[figure] Consider the above sequence of reactions. 151 g of 2-bromopentane is made to react. Yield of major product P is 80\% whereas Q is 100\%. Mass of product Q obtained is g. (Given molar mass in g mol-1 H : 1, C : 12, O : 16, Br : 80)
Answer
184
Solution
Molar mass of 2-bromopentane, C5H11Br = 60 + 11 + 80 = 151 g mol-1
151 g = 1 mol
Dehydrohalogenation with alcoholic KOH follows Saytzeff rule; the more substituted alkene, pent-2-ene, is the major product P:
CH3-CH(Br)-CH2-CH2-CH3 alc. KOH CH3-CH=CH-CH2-CH3
moles of P = 1 0.80 = 0.80 mol
Addition of bromine gives 2,3-dibromopentane (Q):
CH3-CH=CH-CH2-CH3 Br2 CH3-CH(Br)-CH(Br)-CH2-CH3
Molar mass of Q, C5H10Br2 = 60 + 10 + 160 = 230 g mol-1
mass of Q = 0.80 230 = 184 g
Answer: 184