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Wittaler [7]
3 years ago
11

Name a possible product of this reaction in the presence of ether and AlCl3: methylbenzene + 1-chlorodecane.a. 1-methyl-2-decylb

enzeneb. 1-methyl-3-decylbenzenec. 1-methyl-4-decylbenzened. None of these

Chemistry
2 answers:
Vikki [24]3 years ago
8 0

Answer:

None of these

Explanation:

Friedel–Craft reaction is a reaction involves the attachment of substituents to the benzene ring.

Mechanism of the reaction of methylbenzene with 1-chlorodecane in the presence of ether and aluminum chloride :

Step -1 : Generation of stable carbocation.

Aluminium chloride acts as Lewis acid which removes the chloride ion from the alkyl halide forming carbocation. The primary carbocation thus formed gets rearranged to secondary primary carbocation which is more stable due to hyperconjugation.

Step-2: Attack of the ring to the carbocation

The pi electrons of the ring behave as a nucleophile and attacks the carbocation. Since, the group attached on the benzene is methyl (+R effect) , the attack is from the ortho and the para positions. Para product is more stable due to less steric hinderance.

The product formed is shown in mechanism does not mention in any of the options.

So, None of these is the answer

otez555 [7]3 years ago
4 0

Answer:

c. 1-methyl-4-decylbenzene

Explanation:

Hello,

On the attached document you will find the major product for the stated chemical reaction even thought the listed product are possible, nonetheless, by cause of the steric hindrance, the most probable and abundant product is the shown one, 1-methyl-4-decylbenzene, as it has more space for the decyl to become part of the ring. Such reaction is a typical Friedel-Crafts alkylation of an aromatic compound whereas aluminium chloride is used as the catalyst to attach the alkyl chloride to the aromatic ring.

Best regards.

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Given the partial equation:
Nikolay [14]

Answer : The balanced chemical equation in acidic medium will be,

IO_3^-(aq)+2Sn^{2+}(aq)+6H^+(aq)\rightarrow I^-(aq)+2Sn^{4+}(aq)+3H_2O(l)

Explanation :

Redox reaction or Oxidation-reduction reaction : It is defined as the reaction in which the oxidation and reduction reaction takes place simultaneously.

Oxidation reaction : It is defined as the reaction in which a substance looses its electrons. In this, oxidation state of an element increases. Or we can say that in oxidation, the loss of electrons takes place.

Reduction reaction : It is defined as the reaction in which a substance gains electrons. In this, oxidation state of an element decreases. Or we can say that in reduction, the gain of electrons takes place.

Rules for the balanced chemical equation in acidic solution are :

First we have to write into the two half-reactions.

Now balance the main atoms in the reaction.

Now balance the hydrogen and oxygen atoms on both the sides of the reaction.

If the oxygen atoms are not balanced on both the sides then adding water molecules at that side where the less number of oxygen are present.

If the hydrogen atoms are not balanced on both the sides then adding hydrogen ion (H^+) at that side where the less number of hydrogen are present.

Now balance the charge.

The given chemical reaction is,

IO_3^-(aq)+Sn^{2+}(aq)\rightarrow I^-(aq)+Sn^{4+}(aq)

The oxidation-reduction half reaction will be :

Oxidation : Sn^{2+}\rightarrow Sn^{4+}

Reduction : IO_3^-\rightarrow I^-

First balance the main element in the reaction.

Oxidation : Sn^{2+}\rightarrow Sn^{4+}

Reduction : IO_3^-\rightarrow I^-

Now balance oxygen atom on both side.

Oxidation : Sn^{2+}\rightarrow Sn^{4+}

Reduction : IO_3^-\rightarrow I^-+3H_2O

Now balance hydrogen atom on both side.

Oxidation : Sn^{2+}\rightarrow Sn^{4+}

Reduction : IO_3^-+6H^+\rightarrow I^-+3H_2O

Now balance the charge.

Oxidation : Sn^{2+}\rightarrow Sn^{4+}+2e^-

Reduction : IO_3^-+6H^++4e^-\rightarrow I^-+3H_2O

The charges are not balanced on both side of the reaction. Thus, we are multiplying oxidation reaction by 2 and the adding both equation, we get the balanced redox reaction.

Oxidation : 2Sn^{2+}\rightarrow 2Sn^{4+}+4e^-

Reduction : IO_3^-+6H^++4e^-\rightarrow I^-+3H_2O

The balanced chemical equation in acidic medium will be,

IO_3^-(aq)+2Sn^{2+}(aq)+6H^+(aq)\rightarrow I^-(aq)+2Sn^{4+}(aq)+3H_2O(l)

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2 years ago
The elements chlorine and iodine have similar chemical properties because they _____. Group of answer choices have the same numb
worty [1.4K]

Answer:

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Explanation:

Elements in the same group have similar chemical properties because they have the same number of valence electron(s) in their outermost shell.

Chlorine and Iodine have similar chemical properties because they have the same number of valence electron in their outermost shell. This can be seen from their electronic configuration as shown below:

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From the above illustration:

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