reagent] CH3-CH=CH-COOH} $$
18 Convert Benzene to p-Aminobenzoic Acid (PABA)
Step 1: Alkylation

Benzene is reacted with $\ce{CH3Cl / AlCl3}$ (Friedel-Crafts). The methyl group provides the carbon for the future carboxyl group. This yields Toluene.

Step 2: Selective Nitration

Toluene is nitrated ($\ce{HNO3 / H2SO4}$). The methyl group directs ortho/para. Due to sterics, the para isomer dominates. We isolate p-Nitrotoluene.

Step 3: Side-Chain Oxidation

The methyl side-chain is vigorously oxidized using acidified Potassium Permanganate ($\ce{KMnO4 / H+}$, heat). This converts the $-\ce{CH3}$ completely into a $-\ce{COOH}$ group while leaving the nitro group unaffected, yielding p-Nitrobenzoic acid.

Step 4: Reduction

Finally, the nitro group is reduced to an amine. Using Tin and Hydrochloric acid ($\ce{Sn / HCl}$) selectively reduces the $-\ce{NO2}$ to $-\ce{NH2}$ without affecting the carboxylic acid, producing p-Aminobenzoic acid.

$$ \text{Step 1: } \ce{C6H6 + CH3Cl ->[AlCl3] C6H5CH3 + HCl} $$ $$ \text{Step 2: } \ce{C6H5CH3 ->[HNO3/H2SO4] p-CH3-C6H4-NO2} $$ $$ \text{Step 3: } \ce{p-CH3-C6H4-NO2 + 3[O] ->[KMnO4/H+] p-NO2-C6H4-COOH + H2O} $$ $$ \text{Step 4: } \ce{p-NO2-C6H4-COOH + 6[H] ->[Sn/HCl] p-NH2-C6H4-COOH + 2H2O} $$
19 Convert Ethyne to 2-Bromobutane
Step 1: Base De-protonation

Ethyne (Acetylene) is weakly acidic. Bubbling it through a strong base like Sodium amide ($\ce{NaNH2}$) in liquid ammonia extracts a proton, forming the highly nucleophilic Sodium acetylide.

Step 2: Alkylation (Chain Extension)

Sodium acetylide is reacted with Bromoethane ($\ce{CH3CH2Br}$). The $S_N2$ substitution extends the carbon chain from 2 to 4 atoms, yielding the terminal alkyne But-1-yne.

Step 3: Partial Hydrogenation

Complete hydrogenation would yield butane. To stop at the alkene, we use a poisoned catalyst: Hydrogen gas over Lindlar's catalyst ($\ce{H2 / Pd-BaSO4}$, quinoline). This specifically reduces the alkyne to an alkene, yielding But-1-ene.

Step 4: Markovnikov Addition

But-1-ene is reacted with Hydrogen Bromide ($\ce{HBr}$). The electrophilic addition strictly follows Markovnikov's rule, placing the bromine on the secondary carbon to form the final product, 2-Bromobutane.

$$ \text{Step 1: } \ce{HC\equiv CH + NaNH2 -> HC\equiv CNa + NH3} $$ $$ \text{Step 2: } \ce{HC\equiv CNa + CH3CH2Br -> HC\equiv C-CH2CH3 + NaBr} $$ $$ \text{Step 3: } \ce{HC\equiv C-CH2CH3 + H2 ->[Lindlar's Cat.] CH2=CH-CH2CH3} $$ $$ \text{Step 4: } \ce{CH2=CH-CH2CH3 + HBr -> CH3-CH(Br)-CH2CH3} $$
20 Convert Chloroethane to Butane-1,3-diol
Step 1: Nucleophilic Substitution

The 2-carbon halide must be converted to an oxygenated species for condensation. Chloroethane is boiled with aqueous $\ce{KOH}$. This $S_N2$ reaction replaces the chloride with a hydroxyl group, yielding Ethanol.

Step 2: Mild Oxidation

Ethanol is subjected to mild oxidation using Pyridinium Chlorochromate ($\ce{PCC}$) in dichloromethane. This halts the oxidation at the aldehyde stage, providing Ethanal, which possesses the critical $\alpha$-hydrogens.

Step 3: Aldol Addition

Ethanal is treated with cold, dilute $\ce{NaOH}$ (10%). The base creates an enolate which attacks another molecule of ethanal. The dimerized product is the $\beta$-hydroxyaldehyde, 3-Hydroxybutanal (Aldol).

Step 4: Reduction

Without heating (to prevent dehydration), the aldol is directly reduced. Sodium Borohydride ($\ce{NaBH4}$) is a mild reducing agent that exclusively reduces the terminal aldehyde group ($-\ce{CHO}$) to a primary alcohol ($-\ce{CH2OH}$) without affecting the existing secondary alcohol. This yields the final diol, Butane-1,3-diol.

$$ \text{Step 1: } \ce{CH3CH2Cl + KOH(aq) ->[\Delta] CH3CH2OH + KCl} $$ $$ \text{Step 2: } \ce{CH3CH2OH ->[PCC] CH3CHO} $$ $$ \text{Step 3: } \ce{2 CH3CHO ->[dil. NaOH] CH3-CH(OH)-CH2-CHO} $$ $$ \text{Step 4: } \ce{CH3-CH(OH)-CH2-CHO + 2[H] ->[NaBH4] CH3-CH(OH)-CH2-CH2OH} $$

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