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3 years ago

Write the structure of the product that would be formed from the S(

Chemistry

1 answer:

prisoha [69]3 years ago

8 0

Answer:The product formed on reaction with hydroxide ion as nucleophile is 2R-hexane-2-ol.

The product formed on reaction with water would be a 50:50 mixture of

2S-hexane-2-ol. and 2R-hexane-2-ol.

Explanation:

2S-iodohexane on reactiong with hydroxide ion would undergo SN² substitution reaction that is substitution bimolecular. Hydroxide ion has a negative charge and hence it is a quite good nucleophile .

The rate of a SN² reaction depends on both the substrate and nucleophile . Here the substrate is a secondary carbon center having Iodine as a leaving group.SN² reaction takes place here as hydroxide ion is a good nucleophile and it can attack the secondary carbon center from the back side leading to the formation of 2R-hexane-2-ol.

In a SN² reaction since the the nucleophile attacks from the back-side so the product formation takes place with the inversion of configuration.

When the same substrate S-2-iodohexane undergoes a substitution reaction with water as a nucleophile then the reaction occurs through (SN¹) substitution nucleophilic unimolecular mechanism .

The rate of a SN¹ reaction depends only on the nature of substrate and is independent of the nature of nucleophile.

The SN¹ reaction is a 2 step reaction , in the first step leaving group leaves leading to the formation of a carbocation and once the carbocation is formed then any weaker nucleophile or even solvent molecules can attack leading the formation of products.

In this case a secondary carbocation would be generated in the first step and then water will attack this carbocation to form the product in the second step.

The product formed on using water as a nucleophile would be a racemic mixture of R and S isomers of hexane -2-ol in 50:50 ratio. The two products formed would be 2R-hexane-2-ol and 2S-hexane-2-ol.

Kindly refer the attachment for reaction mechanism and structure of products.

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Nucleophilicity is a kinetic property. A higher nucleophilicity indicates that the nucleophile will easily donate its electrons

lord [1]

The rate constant (K) is related to activation energy (Ea), frequency factor (A) and temperature (T) in Kelvin by the equation

R = molar gas constant

K = A(e^(-Ea/RT))

Taking natural log of both sides

In K = In A - (Ea/RT)

In K = (-Ea/R)(1/T) + In A

Comparing this to the equation of a straight line; y = mx + c

y = In K, slope, m = (-Ea/R), x = (1/T) and intercept, c = In A

a) From the question, m = (-Ea/R) = -1.10 × (10^4) K

(-Ea/R) = -1.10 × (10^4) = -11000

R = 8.314 J/K.mol

Ea = -11000 × 8.314 = 91454 J/mol = 91.454 KJ/mol

b) c = In A = 33.5

A = e^33.5 = (3.54 × (10^14))/s

c) K = A(e^(-Ea/RT))

A = (3.54 × (10^14))/s, Ea = 91454 J/mol, T = 25°C = 298.15 K, R = 8.314 J/K.mol

K = (3.54 × (10^14))(e^(-91454/(8.314×298.15))) = 0.0336/s

QED!

3 0

3 years ago

After substantial heating, 6.25 g of iron produced 18.00 g of a compound with chlorine. The empirical formula is:

babunello [35]

Answer:

Option A. FeCl3

Explanation:

The following data were obtained from the question:

Mass of iron (Fe) = 6.25g

Mass of the compound formed = 18g

From the question, we were told that the compound formed contains chlorine. Therefore the mass of chlorine is obtained as follow

Mass of chlorine (Cl) = Mass of compound formed – Mass of iron.

Mass of chlorine (Cl) = 18 – 6.25

Mass of chlorine (Cl) = 11.75g

The compound therefore contains:

Iron (Fe) = 6.25g

Chlorine (Cl) = 11.75g

The empirical formula for the compound can be obtained by doing the following:

Step 1:

Divide by their molar mass

Fe = 6.25/56 = 0.112

Cl = 11.75/35.5 = 0.331

Step 2:

Divide by the smallest

Fe = 0.112/0.112 = 1

Cl = 0.331/0.112 = 3

The empirical formula for the compound is FeCl3

3 0

3 years ago

When metallic aluminum is added to a solution containing iron(II) sulfate a reaction occurs. What species is being oxidized in t

REY [17]

Answer: Aluminium is getting oxidized in the given chemical reaction.

Explanation:

Oxidation reaction is defined as the chemical reaction in which an atom looses its electrons. The oxidation number of the atom gets increased during this reaction.

$X\rightarrow X^{n+}+ne^-$

Reduction reaction is defined as the chemical reaction in which an atom gains electrons. The oxidation number of the atom gets reduced during this reaction.

$X^{n+}+ne^-\rightarrow X$

For the given chemical reaction:

$2Al(s)+3FeSO_4(aq.)\rightarrow Al_2(SO_4)_3(aq.)+3Fe(s)$

The half cell reactions for the above reaction follows:

Oxidation half reaction: $Al\rightarrow Al^{3+}+3e^-$

Reduction half reaction: $Fe^{2+}+2e^-\rightarrow Fe$

As, aluminium is loosing 3 electrons to form aluminium cation. Thus, it is getting oxidized. Iron is gaining 2 electrons to form iron anion. Thus, it is getting reduced.

Hence, the oxidized species of the given reaction is aluminium.

4 0

3 years ago

Describe any new ways you can change the polarity of the three-atom molecule.

yan [13]

- By changing three atom bond angle , we can change molecular polarity. if bond angle is 120° and all atom has same electronegativity, the resultant polarity ( dipole moment ) become zero. we can change bond anngle either less than or greater than 120°, but not exactly 120

- Replacing one or more atom with different atoms having electro negativity values also make molecular poles.

- By placing molecules under external electric field or magnetic field also causes to the molecule.

5 0

3 years ago

Which description identifies the correct charge of each atomic particle? A. electron = positive; proton = negative; neutron = no

stiv31 [10]

B. electron = negative; proton = positive; neutron = no charge

3 0

3 years ago