describe-how-the-ir-spectrum-of-the-products-would-be-used-to-differentiate-them

Question

Describe how the IR spectrum of the products would be used to differentiate them from the reactants in the following three reactions.

Accepted Answer

(a) Reactant = broad peak near 3300 $$cm^{–1}$$, product = strong peak near 1715 $$cm^{–1}(b) $$Reactant = broad peak near 3000 $$cm^{–1}$$, product = Missing peak near 3000 $$cm^{–1}(c) $$Reactant = strong peak near 1715 $$cm^{–1}$$, product = strong peak near 3300 $$cm^{–1}$$

Answer

(a) Reactant = broad peak near 1700 $$cm^{–1}$$, product = strong peak near 3000 $$cm^{–1}(b) $$Reactant = broad peak near 3000 $$cm^{–1}$$, product = Missing peak near 3000 $$cm^{–1}(c) $$Reactant = strong peak near 3300 $$cm^{–1}$$, product = strong peak near 1715 $$cm^{–1}$$

Answer

(a) Reactant = broad peak near 1300 $$cm^{–1}$$, product = strong peak near 1715 $$cm^{–1}(b) $$Reactant = broad peak near 1600 $$cm^{–1}$$, product = Strong peak near 3300 $$cm^{–1}(c) $$Reactant = strong peak near 3300 $$cm^{–1}$$, product = strong peak near 1715 $$cm^{–1}$$

Answer

(a) Reactant = broad peak near 3000 $$cm^{–1}$$, product = strong peak near 1715 $$cm^{–1}(b) $$Reactant = broad peak near 1600 $$cm^{–1}$$, product = Broad peak near 1100 $$cm^{–1}(c) $$Reactant = strong peak near 1750 $$cm^{–1}$$, product = strong peak near 3500 $$cm^{–1}$$