Monocyclic compounds P, Q, R and S are the major products formed in the reaction sequences given below. [figure] The product having the highest number of unsaturated carbon atom(s) is
- P
- Q
- R
- S
Answer
(4) S
Solution
Unsaturated (sp2 / sp) carbons are counted in each product.
P (HVZ reaction – -bromination of the acid):
[figure] [(ii) H2O](i) Br2/Red P [figure] (P)
Unsaturated C = 6 (ring) + 1 (COOH) = 7
Q (cross aldol, then dehydration – Claisen–Schmidt):
[figure] + [figure] aq. NaOH, 293 K [figure] (Q)
Unsaturated C = 6 + 2 + 1 = 9
R (terminal alkyne → acetylide; SN2 on allyl bromide; then Hg2+ hydration of the C≡C):
[figure] NaNH2 [figure] CH2=CH-CH2Br [figure]
Hg2+, H3O+ [figure] (R)
Unsaturated C = 6 + 1 (C=O) + 2 (C=C) = 9
S (ozonolysis opens the five-membered ring to a keto-aldehyde; both C=O groups take one CH3MgBr; acid/heat dehydrates both alcohols):
[figure] O3, Zn-H2O [figure]
CH3MgBr (2 equiv.), then H2O [figure]
H+, (-2H2O) [figure] (S)
Unsaturated C = 6 + 2 + 2 = 10
S has the highest number (10) of unsaturated carbon atoms.
Answer: (D)