1836 United States presidential election in Vermont
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A presidential election was held in Vermont on November 15, 1836 as part of the 1836 United States presidential election.[1] Voters chose seven representatives, or electors to the Electoral College, who voted for President and Vice President.
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November 15, 1836
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 County results
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Vermont voted for Whig candidate William Henry Harrison over Democratic candidate Martin Van Buren. Harrison won Vermont by a margin of 19.86%.
This would be the final time a Democratic candidate would carry Essex County until Franklin D. Roosevelt won it 104 years later in 1940.
1836 would stand as the strongest performance for a Democratic candidate in Vermont until 96 years later in 1932, when Franklin D. Roosevelt performed slightly better with 41.08%.
Harrison would later win Vermont again four years later when he defeated Van Buren.
Results
| 1836 United States presidential election in Vermont[2] | ||||||||
|---|---|---|---|---|---|---|---|---|
| Party | Candidate | Running mate | Popular vote | Electoral vote | ||||
| Count | % | Count | % | |||||
| Whig | William Henry Harrison of Ohio | Francis Granger of New York | 20,994 | 59.93% | 7 | 100.00% | ||
| Democratic | Martin Van Buren of New York | Richard Mentor Johnson of Kentucky | 14,037 | 40.07% | 0 | 0.00% | ||
| Total | 35,031 | 100.00% | 7 | 100.00% | ||||