1836 United States presidential election in Alabama explained

See main article: 1836 United States presidential election.

Election Name:1836 United States presidential election in Alabama
Country:Alabama
Type:presidential
Ongoing:no
Previous Election:1832 United States presidential election in Alabama
Previous Year:1832
Next Election:1840 United States presidential election in Alabama
Next Year:1840
Election Date:November 3 – December 7, 1836
Image1:Martin Van Buren circa 1837 crop.jpg
Nominee1:Martin Van Buren
Party1:Democratic Party (United States)
Home State1:New York
Running Mate1:Richard Johnson
Electoral Vote1:7
Popular Vote1:20,638
Percentage1:55.34%
Nominee2:Hugh White
Party2:Whig Party (United States)
Home State2:Tennessee
Running Mate2:John Tyler
Electoral Vote2:0
Popular Vote2:16,658
Percentage2:44.66%
Map Size:200px
President
Before Election:Andrew Jackson
Before Party:Democratic Party (United States)
After Election:Martin Van Buren
After Party:Democratic Party (United States)
Colour2:800080

The 1836 United States presidential election in Alabama took place between November 3 and December 7, 1836, as part of the 1836 presidential election. Voters chose seven representatives, or electors, to the Electoral College, who voted for President and Vice President.

Alabama voted for the Democratic candidate, Martin Van Buren, over Whig candidate Hugh White. Van Buren won Alabama by a margin of 10.68%.

Results

1836 United States presidential election in Alabama[1]
PartyCandidateVotesPercentageElectoral votes
DemocraticMartin Van Buren20,63855.34%7
WhigHugh White16,65844.66%0
Totals37,296100.00%7

See also

Notes and References

  1. Web site: 1836 Presidential General Election Results - Alabama. U.S. Election Atlas. August 4, 2012.