Question 23Get help answering Molecular Drawing questions Get help answering Molecular Drawing questions.A compound with molecul
ar 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
1 answer:
Answer:
The structure for the compound elucidated (4-Ethylbenzoic acid) is attached below.
Explanation:
Degree of Unsaturation = (where n = no of moles)
=
= 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
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Answer:
Q = 0.50
No
Left
Explanation:
At a generic reversible equation
aA + bB ⇄ cC + dD
The reaction coefficient (Q) is the ratio of the substances concentrations:
Solids and liquid water are not considered in this calculus.
When the reaction achieves equilibrium (concentrations are constant), the Q value is named as Kc, which is the equilibrium constant of the reaction. If Q > Kc, it indicates that the concentration of the products is higher, so, the reaction must progress to the left and form more reactants; if Q < Kc, than the concentrations of the reactants, are higher, so, the reaction progress to the right.
In this case:
Q =
Q = 0.50
So, Q > Kc, the reaction is not at equilibrium and it progresses to the left.