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8090 [49]
3 years ago
7

Which of the following does not correctly describe Sn2 reactions of alkyl halides? A) The mechanism consists of a single step wi

th no intermediates. B) Tertiary halides react faster than secondary halides. C) The transition state species has a pentavalent carbon atom. D) Rate of reaction depends on the concentrations of both the alkyl halide and the nucleophile.
Chemistry
1 answer:
Rashid [163]3 years ago
7 0

Answer:

The correct answer is B the tertiary halides reacts faster than primary halides.

Explanation:

During SN2 reaction the nucleophile attack the alkyl halide from the opposite side resulting in the formation of transition state in which a bond is not completely broken or a new bond is not completely formed.

   After a certain period of time the nucleophile attach with the substrate by substituting the existing nuclophile.

  An increase in the bulkiness in the alkyl halide the SN2 reaction rate of that alkyl halide decreases.This phenomenon is called steric hindrance.

  So from that point of view the that statement tertiary halides reacts faster that secondary halide is not correct.

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Given the E0 values of the following two half-reactions: Zn  Zn2+ + 2e- E0 = 0.763 volt Fe  Fe2+ + 2e- E0 = 0.441 volt a) Writ
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<u>Answer:</u>

<u>For a:</u> The balanced chemical equation is written below.

<u>For b:</u> The corrosion of iron pipe will take place in the presence of zinc.

<u>For c:</u> Zinc will not protect iron pipe from corrosion.

<u>Explanation:</u>

  • <u>For a:</u>

The given half reaction follows:

The substance having highest positive E^o potential will always get reduced and will undergo reduction reaction. Here, zinc will undergo reduction reaction will get reduced.

<u>Oxidation half reaction:</u>  Zn\rightarrow Zn^{2+}+2e^-;E^o_{Zn^{2+}/Zn}=-0.763V

<u>Reduction half reaction:</u>  Fe+2e^-\rightarrow Fe;E^o_{Fe^{2+}/Fe}=-0.441V

The balanced chemical equation follows:

Fe+Zn^{2+}\rightarrow Fe^{2+}+Zn

  • <u>For b:</u>

For a reaction to be spontaneous (thermodynamically feasible) , the standard electrode potential must be positive.

To calculate the E^o_{cell} of the reaction, we use the equation:

E^o_{cell}=E^o_{cathode}-E^o_{anode}

Substance getting oxidized always act as anode and the one getting reduced always act as cathode.

Calculating the E^o_{cell} using above equation, we get:

E^o_{cell}=-0.441-(-0.763)=0.322V

As, the EMF is coming out to be positive, the reaction will be thermodynamically feasible and corrosion of iron pipe will take place in the presence of zinc.

  • <u>For c:</u>

As, the EMF of the cell is positive, the zinc will not protect the iron pipe from corrosion and the reaction will take place.

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