Which combination of reagents used in the indicated order with benzene will give m-nitropropylbenzene? a. 1) CH3CH2CH2Cl/AlCl3, 2) HNO3/H2SO4 b. 1) HNO3/H2SO4, 2) CH3CH2CH2Cl/AlCl3 c. 1) CH3CH2CH2Cl/AlCl3, 2) HNO3/H2SO4, 3) Zn/HCl d. 1) HNO3/H2SO4, 2) CH3CH2CH2Cl/AlCl3, 3) Zn/HCl

1) HNO3/H2SO4, 2) CH3CH2CH2Cl/AlCl3

Explanation:

Benzene is a stable aromatic compound hence it undergoes substitution rather than addition reaction.

When benzene undergoes substitution reaction, the substituent introduced into the ring determines the position of the incoming electrophile.

If I want to synthesize m-nitropropylbenzene, I will first carry out the nitration of benzene using HNO3/H2SO4 since the -nitro group is a meta director. This is now followed by Friedel Craft's alkykation using CH3CH2CH2Cl/AlCl3.

as given the equation is

2al + 6 hcl > 2 alcl3 + 3 h2(g)

as per equation two moles of aluminum reacts with six moles of hcl to give two moles of alcl3 and three moles of hydrogen gas

given: three moles of al are present and 7 moles of hcl

for three moles of aluminum nine moles of hcl will be required

however as 7 moles of hcl are present so the limiting reagent is hcl

hcl is going to decide the number of moles of product formed

when six moles of hcl are present = 3 moles of h2 are formed

when one mole of hcl is present == 3/6 mole sof h2 will be formed

when 7 moles of hcl are present = 3 x 7/ 6 moles of h2 will be formed = 3.5 moles of h2