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.
Answer:
a. AlCl is incorrect because Al has a +3 charge while Cl only has a -1 charge. The correct formula would be AlCl₃. This balances the charges.
b. Na₃SO₄ is incorrect because Na has a charge of +1 and there are three of them so its +3 and SO₄ has a charge of -2. The correct formula would be Na₂SO₄. This balances the charges.
c. BaOH₂ is incorrect because the polyatomic ion OH would not be written that way. It would be written like this Ba(OH)₂. Writing it like BaOH₂ gives the impression that instead of having 2 OH it has 2 H and 1 O.
d. Fe₂O is incorrect because Fe either has a +2 charge or +3 charge while O has a -2 charge. Possible correct answers could be FeO (iron (II) oxide) or Fe₂O₃ (iron (III) oxide).
Answer:
The catalyzed reaction will take 1,41 s
Explanation:
The rate constant for a reaction is:

Assuming frequency factor is the same for both reactions (with and without catalyst) it is possible to obtain:

Replacing:


That means the reaction occurs 5,64x10¹⁰ faster than the uncatalyzed reaction, that is 2537 years / 5,64x10¹⁰ = 4,50x10⁻⁸ years. In seconds:
4,50x10⁻⁸ years×
×
×
=<em> 1,41 s</em>
I hope it helps!
Answer:
22.45g of Fe will be produced
Explanation:
The balanced reaction is:
Fe₂O₃(s) + 3CO(g) → 2Fe(s) + 3CO₂(g)
<em>Where 1 mole of Iron (III) oxide reacts with 2 moles of Iron</em>
<em />
To solve this question we have to find the moles of iron (III) oxide. With these moles and the balanced reaction we can find the moles of iron produced and its mass:
<em>Moles iron (III) oxide -Molar mass: 159.69g/mol-</em>
32.1g Fe₂O₃ * (1mol / 159.69g) = 0.201 moles Fe₂O₃
<em>Moles Iron:</em>
0.201 moles Fe₂O₃ * (2mol Fe / 1mol Fe₂O₃) = 0.402 moles of Fe
<em>Mass Fe -Molar mass: 55.845g/mol-</em>
0.402 moles of Fe * (55.845g/mol) =
<h3>22.45g of Fe will be produced</h3>