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20 Grandmaster 5-Step Organic Conversions

20 Grandmaster 5-Step Organic Conversions | Chemca

Chemca

Grandmaster Level Synthesis

Master 5-Step
Organic Conversions

The pinnacle of synthetic logic and competitive exam preparation. This collection features 20 intricate five-step organic conversions. Learn to sequence step-ups, rigorous protections, meta-directing flips, and functional group migrations. Click to unveil the complete architectural pathways.

1 Convert Methanol to Lactic Acid (2-Hydroxypropanoic acid)
Step 1: Halogenation

Methanol is converted to a reactive alkyl halide using Thionyl chloride ($\ce{SOCl2}$), yielding Chloromethane.

Step 2: Cyanide Step-Up

Chloromethane is heated with alcoholic $\ce{KCN}$. The nucleophilic attack extends the chain to 2 carbons, forming Acetonitrile (Ethanenitrile).

Step 3: Controlled Reduction (Stephen's/DIBAL-H)

The nitrile is partially reduced to an aldehyde. Using Diisobutylaluminum hydride ($\ce{DIBAL-H}$) followed by hydrolysis yields Ethanal.

Step 4: Cyanohydrin Formation (2nd Step-Up)

Ethanal undergoes nucleophilic addition with Hydrogen Cyanide ($\ce{HCN}$), extending the chain to 3 carbons and forming Ethanal cyanohydrin.

Step 5: Acid Hydrolysis

The nitrile group is completely hydrolyzed by boiling with dilute mineral acid ($\ce{H3O+}$), yielding 2-Hydroxypropanoic acid (Lactic Acid).

$$ \text{Step 1: } \ce{CH3OH + SOCl2 -> CH3Cl + SO2 ^ + HCl ^} $$ $$ \text{Step 2: } \ce{CH3Cl + KCN (alc) -> CH3CN + KCl} $$ $$ \text{Step 3: } \ce{CH3CN ->[1. DIBAL-H][2. H2O] CH3CHO} $$ $$ \text{Step 4: } \ce{CH3CHO + HCN -> CH3-CH(OH)-CN} $$ $$ \text{Step 5: } \ce{CH3-CH(OH)-CN + 2H2O ->[H+] CH3-CH(OH)-COOH + NH4+} $$
2 Convert Ethene to 1-Bromobutane
Step 1: Hydration

Ethene (2 carbons) undergoes acid-catalyzed hydration ($\ce{H2O / H+}$) to form Ethanol.

Step 2: Mild Oxidation

Ethanol is oxidized with Pyridinium Chlorochromate ($\ce{PCC}$) to halt at the aldehyde stage, yielding Ethanal.

Step 3: Aldol Condensation

Ethanal is treated with dilute $\ce{NaOH}$ (Aldol addition) then heated ($\Delta$). The dimerization and dehydration form the 4-carbon unsaturated aldehyde, But-2-enal.

Step 4: Catalytic Hydrogenation

But-2-enal is completely reduced using Hydrogen gas over a Nickel catalyst ($\ce{H2 / Ni}$). Both the double bond and the aldehyde are reduced, yielding Butan-1-ol.

Step 5: Halogenation

The primary alcohol is reacted with Phosphorus Tribromide ($\ce{PBr3}$) or $\ce{HBr}$ to substitute the hydroxyl group, yielding 1-Bromobutane.

$$ \text{Step 1: } \ce{CH2=CH2 + H2O ->[H+] CH3CH2OH} $$ $$ \text{Step 2: } \ce{CH3CH2OH ->[PCC] CH3CHO} $$ $$ \text{Step 3: } \ce{2 CH3CHO ->[1. dil NaOH][2. \Delta] CH3-CH=CH-CHO + H2O} $$ $$ \text{Step 4: } \ce{CH3-CH=CH-CHO + 2H2 ->[Ni] CH3CH2CH2CH2OH} $$ $$ \text{Step 5: } \ce{3 CH3CH2CH2CH2OH + PBr3 -> 3 CH3CH2CH2CH2Br + H3PO3} $$
3 Convert Benzene to m-Fluoronitrobenzene
Step 1: First Nitration

Benzene is nitrated using a mixture of concentrated $\ce{HNO3}$ and $\ce{H2SO4}$ at 330 K to yield Nitrobenzene, installing a meta-directing group.

Step 2: Second Nitration

Nitrobenzene requires vigorous conditions (fuming $\ce{HNO3}$, conc. $\ce{H2SO4}$, high heat) to nitrate again. The electrophile attacks the meta position, yielding m-Dinitrobenzene.

Step 3: Selective Reduction

To differentiate the groups, one nitro group is selectively reduced using Ammonium hydrogen sulfide ($\ce{NH4HS}$) or $\ce{Na2S}$. This yields m-Nitroaniline.

Step 4: Diazotization

m-Nitroaniline is diazotized with cold $\ce{NaNO2 / HCl}$ (273 K) to convert the amino group into the excellent diazonium leaving group, forming m-Nitrobenzene diazonium chloride.

Step 5: Balz-Schiemann Reaction

The diazonium salt is treated with Fluoroboric acid ($\ce{HBF4}$) to precipitate the fluoroborate salt. Heating this dry salt decomposes it, yielding m-Fluoronitrobenzene.

$$ \text{Step 1: } \ce{C6H6 ->[HNO3/H2SO4] C6H5NO2} $$ $$ \text{Step 2: } \ce{C6H5NO2 ->[fuming HNO3/H2SO4][\Delta] m-C6H4(NO2)2} $$ $$ \text{Step 3: } \ce{m-C6H4(NO2)2 + 3H2S ->[NH3] m-NO2-C6H4-NH2 + 3S + 2H2O} $$ $$ \text{Step 4: } \ce{m-NO2-C6H4-NH2 ->[NaNO2/HCl, 273K] m-NO2-C6H4-N2+Cl-} $$ $$ \text{Step 5: } \ce{m-NO2-C6H4-N2+Cl- ->[1. HBF4][2. \Delta] m-NO2-C6H4-F + N2 ^ + BF3} $$
4 Convert 1-Bromopropane to 2-Methylpropan-1-ol
Step 1: Dehydrohalogenation

To shift functionalization to the middle carbon, 1-Bromopropane is heated with alcoholic $\ce{KOH}$. The $\beta$-elimination forms Propene.

Step 2: Markovnikov Halogenation

Propene is reacted with $\ce{HBr}$. The electrophilic addition places the bromine on the secondary carbon, yielding 2-Bromopropane.

Step 3: Cyanide Step-Up

2-Bromopropane is heated with alcoholic $\ce{KCN}$. The nucleophilic substitution creates a branch and extends the chain, yielding 2-Methylpropanenitrile (Isopropyl cyanide).

Step 4: Complete Hydrolysis

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

Step 5: Strong Reduction

The carboxylic acid is strongly reduced using Lithium Aluminum Hydride ($\ce{LiAlH4}$), converting the carboxyl group to a primary alcohol, yielding 2-Methylpropan-1-ol.

$$ \text{Step 1: } \ce{CH3CH2CH2Br + KOH(alc) ->[\Delta] CH3CH=CH2 + KBr + H2O} $$ $$ \text{Step 2: } \ce{CH3CH=CH2 + HBr -> CH3-CH(Br)-CH3} $$ $$ \text{Step 3: } \ce{CH3-CH(Br)-CH3 + KCN(alc) -> CH3-CH(CN)-CH3 + KBr} $$ $$ \text{Step 4: } \ce{CH3-CH(CN)-CH3 + 2H2O ->[H+] CH3-CH(COOH)-CH3 + NH4+} $$ $$ \text{Step 5: } \ce{CH3-CH(COOH)-CH3 ->[1. LiAlH4][2. H3O+] CH3-CH(CH2OH)-CH3} $$
5 Convert Benzene to p-Iodotoluene
Step 1: Friedel-Crafts Alkylation

Benzene is reacted with Methyl chloride and $\ce{AlCl3}$. This places a methyl group on the ring, yielding Toluene, an ortho/para-directing activator.

Step 2: Selective Nitration

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

Step 3: Reduction

The nitro group is reduced to a primary amine using Tin and Hydrochloric acid ($\ce{Sn / HCl}$). This yields p-Toluidine (p-Methylaniline).

Step 4: Diazotization

p-Toluidine is diazotized with cold $\ce{NaNO2 / HCl}$ (273K). This converts the amino group into a diazonium group, forming p-Toluenediazonium chloride.

Step 5: Iodination

The diazonium salt is simply warmed with an aqueous solution of Potassium Iodide ($\ce{KI}$). The iodide ion nucleophilically replaces the diazonium group, yielding p-Iodotoluene.

$$ \text{Step 1: } \ce{C6H6 + CH3Cl ->[AlCl3] C6H5CH3 + HCl} $$ $$ \text{Step 2: } \ce{C6H5CH3 + HNO3 ->[H2SO4] p-CH3-C6H4-NO2} \text{ (major)} $$ $$ \text{Step 3: } \ce{p-CH3-C6H4-NO2 + 6[H] ->[Sn/HCl] p-CH3-C6H4-NH2 + 2H2O} $$ $$ \text{Step 4: } \ce{p-CH3-C6H4-NH2 ->[NaNO2/HCl, 273K] p-CH3-C6H4-N2+Cl-} $$ $$ \text{Step 5: } \ce{p-CH3-C6H4-N2+Cl- + KI ->[\Delta] p-CH3-C6H4-I + N2 ^ + KCl} $$
6 Convert Propan-1-ol to 2-Methylbutan-2-ol
Step 1: Dehydration

Propan-1-ol is heated with concentrated $\ce{H2SO4}$ at 443 K. The elimination of water forms Propene.

Step 2: Markovnikov Hydration

Propene is passed through water in the presence of dilute acid ($\ce{H2O / H+}$). Electrophilic addition yields the secondary alcohol, Propan-2-ol.

Step 3: Oxidation to Ketone

Propan-2-ol is oxidized using acidified $\ce{K2Cr2O7}$ or $\ce{PCC}$. The secondary alcohol cleanly oxidizes to a ketone, yielding Propanone (Acetone).

Step 4: Grignard Addition (2-Carbon Step-Up)

Propanone is reacted with Ethylmagnesium bromide ($\ce{CH3CH2MgBr}$) in dry ether. The nucleophilic ethyl group attacks the carbonyl carbon, forming a tertiary magnesium alkoxide intermediate.

Step 5: Acid Hydrolysis

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

$$ \text{Step 1: } \ce{CH3CH2CH2OH ->[conc. H2SO4][443 K] CH3CH=CH2 + H2O} $$ $$ \text{Step 2: } \ce{CH3CH=CH2 + H2O ->[H+] CH3-CH(OH)-CH3} $$ $$ \text{Step 3: } \ce{CH3-CH(OH)-CH3 ->[K2Cr2O7/H+] CH3-CO-CH3 + H2O} $$ $$ \text{Step 4: } \ce{CH3-CO-CH3 + CH3CH2MgBr -> CH3-C(OMgBr)(CH2CH3)-CH3} $$ $$ \text{Step 5: } \ce{CH3-C(OMgBr)(CH2CH3)-CH3 + H3O+ -> CH3-C(OH)(CH2CH3)-CH3} $$
7 Convert Ethene to But-2-yne
Step 1: Halogen Addition

Ethene reacts with Bromine in a non-polar solvent ($\ce{Br2/CCl4}$). The anti-addition across the double bond yields the vicinal dihalide, 1,2-Dibromoethane.

Step 2: Double Dehydrohalogenation

The vicinal dihalide is heated with strong base Sodium amide ($\ce{NaNH2}$) in liquid ammonia. Two successive elimination reactions remove two moles of $\ce{HBr}$, yielding Ethyne (Acetylene).

Step 3: Acetylide Formation

Ethyne is treated with another mole of Sodium amide ($\ce{NaNH2}$). The strongly basic amide ion extracts one weakly acidic terminal proton, forming Sodium acetylide ($\ce{HC\equiv CNa}$).

Step 4: First Alkylation

Sodium acetylide is reacted with Methyl iodide ($\ce{CH3I}$). The $S_N2$ substitution extends the chain by one carbon, yielding Propyne.

Step 5: Second Alkylation

Propyne still has one acidic terminal proton. It is reacted with $\ce{NaNH2}$ to form sodium propynide, which is immediately reacted with another mole of Methyl iodide ($\ce{CH3I}$), yielding the internal alkyne, But-2-yne.

$$ \text{Step 1: } \ce{CH2=CH2 + Br2 ->[CCl4] CH2(Br)-CH2(Br)} $$ $$ \text{Step 2: } \ce{CH2(Br)-CH2(Br) + 2NaNH2 -> HC\equiv CH + 2NaBr + 2NH3} $$ $$ \text{Step 3: } \ce{HC\equiv CH + NaNH2 -> HC\equiv CNa + NH3} $$ $$ \text{Step 4: } \ce{HC\equiv CNa + CH3I -> HC\equiv C-CH3 + NaI} $$ $$ \text{Step 5: } \ce{HC\equiv C-CH3 ->[1. NaNH2][2. CH3I] CH3-C\equiv C-CH3 + NaI} $$
8 Convert Phenol to p-Hydroxyazobenzene (Orange Dye)
Step 1: Deoxygenation

Phenol is heated with Zinc dust. The zinc acts as a reducing agent, extracting the oxygen to form Zinc oxide, leaving pure Benzene.

Step 2: Nitration

Benzene is nitrated using a mixture of concentrated $\ce{HNO3}$ and $\ce{H2SO4}$ at 330 K. The electrophilic attack forms Nitrobenzene.

Step 3: Reduction

Nitrobenzene is reduced to a primary amine using active metal and acid, specifically Tin and Hydrochloric acid ($\ce{Sn / HCl}$). This yields Aniline.

Step 4: Diazotization

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

Step 5: Coupling Reaction

The diazonium salt is reacted with another molecule of Phenol in a mildly alkaline medium (pH 9-10). The diazonium ion attacks the para-position of the activated phenoxide ring, forming the extended conjugated azo system, p-Hydroxyazobenzene.

$$ \text{Step 1: } \ce{C6H5OH + Zn (dust) ->[\Delta] C6H6 + ZnO} $$ $$ \text{Step 2: } \ce{C6H6 + HNO3 ->[H2SO4] C6H5NO2 + H2O} $$ $$ \text{Step 3: } \ce{C6H5NO2 + 6[H] ->[Sn/HCl] C6H5NH2 + 2H2O} $$ $$ \text{Step 4: } \ce{C6H5NH2 + NaNO2 + 2HCl ->[273 K] C6H5N2+Cl- + NaCl + 2H2O} $$ $$ \text{Step 5: } \ce{C6H5N2+Cl- + C6H5OH ->[OH-] C6H5-N=N-C6H4OH + Cl- + H2O} $$
9 Convert Toluene to 1-Phenylethanol
Step 1: Side-Chain Halogenation

Toluene is reacted with Chlorine gas under UV light ($\ce{Cl2/h\nu}$). This directs the free-radical substitution exclusively to the methyl group, producing Benzyl chloride.

Step 2: Hydrolysis

Benzyl chloride is boiled with aqueous Potassium Hydroxide ($\ce{KOH(aq)}$). The chloride is displaced by a hydroxyl group via $S_N2$ mechanism, yielding Benzyl alcohol.

Step 3: Mild Oxidation

Benzyl alcohol is oxidized using Pyridinium Chlorochromate ($\ce{PCC}$) in dichloromethane. This stops the oxidation exactly at the aldehyde stage, yielding Benzaldehyde.

Step 4: Grignard Addition

Benzaldehyde is reacted with Methylmagnesium bromide ($\ce{CH3MgBr}$) in dry ether. The nucleophilic methyl carbanion attacks the carbonyl carbon, forming a magnesium alkoxide complex.

Step 5: Acid Hydrolysis

The alkoxide intermediate is immediately hydrolyzed with dilute acid ($\ce{H3O+}$). Protonation of the oxygen yields the secondary alcohol, 1-Phenylethanol.

$$ \text{Step 1: } \ce{C6H5CH3 + Cl2 ->[h\nu] C6H5CH2Cl + HCl} $$ $$ \text{Step 2: } \ce{C6H5CH2Cl + KOH(aq) ->[\Delta] C6H5CH2OH + KCl} $$ $$ \text{Step 3: } \ce{C6H5CH2OH ->[PCC] C6H5CHO} $$ $$ \text{Step 4: } \ce{C6H5CHO + CH3MgBr -> C6H5-CH(OMgBr)-CH3} $$ $$ \text{Step 5: } \ce{C6H5-CH(OMgBr)-CH3 + H3O+ -> C6H5-CH(OH)-CH3 + Mg(OH)Br} $$
10 Convert Calcium Carbide to Butan-1-ol
Step 1: Hydrolysis

Calcium carbide ($\ce{CaC2}$) reacts with water at room temperature, releasing Ethyne (Acetylene) gas.

Step 2: Kucherov Hydration

Ethyne is hydrated in the presence of Mercuric sulfate and dilute Sulfuric acid ($\ce{HgSO4 / H2SO4}$). The addition of water forms vinyl alcohol, which tautomerizes instantly into Ethanal.

Step 3: Aldol Condensation (Addition)

Ethanal is treated with dilute $\ce{NaOH}$. Two molecules condense (Aldol addition) to form the $\beta$-hydroxyaldehyde, 3-Hydroxybutanal.

Step 4: Dehydration

Heating 3-Hydroxybutanal causes E1cB elimination of water. This creates a highly stable, conjugated $\alpha,\beta$-unsaturated aldehyde, But-2-enal.

Step 5: Complete Catalytic Reduction

But-2-enal is completely reduced using Hydrogen gas over a Nickel or Platinum catalyst ($\ce{H2 / Ni}$). Both the double bond and the carbonyl group are hydrogenated, yielding the primary alcohol, Butan-1-ol.

$$ \text{Step 1: } \ce{CaC2 + 2H2O -> HC\equiv CH + Ca(OH)2} $$ $$ \text{Step 2: } \ce{HC\equiv CH + H2O ->[Hg^2+ / H+] CH3CHO} $$ $$ \text{Step 3: } \ce{2 CH3CHO ->[dil. NaOH] CH3-CH(OH)-CH2-CHO} $$ $$ \text{Step 4: } \ce{CH3-CH(OH)-CH2-CHO ->[\Delta] CH3-CH=CH-CHO + H2O} $$ $$ \text{Step 5: } \ce{CH3-CH=CH-CHO + 2H2 ->[Ni] CH3CH2CH2CH2OH} $$
11 Convert Benzene to m-Bromobenzyl Alcohol
Step 1: Friedel-Crafts Alkylation

Benzene is reacted with Methyl chloride and anhydrous $\ce{AlCl3}$. This places a methyl group on the ring, yielding Toluene.

Step 2: Vigorous Side-Chain Oxidation

Toluene is oxidized with acidified Potassium Permanganate ($\ce{KMnO4 / H+}$, heat). The methyl group is fully oxidized to a carboxyl group, yielding Benzoic acid. The carboxyl group is strongly meta-directing.

Step 3: Meta-Bromination

Benzoic acid is reacted with Bromine and Iron(III) bromide catalyst ($\ce{Br2 / FeBr3}$). The electrophile is directed to the meta position, yielding m-Bromobenzoic acid.

Step 4: Acid Chloride Formation

Direct reduction of a carboxylic acid is difficult and often incompatible with other halogens. The acid is first converted to a highly reactive acid chloride using Thionyl chloride ($\ce{SOCl2}$), yielding m-Bromobenzoyl chloride.

Step 5: Hydride Reduction

The acid chloride is smoothly reduced using Sodium Borohydride ($\ce{NaBH4}$) or mild Lithium Aluminum Hydride ($\ce{LiAlH4}$). This reduces the carbonyl group down to a primary alcohol, yielding m-Bromobenzyl alcohol.

$$ \text{Step 1: } \ce{C6H6 + CH3Cl ->[AlCl3] C6H5CH3 + HCl} $$ $$ \text{Step 2: } \ce{C6H5CH3 + 3[O] ->[KMnO4/H+] C6H5COOH + H2O} $$ $$ \text{Step 3: } \ce{C6H5COOH + Br2 ->[FeBr3] m-Br-C6H4-COOH + HBr} $$ $$ \text{Step 4: } \ce{m-Br-C6H4-COOH + SOCl2 -> m-Br-C6H4-COCl + SO2 ^ + HCl ^} $$ $$ \text{Step 5: } \ce{m-Br-C6H4-COCl ->[1. NaBH4][2. H3O+] m-Br-C6H4-CH2OH} $$
12 Convert Benzene to p-Nitroaniline
Step 1: Nitration

Benzene is treated with a nitrating mixture ($\ce{HNO3 / H2SO4}$) at 330K to yield Nitrobenzene.

Step 2: Reduction

Nitrobenzene is fully reduced using Tin and Hydrochloric acid ($\ce{Sn / HCl}$) to form Aniline, which acts as a strong ortho/para director for the second substituent.

Step 3: Amine Protection (Acetylation)

Aniline is highly susceptible to oxidation during subsequent nitration. It is protected by reacting it with Acetic anhydride ($\ce{Ac2O}$) in pyridine, forming Acetanilide. This also mitigates its activating power.

Step 4: Second Nitration

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

Step 5: Deprotection (Hydrolysis)

The acetyl protecting group is removed by boiling with dilute aqueous acid or base ($\ce{H3O+}$ or $\ce{OH-}$, $\Delta$). This restores the primary amine group, yielding pure p-Nitroaniline.

$$ \text{Step 1: } \ce{C6H6 ->[HNO3/H2SO4] C6H5NO2} $$ $$ \text{Step 2: } \ce{C6H5NO2 + 6[H] ->[Sn/HCl] C6H5NH2 + 2H2O} $$ $$ \text{Step 3: } \ce{C6H5NH2 + (CH3CO)2O ->[Pyridine] C6H5NHCOCH3 + CH3COOH} $$ $$ \text{Step 4: } \ce{C6H5NHCOCH3 + HNO3 ->[H2SO4] p-NO2-C6H4-NHCOCH3 + H2O} $$ $$ \text{Step 5: } \ce{p-NO2-C6H4-NHCOCH3 + H2O ->[H+] p-NO2-C6H4-NH2 + CH3COOH} $$
13 Convert Benzene to 1,3,5-Tribromobenzene
Step 1: Nitration

Benzene is treated with a nitrating mixture ($\ce{HNO3 / H2SO4}$) at 330 K, yielding Nitrobenzene.

Step 2: Reduction

Nitrobenzene is reduced to an amine using active metal and acid ($\ce{Sn / HCl}$). The product is Aniline, which highly activates the aromatic ring.

Step 3: Exhaustive Bromination

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

Step 4: Diazotization

The amino group must now be removed to leave only the bromines in a meta relationship. The tribromoaniline is diazotized using cold $\ce{NaNO2 / HCl}$ at 273 K, yielding 2,4,6-Tribromobenzene diazonium chloride.

Step 5: Deamination

The diazonium salt is reacted with a mild reducing agent, Hypophosphorous acid ($\ce{H3PO2}$) and water. This completely removes the diazonium group, replacing it with a hydrogen atom, yielding 1,3,5-Tribromobenzene.

$$ \text{Step 1: } \ce{C6H6 ->[HNO3/H2SO4] C6H5NO2} $$ $$ \text{Step 2: } \ce{C6H5NO2 + 6[H] ->[Sn/HCl] C6H5NH2 + 2H2O} $$ $$ \text{Step 3: } \ce{C6H5NH2 + 3Br2(aq) -> 2,4,6-C6H2Br3(NH2) v + 3HBr} $$ $$ \text{Step 4: } \ce{2,4,6-C6H2Br3(NH2) ->[NaNO2/HCl, 273 K] 2,4,6-C6H2Br3(N2+Cl-)} $$ $$ \text{Step 5: } \ce{2,4,6-C6H2Br3(N2+Cl-) + H3PO2 + H2O -> 1,3,5-C6H3Br3 + H3PO3 + HCl + N2 ^} $$
14 Convert Methanol to Butan-2-ol
Step 1: Halogenation

Methanol is converted to Chloromethane by reacting with Phosphorus Pentachloride ($\ce{PCl5}$) or Thionyl chloride ($\ce{SOCl2}$).

Step 2: Cyanide Step-Up

Chloromethane is heated with alcoholic $\ce{KCN}$. The $S_N2$ substitution extends the chain to 2 carbons, forming Acetonitrile (Ethanenitrile).

Step 3: Grignard Addition (2-Carbon Step-Up)

Acetonitrile is reacted with Ethylmagnesium bromide ($\ce{CH3CH2MgBr}$). The nucleophilic ethyl group attacks the nitrile carbon, breaking the triple bond to form an unstable Imine magnesium salt.

Step 4: Acid Hydrolysis to Ketone

The imine salt is boiled with dilute mineral acid ($\ce{H3O+}$). The nitrogen is expelled as ammonia, yielding the 4-carbon ketone, Butan-2-one.

Step 5: Hydride Reduction

Butan-2-one is reduced using Sodium Borohydride ($\ce{NaBH4}$). The hydride attacks the carbonyl, forming a secondary alcohol: Butan-2-ol.

$$ \text{Step 1: } \ce{CH3OH + PCl5 -> CH3Cl + POCl3 + HCl} $$ $$ \text{Step 2: } \ce{CH3Cl + KCN (alc) -> CH3CN + KCl} $$ $$ \text{Step 3: } \ce{CH3C\equiv N + CH3CH2MgBr -> CH3-C(CH2CH3)=NMgBr} $$ $$ \text{Step 4: } \ce{CH3-C(CH2CH3)=NMgBr + 2H2O ->[H+] CH3-CO-CH2CH3 + NH3 + Mg(OH)Br} $$ $$ \text{Step 5: } \ce{CH3-CO-CH2CH3 + 2[H] ->[NaBH4] CH3-CH(OH)-CH2CH3} $$
15 Convert Phenol to p-Methoxybenzyl Alcohol
Step 1: Acid-Base Salt Formation

Phenol is reacted with aqueous Sodium Hydroxide ($\ce{NaOH}$) to form Sodium phenoxide, setting up the molecule for Williamson ether synthesis.

Step 2: Williamson Ether Synthesis

Sodium phenoxide is reacted with Methyl iodide ($\ce{CH3I}$). The phenoxide nucleophile displaces the iodide, yielding Methoxybenzene (Anisole). This protects the oxygen and makes the ring highly ortho/para directing.

Step 3: Friedel-Crafts Alkylation

Anisole is reacted with Methyl chloride ($\ce{CH3Cl}$) and anhydrous $\ce{AlCl3}$. The methyl electrophile attacks the less sterically hindered para position, yielding p-Methoxytoluene.

Step 4: Side-Chain Halogenation

p-Methoxytoluene is treated with Chlorine gas under UV light ($\ce{Cl2 / h\nu}$). Substitution occurs selectively at the benzylic position (free radical mechanism), yielding p-Methoxybenzyl chloride.

Step 5: Hydrolysis

The benzylic chloride is boiled with aqueous Potassium Hydroxide ($\ce{KOH(aq)}$). $S_N2$ substitution replaces the chloride with a hydroxyl group, yielding p-Methoxybenzyl alcohol.

$$ \text{Step 1: } \ce{C6H5OH + NaOH -> C6H5ONa + H2O} $$ $$ \text{Step 2: } \ce{C6H5ONa + CH3I -> C6H5OCH3 + NaI} $$ $$ \text{Step 3: } \ce{C6H5OCH3 + CH3Cl ->[AlCl3] p-CH3O-C6H4-CH3 + HCl} $$ $$ \text{Step 4: } \ce{p-CH3O-C6H4-CH3 + Cl2 ->[h\nu] p-CH3O-C6H4-CH2Cl + HCl} $$ $$ \text{Step 5: } \ce{p-CH3O-C6H4-CH2Cl + KOH(aq) ->[\Delta] p-CH3O-C6H4-CH2OH + KCl} $$
16 Convert Toluene to m-Aminobenzyl Alcohol
Step 1: Side-Chain Oxidation (Meta Setup)

To direct an incoming group to the meta position, the methyl group must be oxidized. Toluene is vigorously oxidized with acidified Potassium Permanganate ($\ce{KMnO4 / H+}$, heat) to yield Benzoic acid.

Step 2: Meta-Nitration

Benzoic acid is nitrated ($\ce{HNO3 / H2SO4}$). The carboxyl group strongly directs the electrophile to the meta position, yielding m-Nitrobenzoic acid.

Step 3: Acid Chloride Formation

The carboxylic acid is converted to an acid chloride using Thionyl chloride ($\ce{SOCl2}$). This activates the carbonyl group for easy reduction, yielding m-Nitrobenzoyl chloride.

Step 4: Hydride Reduction

The acid chloride is reduced with Sodium Borohydride ($\ce{NaBH4}$). This mild reducing agent converts the acid chloride to a primary alcohol while leaving the nitro group completely unaffected, yielding m-Nitrobenzyl alcohol.

Step 5: Nitro Group Reduction

The nitro group is finally reduced to an amine using Tin and Hydrochloric acid ($\ce{Sn / HCl}$). This yields the target product, m-Aminobenzyl alcohol.

$$ \text{Step 1: } \ce{C6H5CH3 + 3[O] ->[KMnO4/H+] C6H5COOH + H2O} $$ $$ \text{Step 2: } \ce{C6H5COOH + HNO3 ->[H2SO4] m-NO2-C6H4-COOH + H2O} $$ $$ \text{Step 3: } \ce{m-NO2-C6H4-COOH + SOCl2 -> m-NO2-C6H4-COCl + SO2 ^ + HCl ^} $$ $$ \text{Step 4: } \ce{m-NO2-C6H4-COCl ->[1. NaBH4][2. H3O+] m-NO2-C6H4-CH2OH} $$ $$ \text{Step 5: } \ce{m-NO2-C6H4-CH2OH + 6[H] ->[Sn/HCl] m-NH2-C6H4-CH2OH + 2H2O} $$
17 Convert Propene to Propan-2-amine
Step 1: Markovnikov Halogenation

Propene is reacted with $\ce{HBr}$. The electrophilic addition places the bromine on the secondary carbon, yielding 2-Bromopropane.

Step 2: Cyanide Branching

2-Bromopropane is heated with alcoholic $\ce{KCN}$. The $S_N2$ substitution replaces the bromine, yielding 2-Methylpropanenitrile (Isopropyl cyanide).

Step 3: Complete Hydrolysis

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

Step 4: Amidation

The acid is treated with Ammonia ($\ce{NH3}$) and heated strongly. Dehydration of the ammonium salt yields 2-Methylpropanamide.

Step 5: Hoffmann Degradation (Step-Down)

The amide undergoes Hoffmann bromamide degradation with Bromine and $\ce{KOH}$. The carbonyl carbon is completely removed, yielding the secondary branched primary amine, Propan-2-amine (Isopropylamine).

$$ \text{Step 1: } \ce{CH3CH=CH2 + HBr -> CH3-CH(Br)-CH3} $$ $$ \text{Step 2: } \ce{CH3-CH(Br)-CH3 + KCN(alc) -> CH3-CH(CN)-CH3 + KBr} $$ $$ \text{Step 3: } \ce{CH3-CH(CN)-CH3 + 2H2O ->[H+] CH3-CH(COOH)-CH3 + NH4+} $$ $$ \text{Step 4: } \ce{CH3-CH(COOH)-CH3 + NH3 ->[\Delta] CH3-CH(CONH2)-CH3 + H2O} $$ $$ \text{Step 5: } \ce{CH3-CH(CONH2)-CH3 + Br2 + 4KOH -> CH3-CH(NH2)-CH3 + K2CO3 + 2KBr + 2H2O} $$
18 Convert Methane to Ethanol
Step 1: First Halogenation

Methane is reacted with Chlorine gas under UV light ($\ce{Cl2 / h\nu}$). Free-radical substitution yields Chloromethane.

Step 2: Wurtz Coupling (Step-Up)

Chloromethane is reacted with Sodium metal in dry ether. The symmetrical coupling of two methyl radicals yields the 2-carbon alkane, Ethane.

Step 3: Second Halogenation

Ethane is reactivated using Chlorine gas under UV light ($\ce{Cl2 / h\nu}$). Substitution yields Chloroethane.

Step 4: Amination

Chloroethane is heated with an excess of alcoholic Ammonia ($\ce{NH3}$). Nucleophilic substitution displaces the chloride, yielding the primary amine, Ethanamine.

Step 5: Nitrous Acid Reaction

Ethanamine is reacted with Nitrous acid ($\ce{HNO2}$ from $\ce{NaNO2 + HCl}$). The highly unstable aliphatic diazonium intermediate immediately decomposes with water, yielding Ethanol and nitrogen gas.

$$ \text{Step 1: } \ce{CH4 + Cl2 ->[h\nu] CH3Cl + HCl} $$ $$ \text{Step 2: } \ce{2 CH3Cl + 2Na ->[Dry Ether] CH3-CH3 + 2NaCl} $$ $$ \text{Step 3: } \ce{CH3-CH3 + Cl2 ->[h\nu] CH3CH2Cl + HCl} $$ $$ \text{Step 4: } \ce{CH3CH2Cl + NH3 (excess) -> CH3CH2NH2 + HCl} $$ $$ \text{Step 5: } \ce{CH3CH2NH2 + HNO2 -> CH3CH2OH + N2 ^ + H2O} $$
19 Convert Benzene to p-Fluorobenzoic Acid
Step 1: Alkylation

Benzene is reacted with $\ce{CH3Cl / AlCl3}$ to place an ortho/para directing methyl group on the ring, yielding Toluene.

Step 2: Nitration

Toluene is nitrated ($\ce{HNO3 / H2SO4}$). Due to steric effects, the para isomer is isolated as the major product, yielding p-Nitrotoluene.

Step 3: Reduction

The nitro group is reduced to an amine using $\ce{Sn / HCl}$. This yields p-Toluidine (p-Methylaniline).

Step 4: Balz-Schiemann Reaction

p-Toluidine is diazotized ($\ce{NaNO2 / HCl}$, 273K) and subsequently reacted with Fluoroboric acid ($\ce{HBF4}$). Heating the resulting diazonium fluoroborate salt yields p-Fluorotoluene.

Step 5: Vigorous Oxidation

The methyl side-chain is oxidized using acidified Potassium Permanganate ($\ce{KMnO4 / H+}$, heat). The $-\ce{CH3}$ group converts directly into a $-\ce{COOH}$ group, yielding p-Fluorobenzoic acid.

$$ \text{Step 1: } \ce{C6H6 + CH3Cl ->[AlCl3] C6H5CH3 + HCl} $$ $$ \text{Step 2: } \ce{C6H5CH3 + HNO3 ->[H2SO4] p-CH3-C6H4-NO2} $$ $$ \text{Step 3: } \ce{p-CH3-C6H4-NO2 + 6[H] ->[Sn/HCl] p-CH3-C6H4-NH2 + 2H2O} $$ $$ \text{Step 4: } \ce{p-CH3-C6H4-NH2 ->[1. NaNO2/HCl][2. HBF4, \Delta] p-CH3-C6H4-F + N2 ^ + BF3} $$ $$ \text{Step 5: } \ce{p-CH3-C6H4-F + 3[O] ->[KMnO4/H+] p-F-C6H4-COOH + H2O} $$
20 Convert Benzene to Aspirin (Acetylsalicylic Acid)
Step 1: Sulfonation

Benzene is reacted with Oleum (fuming sulfuric acid, $\ce{H2S2O7}$). The electrophilic $\ce{SO3}$ attacks the ring to form Benzene sulfonic acid.

Step 2: Alkali Fusion

Benzene sulfonic acid is fused with solid $\ce{NaOH}$ at high temperatures (600 K), followed immediately by acidification ($\ce{H+}$). This replaces the sulfonic acid group with a hydroxyl group, yielding Phenol.

Step 3: Salt Formation

Phenol is reacted with aqueous Sodium Hydroxide ($\ce{NaOH}$) to form Sodium phenoxide. This drastically activates the ring for the upcoming weak electrophile.

Step 4: Kolbe's Reaction

Sodium phenoxide is heated with Carbon dioxide ($\ce{CO2}$) at 400 K and 4-7 atm pressure. The electrophile attacks the ortho position. Subsequent acidification ($\ce{H+}$) yields 2-Hydroxybenzoic acid (Salicylic acid).

Step 5: Acetylation

Salicylic acid is treated with Acetic anhydride ($\ce{(CH3CO)2O}$) in the presence of an acid catalyst ($\ce{H+}$). This acetylates the phenolic $-\ce{OH}$ group, yielding Acetylsalicylic acid (Aspirin).

$$ \text{Step 1: } \ce{C6H6 + H2S2O7 -> C6H5SO3H + H2SO4} $$ $$ \text{Step 2: } \ce{C6H5SO3H ->[1. NaOH(fuse), 600K][2. H+] C6H5OH + Na2SO3} $$ $$ \text{Step 3: } \ce{C6H5OH + NaOH -> C6H5ONa + H2O} $$ $$ \text{Step 4: } \ce{C6H5ONa + CO2 ->[400 K][then H+] C6H4(OH)COOH} $$ $$ \text{Step 5: } \ce{C6H4(OH)COOH + (CH3CO)2O ->[H+] C6H4(OCOCH3)COOH + CH3COOH} $$

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