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emmainna [20.7K]
3 years ago
14

(f) Which of the following alkenes is the major product when 2-bromo-2-methylpentane is treated with sodium ethoxide in ethanol?

Chemistry
2 answers:
belka [17]3 years ago
8 0

Answer:

D) 2-methylpent-2-ene

Explanation:

This is an elimination reaction of Halogenoalkane. 2-bromo-2-methylpentane when is heated with NaOH or NaOC2O5( sodium ethoxide) in ethanol will form alkene rather than alcohol.

2-methylpent-1-ene is minor product since double bond form with secondary Carbon rather than primary Carbon.

serious [3.7K]3 years ago
4 0

Answer:

The major product when 2-bromo-2-methylpentane is treated with sodium ethoxide in ethanol is: 2-methylpent-2-ene

Explanation:

The elimination reaction generates the most substituted alkene (Zaitsev's rule)

Hence, the major product when 2-bromo-2-methylpentane is treated with sodium ethoxide in ethanol is 2-methylpent-2-ene.

In the attached picture, you can see that the 2-methylpent-1-ene is less substituted

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Answer:

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Feliz [49]
Number of Atoms in Gold for given mass can be calculated using following formula,

              # of Moles  =  Number of Atoms / 6.022 × 10²³
Or,
             Number of Atoms  =  Moles × 6.022 × 10²³     ------- (1)    

Calculating Moles,
As,
                              Moles  =  Mass / M.mass
So,
                              Moles  =  4.25 g / 196.96 g/mol

                              Moles  =  0.0215

Putting value of mole in eq.1,

             Number of Atoms  =  0.0215 × 6.022 × 10²³

             Number of Atoms  =  1.299 × 10²²

Result:
            4.25 g 
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3 years ago
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If we take the lower value (0.1 g/pea), the number of peas in 454 g is:

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If we take the higher value (0.36 g/pea), the number of peas in 454 g is:

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You can learn more about conversion factors here: brainly.com/question/1844638

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Alexxx [7]

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Justification:


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