Which one of the carbocations from the following is most stable?

Solution

Chemistry ma’am · 4 steps

All four are allyl cations, so ask which end group can push a lone pair straight onto the positive charge.

1. Common skeleton. Every option is CH₂⁺-CH=CH⁻X, an allyl cation. Resonance moves the positive charge to the carbon that carries X. So the group bonded directly to that carbon decides the stability.
2. What helps most. A group joined directly to that carbon with a free lone pair can donate it (+M). The charge then spreads onto that atom as well. The more freely the lone pair is given, the more stable the cation.

   Option	Group on the end carbon	Lone pair donation	Net effect
   (a)	-CH₂-O-CH₃	none, the CH₂ blocks conjugation	only a weak -I pull from O
   (b)	-O-CO-CH₃	weak, O also shares its lone pair with C=O	small +M
   (c)	-O-CH₃	strong, the O lone pair is fully free	large +M, every atom gets an octet
   (d)	-F	weak, F holds its lone pairs tightly	strong -I cancels most of the +M

3. Extra resonance form. In (c) the methoxy oxygen gives the form CH₂=CH-CH=O⁺-CH₃. In this form every carbon and the oxygen have a complete octet, so it adds a lot of stability.
4. So. The cation with the methoxy group directly on the double bond, option (c), is the most stable.

The idea to keep. For a carbocation, the best stabiliser is a lone pair placed directly in conjugation with the positive charge. Check that nothing blocks the conjugation and that nothing else competes for that lone pair.

Answer from the NTA final answer key. Solution by PrepWiser.