1819 French legislative election explained

Election Name:1819 French legislative election
Country:Bourbon Restoration
Type:Parliamentary
Election Date:11 September 1819
Previous Election:1817 French legislative election
Previous Year:1817
Next Election:1820 French legislative election
Next Year:1820
Party1:Liberal Party
Color1:FDEE00
Seats1:35
Party2:Doctrinaires
Color2:9966CC
Seats2:13
Party3:Ultra-Royalists
Color3:000080
Seats3:5
Ongoing:no
Flag Image:Royal flag of France during the Bourbon Restoration.svg

The 1819 French partial election took place on 11 September 1819, during the Second Restoration, to choose delegates to the French Chamber of Deputies. It was the third of three elections (the others coming in 1817 and 1818) under a new law that called for legislative elections to be held annually in one-fifth of the nation's departments.[1]

A total of 53 seats were contested. For the third straight election, the Liberals (left-opposition) made a strong showing, picking up 35 seats.[2] [3] The election of Grégoire caused a scandal, as he was a famous member of the Convention, which forced the government of Decazes to cancel this election. Nevertheless, the liberal group represented nearly one-third of the Lower House after this partial election, whereas the Ultras were reduced to 30 MPs.

Results

PartySeats
 Liberals35
 Doctrinaires13
 Ultra-royalists5

Notes and References

  1. Thomas D. Beck, French Legislators, 1800-1834: A Study in Quantitative History (University of California Press, 1974), pp. 63-71.
  2. Le Constitutionnel, 20 September 1819 (No. 264), p. 2. Accessed at the Gallica Digital Library, 17 April 2014.
  3. Journal des débats politiques et littéraires, 20 September 1819, p. 2. Accessed at the Gallica Digital Library, 17 April 2014.