Constructing ColorData with blue, white and red color

How about

French = Blend[{{0, Blue}, {1/2 - 0.1, White},
     {1/2 + 0.1, White}, {1, Red}}, #1] &;

Then

  French /@ (Range[15]/15.)

ContourPlot[x, {x, 0, 1}, {y, 0, 1}, ColorFunction -> French]

Any variation is possible: e.g.

French2 = Blend[{{0, Darker[Blue, 0.7]}, {0.15, Blue}, {1/2 - 0.05, 
      White},
     {1/2 + 0.05, White}, {0.9, Red}, {1, Darker[Red, 0.5]}}, #1] &;

---

Update

If you want something which looks more closely to brewer-23

col0=List[List[0,RGBColor[List[Rational[103,255],0,Rational[31,255]]]],List[Rational[1,23],RGBColor[List[Rational[8,15],Rational[3,85],Rational[7,51]]]],List[Rational[2,23],RGBColor[List[Rational[167,255],Rational[19,255],Rational[8,51]]]],List[Rational[3,23],RGBColor[List[Rational[63,85],Rational[44,255],Rational[52,255]]]],List[Rational[4,23],RGBColor[List[Rational[41,51],Rational[74,255],Rational[67,255]]]],List[Rational[5,23],RGBColor[List[Rational[44,51],Rational[104,255],Rational[28,85]]]],List[Rational[6,23],RGBColor[List[Rational[232,255],Rational[133,255],Rational[7,17]]]],List[Rational[7,23],RGBColor[List[Rational[244,255],Rational[163,255],Rational[128,255]]]],List[Rational[8,23],RGBColor[List[Rational[50,51],Rational[62,85],Rational[52,85]]]],List[Rational[9,23],RGBColor[List[Rational[254,255],Rational[211,255],Rational[188,255]]]],List[Rational[10,23],RGBColor[List[Rational[254,255],Rational[227,255],Rational[212,255]]]],List[Rational[11,23],RGBColor[List[Rational[251,255],Rational[16,17],Rational[232,255]]]],List[Rational[12,23],RGBColor[List[Rational[82,85],Rational[247,255],Rational[83,85]]]],List[Rational[13,23],RGBColor[List[Rational[77,85],Rational[241,255],Rational[49,51]]]],List[Rational[14,23],RGBColor[List[Rational[43,51],Rational[233,255],Rational[242,255]]]],List[Rational[15,23],RGBColor[List[Rational[193,255],Rational[223,255],Rational[14,15]]]],List[Rational[16,23],RGBColor[List[Rational[167,255],Rational[208,255],Rational[229,255]]]],List[Rational[17,23],RGBColor[List[Rational[46,85],Rational[193,255],Rational[13,15]]]],List[Rational[18,23],RGBColor[List[Rational[104,255],Rational[172,255],Rational[209,255]]]],List[Rational[19,23],RGBColor[List[Rational[5,17],Rational[3,5],Rational[199,255]]]],List[Rational[20,23],RGBColor[List[Rational[56,255],Rational[134,255],Rational[63,85]]]],List[Rational[21,23],RGBColor[List[Rational[14,85],Rational[23,51],Rational[12,17]]]],List[Rational[22,23],RGBColor[List[Rational[29,255],Rational[97,255],Rational[11,17]]]],List[1,RGBColor[List[Rational[6,85],Rational[74,255],Rational[134,255]]]]];

So that

col0//Transpose

Then

col[x_] = Blend[col0, x]

Would work like this:

ContourPlot[Sin[ 10 x y] , {x, 0, 1}, {y, 0, 1}, ColorFunction -> col]

Or

ContourPlot[x y , {x, -1, 1}, {y, -1, 1}, ColorFunction -> col, 
 Contours -> 26, ContourStyle -> None, PlotRange -> Full]

A cheap way to get what you want is to use Lighter[] along with "RedBlueTones" and an appropriate bell-shaped curve:

LinearGradientImage[Function[x, Lighter[ColorData["RedBlueTones", x],
                                        Sech[5 (x - 1/2)]]], {300, 30}]

ContourPlot[x y, {x, -1, 1}, {y, -1, 1}, 
            ColorFunction -> (Lighter[ColorData["RedBlueTones", #], 
                                      Sech[7.2 (# - 0.5)]] &),
            Contours -> 26, ContourStyle -> None, PlotRange -> Full]