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snow_tiger [21]
4 years ago
14

Which compound will have the larger rf value on a si02 tlc plate using 10 % ethyl acetate/ hexane as eluent: 3-decanone or 3-dec

anol, toluene or benzoic acid, cyclooctane or cyclooctanone? why?
Chemistry
1 answer:
Hatshy [7]4 years ago
7 0

The formula of determining the R_f value is:

R_f = \frac{distance travelled by the solute}{distance travelled by the solvent}

where both the distance are measured from the common origin.

10 % ethyl acetate/ hexane means it has 10 % ethyl acetate and 90 % hexane which are non-polar in nature. Hence, the non-polar compound will travel farther in the TLC plate along with the eluent and the polar compound will travel less as they remain attracted to the polar adsorbent SiO_2.

Among 3-decanone and 3-decanol, 3-decanone will have high R_f as it is non-polar.

Among toluene and benzoic acid, toluene will have high R_f as it is non-polar.

Among cyclooctane and cyclooctanone, cyclooctane will have high R_f as it is non-polar.

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Compounds X, C9H19Br, and Y, C9H19Cl, undergo base-promoted E2 elimination to give the same single alkene product, Z. Catalytic
Pavlova-9 [17]

Answer:

Compound X= 4-bromo-2,3,3-trimethylhexane

Compound Y= 5-chloro-2,3,3-trimethylhexane

Explanation:

The first step is set up the problem. That way we can obtain some clues. If we check figure 1 we can obtain some ideas:

-) If we have E2 reaction is not possible a <u>methyl or hydride shift</u>.

-) If we have an E2 reaction we will need an H in <u>anti position</u> to obtain the double bond. Therefore a double bond with the quaternary carbon (the carbon bonded to the 2 methyl groups).

The second step is to solve the alkene structure. We have to put the <u>leaving group</u> near to carbon that has more possible <u>removable hydrogens</u>.  That's why the double bond is put it between carbons 5 and 4 of the alkane (Figure 2).

The third step is the structure of the <u>alkyl bromide</u> structure. To do this we have to check the alcohol produced by the alkene. In the <u>hydration of alkanes</u> reaction we will have a <u>carbocation</u> formation. Therefore we can have for the alkene proposed a methyl shift to obtain the most stable carbocation. With this in mind, we have to do the same for the Alkyl bromide that's why the Br is put it carbon 4 of the alkane. If we put the Br on this carbon we can have the chance of this <u>methyl shift</u> also, to obtain the same alcohol (figure 3).

Finally, for the <u>alkyl chloride</u>, we only have 2 choices because to produce the alkane we have to put the <u>leaving group</u> on one of the 2 carbons of the double bond. If we choose the same carbon on which we put the Br we can have the same behavior of the alkyl bromide (the <u>methyl shift</u>), therefore we have to put in the other carbon.

7 0
4 years ago
What is the molar mass of water in the hydrate ?
Free_Kalibri [48]
The mass of water equals the mass of the hydrate minus the mass of the anhydrite 4.9 G -2.9 G equals 2.0 G of water moles of water equals the mass of water/the molar mass of water 2.0 GH20/18 G/MOLHTO equals 0.11 Moles H20
6 0
3 years ago
A sample of helium, at 20 °c, occupies a volume of 3.17 l at a pressure of 6.21 atm. what volume does this helium sample occupy
Lynna [10]
Boyle's law gives the relationship between pressure of a gas and volume.
It states that at a constant temperature the pressure of gas is inversely proportional to volume of gas.
PV = k
where P - pressure, V - volume and k - constant
P1V1 = P2V2
where parameters for the first instance are given on the left side and parameters for the second instance are given on the right side of the equation.
substituting these values in the equation
6.21 atm x 3.17 L = 5.57 atm x V
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new volume is 3.53 L
8 0
4 years ago
A student pours 10.0 g of salt into a container of water and observes the amount of time it takes for the salt to dissolve. She
9966 [12]

Answer:

It will go faster each time because she is stirring therefore the water can get to the salt faster than it just sitting at the top

Explanation:

6 0
3 years ago
Which of the following groups of elements are likely to have substances that have both metal and non-metallic properties?
Tom [10]

Answer : Group 14 elements are likely to have substances that have both metal and non-metallic properties.

Explanation :

Metalloid : It is a type of element whose properties lies in between the metals and non-metals.

There are commonly six metalloids : Boron(group 13), Silicon & Germanium(group 14), Arsenic & Antimony(group 15), Tellurium(group 16).

Properties of Metalloids :

  • Metalloids look like metals but behave largely like non-metals.
  • Metalloids are solid, brittle and lustrous.
  • Metalloids has ability to conduct electricity.

Therefore, Group 14 elements are likely to have substances that have both metal and non-metallic properties.


4 0
3 years ago
Read 2 more answers
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