Is q"'sym"
too obvious?
c.Expr[Any](q"'mysym")
By obvious, I mean, sorry if this is obviously not what you mean.
Less conveniently,
c.Expr[Any](c.parse("'asym"))
c.Expr[Any](c.parse(s"'$name"))
or maybe painfully,
c.Expr[Any](Apply(c.typeOf[scala.Symbol].typeSymbol, c.literal("mysym").tree))