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Alenkasestr [34]
3 years ago
8

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)

Chemistry
1 answer:
Svet_ta [14]3 years ago
4 0

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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Answer the following questions based on the reaction below: NaOH(aq) + KHP(s) --> NaKP(aq)+H2O(I)
Natali [406]

Answer:

<u />

  • <u>a) 1.44g</u>

<u />

  • <u>b) 77.3%</u>

<u />

Explanation:

<u>1. Chemical balanced equation (given)</u>

       NaOH(aq)+ KHP(s)\rightarrow NaKP(aq)+H_2O(l)

<u>2. Mole ratio</u>

1molNaOH(aq):1molKHP(s)

This is, 1 mol of NaOH will reacts with 1 mol of KHP.

<u />

<u>3. Find the number of moles in 72.14 mL of the base</u>

    Molarity=\text{number of moles of solute}/\text{volume of solution in liters}

    \text{Volume of solution}=72.14mL=0.07214liters

     \text{Number of moles of NaOH}=0.0978M\times 0.07214liter=0.007055mol

<u>4. Find the number of grams of KHP that reacted</u>

The number of moles of KHP that reacted is equal to the number of moles of NaOH, 0.007055 mol

Convert moles to grams:

  • mass = number moles × molar mass = 0.007055mol × 204.23g/mol
  • mass = 1.4408 g.

You have to round to 3 significant figures: 1.44 g (because the molarity is given with 3 significant figures).

<u>5. Find the percentage of KHP in the sample</u>

The percentage is how much of the substance is in 100 parts of the sample.

The formula is:

  • % = (mass of substance / mass of sample) × 100

  • % = (1.4408g/ 1.864g) × 100 = 77.3%
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