Question #33b4d

1 Answer
Jan 16, 2017

The final product is 1-nitropropane.

Explanation:

"HBr"HBr reacts with propene in the presence of organic peroxides by a free radical mechanism to give an anti-Markovnikov addition product.

The steps are:

Initiation: "RO-OR → 2RO·"RO-OR → 2RO⋅

Initiation: "RO·" + "H-Br" → "RO-H" + "Br·"RO⋅+H-BrRO-H+Br⋅

Propagation: underbrace("CH"_3"CH=CH"_2)_color(red)("propene") + "Br·" → "CH"_3 stackrelcolor(blue)(bb·)("C")"H-CH"_2"Br"

Propagation: "CH"_3 stackrelcolor(blue)(bb·)("C")"H-CH"_2"Br" + "H-Br" → underbrace("CH"_3"CH"_2"CH"_2"Br")_color(red)("1-bromopropane") + "Br·"

The product is 1-bromopropane.

Primary alkyl halides react with aqueous silver nitrite in an "S"_"N"2 displacement to form primary nitroalkanes.

Thus,

"CH"_3"CH"_2"CH"_2"Br" + "Ag-O-"stackrelcolor(blue)(. .)("N")"=O" → underbrace("CH"_3"CH"_2"CH"_2"NO"_2)_color(red)("1-nitropropane") + "AgBr"

The product is a nitro compound and not a nitrite because the "Ag" is covalently bound to the "O".

Thus, it is the lone pair on the "N" that attacks the α-carbon of the bromide.