1868 United States presidential election in Oregon explained

See main article: 1868 United States presidential election.

Election Name:1868 United States presidential election in Oregon
Country:Oregon
Flag Year:1868
Type:presidential
Ongoing:no
Previous Election:1864 United States presidential election in Oregon
Previous Year:1864
Next Election:1872 United States presidential election in Oregon
Next Year:1872
Election Date:November 3, 1868
Image1:Hon. Horatio Seymour, N.Y - NARA - 528568 (cropped).jpg
Nominee1:Horatio Seymour
Party1:Democratic Party (United States)
Home State1:New York
Running Mate1:Francis Preston Blair, Jr.
Electoral Vote1:3
Popular Vote1:11,125
Percentage1:50.37%
Nominee2:Ulysses S. Grant
Party2:Republican Party (United States)
Home State2:Illinois
Running Mate2:Schuyler Colfax
Electoral Vote2:0
Popular Vote2:10,961
Percentage2:49.63%
Map Size:375px
President
Before Election:Andrew Johnson
Before Party:Democratic Party (United States)
After Election:Ulysses S. Grant
After Party:Republican Party (United States)

The 1868 United States presidential election in Oregon took place on November 3, 1868, as part of the 1868 United States presidential election. Voters chose three representatives, or electors to the Electoral College, who voted for president and vice president.

Oregon voted for the Democratic nominee, Horatio Seymour over the Republican nominee, Ulysses S. Grant. Seymour won the state by a narrow margin of 0.74%.

As a result of his win, Seymour became the first Democratic presidential candidate to ever win Oregon. Another Democrat would not win Oregon again on a presidential level until Woodrow Wilson won the state in 1912. Democrats wouldn't again win a majority in Oregon until Franklin D. Roosevelt won the state in 1932.

This was the only time a Republican would win a presidential election without Oregon until 1988, 120 years later; every Republican victor since then has lost the state.

See also