The specific rotation of (r)-2-butanol is −13.5°. what specific rotation would be observed for a 3: 1 mixture of (s): (r) 2-butanol?

If the specific rotation of (r)-2-butanol is −13.5° then the specific rotation of (s)-2-butanol is simply the opposite in sign 13.5°. In a 3:1 mixture of (s):(r) 2-butanol, 75% is (s) while 25% is r, so the specific rotation of the mixture is:

 

mixture rotation = 0.75 * (13.5°) + 0.25 * (- 13.5°)

mixture rotation = 6.75°

the catch is that the top numbers are atomic masses the bottom numbers are the numbers of protons and electrons in an uncharged atom. they are the number of that atoms as you would find them on the periodic table. further, the numbers on the left are used to determine the missing numbers on the right

so here we go

atomic mass

239 + 1 = y + 134 + 3 combine like terms

240 = y + 137             subtract 137 from both terms

240 - 137 = y

103 = y

atomic number

94 = z +54 + 0

94 = z + 54               subtract 54 from both sides.

40 = z

answer

; lc