Identifying Incorrect Options

We are given three options and need to determine which one(s) are incorrect. Let's analyze each one step by step.

Step 1: Analyze Option 1

Option 1: Reaction of an alkyl halide with KCN followed by reduction

This reaction sequence is: Alkyl halide (R-X) reacts with KCN to form a nitrile (R-CN), which is then reduced to a primary amine (R-CH₂-NH₂).

This is a standard and correct method for preparing primary amines with one extra carbon atom. Therefore, this option is correct.

Step 2: Analyze Option 2

Me–Cl $\to_{{Ag}_{2} O}^{Moist}$ Me–O–Me (major)

This reaction involves methyl chloride (CH₃Cl) reacting with moist silver oxide (Ag₂O/H₂O).

Moist Ag₂O provides AgOH, which acts as a base. When a methyl halide is treated with a base, it undergoes an S_N2 reaction. For methyl halides, this leads to the formation of methanol (CH₃OH), not dimethyl ether (CH₃-O-CH₃). Dimethyl ether is the major product when the reaction is carried out with dry silver oxide (Ag₂O), which acts as a source of alkoxide ion (CH₃O^-).

Conclusion: The given condition is "Moist", which would produce methanol. The product shown is dimethyl ether, which is incorrect for these conditions. Therefore, this option is incorrect.

Step 3: Analyze Option 3

Option 3: Hofmann Bromamide Reaction

This image depicts the Hofmann Bromamide Reaction. In this reaction, a primary amide (R-CONH₂) is treated with bromine (Br₂) and a strong base (e.g., NaOH or KOH) to form a primary amine (R-NH₂) with one less carbon atom.

This is a well-established and correct reaction for the degradation of amides to amines. Therefore, this option is correct.

Step 4: Analyze Option 4

MeI (excess) + NH₃ $\to_{}^{}$ Me₃N

This reaction involves ammonia (NH₃) reacting with an excess of methyl iodide (CH₃I).

Ammonia undergoes nucleophilic substitution with alkyl halides. Because methyl halides are highly reactive, the reaction does not stop at the first substitution. Even with excess ammonia, the reaction leads to exhaustive alkylation, producing the quaternary ammonium salt, not the tertiary amine.

The product of ammonia with excess alkyl halide is always the quaternary ammonium salt (R₄N⁺X^-). To specifically get a tertiary amine (R₃N), a controlled reaction with a secondary amine (R₂NH) must be used.

Conclusion: The product shown is trimethylamine (Me₃N), but the actual product is tetramethylammonium iodide ((CH₃)₄N⁺I^-). Therefore, this option is incorrect.

Final Answer

The incorrect options are Option 2 and Option 4.

Related Topics & Formulae

1. Nucleophilic Substitution Reactions

Alkyl halides (R-X) undergo substitution reactions where a nucleophile (Nu^-) replaces the halide ion (X^-).

General Reaction: $R - X + {Nu}^{-} \to R - Nu + X^{-}$

Mechanisms: S_N1 (Unimolecular) and S_N2 (Bimolecular). Methyl and primary alkyl halides typically favor the S_N2 mechanism.

2. Preparation of Amines

From Alkyl Halides: Reaction with ammonia gives a mixture of primary, secondary, tertiary amines, and quaternary ammonium salts. $N H_{3} + R X \to R {NH}_{2} + R 2_{2} NH + R 3_{3} N + R 4_{4} N^{+} X^{-}$
Gabriel Phthalimide Synthesis: A method to obtain pure primary amines. $P h t h a l i m i d e + R X \overset{KOH}{\to} N - a l k y l p h t h a l i m i d e \overset{H_{2} O / H +}{\to} R - {NH}_{2}$
Reduction of Nitriles: R-CN $\overset{LiAl H_{4} / H 2, Ni}{\to}$ R-CH₂-NH₂
Hofmann Bromamide Reaction: R-CONH₂ + Br₂ + 4NaOH → R-NH₂ + 2NaBr + Na₂CO₃ + 2H₂O

3. Ether Formation: Williamson Ether Synthesis

The correct way to synthesize an ether (R-O-R') is by reacting an alkoxide ion (RO^-) with a primary alkyl halide (R'X).

General Reaction: $R^{-} O^{-} + R^{'} - X \to R - O - R^{'} + X^{-}$

Dry Ag₂O provides the alkoxide ion. Moist Ag₂O provides hydroxide ion (OH^-), which leads to alcohol formation, especially with methyl halides.