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When Was John Quincy Adams First Election ?
John Quincy Adams was elected as the 6th President of the United States. During the elections of 1824, he ran along with four other candidates. The Speaker of the House, Henry Clay from Kentucky; William H Crawford from Georgia; Andrew Jackson from Tennessee; and John C Calhoun from South Carolina were the four runners for presidency. |
John Quincy Adams was the son of the second president of the United States, John Adams. Besides Crawford, he did not have any real competition among the other presidential hopefuls. However, Crawford suffered from a stroke in the mean time. He had the majority of the electoral votes. Jackson had been more popular than him but due to his health, he lost.
Under the rules of the Twelfth Amendment, the presidential election was left to the House of Representatives to vote for the top candidate. So, it was Jackson, Adams and Crawford in order and Clay was fourth. This made Clay ineligible. Crawford became unviable due to the stroke. So, Adams by default became the president.
Clay disliked John Adams and his take on tariffs. He knew that the south was at loss due to this so he never supported Adams. Adams victory actually left Jackson in a deep shock. He thought of himself as the president, and could not believe that John Adams could overtake him. Adams appointed Clay as the Secretary of State. Due to this, the democrats were outraged. They believed that Adams and Clay had struck a corrupt deal. This contention remained throughout Adams presidency, and is believed to have caused the loss during his second attempted to get elected.
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