1849 Connecticut gubernatorial election

The 1849 Connecticut gubernatorial election was held on April 4, 1849. Former congressman and Whig nominee Joseph Trumbull defeated former congressman and Democratic nominee Thomas H. Seymour as well as former Senator and Free Soil nominee John M. Niles with 49.35% of the vote. Niles had previously been the Democratic nominee for this same office in 1840.

1849 Connecticut gubernatorial election

April 4, 1849

Nominee Joseph Trumbull Thomas H. Seymour John M. Niles Party Whig Democratic Free Soil Electoral vote 122 110 Popular vote 27,800 25,018 3,520 Percentage 49.35% 44.41% 6.25%

Governor before election Clark Bissell Whig Elected Governor Joseph Trumbull Whig

Trumbull won a plurality of the vote, but fell short of a majority. As a result, the Connecticut General Assembly elected the governor, per the state constitution. Trumbull won the vote over Seymour 122 to 110 in the General Assembly, and became the governor.[1] This was the first of six consecutive elections in which the Free Soil Party participated.

General election

Candidates

Major party candidates

• Joseph Trumbull, Whig
• Thomas H. Seymour, Democratic

Minor party candidates

• John M. Niles, Free Soil

Results

1849 Connecticut gubernatorial election[2]

Party		Candidate	Votes	%	±%
	Whig	Joseph Trumbull	27,800	49.35%
	Democratic	Thomas H. Seymour	25,018	44.41%
	Free Soil	John M. Niles	3,520	6.25%
Plurality			2,782
Turnout

1849 Connecticut gubernatorial election, contingent General Assembly election

Party		Candidate	Votes	%	±%
	Whig	Joseph Trumbull	122	52.59%
	Democratic	Thomas H. Seymour	110	47.41%
Majority			12
	Whig hold		Swing

References

1. "Gov. Joseph Trumbull", National Governors Association, retrieved 09-15-2020
2. "Our Campaigns". Retrieved 2020-09-15.