Question

Among the following, identify the compound which will form resonance stabilized carbocations after the leaving group is lost.

Compounds: I, II, III, IV

• A. I & II only
• B. I & III only
• C. II & III only
• D. II & IV only

Hint:

Benzyl and allyl carbocations are resonance stabilized because the positive charge can delocalize into the \(\pi\)-system.

Answer 1

The Correct Option is D

Concept: A carbocation is resonance stabilized when the positive charge is adjacent to a \(\pi\)-bond or aromatic ring, allowing delocalization.

Step 1: Check Compound I Compound I is chlorobenzene. On loss of \(Cl^-\): \[ C_6H_5^+ \] This is a phenyl carbocation. It is highly unstable and does not get resonance stabilization. \[ \boxed{\text{Not correct}} \]

Step 2: Check Compound II Compound II is benzyl chloride. On loss of \(Cl^-\): \[ C_6H_5-CH_2^+ \] This is a benzyl carbocation. Positive charge gets delocalized over the benzene ring. Hence resonance stabilized. \[ \boxed{\text{Correct}} \]

Step 3: Check Compound III Compound III is vinylic bromide. On loss of \(Br^-\): \[ CH_3-CH^+=CH_2 \] This is a vinylic carbocation. Vinylic carbocations are unstable and not resonance stabilized. \[ \boxed{\text{Not correct}} \]

Step 4: Check Compound IV Compound IV is allyl bromide. On loss of \(Br^-\): \[ CH_2=CH-CH_2^+ \] This is an allyl carbocation. Positive charge is delocalized over the double bond. Hence resonance stabilized. \[ \boxed{\text{Correct}} \] Thus the correct compounds are: \[ II \text{ and } IV \] \[ \boxed{\text{II \& IV only}} \] Therefore, the correct answer is: \[ \boxed{(D)} \]