# (a) Your TA will give you a 1H NMR spectrum of the 3-nitroaniline product. Using your NMR knowledge and the special NMR section

###### Question:

## Answers

provided in the answer segment

Explanation:

Below is a step by step process to analyzing this problem

Let us begin;

From 1H-NMR singlet at 5.80 ppm show N-H peak as shown in structure.

Here, H(A) hydrogen has no neighbor hydrogen so it appears integrated singlet at 7.38 ppm.

H(C) hydrogen has the next 2 neighbor hydrogen H(B) and H(D) so it appears as a triplet at region 7.23-7.29 ppm.

H(B) hydrogen has next to one neighbor hydrogen H(C) show doublet at 6.92-6.98 ppm.

H(D) hydrogen has next to one neighbor hydrogen H(C) show doublet at 7.28-7.32 ppm.

(b). From our basic chemistry knowledge, we know that benzene molecule is planer so H(A) is more deshielding because of two substituent groups than H(C), which makes the delta value of H(A) is greater than H(C).

cheers i hope this helped

[tex](a) Your TA will give you a 1H NMR spectrum of the 3-nitroaniline product. Using your NMR knowledge[/tex]

part 1: the half-life of the element is 22 years.

part 2: it will take 132.02 year for the sample to decay from 308.0 g to 4.8125 g.

explanation:

part 1: there are two ways to determine the half life time.

first way: the half life time can be determined directly from its definition as it is the time required for the sample to decay to its half concentration.

from the data the initial concentration was 45.0 g and decayed to its half value (22.5 g) in 22.0 years.

second way: the radio active decay is considered first order reaction so we can use the laws of first order reactions.

k = (1/t) ln (a/a-x), where k is the rate constant of the reaction,

t is the time of the reaction

a is the initial concentration of the reactant

a-x is the remaining concentration at time (t)

a = 45.0 g the initial concentration at t = 0 year.

at t = 22 year, a-x = 22.5 g

then k = (1/22 year) ln (45.0 g/ 22.5 g) = 0.0315 year-1.

the half-life time (t1/2) = ln 2/ k = 0.693/ k =0.693/ 0.0315 = 22 years.

part 2:

using the same law k = (1/t) ln (a/a-x),

using the given data; a = 308.0 g and a-x = 4.8125 g and the calculated k from the part 1 (k = 0.0315 year-1)

t = (1/k) ln (a/a-x) = (1/0.0315) ln (308.0/ 4.8125) = 132.02 years

i don,t know so just answer mine