Question

Carbon-hydrogen bonds exhibit a range of different chemical reactivity that depends on molecular structure. Classify the C-H bonds at the carbons labeled a-c in the structure below. Possible classifications are: primary, secondary, & tertiary or none if there are no hydrogens at the labeled carbon. C-H bond(s) at a C-H bond(s) at b or C-H bond(s) at v primary secondary tertiary 2 item attempts Submit Answer Try Another Version none

Answer 1

"Answer is the carbon is linked to other three carbons a carbon is not attached to any hydrogen – it is tottary (3°) eubon. If the carbon attached to only one hydrogen - ď If the carbon attached to a more hydrogen - 1 a Primary Hatt a- primary - secondary D b - secondary c- primary H. HC c- poinary @ NaBr + HSD -> Hør +NaHSDA -butanol +88 Hromopropane tho Overall rean De Naßr+H50, + Hutonoy -> Hormoponpane + Nahisty +QD moles of Nagr = mass - 11.19 = 0.108 moles mular mass 102.9 mass= density xvolume moles o butanol = 8.10 - 0.109 nolls = 0.810 gut x 104 74.1 = 8.10 g moles y HSD = volume x concentration = 0.106x 18 molet - 1.80 noles .. from the above calculation it is clear that pretende is NaBr the il miting reagent : since Sodium boomide is the limetting reagent theoretical field can be calculate by using it.

since Imole of NaBr produce I mole of thromopropane 0.108 moles of Na Br produse 0.108 moles 9 Houmopoupane 10.108 moles) mass g Hboomopompane în ġ = moles x molar mass = 0.108 mot « 137 g mott - 14.8g © (3 &-methyl-2 butanol & Hi 2-chloro-amethyl butane tho () a moles B A = mass – 8.10 nol - molar mass 88 gmert = 0.09 & moles. mass of a = density x volume = 0.8194 « 10 ml = 8.10g molls of B (HI) - concentration xvolume - 120 x 25x10 L = 0.3 moles. 2-methyl-2-butanol is the limiting reagent. 6 moles of A = moles of 'c' . moles of d-chloro-2-methyl butone - 0.092 moles © Theoretical Yield on the product in grams - moles x molen mass = 0.092 motx 106.5 g mot - 9.80 9.