1832 United States presidential election in Alabama

The 1832 United States presidential election in Alabama took place between November 2 and December 5, 1832, as part of the 1832 United States presidential election. Voters chose seven representatives, or electors, to the Electoral College, who voted for president and vice president.

Main article: 1832 United States presidential election
1832 United States presidential election in Alabama
November 2 – December 5, 1832
 
Nominee Andrew Jackson
Party Democratic
Home state Tennessee
Running mate Martin Van Buren
Electoral vote 7
Popular vote 14,286
Percentage 99.97%
President before election

John Quincy Adams
Democratic-Republican

 Elected President

Andrew Jackson
Democratic

Elections in Alabama
Federal government
U.S. President
U.S. Senate
U.S. House
Government of Alabama
State-wide
Gubernatorial
Alabama House of Representatives
  • 1990
  • 1994
  • 1998
  • 2002
  • 2006
  • 2010
  • 2014
  • 2018
  • 2022
Alabama Senate
  • 1990
  • 1994
  • 1998
  • 2002
  • 2006
  • 2010
  • 2014
  • 2018
  • 2022
Ballot measures
2006
Amendment 774
Birmingham
Mayor
Montgomery
Mayor
Government

Alabama voted for the Democratic candidate, Andrew Jackson, over the National Republican candidate, Henry Clay. Jackson won Alabama by a margin of 99.94%.

Results
1832 United States presidential election in Alabama[1]
Party Candidate Votes Percentage Electoral votes
Democratic Andrew Jackson 14,286 99.97% 7
National Republican Henry Clay 5 0.03% 0
Totals 14,291 100.0% 7

References
  1. "1832 Presidential General Election Results - Alabama". U.S. Election Atlas. Retrieved April 10, 2013.


This article is issued from Wikipedia. The text is licensed under Creative Commons - Attribution - Sharealike. Additional terms may apply for the media files.