Q15-39E

Question

Compound A, C8H10 , yields three substitution products, C8H9Br, on reaction with Br2. Propose two possible structures for A. The 1HNMR spectrum of A shows a complex four-proton multiplet at 7.0 d and a six-proton singlet at 2.30 d. What is the structure of A?

Step-by-Step Solution

Verified
Answer


Compound A

O-xylene

1Step 1: Double bond equivalence

The double bond equivalence of a molecule helps in the determination of the amount of unsaturation present in the molecule. DBE is of great use in the determination of the structure of a compound, given its molecular formula is known. The double bond equivalent is given by the following formula.

DBE=C+1-(H+X-N2)

Here, C is the number of carbon atoms, H is the number of hydrogen atoms, X is the number of halogen atoms, and N is the number of nitrogen atoms.


2Step 2: Compound A




The double bond equivalence for the formula C8H10 can be calculated as follows:

                      DBE=C+1(H+XN2)                              =(8+1(10+00)2)=4

From the value of DBE, which is equal to 4, it can be deduced that the unknown compound must have had three double bonds and a ring. The molecule closest to this approximation is a benzene derivative.

The following benzene derivatives are possible for the molecular formula C8H10are



Possible compounds


From the NMR spectrum data, compound A must have two distinct sets of protons, So, compound I ( three sets of protons) and compound II  (four sets of protons) get eliminated from the list. Therefore, two possible structures for A are as follows:





From the above two possible structures, the compound (II) shows a complex four-proton multiplet at 7.0δ and a six-proton single at 2.30δ. Therefore, the structure of compound A is represented as:




Compound A (o-xylene)


Therefore, the given compound is o-xylene .