Question

Provide a structure for the following compound: C9H10O3; IR: 2400–3200, 1700, 1630 cm–1; 1H NMR: δ 1.53 (3H, t, J = 8 Hz); δ 4.32 (2H, q, J = 8 Hz); δ 7.08, δ 8.13 (4H, pair of leaning doublets, J = 10 Hz); δ 10 (1H, broad, disappears with D2O shake

Answer 1

Answer:

The structure of the compound with H-NMR is 4 -ethoxy benzoic acid.

Explanation:

Degree of Unsaturation:

From the given,

The molecular formula of the compound -$C_{9}H_{10}O_{3}$

$Degree\,\,of\,\,unsaturation=C-\frac{H}{2}-\frac{X}{2}+N+1$

$=9-\frac{10}{2}+1=5$

From the $^{1}H-NMR$ data,

3-H -triplet and 2-H-quartet indicates ethyl group.

δ values of ethyl groups indicates that the ethyl group is attached to the oxygen atom.

hence, one functional group is ethoxy group.

δ= 7.08 value indicates the para substituted benzene.

At δ=10 1-H singlet indicates the presence of acid-H.

From the IR data:

the peak at $1700-1630 \,cm^{-1}$ indicates the presence of carbonyl group.

Therefore, the p-substituted benzene contain ethoxy group and carboxylic group.

Answer 2

Using the IR table of values we decipher the structure of  Ethyl 3-hydroxybenzoate as (picture attached below)

What is the structure of  Ethyl 3-hydroxybenzoate?

Question Parameter(s):

COmpound details

C9H10O3;

IR: 2400–3200, 1700, 1630 cm–1;

1H NMR: δ 1.53 (3H, t, J = 8 Hz);

δ 4.32 (2H, q, J = 8 Hz);

δ 7.08, δ 8.13 (4H,

Generally, Using the IR table of values the detail are given as

2400- 3200 cm-1=Carboxylic acid

17000 cm-1 = C=O bond 1

630 cm-1 = C=C

In conclusion, the H NMR  gives

1.53 = CH3 gives high chemical shift

4.32=CH2 n

7.08, 8.13 = aromatics.

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