The 1836 United States presidential election in Arkansas took place between November 3 and December 7, 1836, as part of the 1836 United States presidential election . Voters chose three representatives, or electors to the Electoral College , who voted for President and Vice President .
Arkansas , having been admitted to the Union as the 25th state on June 15, 1836, voted for the Democratic candidate, Martin Van Buren , over Whig candidate Hugh White during its first presidential election. Van Buren won Arkansas by a margin of 28.16%.
1836 United States Presidential Election in Arkansas (By County)[ 2] County Martin Van Buren Democratic
Hugh Lawson White Whig
Total # % # % Arkansas 38 31.67% 82 68.33% 120 Carroll 73 92.41% 6 7.59% 79 Chicot 43 45.74% 51 54.26% 94 Clark 41 85.42% 7 14.58% 48 Conway 23 32.39% 48 67.61% 71 Crawford 109 47.81% 119 52.19% 228 Crittenden 38 58.46% 27 41.54% 65 Hempstead 110 56.99% 83 43.01% 193 Hot Spring 11 68.75% 5 31.25% 16 Independence 134 54.25% 113 45.75% 247 Izard 87 85.29% 15 14.71% 102 Jackson 56 53.85% 48 46.15% 104 Jefferson 50 51.02% 48 48.98% 98 Johnson 107 72.79% 40 27.21% 147 Lafayette 30 73.17% 11 26.83% 41 Lawrence 82 72.57% 31 27.43% 113 Monroe 17 32.69% 35 67.31% 52 Phillips 96 59.63% 65 40.37% 161 Pike 33 100.00% 0 0.00% 33 Pope 93 66.91% 46 33.09% 139 Pulaski 234 55.06% 191 44.94% 425 Randolph 138 89.03% 17 10.97% 155 Saline 81 60.90% 52 39.10% 133 Sevier 67 66.34% 34 33.66% 101 St. Francis 108 85.71% 18 14.29% 126 Van Buren 19 67.86% 9 32.14% 28 Washington 622 82.38% 133 17.62% 755 White 17 36.96% 29 63.04% 46 Totals 2,380 64.08% 1,334 35.92% 3,714
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