1849 Connecticut gubernatorial election

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.

1849 Connecticut gubernatorial election

← 1848 April 2, 1849 1850 →
 
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%

County results
Trumbull:      40–50%      50–60%
Seymour:      40–50%

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.[2] This was the first of six consecutive elections in which the Free Soil Party participated.

General election

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Candidates

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Major party candidates

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

Minor party candidates

  • John M. Niles, Free Soil

Results

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1849 Connecticut gubernatorial election[3]
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

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  1. ^ "Political Intelligence". The New York herald. New York, N.Y. April 2, 1849. p. 2. Retrieved May 1, 2022.
  2. ^ "Gov. Joseph Trumbull", National Governors Association, retrieved 09-15-2020
  3. ^ "Our Campaigns". Retrieved September 15, 2020.