An alkene (X) on ozonolysis followed by reduction gives following products. H- O \|C-H (2 moles), H- O \|C- O \|C- O \|C-H, CH3- O \|C- O \|C- O \|C-CH3 The alkene (X) is :
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Answer
(4) [figure]
Solution
Reductive ozonolysis converts each >C=C< into two >C=O groups.
C=O groups in the products: 2 (2 HCHO) + 3 (OHC–CO–CHO) + 3 (CH3COCOCOCH3) = 8 ⇒ X has 4 C=C.
Carbon atoms: 2 + 3 + 5 = 10 ⇒ X is a C10 compound.
2 HCHO ⇒ two terminal =CH2 groups.
The two tricarbonyl fragments must come from one six-membered ring with two ring C=C bonds:
CH3-C(=O)-C(=O)-C(=O)-CH3 gives the ring chain C(CH3)-C-C(CH3), and OHC–CO–CHO gives CH-C-CH.
Joining them, each terminal CH of OHC–CO–CHO is double-bonded to a C(CH3) carbon, and both middle carbons carry the =CH2 groups:
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Its ozonolysis: ring C=C bonds → CH3COCOCOCH3 + OHC–CO–CHO; two C=CH2 → 2 HCHO. This is structure (4).
(Structures (1) and (2) have a ring CH2 or CH(CH3) and give different products; (3) has 11 C atoms.)
Answer: (4)