1-Bromopropane is treated with each of the following reagents. Draw the major substitution product if the reaction proceeds in g
ood yield. If no reaction is expected or yields will be poor, draw the starting material in the box. If a charged product is formed, be sure to draw the counterion. a. with H_2 O
b. with H_2 SO_4
c. with 1 equiv of KOH
d. with Csl
e. with NaCN
f. with HCl
g. with (CH_3)_2 S
h. with 1 equiv of NH_3
j. with CI_2
k. with KF
If we look at the structure of 1-Bromopropane; we will see that it is a derivative of alkane family by the the substitution of an alkyl group. The position of the Bromine in the propane is 1, making 1-Bromopropane a primary alkyl-halide.
Primary alkyl - halide undergo SN2 mechanism. This nucleophilic reaction needs to be a strong alkyl halide , such as 1-Bromopropane used otherwise it will result to a reactive mechanism if a weak electrophile is used.
However, the critical and the main objective here is to Draw the major substitution product if the reaction proceeds in good yield. If no reaction is expected or yields will be poor, draw the starting material in the box. If a charged product is formed, be sure to draw the counterion.
The attached diagrams portraying this notions is shown in the attached file below.
First convert 0.163 grams of N2O to mol by dividing it with
the molecular weight. The molecular weight of N2O is 44 grams/mol. The answer
would be 3,79x10^-3. Then multiply it with 2 since there are 2 Nitrogen in one
mole of N2O. Therefore, there are 7.41x10^-3 moles of Nitrogen.