[pfnm]

The question reads:

"Ozonolysis of Compound A yielded compound B as the sole organic product, which upon reaction with two mole equivalents of methyl magnesium iodide with an acidic workup gave compound C. Oxidation of C with PCC produced compound D. Compound D has a molecular formula of C8H12O2. Treatment of D with excess methylamine (under acid catalysis) produced Compound E. Compound E has a molecular formula of C10H20N2. Proton NMR of compound A exhibits a multiplet at 1.6ppm, a multiplet at 2.0ppm, and a singlet at 5.5ppm. The off-resonance decoupled 13 Carbon NMR spectrum of A shows no quartets."

What is Compound A?

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D and E have 2 IHD each.

What I think so far:
-A is a symmetrical alkene (produces only one product on cleavage)  no methyl groups in Compound A (from the off-resonance decoupled cNMR showing no quartets)
- D is a diketone which forms a diimine (E) when reacted with excess CH3-NH2
-A has 6 carbons (if the alkene is in the ring) or 12 carbons (if the alkene is symmetrical and has a ring on either side of the double bond)
- B is a dialdehyde, and reacts with the CH3MgBr twice to form a diol, which is then oxidised to a diketone

A is cyclohexene (it seems to fit the NMR data)

Is this right? I am pretty uncertain,

Thanks

[frisbee]

 Your assignment for A sounds right. The fact that compound D has formula C₈H₁₂O₂ seems wrong. Are you sure there wasn't a typo and it's not C₈H₁₄O₂?

[pfnm]

Yep, thanks for pointing that out, it was a typo, its C8H14O2