Method of preparation of haloalkanes
The hydroxyl group of an alcohol is replaced by halogen on reaction with concentrated halogen acids, phosphorus halides, or thionyl chloride. Thionyl chloride is preferred because in this reaction alkyl halide is formed along with gases SO2 and HCl. The two gaseous products are escapable, hence, the reaction gives pure alkyl halides. The reactions of primary and secondary alcohols with HCl require the presence of a catalyst, ZnCl2. With tertiary alcohols, the reaction is conducted by simply shaking the alcohol with concentrated HCl at room temperature. Constant boiling with HBr (48%) is used for preparing alkyl bromide. Good yields of R-I may be obtained by heating alcohols with sodium or potassium iodide in 95% orthophosphoric acid.
An alkene is converted to corresponding alkyl halide by reaction with hydrogen chloride, hydrogen bromide. Further, addition of bromine in CCl4 to an alkene resulting in discharge of reddish brown colour of bromine constitutes, an important method for the detection of double bond in a molecule.
Alkyl iodides are often prepared by the reaction of alkyl chlorides/ bromides with NaI in dry acetone.
The alkyl fluorides are best prepared by heating an alkyl chlorides/bromides in the presence of a metallic fluoride.
The primary aromatic amine, dissolved or suspended in cold aqueous mineral acid, is treated with sodium nitrite, a diazonium salt is formed. Mixing the solution of freshly prepared diazonium salt with cuprous chloride or cuprous bromide results in the replacement of the diazonium group by -Cl or -Br.
The reactions of haloalkanes may be divided into the following categories: 1. Nucleophilic substitution 2. Elimination reactions 3. Reaction with metals.
In the given reactions, identify A & B:
3ROH + PBr3 → 3RBr + A
ROH + PCl5 → RCl + POCl3 + B
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → A = H3PO3, B = HCl
Alcohols are converted into alkyl halides by phosphorus halides. The –OH group is replaced by the halogen, and the phosphorus ends up in an oxy-acid or oxyhalide:
Phosphorus trihalide (PX3): three R–OH molecules replace the three halogens of one \( \ce{PX3} \). Phosphorus keeps its +3 state and ends up as phosphorous acid, H3PO3.
\( \displaystyle \ce{3ROH + PBr3 -> 3RBr + H3PO3} \)
Phosphorus pentachloride (PCl5): one R–OH reacts; the OH and one Cl combine to form HCl, and \( \ce{PCl5} \) becomes POCl3.
\( \displaystyle \ce{ROH + PCl5 -> RCl + POCl3 + HCl} \)
Balancing the first equation: the three O atoms and three H atoms left over from three ROH molecules combine with the P atom to form \( \ce{H3PO3} \). In the second, the H of ROH and one Cl form HCl.
So A = H3PO3 and B = HCl.
Why the other options are incorrect
Option (1) A = H3PO4, B = HCl — Incorrect. \( \ce{H3PO4} \) has P in the +5 state; \( \ce{PBr3} \) (P +3) is not oxidised in this reaction.
Option (3) A = H3PO3, B = HBr — Incorrect. There is no bromine in the \( \ce{PCl5} \) reaction; the hydrogen halide formed must be HCl.
Option (4) A = H3PO4, B = HBr — Incorrect. \( \ce{H3PO4} \) has P in the +5 state (\( \ce{PBr3} \) is not oxidised), and the \( \ce{PCl5} \) reaction contains no bromine, so HBr cannot form.
NCERT (2026-27 edition):
Methods of Preparation of Haloalkanes
