1849 Connecticut gubernatorial election
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County results Trumbull: 40–50% 50–60% Seymour: 40–50% | |||||||||||||||||||||||||
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Elections in Connecticut |
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The 1849 Connecticut gubernatorial election was held on April 2, 1849.[1] 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.
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.[2] This was the first of six consecutive elections in which the Free Soil Party participated.
General election
[edit]Candidates
[edit]Major party candidates
- Joseph Trumbull, Whig
- Thomas H. Seymour, Democratic
Minor party candidates
- John M. Niles, Free Soil
Results
[edit]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 |
Party | Candidate | Votes | % | ±% | |
---|---|---|---|---|---|
Whig | Joseph Trumbull | 122 | 52.59% | ||
Democratic | Thomas H. Seymour | 110 | 47.41% | ||
Majority | 12 | ||||
Whig hold | Swing |
References
[edit]- ^ "Political Intelligence". The New York herald. New York, N.Y. April 2, 1849. p. 2. Retrieved May 1, 2022.
- ^ "Gov. Joseph Trumbull", National Governors Association, retrieved 09-15-2020
- ^ "Our Campaigns". Retrieved September 15, 2020.