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Maksim231197 [3]
3 years ago
12

Explain why (CH3)2CHCH(Br)CH2CH3 reacts faster than (CH3)2CHCH2CH(Br)CH3 in an E2 reaction, even though both alkyl halides are 2

°. The first structure has more hydrogens attached to the β carbons than the second structure. The major and minor products formed from the first structure have more alkyl groups on the C═C than those formed from the second structure. The major and minor products formed from the second structure have more alkyl groups on the C═C than those formed from the first structure. The second structure has more hydrogens attached to the β carbons than the first structure.
Chemistry
1 answer:
NikAS [45]3 years ago
3 0

Answer:

The major and minor products formed from the first structure have more alkyl groups on the C═C than those formed from the second structure.

The second structure has more hydrogens attached to the β carbons than the first structure.

Explanation:

It is possible to explain the reaction based on Zaitsev’s rule that states that an elimination will normally lead to the most stable alkene as the major product.  This normally translates to it giving the most substituted alkene.

The two adjacent carbons in the first molecule are secondary. That means the two products will produce a disubstituted alkene but in the second molecule just one product will be disubstituted. Thus:

The first structure has more hydrogens attached to the β carbons than the second structure. <em>FALSE</em>. If the structure has more hydrogens will produce an alkene less stable.

The major and minor products formed from the first structure have more alkyl groups on the C═C than those formed from the second structure. <em>TRUE</em>. As the first structure have more alkyl groups the product is most stable.

The major and minor products formed from the second structure have more alkyl groups on the C═C than those formed from the first structure. <em>FALSE</em>. Is the opposite of the last option.

The second structure has more hydrogens attached to the β carbons than the first structure. <em>TRUE</em>. As the second structure has more hydrogens, the alkenes produced will be less substituted being less stable.

I hope it helps!

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Sighting along the C2-C3 bond of 2-methylbutane, the least stable conformation (Newman projection) has a total energy strain of
natima [27]

Answer:

21 KJ/mol

Explanation:

For this question, we have to start with the <u>linear structure</u> of 2-methylbutane. With the linear structure, we can start to propose all the <u>Newman projections</u> keep it in mind that the point of view is between carbons 2 and 3 (see figure 1).

Additionally, we have several <u>energy values for each interaction</u> present in the Newman structures:

-) Methyl-methyl <em>gauche: 3.8 KJ/mol</em>

-) Methyl-H <em>eclipse: 6.0 KJ/mol</em>

-) Methyl-methyl <em>eclipse: 11.0 KJ/mol</em>

-) H-H <em>eclipse:</em> 4.0 KJ/mol

Now, we can calculate the energy for each molecule.

<u>Molecule A</u>

In this molecule, we have 2 Methyl-methyl <em>gauche </em>interactions only, so:

(3.8x2) = 7.6 KJ/mol

<u>Molecule B</u>

In this molecule, we have a Methyl-methyl <em>eclipse </em>interaction a Methyl-H <em>eclipse </em>interaction and an H-H <em>eclipse</em> interaction, so:

(11)+(6)+(4) = 21 KJ/mol

<u>Molecule C</u>

In this molecule, we have 1 Methyl-methyl <em>gauche </em>interaction only, so:

3.8 KJ/mol

<u>Molecule D</u>

In this molecule, we have three Methyl-H <em>eclipse </em>interaction, so:

(6*3) = 18 KJ/mol

<u>Molecule E</u>

In this molecule, we have 1 Methyl-methyl <em>gauche </em>interaction only, so:

3.8 KJ/mol

<u>Molecule F</u>

In this molecule, we have a Methyl-methyl <em>eclipse </em>interaction a Methyl-H <em>eclipse </em>interaction and an H-H <em>eclipse</em> interaction, so:

(11)+(6)+(4) = 21 KJ/mol

The structures with higher energies would be less stable. In this case, structures B and F with an energy value of 21 KJ/mol (see figure 2).

I hope it helps!

3 0
3 years ago
c) What is the pH of the buffer system in part a when 0.030 moles of strong acid are added (without a change in volume)
pishuonlain [190]

Answer:

remain the same

Explanation:

The pH of the buffer system remain the same when 0.030 moles of strong acid are added because buffer system has the property to resist any change in the pH  when acid or base is added to the solution. In buffer system, one molecule is responsible for neutralizing the pH of the solution by giving H+ or OH-.This molecule is known as buffer agent. If more base is added, the molecule provide H+ and when more acid is added to the solution, then the molecule add OH- to the solution.

7 0
3 years ago
How many sig figs does 010.5608 have
katovenus [111]

Answer:

I think it's 6 but I'm not quite sure

5 0
3 years ago
determine the ph at the equivalence point for the titration of a 40.0 ml sample of 0.100 mhno2 by 0.200 mkoh . (the pka of nitro
timama [110]

pH at the equivalence point is 3.60

<h3>Define pH.</h3>

pH is a numerical indicator of how acidic or basic aqueous or other liquid solutions are. The phrase, which is frequently used in chemistry, biology, and agronomy, converts the hydrogen ion concentration, which typically ranges between 1 and 10^{14} gram-equivalents per liter, into numbers between 0 and 14. The hydrogen ion concentration in pure water, which has a pH of 7, is 10^7} gram-equivalents per liter, making it neutral (neither acidic nor alkaline). A solution with a pH below 7 is referred to as acidic, and one with a pH over 7 is referred to as basic, or alkaline.

given that,

HNO2 = 0.100M

KOH= 0.200M

pKa = 3.34

using formula,

pH = pKa + log (salt /acid)

     = 3.34 + log (0.200/0.100)

     = 3.34 + log 2

     = 3.34 + 0.30

     = 3.60

pH at the equivalence point is 3.60

Learn more about hydrogen ion concentration here:-

brainly.com/question/21931436

#SPJ4

8 0
2 years ago
A certain watch’s luminous glow is due to zinc sulfide paint that is energized by beta particles given off by tritium, the radio
sp2606 [1]

Answer:

The watch is 40.9 years old.

Explanation:

To know how many years old is the watch we need to use the following equation:

I_{(t)} = I_{0}e^{-\lambda t}   (1)

Where:

I_{(t)}: is the brightness in a time t = (1/10)I₀

I_{0}: is the initial brightness

λ: is the decay constant of tritium

The decay constant is given by:

\lambda = \frac{ln(2)}{t_{1/2}}   (2)

Where:

t_{1/2}: is the half-life of tritium = 12.3 years

By entering equation (2) into (1)  we have:

I_{(t)} = I_{0}e^{-\lambda t} = I_{0}e^{-\frac{ln(2)}{t_{1/2}}t}

\frac{I_{(t)}}{I_{0}} = e^{-\frac{ln(2)}{t_{1/2}}t}

By solving the above equation for "t" we have:

ln(\frac{I_{(t)}}{I_{0}}) = -\frac{ln(2)}{t_{1/2}}t

t = -\frac{ln(\frac{I_{(t)}}{I_{0}})}{\frac{ln(2)}{t_{1/2}}} = -\frac{ln(\frac{1}{10})}{\frac{ln(2)}{12.3}} = 40.9 y

Therefore, the watch is 40.9 years old.

 

I hope it helps you!

7 0
3 years ago
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