Chemical Reaction

For the reaction given by , the rate of reaction is determined by the rate law: . Here, is the rate constant, is the concentration of , and is the concentration of .

If the volume is doubled, according to the ideal gas law ( ), the pressure is reduced by half. Because concentrations are directly proportional to pressure in gases, each concentration is halved: and .

Substituting these into the rate equation:

Thus, the reaction rate decreases to of its original value, meaning the rate has decreased by eight times.

Aldehydes that lack an alpha hydrogen atom undergo the Cannizzaro reaction. This chemical reaction involves the reduction and oxidation of aldehydes without the presence of an alpha hydrogen. During this reaction, two molecules of the aldehyde react with each other:

Step-by-Step Explanation:

1. Consider an aldehyde, RCHO, without an alpha hydrogen.

2. In the Cannizzaro reaction, two aldehyde molecules undergo a disproportionation reaction, which is a redox process.

3. One molecule of the aldehyde is reduced to form an alcohol (RCH₂OH), while the other is oxidized to form a carboxylate (RCOO^-).

Chemical Reaction:

2RCHO → RCH₂OH + RCOO^-

Thus, aldehydes with no alpha hydrogen atom undergo the Cannizzaro reaction, making the answer: Cannizzaro reaction.

The reaction of aniline with concentrated nitric acid ( ) and concentrated sulfuric acid ( ) at 298 K involves the nitration of the aromatic compound. In the presence of these strong acids, an aniline group initially converts to an anilinium ion due to the acidic environment:

This conversion deactivates the aromatic ring towards electrophilic attack, affecting its reactivity. The strong electron-withdrawing effect of the anilinium ion lowers the reactivity of the ortho and para positions, making the meta position more favorable for substitution. Consequently, nitration predominantly occurs at the meta position, resulting in the formation of m-nitroaniline.

As the final outcome of this reaction in the specified conditions is a predominant formation of m-nitroaniline, 47% of the product will be: m-Nitroaniline.

When ammonia ( ) is added to a solution containing copper ions ( ), a complex ion is formed. Initially, ammonia reacts with the ions to produce a pale blue precipitate of copper(II) hydroxide:

Upon adding excess ammonia, this precipitate dissolves, forming a deep blue complex ion, [ :

The formation of this complex ion is responsible for the deep blue color of the final solution. Therefore, the correct answer is: Deep blue.

You've correctly outlined the steps of the reaction sequence. Here's a summary of the reactions and products:

1. Reduction of Phenol to Benzene (A):

   C₆H₅OH + Zn → C₆H₆ + ZnO

   Phenol reacts with zinc dust to produce benzene.

2. Friedel-Crafts Alkylation of Benzene (A) to Toluene (B):

   C₆H₆ + CH₃Cl → C₆H₅CH₃ + HCl

   Benzene reacts with methyl chloride in the presence of anhydrous AlCl₃ to produce toluene.

3. Oxidation of Toluene (B) to Benzoic Acid (C):

   C₆H₅CH₃ + [O] → C₆H₅COOH

   Toluene is oxidized by potassium dichromate in acidic medium to produce benzoic acid.

4. Nitration of Benzoic Acid (C) to m-Nitrobenzoic Acid (D):

   C₆H₅COOH + HNO₃ → m-C₆H₄(NO₂)COOH + H₂O

   Benzoic acid reacts with a mixture of sulfuric acid and nitric acid to produce m-nitrobenzoic acid.

Therefore, the final product (D) is indeed m-nitrobenzoic acid.

The reaction of (CH₃)₃C − O − CH₃ with HI leads to the formation of CH₃OH and (CH₃)₃CCl instead of CH₃I and (CH₃)₃COH due to the type of mechanism it follows. This reaction proceeds via an SN₁ mechanism. In an SN₁ reaction, the rate-determining step involves the formation of a carbocation intermediate. The stability of the carbocation is crucial for the progression of the reaction. In this case, the (CH₃)₃C⁺ carbocation is more stable compared to any carbocation that would be formed from CH₃ group. Therefore, the continuing steps naturally lead to the products of CH₃OH and (CH₃)₃CCl. This stability plays a key role in favoring the pathway that produces the observed products rather than alternative products like CH₃I and (CH₃)₃COH

Hence, the correct answer is: (B) The reaction follows an SN₁ mechanism