Main article: 1824 United States presidential election
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 County Results
Adams 60-70% 70-80% 80-90% 90-100%
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The 1824 United States presidential election in Connecticut took place between October 26 and December 2, 1824, as part of the 1824 United States presidential election. Voters chose eight representatives, or electors to the Electoral College, who voted for President and Vice President.
During this election, the Democratic-Republican Party was the only major national party, and four different candidates from this party sought the Presidency. Connecticut voted for John Quincy Adams over William H. Crawford, Andrew Jackson, and Henry Clay. Adams won Connecticut by a margin of 51.93%.
Results
| 1824 United States presidential election in Connecticut | |||||
|---|---|---|---|---|---|
| Party | Candidate | Votes | Percentage | Electoral votes | |
| Democratic-Republican | John Quincy Adams | 7,494 | 70.39% | 8 | |
| Democratic-Republican | William H. Crawford | 1,965 | 18.46% | 0 | |
| N/A | Other | 1,188 | 11.16% | 0 | |
| Totals | 10,647 | 100.0% | 8 | ||
See also
References
- "1824 Presidential General Election Results - Connecticut". U.S. Election Atlas. Retrieved February 27, 2013.
| Elections in Connecticut | |
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| General | |
| State Senate | |
| State House | |
| Governor |
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| U.S. President |
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| U.S. Senate |
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| U.S. House |
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| See also: Political party strength in Connecticut | |
| Elections in the United States | |
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| State and district results of the 1824 United States presidential election | ||
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