Organic Reactions

Step 1: Recall reaction type.
Grignard reagents (RMgX) are strong bases and react with acidic hydrogens. Primary amines (RNH ) contain –NH hydrogen.
Step 2: Reaction.
RNH + RMgX R–MgNH + RH.
Thus, an alkane (RH) is formed.
Step 3: Eliminate options.
- Higher amine: Not possible.
- Secondary amine: Not formed.
- None of these: Incorrect.
Step 4: Conclusion.
Thus, the correct product is an alkane.

Step 1: Nature of the reaction.
Zinc in water acts as a reducing system providing nascent hydrogen, which dehalogenates haloforms.
Step 2: Apply to chloroform.
CHCl undergoes stepwise replacement of Cl by H under reducing conditions, ultimately forming CH .
Step 3: Conclusion.
Therefore, chloroform reduces to methane (D).

Step 1: Known conversions of alkyl halides.
Alkyl halides are versatile: (i) Alkanes via Wurtz coupling (R–X + Na → R–R),
(ii) Alkenes via -elimination (alc. KOH), and (iii) Alcohols via nucleophilic substitution (aq. KOH).
Step 2: Conclusion.
Since all three can be prepared from alkyl halides, the answer is (D) All of these.

Step 1: Condition for Cannizzaro reaction.
It occurs with aldehydes that lack -hydrogen. Such aldehydes undergo self-oxidation and self-reduction in strong base to give a salt of a carboxylic acid and an alcohol.
Step 2: Check options for -H.
(A) : has -H gives aldol, not Cannizzaro.
(B) (formaldehyde): has no -H undergoes Cannizzaro.
(C) : not an aldehyde (it is a carboxylic acid).
(D) : a ketone; does not show Cannizzaro.
Step 3: Conclusion.
Thus, is the correct choice (B).

Step 1: Understanding the reaction type.
This reaction involves the substitution of the bromine atom in with the hydroxide ion (OH⁻), which is a nucleophile. The reaction is a typical example of nucleophilic substitution. Step 2: Analysis of options.
(A) Electrophilic substitution: This is incorrect because electrophilic substitution involves an electrophile, and the reaction here involves a nucleophile.
(B) Nucleophilic substitution: Correct — The hydroxide ion (OH⁻) is a nucleophile that replaces the bromine atom.
(C) Both (A) and (B): This is incorrect because the reaction is clearly nucleophilic substitution, not electrophilic.
(D) None of these: This is incorrect because (B) is the correct answer.
Step 3: Conclusion.
The correct answer is (B) because the reaction is a nucleophilic substitution where the nucleophile (OH⁻) replaces the leaving group (Br⁻).

Step 1: Understanding mechanism.
In the mechanism, the rate-determining step is the formation of a carbocation intermediate. Therefore, this mechanism is more likely to occur when a carbocation can be stabilized, such as in the case of tertiary carbocations or allylic/benzylic carbocations. Step 2: Analysis of options.
(A) : Correct — The mechanism is favored for primary carbons with good leaving groups such as X. A carbocation can easily form when the leaving group departs.
(B) : This is incorrect; it would undergo instead of .
(C) : This is incorrect because the alkyl group is primary, which favors the mechanism.
(D) : This is incorrect because this molecule has a methyl group, which does not favor .
Step 3: Conclusion.
The correct answer is (A) because it is the primary alkyl halide that can undergo the mechanism due to carbocation formation.

Step 1: Understanding the reaction with dry silver oxide.
When alkyl halides like ethyl bromide are treated with dry silver oxide, they undergo a nucleophilic substitution, leading to the formation of an ether. Step 2: Analysis of options.
(A) Diethyl ether: Correct — The reaction of ethyl bromide with dry silver oxide produces diethyl ether.
(B) Ethanal: This is incorrect; ethanal is not formed in this reaction.
(C) Ethane: This is incorrect; ethane is not produced when silver oxide is used.
(D) Ethene: This is incorrect; ethene is formed under different conditions (e.g., elimination).
Step 3: Conclusion.
The correct answer is (A) because the reaction of ethyl bromide with dry silver oxide results in the formation of diethyl ether.

Step 1: Understanding the oxidation of aldehydes.
Aldehydes are oxidized to carboxylic acids when they react with oxidizing agents. The oxidation involves the conversion of the aldehyde group (-CHO) to a carboxyl group (-COOH), forming an acid. Step 2: Analysis of options.
(A) an alcohol: Aldehydes are not typically oxidized to alcohols; they are reduced to alcohols instead.
(B) a ketone: Aldehydes are not oxidized to ketones; ketones are already oxidized compounds.
(C) an ether: Aldehydes do not oxidize to form ethers. Ethers are formed by different processes, such as dehydration reactions.
(D) an acid: Correct — Aldehydes are oxidized to carboxylic acids.
Step 3: Conclusion.
The correct answer is (D) because the oxidation of an aldehyde results in the formation of a carboxylic acid.

Step 1: Understanding chloretone formation.
Chloretone is formed by the reaction of chloroform with acetone, resulting in the formation of a compound used as an antiseptic and sedative. This reaction is a halogenation reaction that leads to the formation of chloretone. Step 2: Analysis of options.
(A) Formaldehyde: Chloretone is not formed by the reaction of chloroform and formaldehyde.
(B) Acetaldehyde: Chloretone is not formed by the reaction of chloroform and acetaldehyde.
(C) Acetone: Correct — Chloretone is formed when chloroform reacts with acetone.
(D) Benzaldehyde: Chloretone is not formed by the reaction of chloroform and benzaldehyde.
Step 3: Conclusion.
The correct answer is (C) because chloretone is formed when chloroform reacts with acetone.

Step 1: Understanding the reaction.
Methylamine reacts with chloroform (CHCl₃) in the presence of alcoholic KOH in the formation of isocyanide (also known as isothiocyanate). This is a characteristic reaction of primary amines, also called the isocyanide or carbylamine reaction. Step 2: Analysis of options.
(A) : This is incorrect, as alcohol is not produced in this reaction.
(B) : This is the nitrile group, but the reaction does not give a nitrile here.
(C) : This is an aldehyde, not a product of this reaction.
(D) : This is the correct answer; methylamine with chloroform and alcoholic KOH gives an isocyanide (CH₃NC).
Step 3: Conclusion.
The correct answer is (D) because methylamine forms an isocyanide (CH₃NC) when heated with chloroform and alcoholic KOH.