Question

Question 23Get help answering Molecular Drawing questions Get help answering Molecular Drawing questions.A compound with molecular formula C9H10O2 exhibits a triplet at δ 1.2 (3H), a quartet at δ 2.6 (2H), a doublet at δ 7.3 (2H), a doublet at δ 8.0 (2H) and a singlet at δ 11 (1H) in its 1HNMR spectrum. What is the structure for this compound?MarvinJS Response ImageEdit

Answer 1

Answer:

The structure for the compound elucidated (4-Ethylbenzoic acid) is attached below.

Explanation:

Degree of Unsaturation = $nC - \frac{nH}{2} +\frac{nO}{2} +1$  (where n = no of moles)

= $9-\frac{10}{2} -\frac{2}{2} +1$

= 9 - 6+1  = 4

⇒ 2 ring. 3 double bonds.

From the 1H NMR data given (peaks labeled A-E)

A. 1.2 (3H, t) ⇒  -CH3

B. 2.6 (2H, q) ⇒  -C=C-H or Aromatic-CH3

C. 7.3 (2H, d) ⇒ Aro.-H

D. 8.0 (2H, d) ⇒ Aro.-H

E. 11.0 (1H, s) ⇒ -COOH