Organic Reactions

Step-by-step transformation: 1. Oxidation (CrO /H SO ) converts the primary alcohol to a carboxylic acid: 2. Chlorination using Cl /red P results in substitution at the α-carbon (adjacent to -COOH), forming: Hence, the major product is: α-chloro propionic acid derivative with a correct carbon skeleton as per the original straight chain.

Step-by-step: 1. CH -CH=CH + HBr → CH -CHBr-CH (Markovnikov addition)
2. Dehydrohalogenation with NaOH:
CH -CHBr-CH → CH -C≡CH\ (propyne)
3. Red hot Cu (573 K):
CH -C≡CH → CH -CHO (Acetaldehyde)
4. Ba(OH) , heat (aldol condensation):
2 CH -CHO → CH -CO-CH + H O
Final product is propanone (acetone).

- For , the reaction with forms a carboxylic acid, which corresponds to option III, . - For , reacts with in the presence of , yielding , which corresponds to option IV. - For , reacts with in the presence of , and water, forming formic acid , matching with option II. - For , in the presence of and forms acetic acid, matching with option I. Thus, the correct match is option 2.

In this reaction, the alkyl group will undergo radical substitution in the presence of UV light, and the major product will be a substitution of the hydrogen atom with a bromine atom. The resulting product is the alkyl bromide. Thus, the correct answer is .

The reactions represent common organic reactions: In option (1), the reaction of alkyl halides (CH Br) with zinc leads to the formation of alkene by reduction (dehalogenation), which is correct.
In option (2), the reaction of an alkyne with sodium in liquid ammonia is a reduction that leads to trans-alkene formation, which is correct. In option (3), the reaction of an alkyne with water, mercuric ions (Hg ), and acid leads to the formation of an aldehyde, which is correct.
In option (4), the reaction of an alkyl halide (Ph-CH Br) with sodium in dry ether leads to a coupling reaction forming biphenyl, which is correct.
Thus, option (2) is the correct answer as it involves an incorrect reaction pathway.

- The reaction involves an alkyl halide (chloropropane) with AlCl , followed by treatment with oxygen and water, which is a typical setup for a Friedel-Crafts alkylation followed by oxidation to form a phenol derivative.
- Option (1) correctly represents the reagents needed to achieve the conversion.
Thus, the correct answer is (1).

In the given options, HNO (Nitric acid) and 2[Ag(NH ) ]⁺ are reagents that can oxidize isobutyraldehyde into its corresponding carboxylic acid. The third option is also a known reagent for this transformation.

In the first step, KMnO /KOH acts as an oxidizing agent, converting styrene into a carboxylated product (X). The second step involves a reaction with H O that replaces the halogen (Y) and forms a carboxylic acid.

Step 1: Determine Product X from the first reaction.
The first reaction is the oxidation of ethane (CH₃CH₃) with oxygen (O₂) in the presence of manganese acetate ((CH₃COO)₂Mn) and heat ( ). This is a controlled catalytic oxidation.
Under these conditions, alkanes can be selectively oxidized to carboxylic acids. For ethane, the controlled oxidation leads to the formation of acetic acid. Thus, X = CH₃COOH. Step 2: Determine Product Y from the second reaction.
The second reaction involves the oxidation of propene (CH₃CH=CH₂) with hot, acidic potassium permanganate (KMnO₄/H⁺). Acidic KMnO₄ is a strong oxidizing agent that causes oxidative cleavage of carbon-carbon double bonds. When an alkene is subjected to hot, acidic KMnO₄: \begin{itemize} \item A carbon atom of the double bond that carries at least one hydrogen atom is oxidized to a carboxylic acid. \item A terminal group is completely oxidized to carbon dioxide ( ) and water ( ). \end{itemize} For propene (CH₃CH=CH₂): \begin{itemize} \item The group (terminal carbon of the double bond) will be oxidized to and . \item The group (the other carbon of the double bond with a hydrogen) will be oxidized to a carboxylic acid. The product from this fragment is . \end{itemize} In organic chemistry problems asking for the major organic product of such cleavages, the organic acid (or ketone) formed is usually the expected answer, rather than . Thus, Y = CH₃COOH. Step 3: Match the products with the given options.
Based on our analysis:
X = CH₃COOH
Y = CH₃COOH
Comparing this with the given options:
% Option (1)
% Option (2)
% Option (3)
% Option (4)
Option (4) matches our derived products.