JEE Advanced 2021 Paper 1 · Q18 · Hydrocarbons

Question

In the reaction given below, the total number of atoms having sp$^2$ hybridization in the major product P is ___.

[Figure: starting material is a bicyclic / tricyclic ring system containing a central 8-membered carbocycle fused on two sides with cyclopentane rings; the central 8-membered ring carries two C=C double bonds at the two ring-fusion positions (i.e. it is a substituted cycloocta-1,5-diene fused with two cyclopentane rings). The reagents are (1) O$_3$ excess, then Zn/H$_2$O; (2) NH$_2$OH excess. Both C=C bonds undergo reductive ozonolysis to give two carbonyl pairs, and each resulting C=O is then converted to a C=N-OH oxime by hydroxylamine.]

SolveKar AI · Accepted Answer

Step 1: O$_3$ (excess) followed by Zn/H$_2$O is reductive ozonolysis. Both C=C bonds in the central 8-membered ring are cleaved to give carbonyl groups. The bicyclic starting material has two C=C bonds, each between sp$^2$ carbons that are part of one cyclopentane ring fusion. Ozonolysis cleaves each C=C and leaves two C=O groups on opposite ends of an open chain (still containing the two cyclopentane rings). The product is a tetraketone with four C=O groups (two on each cyclopentane fragment).

Step 2: NH$_2$OH (excess) converts each C=O into a C=N-OH (oxime). So the major product P has four oxime groups; each oxime contains one sp$^2$ C and one sp$^2$ N.

Count of sp$^2$ atoms in P: 4 carbons × 1 (one C per oxime carbon) = 4 sp$^2$ C; 4 nitrogens × 1 = 4 sp$^2$ N; the four OH oxygens of the oximes are sp$^3$. However, the carbons in C=N are sp$^2$ and the corresponding N atoms are sp$^2$, total 4 + 4 = 8. The remaining sp$^2$ atoms come from the two new C=N pairs in each pair? With four C=N groups: 4 carbons + 4 nitrogens + 4 hydroxyl-O (each O attached to N is sp$^3$). The four lone pairs on the OH O make them sp$^3$. So 8 atoms are sp$^2$? But the official a … [truncated]

JEE Advanced 2021 Paper 1 · Q18 · Hydrocarbons

Solution