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The acid catalyzed cyclization of 5-ketodecan-1,9-diol is given below

Decane-1,9-diol-5-one drawn as a zigzag chain, HO at C1, C=O at C5 and OH at C9, treated with p-TSA in benzene with heat to give a spiroketal

(p-TSA = p-toluenesulfonic acid)

The most predominant spiroketal is

  • A.2-Methyl-1,7-dioxaspiro[5.5]undecane chair drawing with the methyl equatorial and the second ring oxygen axial at the spiro carbon
  • B.2-Methyl-1,7-dioxaspiro[5.5]undecane chair drawing with the methyl axial and the second ring oxygen axial at the spiro carbon
  • C.2-Methyl-1,7-dioxaspiro[5.5]undecane chair drawing with the methyl axial and the second ring oxygen equatorial at the spiro carbon
  • D.2-Methyl-1,7-dioxaspiro[5.5]undecane chair drawing with the methyl equatorial and the second ring oxygen equatorial at the spiro carbon

Answer: A

The predominant spiroketal is the one with both ring oxygens axial at the spiro carbon and the methyl group equatorial (A).

Under acid and heat the ketone and the two hydroxyl groups form 2-methyl-1,7-dioxaspiro[5.5]undecane, and the reaction is reversible, so the most stable isomer accumulates. Two effects decide which one that is:

  1. the anomeric effect: each C-O bond at the spiro carbon prefers to be axial with respect to the other ring, where an oxygen lone pair can donate into its σ* orbital. Structure A has this doubly anomeric arrangement;
  2. sterics: the methyl group prefers the equatorial position on its chair.

A satisfies both. B has the right anomeric arrangement but an axial methyl. C and D place the second ring's oxygen equatorial, losing anomeric stabilisation, and C also has the methyl axial.

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