See image — GOC and Organic Chemistry Basics Chemistry Question

💡 Solution & Explanation

# Solution: Reaction of Alkene with Isonitrile ($IN_3^⊖$) ## Mechanism Analysis **Step 1: Electrophilic Addition** The isonitrile carbon ($IN_3^⊖$) acts as a nucleophile with a lone pair. It attacks the $\pi$-electrons of the alkene $R-CH=CH_2$, forming a new C-N bond. **Step 2: Carbocation Intermediate** This generates a carbocation intermediate at the terminal carbon: $$R-\overset{+}{CH}-CH_2-IN_3$$ **Step 3: Nucleophilic Attack by Azide** The azide anion ($N_3^⊖$), generated in situ or present, attacks the carbocation from the back side (Markovnikov's rule preference at the more substituted carbon): $$R-CH-CH_2-IN_3 \rightarrow R-CH(N_3)-CH_2-IN_3$$ However, the key point is that the isonitrile group adds to form an iminodium intermediate, which then gets trapped by azide to give: $$R-\overset{+}{\overset{|}N=}CH-CH_2N_3 \rightarrow R-CH(I)-CH_2N_3$$ where $I$ represents the isonitrile moiety becoming an iodine-like leaving group analog, or more correctly: **the product is $R-CH(IN_3)-CH_2$ with the imine intermediate collapsing**. ## Why Option A is Correct **Option (A)** shows $R-CH-CH_2N_3$ with iodine substitution at the first carbon. This represents the addition of $IN_3$ across the double bond following Markovnikov's rule, with the isonitrile nitrogen group attaching to the secondary carbon and azide to the primary carbon. **Why others are wrong:** - **(B)** incorrectly shows $N_3$ as a substituent without proper bonding - **(C)** forms a 3-membered ring (aziridine), inconsistent with this reaction type - **(D)** forms a 2-membered ring (impossible geometry) and doesn't account for azide