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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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NaCl sodium chloride

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Since Br-80 does not exist, maybe that means Br-79 or Br-81 have very unequal abundances. For example, Br-79 may have 75% abundance whereas Br-81 may have 25% abundance.

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If Liquid 1 has a higher specific heat than Liquid 2, then Liquid 1 will take longer to increase in temperature because the higher specific heat of a liquid needs more thermal energy for heating a liquid.

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A solution is made by dissolving 10.20 grams of glucose (C6H12O6) in 355 grams of water. What is the freezing point depression o
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0.297 °C

Step-by-step explanation:

The formula for the <em>freezing point depression </em>ΔT_f is

ΔT_f = iK_f·b

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For glucose,

       glucose(s) ⟶ glucose(aq)

1 mole glucose ⟶ 1 mol particles     i = 1

Data:

Mass of glucose = 10.20 g

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n = 10.20 g × (1 mol/180.16 g)

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m = 355 g × (1 kg/1000 g)

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(c) <em>Molal concentration </em>

b = moles of solute/kilograms of solvent

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  = 0.1595 mol·kg⁻¹

(d) <em>Freezing point depression </em>

ΔT_f = 1 × 1.86 × 0.1595

        = 0.297 °C

3 0
3 years ago
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