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NCERT Ex 13.5 Solved: Amines & Carboxylic Acid Conversions | ChemCa

NCERT Ex 13.5 Solved: Amines & Carboxylic Acid Conversions | ChemCa

NCERT Exercise 13.5 Solved

Detailed conversion mechanisms for Amines and Carboxylic Acids. Master the techniques of Ascent of Series (Step Up) and Descent of Series (Step Down).

Board Exam Class 12

(i) Ethanoic acid into methanamine

Step Down
$$ \text{CH}_3\text{COOH} \xrightarrow[\Delta]{\text{NH}_3} \text{CH}_3\text{CONH}_2 $$ $$ \text{CH}_3\text{CONH}_2 \xrightarrow{\text{Br}_2/\text{KOH}} \text{CH}_3\text{NH}_2 $$

Strategy: Convert carboxylic acid to amide. Then use the Hoffmann Bromamide Degradation to remove the carbonyl carbon and obtain an amine with one less carbon.

(ii) Hexanenitrile into 1-aminopentane

Step Down
$$ \text{C}_5\text{H}_{11}\text{CN} \xrightarrow[\text{Partial Hydrolysis}]{\text{H}_3\text{O}^+} \text{C}_5\text{H}_{11}\text{CONH}_2 $$ $$ \text{C}_5\text{H}_{11}\text{CONH}_2 \xrightarrow{\text{Br}_2/\text{KOH}} \text{C}_5\text{H}_{11}\text{NH}_2 $$

Strategy: Hexanenitrile (6C) needs to become pentanamine (5C). Hydrolyze the nitrile to amide, then degrade it using Hoffmann Bromamide reaction.

(iii) Methanol to ethanoic acid

Step Up
$$ \text{CH}_3\text{OH} \xrightarrow{\text{PCl}_5} \text{CH}_3\text{Cl} \xrightarrow{\text{KCN(alc)}} \text{CH}_3\text{CN} $$ $$ \text{CH}_3\text{CN} \xrightarrow{\text{H}_3\text{O}^+} \text{CH}_3\text{COOH} $$

Strategy: To add a carbon, convert alcohol to alkyl halide, then react with KCN to form a Nitrile. Complete hydrolysis of nitrile yields the acid.

(iv) Ethanamine into methanamine

Step Down
$$ \text{C}_2\text{H}_5\text{NH}_2 \xrightarrow{\text{HNO}_2} \text{C}_2\text{H}_5\text{OH} \xrightarrow{\text{KMnO}_4} \text{CH}_3\text{COOH} $$ $$ \xrightarrow[\Delta]{\text{NH}_3} \text{CH}_3\text{CONH}_2 \xrightarrow{\text{Br}_2/\text{KOH}} \text{CH}_3\text{NH}_2 $$

Strategy: Convert amine to alcohol (HONO), oxidize to acid, convert to amide, and finally use Hoffmann degradation to step down.

(v) Ethanoic acid into propanoic acid

Step Up
$$ \text{CH}_3\text{COOH} \xrightarrow{\text{LiAlH}_4} \text{CH}_3\text{CH}_2\text{OH} \xrightarrow{\text{PCl}_5} \text{CH}_3\text{CH}_2\text{Cl} $$ $$ \text{CH}_3\text{CH}_2\text{Cl} \xrightarrow{\text{KCN}} \text{CH}_3\text{CH}_2\text{CN} \xrightarrow{\text{H}_3\text{O}^+} \text{CH}_3\text{CH}_2\text{COOH} $$

Strategy: Reduce acid to alcohol, convert to halide, use KCN to add a carbon (Step Up), then hydrolyze.

(vi) Methanamine into ethanamine

Step Up
$$ \text{CH}_3\text{NH}_2 \xrightarrow{\text{HNO}_2} \text{CH}_3\text{OH} \xrightarrow{\text{PCl}_5} \text{CH}_3\text{Cl} $$ $$ \text{CH}_3\text{Cl} \xrightarrow{\text{KCN}} \text{CH}_3\text{CN} \xrightarrow{\text{LiAlH}_4} \text{CH}_3\text{CH}_2\text{NH}_2 $$

Strategy: Amine $\to$ Alcohol $\to$ Halide $\to$ Nitrile (add carbon) $\to$ Reduction to Amine (Mendius reaction).

(vii) Nitromethane into dimethylamine

Isocyanide Route
$$ \text{CH}_3\text{NO}_2 \xrightarrow{\text{Sn/HCl}} \text{CH}_3\text{NH}_2 $$ $$ \text{CH}_3\text{NH}_2 \xrightarrow[\text{KOH, }\Delta]{\text{CHCl}_3} \text{CH}_3\text{NC} \xrightarrow{\text{LiAlH}_4} \text{CH}_3\text{NHCH}_3 $$

Strategy: Reduce nitro to primary amine. Use Carbylamine Reaction to form Isocyanide. Reduction of isocyanide yields a secondary amine.

(viii) Propanoic acid into ethanoic acid

Step Down
$$ \text{C}_2\text{H}_5\text{COOH} \xrightarrow[\Delta]{\text{NH}_3} \text{C}_2\text{H}_5\text{CONH}_2 \xrightarrow{\text{Br}_2/\text{KOH}} \text{C}_2\text{H}_5\text{NH}_2 $$ $$ \text{C}_2\text{H}_5\text{NH}_2 \xrightarrow{\text{HNO}_2} \text{C}_2\text{H}_5\text{OH} \xrightarrow{\text{KMnO}_4} \text{CH}_3\text{COOH} $$

Strategy: Standard degradation: Acid $\to$ Amide $\to$ Amine (less carbon) $\to$ Alcohol $\to$ Acid.

Reaction Cheat Sheet

Br₂ / KOH Hoffmann Bromamide (Step Down)
KCN (alc) Nitrile Synthesis (Step Up)
HNO₂ (HONO) Amine $\to$ Alcohol
CHCl₃ / KOH Carbylamine (Isocyanide Test)

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NCERT Ex 12.15 Solved: Aldehydes & Ketones Conversions | ChemCa

NCERT Ex 12.15 Solved: Aldehydes & Ketones Conversions | ChemCa

NCERT Exercise 12.15 Solved

Detailed "2-step" conversion strategies for Aldehydes, Ketones, and Carboxylic Acids. Master reagents like Grignard, Rosenmund Catalyst, and Aldol Condensation.

Board Exam Important Class 12

(i) Propanone to Propene

$$ \text{CH}_3\text{COCH}_3 \xrightarrow{\text{NaBH}_4} \text{CH}_3\text{CH(OH)CH}_3 $$ $$ \xrightarrow[\Delta]{\text{conc. H}_2\text{SO}_4} \text{CH}_3\text{CH}=\text{CH}_2 $$

Strategy: Reduce ketone to secondary alcohol, then dehydrate to alkene.

(ii) Benzoic acid to Benzaldehyde

$$ \text{Ph-COOH} \xrightarrow{\text{SOCl}_2} \text{Ph-COCl} $$ $$ \xrightarrow[\text{Pd-BaSO}_4]{\text{H}_2} \text{Ph-CHO} $$

Strategy: Convert acid to acid chloride, then use Rosenmund Reduction to stop reduction at aldehyde stage.

(iii) Ethanol to 3-Hydroxybutanal

$$ \text{CH}_3\text{CH}_2\text{OH} \xrightarrow[\text{573 K}]{\text{Cu}} \text{CH}_3\text{CHO} $$ $$ \xrightarrow{\text{dil. NaOH}} \text{CH}_3\text{CH(OH)CH}_2\text{CHO} $$

Strategy: Mild oxidation to ethanal (acetaldehyde), followed by Aldol Condensation.

(iv) Benzene to m-Nitroacetophenone

$$ \text{C}_6\text{H}_6 \xrightarrow[\text{AlCl}_3]{\text{CH}_3\text{COCl}} \text{Ph-COCH}_3 $$ $$ \xrightarrow{\text{conc. HNO}_3/\text{H}_2\text{SO}_4} \text{m-NO}_2\text{-C}_6\text{H}_4\text{-COCH}_3 $$

Strategy: Friedel-Crafts Acylation first. The -COCH3 group is meta-directing, guiding the nitro group to the correct position.

(v) Benzaldehyde to Benzophenone

$$ \text{Ph-CHO} \xrightarrow[\text{2. H}_3\text{O}^+]{\text{1. PhMgBr}} \text{Ph-CH(OH)-Ph} $$ $$ \xrightarrow{\text{CrO}_3 \text{ or PCC}} \text{Ph-CO-Ph} $$

Strategy: Grignard reaction (Phenyl Magnesium Bromide) creates a secondary alcohol. Oxidation converts it to a ketone.

(vi) Bromobenzene to 1-Phenylethanol

$$ \text{Ph-Br} \xrightarrow[\text{Ether}]{\text{Mg}} \text{PhMgBr} $$ $$ \xrightarrow[\text{2. H}_3\text{O}^+]{\text{1. CH}_3\text{CHO}} \text{Ph-CH(OH)-CH}_3 $$

Strategy: Convert to Grignard reagent. Reacting Grignard with Acetaldehyde yields a secondary alcohol.

(vii) Benzaldehyde to 3-Phenylpropan-1-ol

$$ \text{PhCHO} + \text{CH}_3\text{CHO} \xrightarrow[\Delta]{\text{dil. NaOH}} \text{PhCH=CHCHO} $$ $$ \xrightarrow{\text{H}_2/\text{Ni}} \text{Ph-CH}_2\text{CH}_2\text{CH}_2\text{OH} $$

Strategy: Cross-Aldol condensation with acetaldehyde gives Cinnamaldehyde. Catalytic hydrogenation reduces both the double bond and the aldehyde to alcohol.

(viii) Benzaldehyde to $\alpha$-Hydroxyphenylacetic acid

$$ \text{Ph-CHO} \xrightarrow{\text{HCN}} \text{Ph-CH(OH)-CN} $$ $$ \xrightarrow{\text{H}_3\text{O}^+} \text{Ph-CH(OH)-COOH} $$

Strategy: Nucleophilic addition of HCN forms a Cyanohydrin. Hydrolysis of the -CN group converts it to -COOH.

(ix) Benzoic acid to m-Nitrobenzyl alcohol

$$ \text{Ph-COOH} \xrightarrow{\text{HNO}_3/\text{H}_2\text{SO}_4} \text{m-NO}_2\text{-C}_6\text{H}_4\text{-COOH} $$ $$ \xrightarrow{\text{B}_2\text{H}_6 \text{ or LiAlH}_4} \text{m-NO}_2\text{-C}_6\text{H}_4\text{-CH}_2\text{OH} $$

Strategy: Nitration first (-COOH is meta directing). Then reduce the carboxylic acid to alcohol (Diborane is preferred as it doesn't reduce the nitro group easily).

Quick Reagent Guide

Rosenmund (Pd-BaSO₄) Acid Chloride → Aldehyde
Grignard (RMgX) Increases Carbon Chain
dil. NaOH Aldol Condensation
HCN / H₃O⁺ Aldehyde → Hydroxy Acid

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NCERT Ex 10.11 Solved: Organic Conversions Guide | ChemCa

NCERT Ex 10.11 Solved: Organic Conversions Guide | ChemCa

NCERT Exercise 10.11 Solved

Step-by-step mechanisms for the 10 important organic conversions from the Haloalkanes & Haloarenes chapter.

Board Exam Important NEET / JEE

(i) Ethanol to But-1-yne

Ascent of Chain
$$ \text{CH}_3\text{CH}_2\text{OH} \xrightarrow{\text{SOCl}_2} \text{CH}_3\text{CH}_2\text{Cl} $$ $$ \text{HC}\equiv\text{CH} \xrightarrow{\text{NaNH}_2} \text{HC}\equiv\text{C}^-\text{Na}^+ $$ $$ \text{CH}_3\text{CH}_2\text{Cl} + \text{NaC}\equiv\text{CH} \to \text{CH}_3\text{CH}_2\text{C}\equiv\text{CH} $$

Concept: Convert alcohol to alkyl halide. React with Sodium Acetylide (from Ethyne) to add two carbons and a triple bond.

(ii) Ethane to Bromoethene

Double Elimination
$$ \text{CH}_3\text{-CH}_3 \xrightarrow[\text{h}\nu]{\text{Br}_2} \text{CH}_3\text{CH}_2\text{Br} \xrightarrow{\text{alc. KOH}} \text{CH}_2=\text{CH}_2 $$ $$ \text{CH}_2=\text{CH}_2 \xrightarrow{\text{Br}_2/\text{CCl}_4} \text{Br-CH}_2\text{-CH}_2\text{-Br} \xrightarrow{\text{alc. KOH}} \text{CH}_2=\text{CH-Br} $$

Concept: Convert alkane to alkene, halogenate to get vicinal dihalide, then partial dehydrohalogenation gives vinyl bromide.

(iii) Propene to 1-Nitropropane

Anti-Markovnikov
$$ \text{CH}_3\text{CH}=\text{CH}_2 \xrightarrow{\text{HBr/Peroxide}} \text{CH}_3\text{CH}_2\text{CH}_2\text{Br} $$ $$ \text{CH}_3\text{CH}_2\text{CH}_2\text{Br} \xrightarrow{\text{AgNO}_2} \text{CH}_3\text{CH}_2\text{CH}_2\text{NO}_2 $$

Concept: Peroxide effect adds Br to the terminal carbon. AgNO2 is used specifically to get the Nitro alkane (KNO2 would give alkyl nitrite).

(iv) Toluene to Benzyl Alcohol

Side Chain
$$ \text{C}_6\text{H}_5\text{CH}_3 \xrightarrow[\text{h}\nu]{\text{Cl}_2} \text{C}_6\text{H}_5\text{CH}_2\text{Cl} \xrightarrow{\text{aq. KOH}} \text{C}_6\text{H}_5\text{CH}_2\text{OH} $$

Concept: Free radical substitution attacks the benzylic position. Nucleophilic substitution ($S_N2$) with aqueous KOH yields the alcohol.

(v) Propene to Propyne

Unsaturation
$$ \text{CH}_3\text{CH}=\text{CH}_2 \xrightarrow{\text{Br}_2/\text{CCl}_4} \text{CH}_3\text{CH(Br)CH}_2\text{Br} $$ $$ \text{CH}_3\text{CH(Br)CH}_2\text{Br} \xrightarrow{\text{alc. KOH}} \xrightarrow{\text{NaNH}_2} \text{CH}_3\text{C}\equiv\text{CH} $$

Concept: Convert alkene to vicinal dihalide. Strong bases (KOH then NaNH2) perform double elimination to create the triple bond.

(vi) Ethanol to Ethyl Fluoride

Swarts Reaction
$$ \text{CH}_3\text{CH}_2\text{OH} \xrightarrow{\text{SOCl}_2} \text{CH}_3\text{CH}_2\text{Cl} $$ $$ \text{CH}_3\text{CH}_2\text{Cl} \xrightarrow{\text{AgF or Hg}_2\text{F}_2} \text{CH}_3\text{CH}_2\text{F} $$

Concept: Direct fluorination is explosive. We use Swarts Reaction (Halide exchange with heavy metal fluorides) to synthesize alkyl fluorides.

(vii) Bromomethane to Propanone

Grignard
$$ \text{CH}_3\text{Br} \xrightarrow{\text{KCN}} \text{CH}_3\text{CN} $$ $$ \text{CH}_3\text{CN} + \text{CH}_3\text{MgBr} \to \text{Complex} \xrightarrow{\text{H}_3\text{O}^+} \text{CH}_3\text{COCH}_3 $$

Concept: Convert to nitrile (2 carbons). React with methyl Grignard reagent (provides 3rd carbon) to form a ketone upon hydrolysis.

(viii) But-1-ene to But-2-ene

Saytzeff Rule
$$ \text{CH}_3\text{CH}_2\text{CH}=\text{CH}_2 \xrightarrow{\text{HBr}} \text{CH}_3\text{CH}_2\text{CH(Br)CH}_3 $$ $$ \text{CH}_3\text{CH}_2\text{CH(Br)CH}_3 \xrightarrow{\text{alc. KOH}} \text{CH}_3\text{CH}=\text{CHCH}_3 $$

Concept: Add HBr (Markovnikov) to move the leaving group to the secondary carbon. Elimination yields the more substituted, stable alkene (But-2-ene) as the major product.

(ix) 1-Chlorobutane to n-Octane

Wurtz Reaction
$$ 2 \text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{Cl} + 2\text{Na} \xrightarrow{\text{Dry Ether}} \text{n-Octane} + 2\text{NaCl} $$

Concept: Classic Wurtz reaction couples two alkyl halide molecules to double the carbon chain.

(x) Benzene to Biphenyl

Fittig Reaction
$$ \text{C}_6\text{H}_6 \xrightarrow{\text{Cl}_2/\text{FeCl}_3} \text{C}_6\text{H}_5\text{Cl} $$ $$ 2 \text{C}_6\text{H}_5\text{Cl} + 2\text{Na} \xrightarrow{\text{Dry Ether}} \text{C}_6\text{H}_5\text{-C}_6\text{H}_5 + 2\text{NaCl} $$

Concept: Halogenate benzene first. The Fittig Reaction couples two aryl halides using Sodium metal.

Essential Reagents Cheat Sheet

SOCl₂ Alcohol → Chloride
alc. KOH Elimination (-HX)
HBr/Peroxide Anti-Markovnikov
Na/Ether Wurtz/Fittig Coupling

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Top 20 Organic Conversions Solved: Class 12 Chemistry Guide | ChemCa

Top 20 Organic Conversions Solved: Class 12 Chemistry Guide | ChemCa

Mastering Organic Conversions

Comprehensive step-by-step solutions to NCERT Exercise 10.19 (Haloalkanes and Haloarenes). Learn how to interconvert functional groups effectively.

(i) Propene to Propan-1-ol

Hydroboration
$$ \text{CH}_3\text{CH}=\text{CH}_2 \xrightarrow[\text{H}_2\text{O}_2/\text{OH}^-]{\text{B}_2\text{H}_6} \text{CH}_3\text{CH}_2\text{CH}_2\text{OH} $$

Strategy: Direct hydration follows Markovnikov's rule (giving propan-2-ol). To get primary alcohol (Anti-Markovnikov product), use Hydroboration-Oxidation.

(ii) Ethanol to But-1-yne

Step-up
$$ \text{CH}_3\text{CH}_2\text{OH} \xrightarrow{\text{SOCl}_2} \text{CH}_3\text{CH}_2\text{Cl} $$ $$ \text{HC}\equiv\text{CH} \xrightarrow{\text{NaNH}_2} \text{HC}\equiv\text{C}^-\text{Na}^+ $$ $$ \text{CH}_3\text{CH}_2\text{Cl} + \text{NaC}\equiv\text{CH} \to \text{CH}_3\text{CH}_2\text{C}\equiv\text{CH} $$

Strategy: Convert ethanol to halide. React with Sodium Acetylide (from acetylene) to increase carbon chain.

(iii) 1-Bromopropane to 2-Bromopropane

Rearrangement
$$ \text{CH}_3\text{CH}_2\text{CH}_2\text{Br} \xrightarrow{\text{alc. KOH}} \text{CH}_3\text{CH}=\text{CH}_2 \xrightarrow{\text{HBr}} \text{CH}_3\text{CH(Br)}\text{CH}_3 $$

Strategy: Eliminate HBr to form Propene, then add HBr. Addition follows Markovnikov's Rule to put Br on the secondary carbon.

(iv) Toluene to Benzyl alcohol

Side-chain Sub
$$ \text{C}_6\text{H}_5\text{CH}_3 \xrightarrow[\text{h}\nu]{\text{Cl}_2} \text{C}_6\text{H}_5\text{CH}_2\text{Cl} \xrightarrow{\text{aq. KOH}} \text{C}_6\text{H}_5\text{CH}_2\text{OH} $$

Strategy: Free radical chlorination attacks the side chain (benzylic position). Then simple nucleophilic substitution with aqueous KOH.

(v) Benzene to 4-Bromonitrobenzene

EAS
$$ \text{C}_6\text{H}_6 \xrightarrow{\text{Br}_2/\text{FeBr}_3} \text{C}_6\text{H}_5\text{Br} \xrightarrow{\text{HNO}_3/\text{H}_2\text{SO}_4} \text{p-BrC}_6\text{H}_4\text{NO}_2 $$

Strategy: Brominate first. -Br is ortho/para directing. Nitration then yields the para product as major isomer.

(vi) Benzyl alcohol to 2-Phenylethanoic acid

Step-up
$$ \text{PhCH}_2\text{OH} \xrightarrow{\text{SOCl}_2} \text{PhCH}_2\text{Cl} \xrightarrow{\text{KCN}} \text{PhCH}_2\text{CN} \xrightarrow{\text{H}_3\text{O}^+} \text{PhCH}_2\text{COOH} $$

Strategy: Use KCN to add a carbon atom (Step-up reaction), then hydrolyze the nitrile to carboxylic acid.

(vii) Ethanol to Propanenitrile

$$ \text{C}_2\text{H}_5\text{OH} \xrightarrow{\text{PCl}_5} \text{C}_2\text{H}_5\text{Cl} \xrightarrow{\text{alc. KCN}} \text{C}_2\text{H}_5\text{CN} $$

Strategy: Convert alcohol to halide, then substitute with Cyanide ion.

(viii) Aniline to Chlorobenzene

Sandmeyer
$$ \text{Ph-NH}_2 \xrightarrow[\text{0-5}^\circ\text{C}]{\text{NaNO}_2 + \text{HCl}} \text{Ph-N}_2^+\text{Cl}^- \xrightarrow{\text{Cu}_2\text{Cl}_2} \text{Ph-Cl} $$

Strategy: Diazotization followed by Sandmeyer's reaction.

(ix) 2-Chlorobutane to 3,4-Dimethylhexane

Wurtz
$$ 2 \text{CH}_3\text{CH(Cl)C}_2\text{H}_5 \xrightarrow[\text{Dry Ether}]{\text{2Na}} \text{CH}_3\text{CH}_2\text{CH(CH}_3)\text{-CH(CH}_3)\text{CH}_2\text{CH}_3 $$

Strategy: Wurtz reaction doubles the carbon chain and mirrors the structure.

(x) 2-Methyl-1-propene to 2-Chloro-2-methylpropane

$$ (\text{CH}_3)_2\text{C}=\text{CH}_2 + \text{HCl} \to (\text{CH}_3)_3\text{C-Cl} $$

Strategy: Simple Markovnikov addition of HCl. The Cl attacks the tertiary carbocation.

(xi) Ethyl chloride to Propanoic acid

$$ \text{C}_2\text{H}_5\text{Cl} \xrightarrow{\text{KCN}} \text{C}_2\text{H}_5\text{CN} \xrightarrow{\text{H}_3\text{O}^+} \text{C}_2\text{H}_5\text{COOH} $$

Strategy: Step-up reaction using KCN, followed by complete hydrolysis.

(xii) But-1-ene to n-Butyliodide

$$ \text{But-1-ene} \xrightarrow{\text{HBr/Peroxide}} \text{1-Bromobutane} \xrightarrow{\text{NaI/Acetone}} \text{1-Iodobutane} $$

Strategy: Anti-Markovnikov addition (Peroxide effect) to get terminal bromide. Then Finkelstein reaction to swap Br for I.

(xiii) 2-Chloropropane to 1-Propanol

Isomerization path
$$ \text{2-Chloropropane} \xrightarrow{\text{alc. KOH}} \text{Propene} \xrightarrow{\text{HBr/Peroxide}} \text{1-Bromopropane} \xrightarrow{\text{aq. KOH}} \text{1-Propanol} $$

Strategy: Eliminate to alkene, then Anti-Markovnikov addition to move functional group to the end.

(xiv) Isopropyl alcohol to Iodoform

$$ (\text{CH}_3)_2\text{CHOH} \xrightarrow{\text{I}_2 + \text{NaOH}} \text{CHI}_3 \downarrow + \text{CH}_3\text{COONa} $$

Strategy: This is the classic Haloform Reaction. Isopropyl alcohol contains the $\text{CH}_3\text{CH(OH)-}$ group required for a positive test.

(xv) Chlorobenzene to p-Nitrophenol

$$ \text{Ph-Cl} \xrightarrow{\text{HNO}_3/\text{H}_2\text{SO}_4} \text{p-Cl-C}_6\text{H}_4\text{NO}_2 \xrightarrow[\text{443K, H}^+]{\text{NaOH}} \text{p-HO-C}_6\text{H}_4\text{NO}_2 $$

Strategy: Nitration first (ortho/para). The presence of $-\text{NO}_2$ at para position activates the ring for Nucleophilic substitution of Cl by OH.

(xvi) 2-Bromopropane to 1-Bromopropane

$$ \text{CH}_3\text{CH(Br)CH}_3 \xrightarrow{\text{alc. KOH}} \text{Propene} \xrightarrow{\text{HBr/Peroxide}} \text{CH}_3\text{CH}_2\text{CH}_2\text{Br} $$

Strategy: Elimination followed by Anti-Markovnikov addition.

(xvii) Chloroethane to Butane

$$ 2 \text{C}_2\text{H}_5\text{Cl} + 2\text{Na} \xrightarrow{\text{Dry Ether}} \text{C}_4\text{H}_{10} $$

Strategy: Classic Wurtz Reaction to double the carbon chain.

(xviii) Benzene to Diphenyl

Fittig
$$ \text{C}_6\text{H}_6 \xrightarrow{\text{Cl}_2/\text{FeCl}_3} \text{C}_6\text{H}_5\text{Cl} $$ $$ 2 \text{C}_6\text{H}_5\text{Cl} + 2\text{Na} \xrightarrow{\text{Dry Ether}} \text{Ph-Ph} $$

Strategy: Convert benzene to chlorobenzene. Then use Fittig Reaction to couple two aryl rings.

(xix) tert-Butyl bromide to Isobutyl bromide

$$ (\text{CH}_3)_3\text{CBr} \xrightarrow{\text{alc. KOH}} (\text{CH}_3)_2\text{C}=\text{CH}_2 $$ $$ (\text{CH}_3)_2\text{C}=\text{CH}_2 \xrightarrow{\text{HBr/Peroxide}} (\text{CH}_3)_2\text{CH-CH}_2\text{Br} $$

Strategy: Dehydrohalogenation gives isobutylene. Peroxide effect adds HBr to the less substituted carbon (primary bromide).

(xx) Aniline to Phenylisocyanide

Carbylamine
$$ \text{Ph-NH}_2 + \text{CHCl}_3 + 3\text{KOH(alc)} \xrightarrow{\Delta} \text{Ph-NC} + 3\text{KCl} + 3\text{H}_2\text{O} $$

Strategy: This is the Carbylamine Reaction, a test for primary amines resulting in a foul-smelling isocyanide.

Key Reaction Mechanisms Used

Markovnikov's Rule Anti-Markovnikov (Peroxide) Wurtz & Fittig Sandmeyer Nucleophilic Substitution

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pH of Polyprotic Acids: H2S, H3PO4 & H2SO4 Guide | ChemCa

pH of Polyprotic Acids: H2S, H3PO4 & H2SO4 Guide | ChemCa

pH of Polyprotic Acids

Step-by-step strategies to calculate pH and ion concentrations for Diprotic and Triprotic acids like $H_2S$, $H_3PO_4$, and $H_2SO_4$.

Ionic Equilibrium pH Calculation

The Dominant Step Principle

Polyprotic acids dissociate in steps. For most weak polyprotic acids (like $H_2CO_3, H_2S, H_3PO_4$), the first dissociation constant ($K_{a1}$) is significantly larger than the second ($K_{a2}$).

Typically: $K_{a1} \gg K_{a2} \gg K_{a3}$ (often by factors of $10^4$ to $10^6$).

Key Approximation: Since the first step produces the vast majority of $H^+$, we can calculate the pH by treating the acid as monoprotic using only $K_{a1}$. The subsequent steps contribute negligible $H^+$.

Stepwise Dissociation ($H_2A$)

$$ \text{Step 1: } H_2A \rightleftharpoons H^+ + HA^- \quad (K_{a1}) $$
$$ \text{Step 2: } HA^- \rightleftharpoons H^+ + A^{2-} \quad (K_{a2}) $$

*Step 2 is suppressed by the Common Ion Effect ($H^+$ from Step 1).


Case 1: Weak Diprotic Acid ($K_{a1} \gg K_{a2}$)

Solved Example: Carbonic Acid

Calculate the pH and $[CO_3^{2-}]$ concentration in a $0.05 M$ solution of $H_2CO_3$.
Given: $K_{a1} = 4 \times 10^{-7}$, $K_{a2} = 5 \times 10^{-11}$.

Step 1: Determine pH (First Dissociation)

Since $K_{a1} \gg K_{a2}$, we treat it as a monoprotic acid. Check if we can neglect $\alpha$ ($C/K_{a1} > 100$).

$$ [H^+] \approx \sqrt{K_{a1} \cdot C} = \sqrt{4 \times 10^{-7} \times 0.05} $$ $$ [H^+] = \sqrt{20 \times 10^{-9}} = \sqrt{2 \times 10^{-8}} \approx 1.41 \times 10^{-4} M $$

$pH = -\log(1.41 \times 10^{-4}) = 4 - 0.15 = \mathbf{3.85}$

Step 2: Concentration of Second Ion ($A^{2-}$)

This is a standard result for weak polyprotic acids. From Step 1, $[H^+] \approx [HA^-]$. Now write the Step 2 equilibrium:

$$ K_{a2} = \frac{[H^+][CO_3^{2-}]}{[HCO_3^-]} $$

Substituting $[H^+] \approx [HCO_3^-]$ (from Step 1):

$$ K_{a2} \approx \frac{[H^+][CO_3^{2-}]}{[H^+]} \implies [CO_3^{2-}] \approx K_{a2} $$ $$ \therefore [CO_3^{2-}] = 5 \times 10^{-11} M $$

*Note: The concentration of the divalent ion equals the second dissociation constant.

Case 2: The $H_2SO_4$ Exception

Sulfuric Acid ($H_2SO_4$) is unique because the first step is Strong (Complete dissociation) while the second step is Weak.

Step 1 (Strong)
$$ H_2SO_4 \longrightarrow H^+ + HSO_4^- $$ $$ [H^+]_1 = C $$
Step 2 (Weak)
$$ HSO_4^- \rightleftharpoons H^+ + SO_4^{2-} $$ $$ \text{Initial: } C \quad C \quad 0 $$ $$ \text{Eq: } C-x \quad C+x \quad x $$

Example: 0.01M $H_2SO_4$ ($K_{a2} = 1.2 \times 10^{-2}$)

$$ K_{a2} = \frac{(C+x)(x)}{C-x} $$ $$ 1.2 \times 10^{-2} = \frac{(0.01+x)x}{0.01-x} $$

Since $K_{a2}$ is comparable to $C$, we cannot ignore x. We must solve the quadratic equation.
Solving gives $x \approx 0.0045 M$.

Total $[H^+]$:

$$ [H^+]_{total} = C + x = 0.01 + 0.0045 = 0.0145 M $$ $$ pH = -\log(0.0145) \approx 1.84 $$

Summary Formulas

Case Type Condition Calculation
Weak Diprotic ($H_2S$) $K_{a1} \gg K_{a2}$ $[H^+] \approx \sqrt{K_{a1}C}$
$[A^{2-}] \approx K_{a2}$
Strong-Weak ($H_2SO_4$) Step 1 Complete $[H^+] = C + x$
Solve Quad for x
Triprotic ($H_3PO_4$) $K_{a1} \gg K_{a2} \gg K_{a3}$ $[H^+] \approx \sqrt{K_{a1}C}$
$[HPO_4^{2-}] \approx K_{a2}$

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