Sodium Iodide → HCID 67530671, compound

Participants

Reaction Steps

This compound was prepared in a manner analogous to that of Step E of Example 2, by 1) the reaction of 0.9 gram (0.0046 mole) of 6-methoxy-7-methylchroman-4-one (ii) with 1.40 grams (0.0138 mole) of trimethylsilyl cyanide, in the presence of 0.1 gram (catalyst) of aluminum chloride in 30 mL of toluene, affording an intermediate product, namely: 6-methoxy-7-methyl-4-(1,1-dimethyl-1-silaethoxy)chromane-4-carbonitrile (cyano-silyl intermediate), then 2) the reaction of the I-silaethoxy intermediate with 2.35 mL (0.0184 mole) of trimethylsilyl chloride, 2.8 grams (0.0184 mole) of sodium iodide, and 0.12 mL of water in 30 mL of acetonitirile, yielding 0.6 gram of subject compound. Contrary to Step E of Example 2, the hydrogenation step was not necessary to obtain the subject compound. The NMR spectrum was consistent with the proposed structure.

Provenance & Evidence