Animated Solution for Chemistry - Organic Chemistry: Among the given species the resonance stabilised carbocations are

When evaluating the stability of reactive intermediates like carbocations, resonance is one of the most powerful stabilizing effects. For resonance to occur, there must be a continuous path of overlapping p-orbitals. In simpler terms, the positive charge must be in conjugation with a $\pi$ bond (like a double or triple bond) or a lone pair of electrons. Let's analyze the four given structures to see which ones meet this critical condition.

Structure (A) is a benzyl carbocation. Here, the positively charged $C{H}_2$ group is attached directly to a benzene ring. If we look closely, the carbon bearing the positive charge has an empty p-orbital that is immediately adjacent to the $\pi$ system of the aromatic ring.

Because there is no $s{p}^3$ hybridized carbon interrupting the path, the $\pi$ electrons from the ring can easily delocalize. They shift towards the positive charge, forming an exocyclic double bond and moving the positive charge into the ring. This extensive delocalization spreads the electron deficiency over multiple atoms, making the benzyl carbocation highly stable.

Structure (B) features a cyclohexenyl ring with a $C{H}_2^{+}$ group attached. Crucially, the double bond inside the ring is located exactly one single bond away from the positive charge. This specific arrangement is known as an allylic system.

Just like in the benzyl case, the $\pi$ electrons of the $C=C$ double bond are in perfect conjugation with the empty p-orbital of the carbocation. The electrons can swing over to form a new double bond between the ring and the $C{H}_2$ group, shifting the positive charge to the adjacent carbon within the ring. Thus, structure (B) is also stabilized by resonance.

Structure (C) is a cyclohexylmethyl carbocation. Take a look at the six-membered ring—it contains only single bonds. Every carbon atom in the ring is $s{p}^3$ hybridized, meaning there are no unhybridized p-orbitals available to share electrons.

Without any adjacent $\pi$ bonds or lone pairs, there is absolutely no possibility for conjugation. The positive charge remains localized on the external carbon atom. Therefore, structure (C) does not exhibit resonance stabilization.

Structure (D) is an open-chain carbocation containing a carbonyl ($C=O$) group. While there is a $\pi$ bond present in the molecule, we must check its position relative to the positive charge.

Between the carbonyl carbon and the positively charged carbon, there is an intervening $C{H}_2$ group. This middle carbon is $s{p}^3$ hybridized. It acts as an "insulator," completely breaking the continuous overlap of p-orbitals required for electron delocalization. Because the conjugation path is broken, the $\pi$ electrons of the carbonyl group cannot interact with the empty p-orbital. Consequently, structure (D) is not stabilized by resonance.

After carefully examining all four species, we find that only the benzyl carbocation (A) and the allylic carbocation (B) possess the necessary conjugation for resonance stabilization. Therefore, the correct answer is (c) (A) and (B) only.