1812 United States presidential election in New York
The 1812 United States presidential election in New York took place between October 30 and December 2, 1812, as part of the 1812 United States presidential election. The state legislature chose 29 representatives, or electors to the Electoral College, who voted for President and Vice President.
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During this election, New York cast its 29 electoral votes to Independent Democratic Republican and Federalist supported candidate DeWitt Clinton, who was then currently serving as the Mayor of New York City and the Lieutenant Governor of New York.
The election ultimately hinged on both New York and neighboring Pennsylvania,[1] and while Clinton was able to take his home state, he failed to take Pennsylvania and thus lost the election to traditional Democratic Republican candidate and incumbent President James Madison won by a narrow margin. This would be the first time New York would vote for a losing presidential candidate. It would also be the only time that happened until 1856.
Notes
- While commonly labeled as the Federalist candidate, Clinton technically ran as a Democratic-Republican and was not nominated by the Federalist party itself, the latter simply deciding not to field a candidate. This did not prevent endorsements from state Federalist parties (such as in Pennsylvania), but he received the endorsement from the New York state Democratic-Republicans as well.
References
- Sabato, Larry; Ernst, Howard (1 January 2009). Encyclopedia of American Political Parties and Elections. Infobase Publishing. pp. 303–304. ISBN 9781438109947. Retrieved 19 December 2018.
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| Other 1812 elections | ||