The 1860 United States presidential election in Missouri took place on November 2, 1860, as part of the 1860 United States presidential election. Voters chose nine representatives, or electors to the Electoral College, who voted for president and vice president.

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1860 United States presidential election in Missouri

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Nominee Stephen A. Douglas John Bell Party Democratic Constitutional Union Home state Illinois Tennessee Running mate Herschel V. Johnson Edward Everett Electoral vote 9 0 Popular vote 58,801 58,372 Percentage 35.52% 35.26%   Nominee John C. Breckinridge Abraham Lincoln Party Southern Democratic Republican Home state Kentucky Illinois Running mate Joseph Lane Hannibal Hamlin Electoral vote 0 0 Popular vote 31,362 17,028 Percentage 18.94% 10.28%

County Results[lower-alpha 1] Douglas   40-50%   50-60%   60-70% Bell   40-50%   50-60%   60-70% Breckinridge   40-50%   50-60%   60-70%   80-90% Lincoln   40-50%   50-60% Unknown/No Vote

President before election James Buchanan Democratic Elected President Abraham Lincoln Republican

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Missouri was won by Democratic candidate Stephen A. Douglas, by a very narrow margin of 0.26%. The state was the only one to fully give its votes to Douglas, though he won the popular vote and three of the seven electoral votes from New Jersey under a fusion ticket.

As of the 2020 presidential election^[update], this is the last occasion when Putnam County, Ozark County, and Taney County voted for a candidate running as a Democrat, with the former voting for Douglas and the latter two voting for Breckinridge. This was the first time in American history that four candidates each won at least one county in the same state, something that has only been repeated in 1912.^[1]

• United States presidential elections in Missouri