Draw a structure for a compound that meets the following description: An optically active compound, C5H10O with an IR absorption
at 1730 cm-1.
1 answer:
Question: A optically active compound, C5H10O, exhibits IR absorption at 1730 cm-1.
Its carbon NMR shifts are given below. The number of hydrogen's at each carbon, determined by DEPT, is given in parentheses after the chemical shift.
13C NMR: δ 22.6 (3), 23.6 (1), 52.8 (2), 202.4 (1)
Draw the structure of this compound in the window below
Explanation:
3-methylbutanal is a butanal substituted by the methyl group at the 3rd position. It is a volatile constituent in the olive. Also, it is used as a flavoring agent and a plant metabolite, it is also a Saccharomyces cerevisiae metabolite. It is also called as the Isovaleraldehyde organic compound. The liquid is colorless at STP, and also found in very low concentrations. It is also seen to be produced commercially for different use. Mostly used compound reagent in the preparation of pharmaceuticals and pesticides.
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Answer:
=> 1366.120 g/mL.
Explanation:
To determine the formula to use in solving such a problem, you have to consider what you have been given.
We have;
mass (m) = 25 Kg
Volume (v) = 18.3 mL.
From our question, we are to determine the density (rho) of the rock.
The formula:
First let's convert 25 Kg to g;
1 Kg = 1000 g
25 Kg = ?
= 25000 g
Substitute the values into the formula:
= 1366.120 g/mL.
Therefore, the density (rho) of the rock is 1366.120 g/mL.
Explanation:
Molarity of solution = 1.00 M = 1.00 mol/L
In 1 L of solution 1.00 moles of calcium chloride is present.
Mass of solute or calcium chloride = m
Mass of solution = M
Volume of solution = V = 1L = 1000 mL
Density of solution , d= 1.07 g/mL
1) The value of %(m/M):
2) The value of %(m/V):
n = Equivalent mass
n =
3) Normality of calcium ions:
Moles of calcium ion = 1 mol (1 mole has 1 mole of calcium ion)
4) Normality of chlorine ions:
Moles of chlorine ion = 2 mol (1 mole has 2 mole of chlorine ion)
Moles of calcium chloride = 1.00 mol
Mass of solvent = Mass of solution - mass of solute
= 1,070 g - 111 g = 959 g = 0.959 kg ( 1 g =0.001 kg)
5) Molality of the solution :
Moles of calcium chloride =
Mass of solvent = 959 g
Moles of water =
Mass of solvent = 959 g
6) Mole fraction of calcium chloride =
7) Mole fraction of water =
8) Mass of solution = m'
Volume of the solution= v = 100 mL
Density of solution = d = 1.07 g/mL
Mass of 100 mL of this solution 107 grams of solution.
9) Volume of solution = V = 100 mL
Mass of solution = M'' = 107 g
Mass of solute = m
The value of %(m/V) of solution = 11.1%
m = 11.1 g
Mass of solvent = M''- m = 107 g -11.1 g = 95.9 g
95.9 grams of water was present in 100 mL of given solution.