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torisob [31]
4 years ago
13

In each of the following indicate which reaction will occur faster. Explain your reasoning. (a) CH3CH2CH2CH2Br or CH3CH2CH2CH2I

with sodium cyanide in dimethyl sulfoxide (b) 1-Chloro-2-methylbutane or 1-chloropentane with sodium iodide in acetone (c) Hexyl chloride or cyclohexyl chloride with sodium azide in aqueous ethanol (d) Solvolysis of 1-bromo-2,2-dimethylpropane or tert-butyl bromide in ethanol (e) Solvolysis of isobutyl bromide or sec-butyl bromide in aqueous formic acid (f) Reaction of 1-chlorobutane with sodium acetate in acetic acid or with sodium methoxide in methanol (g) Reaction of 1-chlorobutane with sodium azide or sodium p-toluenesulfonate in aqueous ethanol
Chemistry
1 answer:
Elena L [17]4 years ago
5 0

Answer:

A. Butyliodide will react faster than butylbromide, this is because iodide is a better leaving group in substitution reaction than bromide. B. 1-chloropentane is more reactive than 1-chloro2methylbutane, this is because Alkyl groups at the carbon atom decrease the reaction rate of SN2 reaction. C. Chloro- hexanechloride is more reactive than hexyl chloride because determining the rate of reaction in SN1 reaction of formation of carbocation is more stable than the primary one. SN1 reaction is a secondary carbocation reaction. D. Terri-butylbromide is more reactive than 1-bromo 2,2 dimethylpropane, this is because Tertiary carbocation is stable compared to the primary, this is because the rate of determining step involved in SN1 reaction. E. Sec-butyl bromide is more reactive than isobutylbromide, this is because Sec-butyl bromide belong to the secondary all halide why the other belongs to primary class. F. The reaction is faster in sodium methoxide in methanol than sodium acetate in acetic acid, this is because 1-chlorobutane is a primary alklyl- chloride than and the strength of nuclephile determines the rate of reaction. G. when sodium azide mixed with ethanol is used, the reaction will be faster, because strong acid form weak conjuage base and also azide is a better nuclephile.

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a model of an atom shows eight electrons in rings that represent different energy levels. how many electrons are in each energy
vlada-n [284]

Answer:

1. four in the first energy level, four in the second energy level

2. eight in the first energy level, zero in the second energy level

3. zero in the first energy level, eight in the second energy level

4. two in the first energy level, six in the second energy level *

Explanation:

1. four in the first energy level, four in the second energy level

2. eight in the first energy level, zero in the second energy level

3. zero in the first energy level, eight in the second energy level

4. two in the first energy level, six in the second energy level *

4 0
3 years ago
If a stock solution of 20.0 g of sugar per 100. ml of solution is available, how much of this stock is needed to prepare 10.0 ml
Vladimir79 [104]

You must use 2.50 mL of the concentrated solution to make 10.0 mL of the dilute solution.

We can use the dilution formula

<em>V</em>_1<em>C</em>_1 = <em>V</em>_2<em>C</em>_2

where

<em>V</em> represents the volumes and

<em>C</em> represents the concentrations

We can rearrange the formula to get

<em>V</em>_2 = <em>V</em>_1 × (<em>C</em>_1/<em>C</em>_2)

<em>V</em>_1 = 10.0 mL; <em>C</em>_1 = 5.00 g/100. mL

<em>V</em>_2 = ?; ____<em>C</em>_2 = 20.0 g/100. mL

∴ <em>V</em>_2 = 10.0 mL × [(5.00 g/100. mL)/(20.0 g/100. mL)] = 10.0 mL × 0.250

= 2.50 mL

3 0
3 years ago
What types of mixture can be separated by chromatography​
ser-zykov [4K]

Answer:

mixture is amino acid, peptides, carbohydrates and other simple organic compounds can be separated by paper chromatography.

4 0
3 years ago
Carbon disulfide is prepared industrially by reacting carbon with sulfur dioxide according to the above equation. If 70.8 g of c
Rus_ich [418]

Answer:

1.18 moles of CS₂ are produced by the reaction.

Explanation:

We present the reaction:

5C + 2SO₂  →  CS₂  +  4CO

5 moles of carbon react to 2 moles of sulfur dioxide in order to produce 1 mol of carbon disulfide and 4 moles of carbon monoxide.

As we do not have data from the SO₂, we assume this as the excess reagent. We convert the mass of carbon to moles:

70.8 g / 12 g/mol = 5.9 moles

Ratio is 5:1, so 5 moles of carbon react to produce 1 mol of CS₂

Then, 5.9 moles will produce (5.9 . 1) / 5 = 1.18 moles

8 0
3 years ago
I would like some help por favor / please
CaHeK987 [17]
What do you need help on
7 0
3 years ago
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