Question

Write the reagent for each of the following reactions:
i) R–OH + …… → R–Cl
ii) R–OH + …… → R–Br
iii) CH₂=CH₂ + …… → BrCH₂–CH₂Br
iv) C₆H₅–CH=CH₂ + …… → C₆H₅–CH₂–CH₂Br

Hint:

OH→Cl uses SOCl₂; OH→Br uses PBr₃; C=C + Br₂ gives a vicinal dibromide; and anti-Markovnikov (Br on the far carbon of styrene) needs HBr with a peroxide.

Answer 1

Each blank asks for the reagent that converts the given substrate into the product shown.

Step 1 (Part i): An alcohol R–OH is turned into an alkyl chloride R–Cl by replacing –OH with –Cl. The best reagent is thionyl chloride, SOCl₂ (the Darzen method), because the by-products SO₂ and HCl are gases and escape, giving a pure product.
\[ R\text{-}OH + SOCl_2 \rightarrow R\text{-}Cl + SO_2\uparrow + HCl\uparrow \]
PCl₅, PCl₃ or conc. HCl with anhydrous ZnCl₂ (Lucas reagent) also work.

Step 2 (Part ii): An alcohol R–OH is converted to an alkyl bromide R–Br using phosphorus tribromide, PBr₃ (usually made in situ from red phosphorus and bromine, P + Br₂).
\[ 3\,R\text{-}OH + PBr_3 \rightarrow 3\,R\text{-}Br + H_3PO_3 \]
Constant-boiling HBr (NaBr + conc. H₂SO₄) is an alternative.

Step 3 (Part iii): Ethene adds bromine across its double bond to give 1,2-dibromoethane. The reagent is bromine, Br₂, dissolved in an inert solvent such as CCl₄. The red-brown colour of bromine is discharged, a test for unsaturation.
\[ CH_2{=}CH_2 + Br_2 \rightarrow BrCH_2\text{-}CH_2Br \]

Step 4 (Part iv): The product C₆H₅–CH₂–CH₂Br has Br on the terminal carbon, i.e. anti-Markovnikov addition. Styrene reacts with HBr in the presence of an organic peroxide (benzoyl peroxide); this is the peroxide or Kharasch effect and gives the anti-Markovnikov product through a free-radical chain.
\[ C_6H_5\text{-}CH{=}CH_2 + HBr \xrightarrow{\text{peroxide}} C_6H_5\text{-}CH_2\text{-}CH_2Br \]

Answer: i) SOCl₂ ii) PBr₃ (P + Br₂) iii) Br₂/CCl₄ iv) HBr with peroxide.