1844 Connecticut gubernatorial election
From Wikipedia, the free encyclopedia
The 1844 Connecticut gubernatorial election was held on April 1, 1844.[1][2] Former state legislator, Amistad lawyer and Whig nominee Roger Sherman Baldwin was elected, defeating incumbent governor and Democratic nominee Chauncey Fitch Cleveland with 49.41% of the vote.
April 1, 1844
| ||||||||||||||||||||
| ||||||||||||||||||||
Baldwin: 40â50% 50â60% 60â70% 70â80% Cleveland: 40â50% 50â60% 60â70% 70â80% Tie: 50% | ||||||||||||||||||||
| ||||||||||||||||||||
Although Baldwin won a plurality of the vote, he fell short of a majority. The state constitution at the time required that in such a case, the Connecticut General Assembly decides the election. The state legislature voted for Baldwin, 116 to 93, and Baldwin became the governor.[3]
General election
Candidates
Major party candidates
- Roger Sherman Baldwin, Whig
- Chauncey Fitch Cleveland, Democratic
Minor party candidates
- Francis Gillette, Liberty
Results
| Party | Candidate | Votes | % | ±% | |
|---|---|---|---|---|---|
| Whig | Roger Sherman Baldwin | 30,093 | 49.41% | ||
| Democratic | Chauncey Fitch Cleveland (incumbent) | 28,846 | 47.36% | ||
| Liberty | Francis Gillette | 1,971 | 3.24% | ||
| Plurality | 1,247 | ||||
| Turnout | |||||
| Party | Candidate | Votes | % | ±% | |
|---|---|---|---|---|---|
| Whig | Roger Sherman Baldwin | 116 | 55.50% | ||
| Democratic | Chauncey Fitch Cleveland (incumbent) | 93 | 44.50% | ||
| Majority | 23 | ||||
| Whig gain from Democratic | Swing | ||||