Question

please help!!!

6. First, draw all three monobromination products for the reaction below. notes, determine the product distribution (percentage of each product). Show W. Second, using the relative rates of halogenation in the Se of each product). Show all work. Bra H₂C- CH₂ light (hv)

Answer 1

b b There are three types of Hydrogens are present. So three mono brominated products are formed Br2 -> H2C- -CH2-Br hD hu bb Br2 → H2C- CH b hv Br2 → H3C hv

Relative rates of mono bromination for 1°, 2° and 3° are 1, 82 and 1640 respectively. Number of a type Hydrogens (primary) = 6 Number of b type Hydrogens (secondary) = 8 Number of c type Hydrogens (tertiary) = 2 Relative Amount of a brominated product =no. of Hydrogens * reactivity = 6*1.0 = 6 Relative Amount of b brominated product = 8*82 = 656 Relative Amount of b brominated product = 2* 1640 = 3280 Percent Yield of a brominated product = 6/(6 + 656 + 3280) *100 = 0.15 % Percent Yield of b brominated product = 656 /(6 +656 + 3280) *100 = 16.64 % Percent Yield of c brominated product = 3280 /(6 + 656 + 3280) *100 = 83.21 %