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 →
  Joseph Trumbull Connecticut Governor.jpg ThomasSeymour.png JohnMiltonNiles.jpg
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

Elected Governor

Joseph Trumbull

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 electionEdit


Major party candidates

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

Minor party candidates

  • John M. Niles, Free Soil


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
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


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