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sleet_krkn [62]
1 year ago
13

What is the IUPAC name of the following compound a. (R)-3-chloro-1-methylcyclohexene b. (S)-3-chloro-1-methylcyclohexene c. (R)-

1-chloro-3-methyl-2-cyclohexene d. (S)-1-chloro-3-methyl-2-cyclohexene
Chemistry
1 answer:
nikklg [1K]1 year ago
7 0

The answer is: (R)-1-chloro-3-methyl-2-cyclohexene.

R- and S- configuration: The nomenclature "right-handed" and "left-handed" is used to name the enantiomers of a chiral compound. Stereocenters are indicated as R or S.

How can the R- and S- configuration is assigned to a structure.

  • We must first identify the carbon(s) with the four distinct groups (atoms) connected in order to designate an absolute configuration. These are called centers of chirality.
  • Give each atom attached to a chiral center a priority based on its atomic number. The higher the atomic number, the higher the priority.
  • Draw an arrow starting at priority one and going to priority two and then to priority 3:
  • If the arrow goes clockwise, as in this case, the absolute configuration is R configuration.
  • In contrast, if the arrow goes counter-clockwise, then the absolute configuration is S configuration.
  • Now, the given structure is (shown in image). The priority is given as ( image 2). As the priority is going clockwise, thus the configuration is 'R'.
  • Hence, the IUPAC name of compound is-  (R)-1-chloro-3-methyl-2-cyclohexene.

To learn more about R- and S- configuration visit: brainly.com/question/16812236

#SPJ4

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Answer:

The answer to your question is below

Explanation:

H₂SO₄ = sulphuric acid

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This is a neutralization reaction

                       H₂SO₄   +  2 KOH   ⇒     K₂SO₄   +  2H₂O

                                 2 -----------   K  ----------- 2

                                 1 -----------   S  ----------  1

                                 4 ----------   H  ---------   4

                                6 ------------   O ----------  6

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Explanation:

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Explanation:

It is known that pK_{a} of propionic acid = 4.87

And, initial concentration of  propionic acid = \frac{0.19}{1.20}

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Concentration of sodium propionate = \frac{0.26}{1.20}[/tex]

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Now, in the given situation only propionic acid and sodium propionate are present .

Hence,      pH = pK_{a} + log(\frac{[salt]}{[acid]})

                       = 4.87 + log \frac{0.216}{0.158}

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                        = 5.00

  • Therefore, when 0.02 mol NaOH is added  then,

     Moles of propionic acid = 0.19 - 0.02

                                              = 0.17 mol

Hence, concentration of propionic acid = \frac{0.17}{1.20 L}

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and,      moles of sodium propionic acid = (0.26 + 0.02) mol

                                                                  = 0.28 mol

Hence, concentration of sodium propionic acid will be calculated as follows.

                        \frac{0.28 mol}{1.20 L}

                           = 0.23 M

Therefore, calculate the pH upon addition of 0.02 mol of NaOH as follows.

             pH = pK_{a} + log(\frac{[salt]}{[acid]})

                       = 4.87 + log \frac{0.23}{0.14}

                        = 4.87 + log (1.64)

                        = 5.08

Hence, the pH of the buffer after the addition of 0.02 mol of NaOH is 5.08.

  • Therefore, when 0.02 mol HI is added  then,

     Moles of propionic acid = 0.19 + 0.02

                                              = 0.21 mol

Hence, concentration of propionic acid = \frac{0.21}{1.20 L}

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and,      moles of sodium propionic acid = (0.26 - 0.02) mol

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                        = 4.87 + log (0.114)

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Hence, the pH of the buffer after the addition of 0.02 mol of HI is 4.98.

7 0
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