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alexira [117]
3 years ago
11

Draw the structures of organic compounds A and B. Indicate stereochemistry where applicable The starting material is ethyne, a c

arbon carbon triple bond where each carbon is bonded to a hydrogen. Step 1 is N a N H 2 followed by 1 equivalent of C H 3 C H 2 C H 2 B r to form compound A. Compound A reacts with hydrogen and lindlar's catalyst to form compound B. COmpound B reacts with H 2 O and H 3 O Plus to form a 5 carbon chain with a hydroxy substituent on carbon 2.
Chemistry
1 answer:
k0ka [10]3 years ago
3 0

Answer:

See explanation below

Explanation:

In this case we have the starting reactant which is the ethine, In the first step reacts with NaNH₂, a strong base. This base will substract the hydrogen from one of the carbon of the ethine, and form a carbanion. This will react with the propane bromide, displacing the bromine and forming a 5 carbon chain with the triple bond on the carbon 1 and 2.

In the second step, reacts with the lindlar catalyst to do a reduction, and form a double bond between carbon 1 and 2. In essence, compound A is similar to compound B.

Finally B reacts with water in acid and makes a addition reaction, and form an alcohol.

The whole process can be seen in the picture below.

Hope this helps

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A chemist prepares a solution of copper(l) sulfate (CuSo4) by measuring out 11.7 g of copper(II) sulfate into a 350. mL volumetr
Ksju [112]

Answer:

0.209 mol/L

Explanation:

Given data

  • Mass of copper(lI) sulfate (solute): 11.7 g
  • Volume of solution: 350 mL = 0.350 L

The molar mass of copper(Il) sulfate is 159.61 g/mol. The moles corresponding to 11.7 grams are:

11.7 g × (1 mol/159.61 g) = 0.0733 mol

The molarity of copper(Il) sulfate is:

M = moles of solute / liters of solution

M = 0.0733 mol / 0.350 L

M = 0.209 mol/L

6 0
3 years ago
Convert 32.56 km/hr into ft/hr
Gekata [30.6K]
Note that
1 m = 3.2808 ft

Therefore
1 km = 3280.8 ft
and
32.56 \,  \frac{km}{h} = (32.56 \,  \frac{km}{h})*(3280.8 \,  \frac{ft}{km}) =  1.0682 \, \times 10^{5} \, \frac{ft}{h}

Answer: 1.0682 x 10⁵ ft/hr

8 0
3 years ago
The rate constant of a certain reaction is known to obey the Arrhenius equation, and to have an activation energy . If the rate
vovangra [49]

Answer:

K2 = 61.2 M^-1.S^-1

Explanation:

We complete the question fully:

The rate constant of a certain reaction is known to obey the Arrhenius equation, and to have an activation energy Ea = 71.0kJ/mol . If the rate constant of this reaction is 6.7M^(-1)*s^(-1) at 244.0 degrees Celsius, what will the rate constant be at 324.0 degrees Celsius?

Answer is as follows:

The question asks us to calculate the value of the rate constant at a certain temperature, given that it is at a particular value for a particular temperature. We solve the question as follows:

According to Arrhenius equation, the relationship between temperature and activation energy is as follows:

            k = Ae^-(Ea/RT)

where,   k = rate constant

              A = pre-exponential factor

          Ea  = activation energy

             R = gas constant

              T = temperature in kelvin

From the equation, the following was derived for a double temperature problem:

ln(k2/k1) = (-Ea/R) * (1/T1 - 1/T2)

We list out the parameters as follows:

         

      T1= (244 + 273.15) K = 517.15 K

      T2= (324+ 273.15) K =597.15 K

    K1  = 6.7 ,     K2 = ?

         R = 8.314 J/mol K

     Ea = 71.0 kJ/mol = 71000 J/mol

Putting the given values into the above formula as follows:

ln(k2/6.7) = (-71000/8.314) * (1/517.15 - 1/597.15)

lnk2 - 1.902 = 8539.8 * 0.000259

lnK2 = 1.902 + 2.21

lnK2 = 4.114

K2 = e^(4.114)

K2 = 61.2

Hence, K2 = 61.2 (M.S)^-1

7 0
3 years ago
Read 2 more answers
Ayye yoo who want to answer this question ill give brainlist :)
frez [133]

Answer:

noble-gas notation

Explanation:

5 0
2 years ago
Hi can someone please help me!
Ratling [72]

Answer:

a) 2

b) 2

c) 5

d) 5

e) 5

Explanation:

a) There is 1 Ag atom and 1 Cl atom. <em>When there's no subscript number next to an element, it means there is only one.</em>

b) There is 1 Ca atom and 1 O atom.

c) There are 3 Mg atoms (there's a subscript 3 next to Mg) and 2 N atoms.

d) There are 2 Al atoms and 3 O atoms.

e) There are 2 Sc atoms and 3 S atoms.

7 0
3 years ago
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