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%.
| 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 | ||