Exams › JEE Advanced › Chemistry

In the reaction between C6H5Br and CH3CH2OH, C6H5Br remains unreactive. However, when C6H5OH reacts with MeCBrMe2, the tert-halide group in CBr undergoes elimination instead of substitution. Therefore, only certain combinations are suitable for ether synthesis.

1. C6H5OH reacts with (CH3)2SO4 to form C6H5OCH3.
2. p-NO2C6H4Br reacts with CH3CH2OH to produce p-NO2C6H4OCH2CH3.
3. C6H5OH reacts with MeCBrMe2 to yield C6H5OCHMe2.
4. C6H5Br reacts with CH3CH2OH but no reaction occurs.

Correct answer: C6H5OH reacts with MeCBrMe2 to yield C6H5OCHMe2.

Solution

C6H5OH reacts with MeCBrMe2 to yield C6H5OCHMe2 because the tert-halide group in CBr undergoes elimination instead of substitution, making it a suitable combination for ether synthesis.

Related JEE Advanced Chemistry questions

• Which compound listed below cannot be synthesized using the standard Williamson ether synthesis?
• When CH3CH2CH2CH2OH is heated with Al2O3 at 350°C, what is the main product formed?
• The compound that does NOT liberate CO₂ on treatment with aqueous sodium bicarbonate solution, is -
• Which type of solution is required to perform the dye test for identifying β-naphthol?
• Among the following alcohols, how many will produce an immediate white turbidity (precipitate) when treated with Lucas reagent (anhydrous ZnCl2 / conc. HCl) at room temperature? The compounds are: (i) 1-butanol, (ii) 2-butanol, (iii) 2-methyl-2-propanol (tert-butanol), (iv) benzyl alcohol.
• Which one of the following compounds does NOT react with aqueous sodium hydrogen carbonate (NaHCO3) solution to produce a gas?

⚔️ Practice JEE Advanced Chemistry free + battle 1v1 →