See main article: 1840 United States presidential election.
| Election Name: | 1840 United States presidential election in Alabama |
| Country: | Alabama |
| Type: | presidential |
| Ongoing: | no |
| Previous Election: | 1836 United States presidential election in Alabama |
| Previous Year: | 1836 |
| Next Election: | 1844 United States presidential election in Alabama |
| Next Year: | 1844 |
| Election Date: | October 30 – December 2, 1840 |
| Image1: | Martin Van Buren circa 1837 crop.jpg |
| Nominee1: | Martin Van Buren |
| Party1: | Democratic Party (United States) |
| Home State1: | New York |
| Running Mate1: | Richard Mentor Johnson |
| Electoral Vote1: | 7 |
| Popular Vote1: | 33,996 |
| Percentage1: | 54.38% |
| Nominee2: | William Henry Harrison |
| Party2: | Whig Party (United States) |
| Home State2: | Ohio |
| Running Mate2: | John Tyler |
| Electoral Vote2: | 0 |
| Popular Vote2: | 28,518 |
| Percentage2: | 45.62% |
| President | |
| Before Election: | Martin Van Buren |
| Before Party: | Democratic Party (United States) |
| After Election: | William Henry Harrison |
| After Party: | Whig Party (United States) |
| Map Size: | 320px |
The 1840 United States presidential election in Alabama took place between October 30 and December 2, 1840, as part of the 1840 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 William Henry Harrison. Van Buren won Alabama by a margin of 8.76%. As of 2020, this remains the only time in American history that Alabama has voted for a different presidential candidate than neighboring Mississippi.