1836 United States presidential election in Alabama
Appearance
![]() | ||||||||||||||||||||||||||
| ||||||||||||||||||||||||||
| ||||||||||||||||||||||||||
|
Elections in Alabama |
---|
![]() |
![]() |
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
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
- ^ "1836 Presidential General Election Results - Alabama". U.S. Election Atlas. Retrieved 4 August 2012.