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Vikki [24]
2 years ago
11

Draw a structure for the product of nucleophilic substitution obtained on solvolysis of tert−butyl bromide in methanol, and arra

nge the correct mechanism for its formation. be sure to answer all part

Chemistry
2 answers:
dem82 [27]2 years ago
5 0

The mechanism will include

a) Slow step: rate determining step

In this step the leaving group (chloride ion) will leave the substrate and will form carbocation

b) Fast step:

The cation will be attacked by nucleophile and will give the substituted product

the mechanism is shown as :


Solnce55 [7]2 years ago
4 0

Final product of the solvolysis of tert-butyl bromide in methanol is 2-methoxy-2-methylpropane.

Further Explanation:

Nucleophilic substitution reaction:

Nucleophilic substitution reaction is the chemical reaction where electron rich nucleophile replaces the leaving group from the substrate. Nucleophilic reactions are widely used in the conversion of one functional group to another.

Nucleophilic substitution reactions are two types:

1. The nucleophilic substitution reaction {{\text{S}}_{\text{N}}}1 is a two-step substitution reaction. In the first step, the removal of the leaving group takes place in order to form the carbocation intermediate. The second step involves the formation of the product by the attack of the incoming nucleophile on the carbocation intermediate.

2. Nucleophilic substitution reaction {{\text{S}}_{\text{N}}}2 is a single step substitution reaction in which electron-rich species substitute leaving group. The attack takes place from the backside of the electrophilic center thus an inversion of the configuration takes place.

The mechanism of solvolysis of tert-butyl bromide in methanol is {{\text{S}}_{\text{N}}}1  type (refer to the image attached), therefore

Step 1: Formation of carbocation intermediate from tert-butyl bromide after leaving of bromide ion.

Step 2: Then methanol act as the nucleophile and attack on the carbocation intermediate.

Step 3: The last step is deprotonation to produce the final product is 2-methoxy-2-methylpropane.

Learn more:

1. Product formed when 2-propanol react with NaH: brainly.com/question/5045356

2. Stereoisomer for 1,2-difluoro-1,2-dimethylcyclopentane: brainly.com/question/8803065

Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Nucleophilic substitution reactions

Keywords: Nucleophilic substitution reaction, SN1, SN2, 2-methoxy-2-methylpropane, mechanism, solvolysis, tert-butyl bromide, methanol, steps of mechanism.

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Answer:

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Explanation:

The AP Classroom will not count your answer to this question as correct unless it includes at least one of the answers listed above. If you say that theanswer to this question is density, it will be marked as incorrect, I found that out the hard way when I used the answers that brainly gave me.

Good luck,

I applaud you for using the sources avalible to you, which is /definetly not/ cheeting.

8 0
1 year ago
James was given a sample of sodium bicarbonate NaHCO3 in a weighing dish to analyze. His teacher asked him to find the percent o
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This question could be answered easily if the results of the abundance of the other elements are given. You will just have to subtract the sum of all their abundances to 100. Since it's not given, the solution would just be:

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Draw all of the constitutional isomers of the molecule with formula C3H5Br. Ignore geometric and stereoisomers.
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Answer:

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Explanation:

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I.D.H~=~\frac{(2C)+2+(N)-(H)-(X)}{2}

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I.D.H~=~\frac{(2*3)+2+(0)-(5)-(1)}{2}~=~1

We have an I.D.H value of one. This indicates that we can have a cyclic structure or a double bond.

We can start with a linear structure with 3 carbon with a double bond in the first carbon and the Br atom also in the first carbon (<u>1-bromoprop-1-ene</u>). In the second structure, we can move the Br atom to the second carbon (<u>2-bromoprop-1-ene</u>), in the third structure we can move the Br to carbon 3 (<u>3-bromoprop-1-ene</u>). Finally, we can have a cyclic structure with a Br atom (<u>bromocyclopropane</u>).

See figure 1

I hope it helps!

7 0
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