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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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How many molecules of SrCrO4 are in a sample of SrCrO4 0.556 moles?
Amiraneli [1.4K]

Answer:

3.35*10^{23}\ SrCrO_4\ molecules

Explanation:

We\ are\ given\ that,\\No.\ of\ moles\ of\ SrCrO_4=0.556\\Hence,\\As\ we\ know\ that,\\No.\ of\ particles=Avagadro's\ Constant*No.\ of\ moles\\We\ already\ know\ that\ Avagadro's\ Constant=6.022*10^{23}\\Here,\\No.\ of\ SrCrO_4\  molecules= 6.022*10^{23}*0.556\\Hence,\\No.\ of\ SrCrO_4\  molecules=3.348*10^{23} molecules\ \approx 3.35*10^{23}\ SrCrO_4\ molecules

7 0
3 years ago
A gas cylinder is filled with argon at a pressure of 177 atm and 25°C. What is the gas pressure when the temperature of the cyli
Elena L [17]

277.79 atm is the calculated gas pressure.

The ideal gas is a fictitious concept used to study how real gases behave by comparing them to their deviations. The pressure-temperature rules are followed by an ideal gas.

177 atm is the initial pressure. The starting temperature is 298 K (25 °C = 25 + 273 °C).

195°C = 195+273

= 468K is the final temperature.

The pressure temperature relation illustrated below can be used to get the final pressure.

P1/T1 = P2/T1

= P1T2/T1

= 177 atm 468 K /298 K

= 277.97 atm

The final pressure is therefore 277.97 atm.

Learn more about Pressure here-

brainly.com/question/4578923

#SPJ4

5 0
2 years ago
Balance the following Equation Show your work.
7nadin3 [17]

Answer:

1) alr balanced

2) 2 2 1

3) 2 1 1 2

4) 2 3 2

5) 2 1 2 1

6) 1 6 2 3

7) 2 2 1

8) 1 2 1 2

9) 2 5 4 6

10) 3 1 2

3 0
3 years ago
Solve the problem. Express your answer to the correct number of significant figures
Shalnov [3]

Answer:

Explanation:

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4 0
3 years ago
Nuclear decay<br>27Al + He 》 39P<br>​
dybincka [34]

Answer:

_{13}^{27}\text{Al} + \rm _{2}^{4}\text{He} \longrightarrow \, _{15}^{31}\text{P}

Explanation:

The unbalanced nuclear equation is

^{27}\text{Al} + \rm \text{He} \longrightarrow \, ^{31}\text{P}

We can insert the subscripts, because these are the atomic numbers of the elements

_{13}^{27}\text{Al} + \, \rm _{2}\text{He} \longrightarrow \, _{15}^{31}\text{P}

That leaves only the superscript of He to be determined,

The main point to remember in balancing nuclear equations is that the sums of the superscripts must be the same on each side of the equation.  

Then

27 + x = 31, so x = 31 - 27 = 4

Then, your nuclear equation becomes

_{13}^{27}\text{Al} + \, \rm _{2}^{4}\text{He} \longrightarrow \, _{15}^{31}\text{P}

6 0
3 years ago
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