Exams › JEE Main › Chemistry

Arrange the following carbocations in decreasing order of stability: I. CH2=CH–CH2+ II. CH3–CH2–CH2+ III. C6H5–CH2+

1. I > II > III
2. II > III > I
3. III > II > I
4. III > I > II

Correct answer: III > I > II

Solution

C6H5-CH2+ (benzyl) is most stabilised by ring resonance, CH2=CH-CH2+ (allyl) is resonance-stabilised, and CH3-CH2-CH2+ is an unstabilised primary cation. Order: III > I > II (index 3).

Related JEE Main Chemistry questions

• Lactic acid loses its optical activity when the hydroxyl group is replaced by hydrogen because
• Which is the most suitable technique for separating a mixture of sugars?
• Which statement correctly compares the staggered and eclipsed conformations of ethane?
• Which statement correctly describes a common feature shared by optical and geometrical isomerism?
• Which substituent lowers the reactivity of a benzene ring toward electrophilic substitution, yet still mainly directs a new group to the ortho and para positions?
• Among the following substances, which has the greatest affinity for a proton?

⚔️ Practice JEE Main Chemistry free + battle 1v1 →