Acid catalyzed hydrolysis of HOCH2CH2C(CH3)2CN forms compound A (C6H10O2). A shows a strong peak in its IR spectrum at 1770 cm−1 and the following signals in its 1 H NMR spectrum: 1.27 (singlet, 6 H), 2.12 (triplet, 2 H), and 4.26 (triplet, 2 H) ppm. Draw the structure for A and select the correct answers to complete a mechanism that accounts for its formation.
Answers
Answer:
3,3-dimethyldihydrofuran-2(3H)-one
Explanation:
The simulated spectrum of molecule => Figure 1
Mechanism=>Figure2
Answer:
From the given problem statement we need to hydrolyses HOCH2CH2C(CH3)2CN structure that forms (C6H10O2)
Peak of IR spectrum=1770 cm^-1
This structure for A involves 6 carbon atoms,four hydrogen atoms and 6 oxygen atoms for a complete reaction to occur.
Related Questions
Which of the following reactions would have the smallest value of K at 298 K? Which of the following reactions would have the smallest value of K at 298 K? A + B → 2 C; E°cell = -0.030 V A + 2 B → C; E°cell = +0.98 V A + B → C; E°cell = +1.22 V A + B → 3 C; E°cell = +0.15 V More information is needed to determine.
The rate constant for the second-order reaction 2NOBr(g) ¡ 2NO(g) 1 Br2(g) is 0.80/M ? s at 108C. (a) Starting with a concentration of 0.086 M, calculate the concentration of NOBr after 22 s. (b) Calculate the half-lives when [NOBr]0 5 0.072 M and [NOBr]0 5 0.054 M.
What is the symbol of the element that is classified as an alkali metal and is in period 4?
Molecular compounds result from covalent bonding which are called _______.This is for high school physical science
Answers
liquid 1 and 2 have the same color and mass so the answer would be liquid 1 and 2
Explanation:
hope this is helpful
Answers
Answer:
0.113 M
Explanation:
Since B and D are on opposite sides of the reaction, the concentration of D increases when the concentration of B decreases. The amount by which D increases is determined by the coefficients of B and D in the balanced chemical equation:
[D]=(0.045 M)=0.113 M.
Answers
Answer:
The freezing point of a solution is lowered compared to the freezing point of the pure solvent. The amount of depression of the freezing point is proportional to the molality of the solute. The greater the molality of a solution, the lower its freezing point. To compare the freezing points of these solutions, we need to determine which one has the highest molality.
First, we need to determine the number of particles that each solute will produce in solution, as this affects the amount of depression of the freezing point.
KNO3 dissociates into two ions: K+ and NO3-, so it will produce two particles per formula unit.
BaCl2 dissociates into three ions: Ba2+ and two Cl-, so it will produce three particles per formula unit.
Ethylene glycol does not dissociate in solution, so it will produce one particle per molecule.
Na3PO4 dissociates into four ions: three Na+ and one PO43-, so it will produce four particles per formula unit.
Now, we can calculate the molality (moles of solute per kilogram of solvent) for each solution:
For 0.10 m KNO3: molality = 0.10 mol / 1 kg = 0.10 m
For 0.10 m BaCl2: molality = 0.10 mol x 3 particles / 1 kg = 0.30 m
For 0.10 m ethylene glycol: molality = 0.10 mol / 1 kg = 0.10 m
For 0.10 m Na3PO4: molality = 0.10 mol x 4 particles / 1 kg = 0.40 m
So, the solutions in order of decreasing freezing points are:
0.10 m Na3PO4 (highest molality)
0.10 m BaCl2
0.10 m KNO3 and 0.10 m ethylene glycol (same molality, but KNO3 has a smaller van't Hoff factor than ethylene glycol, so it will have a slightly higher freezing point)
Explanation:
Onoble gases
O halogens
O transitional metals
Answers
Final answer:
Group/Family 18 on the periodic table is called the noble gases.
Explanation:
Group/Family 18 on the periodic table is called the noble gases. The noble gases are a group of chemical elements that have full valence electron shells, which makes them stable and nonreactive. This group includes elements like helium, neon, argon, krypton, xenon, and radon.
Learn more about Noble gases on the periodic table here:
Answers
Answer:
It produces water.
Explanation:
H+ + OH- produces H2O.
It is a type of Neutralization reaction.
b. 72.8 g c2h6o in 2.34 l of solution
c. 12.87 mg ki in 112.4 ml of solution
Answers
molarity is defined as the number of moles of solute in 1 L solution
the number of moles of LiNO₃ - 0.38 mol
volume of solution - 6.14 L
since molarity is number of moles in 1 L
number of moles in 6.14 L - 0.38 mol
therefore number of moles in 1 L - 0.38 mol / 6.14 L = 0.0619 mol/L
molarity of solution is 0.0619 M
Q2)
the mass of C₂H₆O in the solution is 72.8 g
molar mass of C₂H₆O is 46 g/mol
number of moles = mass present / molar mass of compound
the number of moles of C₂H₆O - 72.8 g / 46 g/mol
number of C₂H₆O moles - 1.58 mol
volume of solution - 2.34 L
number of moles in 2.34 L - 1.58 mol
therefore number of moles in 1 L - 1.58 mol / 2.34 L = 0.675 M
molarity of C₂H₆O is 0.675 M
Q3)
Mass of KI in solution - 12.87 x 10⁻³ g
molar mass - 166 g/mol
number of mole of KI = mass present / molar mass of KI
number of KI moles = 12.87 x 10⁻³ g / 166 g/mol = 0.0775 x 10⁻³ mol
volume of solution - 112.4 mL
number of moles of KI in 112.4 mL - 0.0775 x 10⁻³ mol
therefore number of moles in 1000 mL- 0.0775 x 10⁻³ mol / 112.4 mL x 1000 mL
molarity of KI - 6.90 x 10⁻⁴ M
The molarities of the given solutions: (a). 0.38 mol of LiNO₃ in 6.14 L of solution has a molarity of 0.062 M. (b). 72.8 g of C₂H₆O in 2.34 L of solution has a molarity of 0.675 M. (c). 12.87 mg of KI in 112.4 mL of solution has a molarity of 0.000688 M.
To calculate the molarity (M) of a solution, you can use the formula:
Molarity (M) = moles of solute / volume of solution (in liters)
a. 0.38 moles of LiNO₃ in 6.14 L of solution:
Molarity (M) = 0.38 moles / 6.14 L = 0.062 M
b. 72.8 grams of C₂H₆O (ethyl alcohol) in 2.34 L of solution:
First, you need to convert grams to moles using the molar mass of C₂H₆O.
Molar mass of C₂H₆O = 2(12.01 g/mol) + 6(1.01 g/mol) + 1(16.00 g/mol) = 46.08 g/mol
Now, calculate moles of C₂H₆O:
moles = 72.8 g / 46.08 g/mol = 1.58 moles
Molarity (M) = 1.58 moles / 2.34 L = 0.675 M
c. 12.87 mg of KI in 112.4 mL of solution:
First, convert milligrams to grams (1 g = 1000 mg):
12.87 mg = 12.87 g (since 12.87 mg / 1000 = 0.01287 g)
Now, convert mL to liters (1 L = 1000 mL):
112.4 mL = 0.1124 L
Calculate moles of KI:
Molar mass of KI = 39.10 g/mol (for K) + 126.90 g/mol (for I) = 166.00 g/mol
moles = 0.01287 g / 166.00 g/mol = 7.75 × 10⁻⁵ moles
Molarity (M) = (7.75 × 10⁻⁵ moles) / 0.1124 L = 0.000688 M
So, the molarities of the solutions are as follows:
a. 0.062 M
b. 0.675 M
c. 0.000688 M
To know more about moles:
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