Consider the following statements about some conversions.
(A) Ethanol can be converted into fluoroethane by heating it with constant boiling HBr and then heating the bromoethane formed with AgF.
(B) Benzene can be converted into biphenyl by brominating it in the presence of iron and then treating the bromobenzene formed with sodium in dry ether.
(C) 1-Chlorobutane can be converted into n-octane by treating it with magnesium in dry ether.
(D) Benzenediazonium chloride can be converted into iodobenzene by mixing its solution with cuprous chloride.
Which of the statements given above are incorrect? Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (C) and (D) only
Each conversion is judged step by step: the reagent of every step must give the product that the next step needs.
(A) Ethanol can be converted into fluoroethane by heating it with constant boiling HBr and then heating the bromoethane formed with AgF. — Correct statement. HBr replaces –OH by Br, and heating the alkyl bromide with a metallic fluoride such as AgF replaces Br by F (Swarts reaction):
\( \displaystyle \ce{C2H5OH + HBr} \xrightarrow{\Delta} \ce{C2H5Br + H2O} \)
\( \displaystyle \ce{C2H5Br + AgF} \xrightarrow{\Delta} \ce{C2H5F + AgBr} \)
(B) Benzene can be converted into biphenyl by brominating it in the presence of iron and then treating the bromobenzene formed with sodium in dry ether. — Correct statement. Iron acts as the Lewis acid catalyst for bromination, and sodium in dry ether then joins two aryl groups (Fittig reaction):
\( \displaystyle \ce{C6H6 + Br2} \xrightarrow{\ce{Fe}} \ce{C6H5Br + HBr} \)
\( \displaystyle \ce{2C6H5Br + 2Na} \xrightarrow{\text{dry ether}} \ce{C6H5-C6H5 + 2NaBr} \)
(C) 1-Chlorobutane can be converted into n-octane by treating it with magnesium in dry ether. — Incorrect statement. Doubling the carbon chain needs sodium in dry ether (Wurtz reaction). Magnesium in dry ether gives the Grignard reagent, butylmagnesium chloride, not octane.
\( \displaystyle \ce{2CH3CH2CH2CH2Cl + 2Na} \xrightarrow{\text{dry ether}} \ce{CH3(CH2)6CH3 + 2NaCl} \)
(D) Benzenediazonium chloride can be converted into iodobenzene by mixing its solution with cuprous chloride. — Incorrect statement. Cuprous chloride replaces the diazonium group by –Cl and gives chlorobenzene. Iodobenzene is made by shaking the diazonium salt with potassium iodide; no cuprous halide is needed.
\( \displaystyle \ce{C6H5N2^+Cl^- + KI -> C6H5I + KCl + N2 ^} \)
NCERT (2026-27 edition):
6.4.1 From Alcohols – “…Constant boiling with HBr (48%) is used for preparing alkyl bromide.”
6.4.3 Halogen Exchange – “The synthesis of alkyl fluorides is best accomplished by heating an alkyl chloride/bromide in the presence of a metallic fluoride such as AgF, Hg2F2, CoF2 or SbF3. The reaction is termed as Swarts reaction.”
6.5 Preparation of Haloarenes – “Aryl chlorides and bromides can be easily prepared by electrophilic substitution of arenes with chlorine and bromine respectively in the presence of Lewis acid catalysts like iron or iron(III) chloride.” … “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.”
“Replacement of the diazonium group by iodine does not require the presence of cuprous halide and is done simply by shaking the diazonium salt with potassium iodide.”
6.7.1 Reactions of Haloalkanes – “…These reagents are obtained by the reaction of haloalkanes with magnesium metal in dry ether.” … “Alkyl halides react with sodium in dry ether to give hydrocarbons containing double the number of carbon atoms present in the halide. This reaction is known as Wurtz reaction.”
6.7.2 Reactions of Haloarenes – “Aryl halides also give analogous compounds when treated with sodium in dry ether, in which two aryl groups are joined together. It is called Fittig reaction.”