Question

Determine the unsaturation number (UN#) for the unknown compounds using the molecular formula. Determine the functional groups present in the IR spectrum

Determine the final structure of the unknown

C7H12 90 80 2126 70 60 50 2879 40 3324 30 20 2971 10 2000 4000 Wavenumbers (cm-1) 9H 1H 2H PPM % Transmittance

Answer 1

Unsaturation number is also known as degree of unsaturation or double bond equivalent or index of hydrogen deficiency

                 …………………………………………………………………..(1)

U (n r2) 2

The base formula for organic compounds = C_nH_n+r

So according to the question the base formula is C₇H₁₂

There is another rule in organic chemistry which is known as Rule of Thirteen. It can be represented as :

…………………………………………………….(2)

м —пtr/13 13

From the molecular formula given in the question C₇H_{12 we can calculate the molecular mass which comes out to be 96 g/mol. We put M = 96 and n = 7 in equation 2 to calculate the value of r from there:}

96 = 7 +r/13 13

We get the value r = 5 from here. We can also calculate the value of r from C_nH_{n+r as well.}

_{Now put the value of n = 7 and r = 5 in equation 1, we can calculate the value of}

U (7 5 2

U = 2

The compound having C₇H₁₂ formula and U = 12 must have either a double bond or a ring or two double bonds or a triple bond.

The 3324 cm^-1 frequency band can be attributed to alkene stretching frequency

The 2971 cm^-1 frequency corresponds to Asymmetric methyl stretching while the frequency at 2879 cm^-1 corresponds the symmetric methyl stretching.

The name and structure of the compound is

The compound is 3,4-Dimethyl-1,3-pentadiene. The 3 methyl groups have the same environment i.e. they are linked to the doubly bonded carbons hence will be deshielded and will not appear at 0.9 ppm. So as we can see in the NMR spectra the methyl carbons appear little downfield The terminal double-bonded carbon has 2 hydrogens and is attached to CH group which will experience most deshielding due to the influence of 2 double bonds.