| Elections in Alabama |
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The 1824 United States presidential election in Alabama took place between October 26 and December 2, 1824, as part of the 1824 presidential election. Voters chose five representatives, or electors, to the Electoral College, who voted for President and Vice President.
During this election, the Democratic-Republican Party was the only major national party, and four different candidates from this party sought the Presidency. Alabama voted for Andrew Jackson over John Quincy Adams, William H. Crawford and Henry Clay. Jackson won Alabama by a margin of 51.52%. This was the first time since achieving statehood in 1819 that Alabama backed the losing candidate in a presidential election.
Results
| 1824 United States presidential election in Alabama[1] | |||||
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| Party | Candidate | Votes | Percentage | Electoral votes | |
| Democratic-Republican | Andrew Jackson | 9,429 | 69.32% | 5 | |
| Democratic-Republican | John Quincy Adams | 2,422 | 17.80% | 0 | |
| Democratic-Republican | William H. Crawford | 1,656 | 12.17% | 0 | |
| Democratic-Republican | Henry Clay | 96 | 0.71% | 0 | |
| Totals | 13,423 | 100.00% | 5 | ||
See also
References
- ^ "1824 Presidential General Election Results - Alabama". U.S. Election Atlas. Retrieved February 27, 2013.
State and district results of the 1824 United States presidential election | ||
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