The stability of carbocations depends on the resonance stabilization and inductive effects from substituents on the benzene ring. In this question, we are asked to identify the most stable carbocation among the following options.
- Option 1: \(\text{C}_6\text{H}_5\text{CH}_3^+ \): The methyl group \( \text{CH}_3 \) is an electron-donating group through inductive effects, which tends to destabilize carbocations.
- Option 2: \(\text{C}_6\text{H}_5\text{CH}_2^+ \): The benzylic carbocation is stabilized by resonance, as the positive charge can delocalize into the aromatic ring, making it more stable than an alkyl carbocation.
- Option 3: \(\text{C}_6\text{H}_4\text{CH}_3^+ \): Similar to option 1, the methyl group donates electrons inductively, which destabilizes the carbocation.
- Option 4: \(\text{C}_6\text{H}_5\text{OCH}_3^+ \): The oxygen atom in the methoxy group \( \text{OCH}_3 \) is highly electronegative and donates electron density via resonance into the ring, stabilizing the carbocation formed at the para-position. This makes this carbocation the most stable among the given options. Therefore, the most stable carbocation is \( \boxed{\text{C}_6\text{H}_5\text{OCH}_3^+} \).
0.1 mole of compound S will weigh ....... g,
(given the molar mass in g mol\(^{-1}\) \( {C} = 12, \, {H} = 1, \, {O} = 16 )\):
If \( {A}_2{B} \) is 30% ionised in an aqueous solution, then the value of van't Hoff factor \( i \) is:
Match List - I with List - II:
Choose the correct answer from the options given below: