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.
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Elections in Alabama |
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Government |
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] | |||||
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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 | ||
References
- "1836 Presidential General Election Results - Alabama". U.S. Election Atlas. Retrieved August 4, 2012.
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