Exams › JEE Advanced › Chemistry
Arrange the following compounds in increasing order of their boiling points: (I) C2H5OH (II) C2H5Cl (III) C2H5CH3 (i.e. propane) (IV) C2H5OCH3 (methoxyethane)
- (III) < (II) < (I) < (IV)
- (II) < (III) < (IV) < (I)
- (IV) < (III) < (I) < (II)
- (III) < (IV) < (II) < (I)
Correct answer: (III) < (IV) < (II) < (I)
Solution
Propane (only weak dispersion) is lowest, then methoxyethane (weak dipole, no H-bond), then chloroethane (larger dipole/heavier), and ethanol highest due to strong O-H hydrogen bonding.
Related JEE Advanced Chemistry questions
- Which compound listed below cannot be synthesized using the standard Williamson ether synthesis?
- 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.
- 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.
⚔️ Practice JEE Advanced Chemistry free + battle 1v1 →