1836 United States presidential election in Alabama

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.

1836 United States presidential election in Alabama

November 3 – December 7, 1836
 
Nominee Martin Van Buren Hugh White
Party Democratic Whig
Home state New York Tennessee
Running mate Richard Johnson John Tyler
Electoral vote 7 0
Popular vote 20,638 16,658
Percentage 55.34% 44.66%

President before election

Andrew Jackson
Democratic

Elected President

Martin Van Buren
Democratic

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]
Party Candidate Votes Percentage Electoral votes
Democratic Martin Van Buren 20,638 55.34% 7
Whig Hugh White 16,658 44.66% 0
Totals 37,296 100.00% 7

See also

References

  1. "1836 Presidential General Election Results - Alabama". U.S. Election Atlas. Retrieved August 4, 2012.


This article is issued from Wikipedia. The text is licensed under Creative Commons - Attribution - Sharealike. Additional terms may apply for the media files.