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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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Ammonia, NH3 is a common base with Kb of 1.8 X 10-5. For a solution of 0.150 M NH3:
Vesnalui [34]

The concentrations : 0.15 M

pH=11.21

<h3>Further explanation</h3>

The ionization of ammonia in water :

NH₃+H₂O⇒NH₄OH

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The concentrations of all species present in the solution = 0.15 M

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M=0.15

\tt [OH^-]=\sqrt{Kb.M}\\\\(OH^-]=\sqrt{1.8\times 10^{-5}\times 0.15}\\\\(OH^-]=\sqrt{2.7\times 10^{-6}}=1.64\times 10^{-3}

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