Question

c) Using your answers to question 5 and 6a, what percent of the mixture obtained in reaction 1 is the ortho isomer?

d) What percent of the mixture obtained in reaction 1 is the meta isomer?

e) What percent of the mixture obtained in reaction 1 is the para isomer?

f) Could we have used Mass Spectrometry after the GC analysis (GC-MS) to identify which peak on the GC was which isomer? Why or why not?

g) Could we have used IR to differentiate between the ortho, meta, and para isomers? Why or why not?

5. calculate the percent composition for each product (A, B and C). Be sure to show your work.

Area of peak A = 125 x 8 = 1000mm²

Area of peak B = 72 x 9 = 648mm²

Area of peak C = 10 x 6 = 60mm²

Total peak area = 1708mm²

Percent composition of A = (1000/1708) x 100% = 58.5%

Percent composition of B = (648/1708) x 100% = 37.9%

Percent composition of C = (60/1708) x 100% = 3.5 %

6a)

Spectrum For Peak A on GC For Peak B on GC For Peak Con GC Isomer Ortho Nitro Toluene Para Nitro Toluene Meta Nitro Toluene

Spectra:

Answer 1

Solution:

c) ortho isomer: 58.5%

d) para isomer : 37.9%

e) meta isomer : 3.5 %

f) since these are structural isomers there no mass difference thus, difficult to judge the mass fragmentation hence difficult to identify the isomers. So taking mass will not be helpful.

g) from FTIR it is easy to distinguish these isomers.

For ortho isomer: 770-735 cm-1 band will appear for CH wag

For meta isomer: 810-750 cm-1 band will appear for CH wag.

For para isomer: 860-790 cm-1 band will appear for CH wag.