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 4.3
2 (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)
1 answer:
Answer:
The answer is Ethyl 3-hydroxybenzoate (structure attached).
Explanation:
The following can be elucidated from IR table of values:
2400- 3200 cm-1 ⇒ Carboxylic acid
17000 cm-1 ⇒ C=O bond (aldehyde, ketone)
1630 cm-1 ⇒ C=C (alkene) or aromatics
H NMR values:
1.53 ⇒ CH3 next to an electronegative, hence the high chemical shift (deshielded)
4.32⇒CH2 next to a CH3 and an R group (possible deshielding)
7.08, 8.13 ⇒ aromatics.
The structure is attached.
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The complete table is inserted.
A table is given,
Formulas used:
pH= -log(H⁺)
pOH= -log(OH⁻)
pH+ pOH=14
Calculations:
For A: (H⁺)=2×10⁻⁸M
Using the pH formula:
pH= -log(H⁺)=-log(2×10⁻⁸)=7.69
pOH=14 - 7.69=6.3
Calculating OH concentration,
pOH= -log(OH⁻)
6.3= -log(OH⁻)
(OH⁻)=5.011×10⁻⁷M
Hence, the nature of A is basic.
Similarily,
For B,
(OH⁻)=1×10⁻⁷
Using the pH formula:
pOH= -log(OH⁻)= -log(1×10⁻⁷)=7
pH=14-7=7
Calculating H concentration,
pH= -log(H⁺)
7= -log(H⁺)
(H⁺)=1×10⁻⁷M
Hence, the nature of B is neutral.
Similarily,
For C,
pH=12.3
Using the pH formula:
pOH=14-12.3=1.7
Calculating H concentration,
pH= -log(H⁺)
12.3= -log(H⁺)
(H⁺)=5.011×10⁻¹³M
Calculating OH concentration,
pOH= -log(OH⁻)
1.7= -log(OH⁻)
(OH⁻)=1.99×10⁻²M
Hence, the nature of C is Basic.
Similarily,
For D,
pOH=6.8
Using the pH formula:
pH=14-6.8=7.2
Calculating H concentration,
pH= -log(H⁺)
7.2= -log(H⁺)
(H⁺)=6.309×10⁻⁸M
Calculating OH concentration,
pOH= -log(OH⁻)
6.8= -log(OH⁻)
(OH⁻)=1.58×10⁻⁷M
Hence, the nature of D is basic.
Learn more about the acid and bases here:
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