1836 United States presidential election in Alabama

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United States presidential election in Alabama, 1836

← 1832 November 3 – December 7, 1836 1840 →
  Van Buren.jpg HLWhite.jpg
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

The 1836 United States presidential election in Alabama took place between November 3 and December 7, 1836, as part of the 1836 United States 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[edit]

United States presidential election in Alabama, 1836[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 37,296 100.0% 7

References[edit]

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