1988 United States presidential election in Oregon

1988 United States presidential election in Oregon
Dukakis 40–50% 50–60% 60–70%	Bush 40–50% 50–60% 60–70%
President before election; Ronald Reagan; Republican	Elected President ; George H. W. Bush; Republican

The 1988 United States presidential election in Oregon took place on November 8, 1988, as part of the 1988 United States presidential election. Voters chose seven electors of the Electoral College, who voted for president and vice president.

Oregon was won by Democratic nominee Massachusetts governor Michael Dukakis over Republican nominee Vice President George H. W. Bush. Oregon was one of just ten states won by Dukakis in an election overwhelmingly won by Bush. It also marked the first victory by a Democratic presidential candidate in Oregon since 1964; Democrats have won every presidential election in Oregon since then, although Bush's son and later Republican candidate George W. Bush came extremely close to winning the state in 2000 and only lost it by a somewhat slender margin in 2004.^[1] Bush's loss marked the first time that a Republican was elected President while losing Oregon since Ulysses S. Grant in 1868.

As of the 2020 presidential election^[update], this is the last occasion Washington County has voted for a Republican presidential nominee,^[2] the only time since 1948 where Oregon has not voted for the same candidate as neighboring California, and the last time that Oregon voted to the left of neighboring Washington.

^ "Oregon Presidential Election Voting History". 270ToWin.com. Retrieved June 12, 2012.
^ Sullivan, Robert David; 'How the Red and Blue Map Evolved Over the Past Century'; America Magazine in The National Catholic Review; June 29, 2016

[1] "Oregon Presidential Election Voting History". 270ToWin.com. Retrieved June 12, 2012.

[How-2] Sullivan, Robert David; 'How the Red and Blue Map Evolved Over the Past Century'; America Magazine in The National Catholic Review; June 29, 2016

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