1849 Connecticut gubernatorial election
 | |||||||||||||||||||||||||
| |||||||||||||||||||||||||
| |||||||||||||||||||||||||
 County results Trumbull: 40–50% 50–60% Seymour: 40–50% | |||||||||||||||||||||||||
| |||||||||||||||||||||||||
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
Candidates
Major party candidates
- Joseph Trumbull, Whig
- Thomas H. Seymour, Democratic
Minor party candidates
- John M. Niles, Free Soil
Results
| 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
| |||
|---|---|---|---|
| General | |||
| State Senate | |||
| State House | |||
| Governor |
| ||
| U.S. President | |||
| U.S. Senate |
| ||
| U.S. House |
| ||
| |||
