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jeyben [28]
3 years ago
6

Draw the alkane formed when 4,5,5-trimethyl-1-hexyne is treated with two equivalents of hbr.

Chemistry
2 answers:
Doss [256]3 years ago
5 0
When an alkyne is reacted with 2 moles of HBr, the Markovnikov rule applies. The H-atoms of the 2HBr attaches to the C atom with more H atoms, while the 2-Br atoms, attaches to the C atom with less H-atoms. The product would be 2,2, dibromo - 4,5,5 - trimethylhexane.

AveGali [126]3 years ago
3 0
<h3>Answer:</h3>

             The Alkane formed is 5,5-dibromo-2,2,3-trimethylhexane. as shown below in attached scheme (Green Color).

<h3>Explanation:</h3>

                        Alkynes like Alkenes undergo <em>Electrophillic Addition Reactions</em>. The reaction given is a two step reaction. In step 1, the Alkyne adds first equivalent of HBr obeying <em>Markovnikov's rule</em> (i.e. Bromine will add to carbon containing less number of hydrogen atoms) and forms <em>2-bromo-4,5,5-trimethylhex-1-ene</em>. In step 2, the alkene formed in first step (2-bromo-4,5,5-trimethylhex-1-ene) undergoes addition reaction with the second equivalent of HBr via Markovnikov's rule to produce <em>5,5-dibromo-2,2,3-trimethylhexane</em>.

The scheme is attached below, Blue color is assigned to starting Alkyne, Red color is assigned to intermediate Alkene and Green color is assigned to product Alkane respectively.

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The low-grade iron ore taconite, which contains Fe3O4, is concentrated and made into pellets for processing. If one ton of tacon
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<u>Answer:</u> The percent composition of Fe_3O_4 in taconite is 37.6 %.

<u>Explanation:</u>

We are given:

Mass of taconite pellets = 1 ton = 907185 g      (Conversion factor: 1 ton = 907185 g)

Mass of iron produced = 545 lb = 247212 g    (Conversion factor:  1 lb = 453.6 g )

We know that:

Molar mass of iron = 55.85 g/mol

Molar mass of Fe_3O_4 = 231.53 g/mol

1 mole of Fe_3O_4 contains 3 moles of iron atom and 4 moles of oxygen atom

(3 × 55.85) = 167.55 g of iron is produced from 231.53 grams of Fe_3O_4

So, 247212 grams of iron will be produced from = \frac{231.53}{167.55}\times 247212=341611.43g of Fe_3O_4

To calculate the percentage of Fe_3O_4 in taconite, we use the equation:

\%\text{ composition of }Fe_3O_4=\frac{\text{Mass of }Fe_3O_4}{\text{Mass of taconite}}\times 100

Mass of taconite = 907185 g

Mass of Fe_3O_4 = 341611.43 g

Putting values in above equation, we get:

\%\text{ composition of }Fe_3O_4=\frac{341611.43g}{907185g}\times 100=37.6\%

Hence, the percent composition of Fe_3O_4 in taconite is 37.6 %.

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