Search This Blog

Home

Chemca - Chemistry Made Easy | JEE, NEET & Board Exam Prep

Chemca - Chemistry Made Easy | JEE, NEET & Board Exam Prep ...

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

© 2025 ChemCa Education. All rights reserved.

No comments:

Post a Comment