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

Read every arrow as "which chemical does this job?" and match it to the standard reagent.

Step 1: Swapping OH for Cl on an alcohol is a substitution of the hydroxyl group. Thionyl chloride is preferred because it leaves no liquid impurity: \(R\text{-}OH + SOCl_2 \rightarrow R\text{-}Cl + SO_2 + HCl\). (Phosphorus chlorides PCl₅/PCl₃, or Lucas reagent conc. HCl + ZnCl₂, are acceptable too.)

Step 2: To install Br in place of OH, use phosphorus and bromine, which generate PBr₃: \(3R\text{-}OH + PBr_3 \rightarrow 3R\text{-}Br + H_3PO_3\). So the blank is P + Br₂ (i.e. PBr₃).

Step 3: Turning a symmetrical alkene into a vicinal dihalide (Br on each of the two carbons) is simple electrophilic addition of molecular bromine: \(CH_2{=}CH_2 + Br_2 \rightarrow BrCH_2CH_2Br\). Reagent = Br₂ in CCl₄.

Step 4: Here Br lands on the CH₂ that is far from the ring (the less substituted carbon). Markovnikov addition of HBr would put Br on the benzylic carbon; to reverse the selectivity we add a peroxide, which switches the mechanism to free radicals. Reagent = HBr in the presence of a peroxide (Kharasch/peroxide effect): \(C_6H_5CH{=}CH_2 + HBr \xrightarrow{\text{peroxide}} C_6H_5CH_2CH_2Br\).

Reagents: (i) SOCl₂, (ii) PBr₃, (iii) Br₂, (iv) HBr + peroxide.