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Elis [28]
3 years ago
13

A single alkyl bromide reactant theoretically yields either of the given products, depending on the reaction conditions. Draw th

e structure of the alkyl bromide compound.

Chemistry
2 answers:
Ierofanga [76]3 years ago
8 0

Answer:

H_{3}C--H_{3}C--CH_{3}--Br--CH_{3}

Explanation:

The steps involved in predicting the structure of the alkyl bromide compound are outlined below.

1) An examination of the product shows that the product could only be formed by a substitution reaction.

2) The structure of the alkyl bromide compound can be then predicted by replacing the methoxide group in the product after the substitution of bromine atom. This is because the methoxide ion acts as a strong nucleophile.

Therefore, by consideration the reaction mechanisms of reactions 1 and 2, it can be predicted that the structure of the alkyl bromide compound is H_{3}C--H_{3}C--CH_{3}--Br--CH_{3}. A pictorial diagram of the alkyl bromide compound is also attached.

Sergeu [11.5K]3 years ago
5 0

Answer:

Yields an ether either with one step reaction or may undergo a bimolecular reaction

Explanation:

Here the substitution reactions of alkyl halides: explores two mechanisms.

Step 1:

 The reaction takes place in a single step- direction, bond-breaking and bond-forming occur simultaneously this is a dissociative and produces carbonation. SN1 Reaction

Or

Step 2: bimolecular reaction and it is stereoselective. SN2

Note: As a rule, nucleophilic substitutions occur at sp3-hybridized carbons, and not where the leaving group is attached to an sp2-hybridized carbon.

Bonds on sp2-hybridized carbons are inherently shorter and stronger than bonds on sp3 - hybridized carbons, meaning that it is harder to break the C-X (Halide) bond in these substrates.

Where as SN2 reactions of this type are unlikely also because the (hypothetical) electrophilic carbon is protected from nucleophilic attack by electron density in the p bond.

SN1 reactions are highly unlikely, because the resulting carbocation intermediate, which would be sp-hybridized, would be very unstable .

So this above births the expression theoretical yield. Here is a question and the reactions.

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1. Calculate the momentum of a 1,500 kg car traveling at 6 m/s.
ziro4ka [17]

Answer:

1.9000kgm/s

2.258,000kgm/s

Explanation:

1.Momentum is given as the product of mass by velocity of an object.

Momentum,p=mv

m=1,500kh, v=6m/s

p=1500*6\\p=9000kgm/s

2.Momentum,p=mv

m=7800kg, v=30m/s

p_1=7800*30\\p_1=234,000kgm/s\\

new mass=7800+800=8600

As mass is increased, so does the resultant velocity as mass is directly proportional to velocity.

p_2=8600*30\\p_2=258,000kgm/s

8 0
4 years ago
2. As NH4OH is added to an HCl solution, the pH of the solution
chubhunter [2.5K]

Answer:

c

Explanation:

Nh4OH + HCL ---> NH4Cl + H3O

so ph decreases as H3O increases

and OH also decreases

5 0
3 years ago
Read 2 more answers
Given the standard enthalpy changes for the following two reactions:
lawyer [7]

Answer:

ΔH° = -186.2 kJ

Explanation:

Hello,

This case in which the Hess method is applied to compute the required chemical reaction. Thus, we should arrange the given first two reactions as:

(1) it is changed as:

SnCl2(s) --> Sn(s) + Cl2(g)...... ΔH° = 325.1 kJ

That is why the enthalpy of reaction sign is inverted.

(2) remains the same:

Sn(s) + 2Cl2(g) --> SnCl4(l)......ΔH° = -511.3 kJ

Therefore, by adding them, we obtain the requested chemical reaction:

(3) SnCl2(s) + Cl2(g) --> SnCl4(l)

For which the enthalpy change is:

ΔH° = 325.1 kJ - 511.3 kJ

ΔH° = -186.2 kJ

Best regards.

7 0
3 years ago
Read 2 more answers
The PH of a solution of Hcl is 2.find out the amount of acid present in a litre of the solution ​
Scrat [10]

Answer:

The solution is 10^-2 or 0.01M in HCl.

Explanation:

meaning of pH is "power of hydrogen".

what is the molar concentration of a HCl solution with pH=2?

Let say pH=2

[H+]=10^-2M

HCL is a strong acid that dissociates completely:

[H+]=[HCL]

Therefore solution is 10^-2 or 0.01M in HCL.

5 0
3 years ago
1.How many mL of 0.401 M HI are needed to dissolve 5.97 g of BaCO3?
garri49 [273]

Answer:

The answer to your question is:

1.- volume = 0.151 l or 151 ml

2.- 0.241 l  or 241 ml of NaOH

Explanation:

1.-

Data

V = ? HI = 0.401 M

BaCO3 = 5.97 g

                     2HI(aq)    +    BaCO3(s)   ⇒   BaI2(aq) + H2O(l) + CO2(g)

MW BaCO3 = 137 + 12 + 48 = 197 g

                     197 g of BaCO3 ----------------- 1 mol

                     5.97 g                -----------------   x

                     x = (5.97 x 1) /197

                    x = 0.03 mol of BaCO3

                    2 moles of HI ----------------  1 mol of BaCO3

                    x                     ----------------  0.03 mol of BaCO3

                    x = (0.03 x 2) / 1

                   x = 0.060 mol of HI

Molarity = moles / volume

volume = moles / molarity

volume = 0.060 / 0.401

volume = 0.151 l or 151 ml

2.-

V = ?    NaoH 0.757 M

Co⁺² Volume = 167 ml   0.548 M

             CoSO4(aq) + 2NaOH(aq)   ⇒   Co(OH)2(s) + Na2SO4(aq)

Moles of Co = Molarity x  volume

Moles of Co = 0.548 x 0.167

Moles of Co = 0.092

                                 1 mol of CoSO4 -------------- 2 moles of NaOH

                                0.092 moles      ---------------   x

                                x = (0.092 x 2) /1

                               x = 0.183 moles of NaOH

Volume of NaOH = moles / molarity

                             = 0.183 / 0.757

                            = 0.241 l  or 241 ml of NaOH

6 0
3 years ago
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