1836 United States presidential election in Alabama

Election in Alabama

1836 United States presidential election in Alabama

← 1832 November 3 – December 7, 1836 1840 →
 
Nominee Martin Van Buren Hugh White
Party Democratic Whig
Home state New York Tennessee
Running mate Richard Johnson John Tyler
Electoral vote 7 0
Popular vote 20,638 16,658
Percentage 55.34% 44.66%

County Results

Van Buren

  50-60%
  60-70%
  70–80%
  80–90%
  90-100%

White

  50-60%
  60-70%
  70–80%
  80–90%

Unknown/No Vote

  


President before election

Andrew Jackson
Democratic

Elected President

Martin Van Buren
Democratic

Elections in Alabama
Presidential elections
Presidential primaries
Democratic
2000
2004
2008
2016
2020
2024
Republican
2004
2008
2012
2016
2020
2024
U.S. Senate elections
U.S. House of Representatives elections
Government
  • v
  • t
  • e

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%.

Results

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

See also

References

  1. ^ "1836 Presidential General Election Results - Alabama". U.S. Election Atlas. Retrieved August 4, 2012.


Stub icon 1 Stub icon 2

This Alabama elections-related article is a stub. You can help Wikipedia by expanding it.

  • v
  • t
  • e