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charle [14.2K]
3 years ago
13

Give the mechanistic symbols (SN1, SN2) that are most consistent with each other the following statements:a) Methyl halides reac

t with sodium ethoxide in ethanol only by this mechanism.b) Unhindered primary halides react with sodium ethoxide in ethanol mainly by this mechanis.c) The substitution product obtained by solvolysis of tert- butyl bromide in ethanol arises by this mechanism.d) Reactions proceeding by this mechanism are stereospecfic.e) Reactions proceeding by this mechanism involve carbocation intermediates.f) this mechanism is most likely to have been involved when the products are found to have a different carbon skeleton from the substrate.g) Alkyl iodides react faster than alkyl bromides in reactions that proceed by these mechanisms.
Chemistry
1 answer:
WARRIOR [948]3 years ago
5 0

Answer:

very comfusing

Explanation:

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a race car at the Daytona 500 zips around the track .its travel the first 80 kilometers in 0.4 hours. the next 114 kilometers ta
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The answer to this should be around 40
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3 years ago
A gaseous mixture contains 434.0 Torr H 2 ( g ) , 434.0 Torr H2(g), 389.9 Torr N 2 ( g ) , 389.9 Torr N2(g), and 77.9 Torr Ar (
weqwewe [10]

Answer:

H₂: 0.48,  N₂: 0.43;  Ar: 0.09

Explanation:

First of all, sum all the pressures to know the total pressure in the mixture.

434 Torr + 389.9 Torr + 77.9 Torr = 901.8 Torr

Mole fraction = Pressure gas / Total Pressure

Mole Fraction H₂: 434 Torr /901.8 Torr = 0.48

Mole Fraction N₂: 389.9 /901.8 Torr =0.43

Mole Fraction Ar:  77.9 /901.8 Torr = 0.09

Remember: <u>SUM OF MOLE FRACTION = 1</u>

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3 years ago
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6 0
3 years ago
Read 2 more answers
Show how you might synthesize this compound from an alkyl bromide and a nucleophile in an SN2 reaction. Use the wedge/hash bond
PolarNik [594]

Answer:

See the image 1

Explanation:

If you look carefully at the progress of the SN2 reaction, you will realize something very important about the outcome. The nucleophile, being an electron-rich species, must attack the electrophilic carbon from the back side relative to the location of the leaving group. Approach from the front side simply doesn't work: the leaving group - which is also an electron-rich group - blocks the way. (see image 2)

The result of this backside attack is that the stereochemical configuration at the central carbon inverts as the reaction proceeds. In a sense, the molecule is turned inside out. At the transition state, the electrophilic carbon and the three 'R' substituents all lie on the same plane. (see image 3)

What this means is that SN2 reactions whether enzyme catalyzed or not, are inherently stereoselective: when the substitution takes place at a stereocenter, we can confidently predict the stereochemical configuration of the product.

6 0
3 years ago
Determine the partial negative charge on the bromine atom in a c−br bond. the bond length is 1.93 å and the bond dipole moment i
vaieri [72.5K]

Answer:

The value is x  =  0.151  \ e

Explanation:

From the question we are told that

The bond length is l  =  1.93\  \r  a =  1.93 *1 *10^{-10}  =1.93 *10^{-10}\  m

The bond dipole moment is \mu  = 1.40 d  = 1.40 *  3.33564 *10^{-30}  =  4.6699 *10^{-30} \  C \cdot m

Generally the dipole moment is mathematically represented as

\mu  =  Q *  l

Here Q is the partial negative charge on the bromine atom

So

Q =  \frac{\mu}{ l}

=> Q =  \frac{4.6699 *10^{-30}}{ 1.93 *10^{-10} }

=> Q = 2.42 *10^{-20} C

Generally

1 electronic charge(e) is equivalent to 1.60*10^{-19} C

So x electronic charge(e) is equivalent to Q = 2.42 *10^{-20} C

=> x  =  \frac{2.42 *10^{-20}}{1.60*10^{-19} }

=>     x  =  0.151  \ e

3 0
3 years ago
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