Key Organic Reactions: Fittig, Wurtz, Swarts, Kolbe‑Schmitt, Williamson

Aryl halide to diphenyl

Aryl halide reacts with sodium metal in the presence of dry ether to form diphenyl.

Name: Fittig reaction

Equations for SN1 step: tert‑butyl bromide to tert‑butyl alcohol

Equations for the step in SN1 of the conversion of tert‑butyl bromide into tert‑butyl alcohol.

IUPAC name for product from 2‑bromopentane with alcoholic KOH

IUPAC name for the major product obtained when 2‑bromopentane is heated with alcoholic KOH. Equation and name:

ANS: pent-2-ene

CH3-CH2-CH-CH3 ------> CH3-CH2-CH=CH-CH3

                     I

                   Br

NAME: elimination reaction

Why aryl halides are less reactive toward nucleophilic substitution

Aryl halides are less reactive towards nucleophilic substitution compared to alkyl halides. Two reasons:

• Instability of the phenyl cation: the phenyl cation is not resonance stabilized.
• Electron repulsion: possible repulsion between the electron‑rich nucleophile and the electron‑rich aromatic ring.

Wurtz and Swarts reactions with examples

1) Wurtz / Fittig-type reaction

Aryl halide reacts with an alkyl halide in the presence of sodium in dry ether to form an alkylbenzene.

C2H5Br + C6H5Br ------> C2H5C6H5 + NaBr

                               ether

2) Swarts reaction (example)

An alkyl fluoride is obtained by treating an alkyl chloride or bromide with silver fluoride.

CH3Br + AgF ----> CH3F + AgBr

Kolbe‑Schmitt and Williamson ether synthesis

1) Kolbe‑Schmitt reaction

Sodium phenoxide undergoes electrophilic carboxylation with CO2 to give, after acidification, ortho‑hydroxybenzoic acid (salicylic acid) as the main product.

2) Williamson ether synthesis

An alkyl halide reacts with a sodium alkoxide to form an ether:

R‑X + Na‑O‑R' -----> R‑O‑R' + NaX

Dehydration of ethanol to ethene

Step 1: protonation of the alcohol

H H H H

I I I I

H‑C‑C‑O‑H + H+ <----> H‑C‑C‑O‑H ---->

I I fast I I I

H H H H H

Step 2: formation of carbocation by loss of water

H H H H

I I I I

H‑C‑C‑O‑H <--> H‑C‑C+ + H2O ---->

I I I I I

H H H H H

STEP 3: (PTO)

Reimer‑Tiemann and dehydration of a primary alcohol

1) Reimer‑Tiemann reaction

Phenol reacts with chloroform in the presence of NaOH to form a product which on acidification gives salicylaldehyde.

2) Dehydration / dehydrogenation of a primary alcohol

Ethyl alcohol when heated with concentrated sulfuric acid dehydrates / dehydrogenates to form ethanal:

CH3CH2OH ------------> CH3CHO

                          conc. H2SO4 (heat)