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professor190 [17]
3 years ago
7

Consider the following statements in reference to SN1, SN2, E1, and E2 reactions of haloalkanes. To which mechanism(s), if any,

does each statement apply?
1. Involves a carbocation intermediate
2. Is first order in haloalkane and first order in base.
3. nvolves inversion of configuration at the site of substitution
4. Involves retention of configuration at the site of substitution
5. Substitution at a stereocenter gives predominantly a racemic product
6. Is first-order in alkyl halide and zero-order in base
7. Is first-order in alkyl halide and first order in base
8. Is greatly accelerated in protic solvents of increasing polarity.
9. Rearrangements are common
10. Order of reactivity is 3°> 2° > 1° > methyl
11. Order of reactivity of haloalkanes is methyl > 1° > 2° > 3°.
Chemistry
1 answer:
Rudik [331]3 years ago
6 0

Answer:

1. Involves a carbocation intermediate- SN1, E1

2. Is first order in haloalkane and first order in base- E2

3. involves inversion of configuration at the site of substitution- SN2

4. Involves retention of configuration at the site of substitution- SN1

5. Substitution at a stereocenter gives predominantly a racemic product- SN1

6. Is first-order in alkyl halide and zero-order in base- E1

7. Is first-order in alkyl halide and first order in base- E2

8. Is greatly accelerated in protic solvents of increasing polarity- SN1,E1

9. Rearrangements are common- SN1, E1

10. Order of reactivity is 3°> 2° > 1° > methyl- SN1

11. Order of reactivity of haloalkanes is methyl > 1° > 2° > 3°- SN2

Explanation:

SN1 is mechanism of nucleophilic substitution in which a nucleophile reacts with an alkyl halide by first order mechanism. The reaction is first order in the alkyl halide and zero order in the nucleophile. The rate determining step is the formation of a carbocation. The ease of occurrence of SN1 reaction depends on the ease of formation of a stable carbocation. The order of carbocation stability is; 3°> 2° > 1° > methyl. SN1 mechanism leads to retention of configuration at the site of substitution and a racemization when substitution occurs at a chiral carbon atom.

SN2 is a bimolecular nucleophilic substitution mechanism which is first order in both nucleophile and alkyl halide. It is favoured by aprotic solvents which do not solvate the nucleophile significantly. The order of alkyl halide reactivity by SN2 mechanism is; methyl > 1° > 2° > 3°.

E1 mechanism refers to elimination reaction in which the rate determining step is the formation of a carbocation. It is common for tertiary alkyl halides. It mostly occurs with weak bases. The reaction is first order in the alkyl halide but zero order in the base.

E2 is an elimination mechanism which is first order in both alkyl halide and base. It is favoured by strong and bulky bases. Primary alkyl halides more commonly react by E2 mechanism.

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Thank you for posting your question here. Below is the solution:

HNO3 --> H+ + NO3- 
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<span>** one doesn't need to find the molarity of water since it is the solvent </span>

<span>0M HNO3 </span>
<span>1x10^-6M H3O+ </span>
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<span>1x10^-8M OH-.....the Kw = 1x10^-14 = [H+][OH-] </span>
<span>you have 1x10^-6M H+ so, 1x10^-14 / 1x10^-6 = 1x10^-8M OH- </span>


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4 years ago
Write a balanced nuclear equation for the formation of 28 Si 14 from beta-minus emission.
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Answer:

Phosphorus-28 undergoes beta-minus decay to produce

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\rm ^{28}_{13} Al \to ^{28}_{14} Si + ^{\phantom{-}0}_{-1}e + \bar{\mathnormal{v}}_{\rm e}

Explanation:

In simple words, when a nucleus undergoes beta-minus decay, a neutron is converted to a proton. An electron and an electron antineutrino will be released.

\rm ^{1}_{0}n^{0} \to ^{1}_{1}p^{+} + ^{\phantom{-}0}_{-1}e^{-} + \bar{\mathnormal{v}}_{\rm e}.

One way to tell whether a neutron is converted to a proton, but not vice versa, is to check the sum charges on the two sides of this equation.

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The charges on the two sides of this equation is the same. Hence this nuclear equation is possible (but not necessarily correct; however, if the proton and the neutron are in the wrong place the charge won't even be the same.)

Since the mass number of a proton and a neutron are both 1, the overall mass number of the atom will stay the same.

The atomic number is the number of protons in each atom. That number determines the symbol and the chemical properties of the atom. When one neutron in an atom is converted to a proton, the atomic number of the atom will increase by 1.

The atomic number of the daughter nucleus, silicon, is 14. It takes a parent nucleus with atomic number 14 - 1 = 13 to produce a silicon atom. Refer to a modern periodic table. Atomic number 13 corresponds to the element aluminum.

Also, the mass number of the daughter nucleus is 28. Since the mass number would stay the same in a beta decay, the mass number of the parent nucleus would also be 28. In other words, it takes an aluminum-28 atom to undergo beta-decay to produce a silicon-28 atom.

Complete the other details (electron and electron antineutrino) to obtain the equation

\rm ^{28}_{13} Al \to ^{28}_{14} Si + ^{\phantom{-}0}_{-1}e + \bar{\mathnormal{v}}_{\rm e}.

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