1836 United States presidential election in Vermont

Main article: 1836 United States presidential election

1836 United States presidential election in Vermont

←  1832 November 15, 1836 1840  →

←  MA NC  →

Nominee William Henry Harrison Martin Van Buren Party Whig Democratic Home state Ohio New York Running mate Francis Granger Richard M. Johnson Electoral vote 7 0 Popular vote 20,994 14,037 Percentage 59.93% 40.07%

County results Harrison   50–60%   60–70%   70–80% Van Buren   50–60% Unknown/No vote

President before election Andrew Jackson Democratic Elected President Martin Van Buren Democratic

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.

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 M. Johnson of Kentucky	14,037	40.07%	0	0.00%
Total				35,031	100.00%	7	100.00%

See also

• United States presidential elections in Vermont

References

1. "Presidential Elections". Weekly Messenger. November 12, 1836.
2. "1836 Presidential General Election Results - Vermont". U.S. Election Atlas. Retrieved December 23, 2013.