United States presidential election in Alabama, 1836

United States presidential election in Alabama, 1836

November 3 – December 7, 1836

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%

President before election Andrew Jackson Democratic Elected President Martin Van Buren Democratic

The 1836 United States presidential election in Alabama took place between November 3 and December 7, 1836, as part of the 1836 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 Hugh White. Van Buren won Alabama by a margin of 10.68%.

Results

United States presidential election in Alabama, 1836[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	100.0%	7

References

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