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ludmilkaskok [199]
2 years ago
12

Cis-4-tertButylcyclohexyl bromide (compound 1) and Trans-4 tert Butylcyclohexylbromide (compound 2) are reacted with Potassium T

ertiary butoxide in Tertiary butanol to produce 4-tertbutylcyclohexene. The following statement is completely true?A. In compound 1 the Tert butyl group occupies the equatorial position and the Bromine occupies the axial position and in compound 2 both the Tert butyl and the bromine occupy equatorial positions. Compound 1 reacts faster than compound 2.
B. In compound 1 the Tert butyl group occupies the axial position and the Bromine occupies the axial position and in compound 2 both the Tert butyl and the bromine occupy equatorial positions. Compound 1 reacts faster than compound 2.
C. In compound 1 the Tert butyl group occupies the equatorial position and the Bromine occupies the equatorial position and in compound 2 both the Tert butyl and the bromine occupy equatorial positions. Compound 1 reacts faster than compound 2.
D. In compound 1 the Tert butyl group occupies the equatorial position and the Bromine occupies the axial position and in compound 2 the Tert butyl occupies the axial and the bromine occupies equatorial positions. Compound 1 reacts faster than compound 2.
E. In compound 1 the Tert butyl group occupies the equatorial position and the Bromine occupies the axial position and in compound 2 both the Tert butyl occupies the equatorial and the bromine occupies axial position. Compound 1 reacts faster than compound 2.
F. In compound 1 the Tert butyl group occupies the equatorial position and the Bromine occupies the axial position and in compound 2 both the Tert butyl and the bromine occupy equatorial positions. Compond 2 reacts faster than compound 1.
Chemistry
1 answer:
Mkey [24]2 years ago
4 0

Answer:

In compound 1 the Tert butyl group occupies the equatorial position and the Bromine occupies the axial position and in compound 2 the Tert butyl occupies the axial and the bromine occupies equatorial positions. Compound 1 reacts faster than compound 2.

Explanation:

In cyclic organic compounds, substituents may occupy the axial or equatorial positions. The axial positions are aligned parallel to the symmetry axis of the ring while the equatorial positions are around the plane of the ring.

Bulky substituents have more room in the equatorial than in the axial position. This means that compound 1 is more stable than compound 2.

This is clear on the basis of stability of the molecules because compound 1 will react faster than compound 2 since the bulky tertiary butyl group in compound 1 occupy equatorial and not axial positions.

You might be interested in
How many double bonds does CCL2H2 have?
galben [10]
None. Both chlorines and both hydrogens are single-bonded to the central carbon atom; the molecule is comprised of four single bonds and no double bonds.

Hope this helps!
4 0
2 years ago
Stacy reads that a 50:50 mixture of methanol and water is best for keeping a car’s radiator from freezing when the temperature g
salantis [7]
1. The hypothesis for this is experiment is that the 50:50 of methanol-water mixture will not turn to solid when the temperature reaches to -40°C.

2. The procedure for this is measuring equal volumes of water and methanol using the graduated cylinder. You can measure 100 mL of water and 100 mL of methanol using the graduated cylinder. Then, mix them in the beaker. Next, measure 200 mL of water, and another 200 mL of methanol. Don't mix them. Also, make a 60:40 mixture by measuring 120 mL of water and 80 mL of methanol, then mix them together. Place them all in the refrigerator at the same time. Record the time when they would freeze to solid.

3. The controls for this experiment are the 200 mL water alone, and the 200 mL methanol alone.

4. The independent variable in here is the time, while the dependent variable is the temperature of the mixtures.

5. If the hypothesis turns out to be true, then all the mixtures prepared should freeze and become solid after a certain period of time, with the exception of the 50:50 mixture. The 50:50 mixture should still remain as a liquid even when left overnight.
5 0
3 years ago
Read 2 more answers
A
amm1812

Solution:

1) Separate out the half-reactions. The only issue is that there are three of them.

<span>Fe2+ ---> Fe3+ 
S2¯ ---> SO42¯ 
NO3¯ ---> NO</span>

How did I recognize there there were three equations? The basic answer is "by experience." The detailed answer is that I know the oxidation states of all the elements on EACH side of the original equation. By knowing this, I am able to determine that there were two oxidations (the Fe going +2 to +3 and the S going -2 to +6) with one reduction (the N going +5 to +2).

Notice that I also split the FeS apart rather than write one equation (with FeS on the left side). I did this for simplicity showing the three equations. I know to split the FeS apart because it has two "things" happening to it, in this case it is two oxidations.

Normally, FeS does not ionize, but I can get away with it here because I will recombine the Fe2+ with the S2¯ in the final answer. If I do everything right, I'll get a one-to-one ratio of Fe2+ to S2¯ in the final answer.

2) Balancing all half-reactions in the normal manner.

<span>Fe2+ ---> Fe3+ + e¯ 
4H2O + S2¯ ---> SO42¯ + 8H+ + 8e¯ 
3e¯ + 4H+ + NO3¯ ---> NO + 2H2O</span>

3) Equalize the electrons on each side of the half-reactions. Please note that the first two half-reactions (both oxidations) total up to nine electrons. Consequently, a factor of three is needed for the third equation, the only one shown below:

<span>3 [3e¯ + 4H+ + NO3¯ ---> NO + 2H2O]</span>

Adding up the three equations will be left as an exercise for the reader. With the FeS put back together, the sum of all the coefficients (including any that are one) in the correct answer is 15.

Problem #2: CrI3 + Cl2 ---> CrO42¯ + IO4¯ + Cl¯ [basic sol.]

Solution:

Go to this video for the solution

Problem #3: Sb2S3 + Na2CO3 + C ---> Sb + Na2S + CO

Solution:

1) Remove all the spectator ions:

<span>Sb26+ + CO32- + C ---> Sb + CO</span>

Notice that I did not write Sb3+. I did this to keep the correct ratio of Sb as reactant and product. It also turns out that it will have a benefit when I select factors to multiply through some of the half-reactions. I didn't realize that until after the solution was done.

2) Separate into half-reactions:

<span>Sb26+ ---> Sb 
CO32- ---> CO 
C ---> CO</span>

3) Balance as if in acidic solution:

<span>6e¯ + Sb26+ ---> 2Sb 
2e¯ + 4H+ + CO32- ---> CO + 2H2O 
H2O + C ---> CO + 2H+ + 2e¯Could you balance in basic? I suppose, but why?</span>

4) Use a factor of three on the second half-reaction and a factor of six on the third.

<span>6e¯ + Sb26+ ---> 2Sb 
3 [2e¯ + 4H+ + CO32- ---> CO + 2H2O] 
6 [H2O + C ---> CO + 2H+ + 2e¯]The key is to think of 12 and its factors (1, 2, 3, 4, 6). You need to make the electrons equal on both sides (and there are 12 on each side when the half-reactions are added together). You get 12 H+ on each side (3 x 4 in the second and 6 x 2 in the third). You get six waters with 3 x 2 in the second and 6 x 1 in the third.Everything that needs to cancel gets canceled!</span>

5) The answer (with spectator ions added back in):

<span>Sb2S3 + 3Na2CO3 + 6C ---> 2Sb + 3Na2S + 9CO</span>

6) Here's a slightly different take on the solution just presented.

<span>a) Write the net ionic equation:<span>Sb26+ + CO32- + C ---> Sb + CO</span>b) Notice that charges must be balanced and that we have zero charge on the right. So, do this:<span>Sb26+ + 3CO32- + C ---> Sb + CO</span>c) Now, balance for atoms:<span>Sb26+ + 3CO32- + 6C ---> 2Sb + 9CO</span>d) Add back the sodium ions and sulfide ions to recover the molecular equation.<span>Sb2S3 + 3Na2CO3 + 6C ---> 2Sb + 3Na2S + 9CO</span></span>

7) Here's a discussion of a wrong answer to the above problem.

However, after reading the above wrong answer example, look at problem #10 below for an instance of having to add in a substance not included in the original reaction.

Problem #4: CrI3 + H2O2 ---> CrO42¯ + IO4¯ [basic sol.]

Solution:

1) write the half-reactions:

<span>Cr3+ ---> CrO42¯ 
I33¯ ---> IO4¯ 
H2O2 ---> H2O</span>

I wrote the iodide as I33¯ to make it easier to recombine it with the chromium ion at the end of the problem.

2) Balance as if in acidic solution:

<span>4H2O + Cr3+ ---> CrO42¯ + 8H+ + 3e¯ 
12H2O + I33¯ ---> 3IO4¯ + 24H+ + 24e¯ 
2e¯ + 2H+ + H2O2 ---> 2H2O</span>

I used water as the product for the hydrogen peroxide half-reaction because that gave me a half-reaction in acid solution. It will all go back to basic at the end of the problem.

3) Recover CrI3 by combining the first two half-reactions from just above:

<span>16H2O + CrI3 ---> 3IO4¯ + CrO42¯ + 32H+ + 27e¯</span>

4) Equalize the electrons:

<span>2 [16H2O + CrI3 ---> 3IO4¯ + CrO42¯ + 32H+ + 27e¯] 
27 [2e¯ + 2H+ + H2O2 ---> 2H2O]leads to:32H2O + 2CrI3 ---> 6IO4¯ + 2CrO42¯ + 64H+ + 54e¯ 
54e¯ + 54H+ + 27H2O2 ---> 54H2O</span>

5) Add the half-reactions together. Strike out (1) electrons, (2) hydrogen ion and (3) water. The result:

<span>2CrI3 + 27H2O2 ---> 2CrO42¯ + 6IO4¯ + 10H+ + 22H2O</span>

6) Add 10 hydroxides to each side. This makes 10 more waters on the right, so combine with the water alreadyon the right-hand side to make 32:

<span>2CrI3 + 27H2O2 + 10OH¯ ---> 2CrO42¯ + 6IO4¯ + 32H2O</span>



3 0
3 years ago
4. Which of the following NOT TRUE about real gas
Semenov [28]

Answer:

The answer is D.

Explanation:

Intermolecular force are negligible

When the distance between molecules decrease,

the attraction or repulsion become greater

7 0
2 years ago
What is the volume of 3.00 mole of ideal gas at 100.0 C and 2.00 kPa
Novosadov [1.4K]

Answer:

The volume for the ideal gas is: 4647.5 Liters

Explanation:

Formula for the Ideal Gases Law must be applied to solve this question:

P . V = n .  R . T

We convert the T° to K → 100°C + 273 = 373 K

We convert pressure value from kPa to atm.

2 kPa . 1atm/101.3 kPa = 0.0197 atm

We replace data in the formula.

V = ( n . R . T) / P → (3 mol . 0.082 . 373K) / 0.0197 atm =

The volume for the ideal gas is: 4647.5 Liters

8 0
2 years ago
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