Chemistry MCQs for NEET — Practice Questions with Answers

The Wurtz reaction is a coupling reaction in organic chemistry, typically used for coupling two alkyl halides to form a longer alkane chain. It is not used for the synthesis or separation of amines. The other methods listed (Hinsberg, Carbylamine, and Hofmann methods) are specifically related to the synthesis or separation of amines.

Ethylamine reacts with $HNO_2$ to form an alcohol, but aniline reacts with $HNO_2$ to form a diazonium salt, not a hydroxy compound. Therefore, the statement 'Ethylamine and aniline both react with $HNO_2$ to give hydroxy compounds' is incorrect.

$ -NO_2 $ ,group is electron with drawing group.

The Gabriel phthalimide reaction is used for the preparation of primary aliphatic amines. In this reaction, phthalimide is reacted with an alkyl halide to form N-alkylphthalimide, which on hydrolysis gives a primary aliphatic amine.

Benzene diazonium chloride is the most versatile compound in the synthesis of aromatic compounds. It can undergo a variety of reactions such as substitution, coupling, and reduction, making it a key intermediate in the synthesis of many aromatic derivatives.

The sequence of reactions given involves the following steps: 1) Nitrobenzene is reduced by Sn/HCl to form aniline (A). 2) Aniline reacts with excess bromine to form 2, 4, 6-tribromoaniline (B). 3) 2, 4, 6-tribromoaniline is then diazotized with NaNO2/HCl to form 2, 4, 6-tribromo benzene diazonium chloride (C). 4) Finally, the diazonium compound is reduced by H3PO2/H2O to form 1, 3, 5-tribromobenzene (D).

The reaction sequence is as follows:

1. $CH_3CH_2CH_2NH_2$ (Propylamine) reacts with $HONO$ (Nitrous acid) to form $CH_3CH_2CH_2OH$ (n-Propanol).
2. n-Propanol reacts with $PCl_5$ (Phosphorus pentachloride) to form $CH_3CH_2CH_2Cl$ (n-Propyl chloride).
3. n-Propyl chloride reacts with $KCN$ (Potassium cyanide) to form $CH_3CH_2CH_2CN$ (n-Propyl cyanide).
4. n-Propyl cyanide is reduced by $Na/C_2H_5OH$ (Sodium in ethanol) to form $CH_3CH_2CH_2CH_2NH_2$ (n-Butylamine). So, the correct option is $[A] = CH_3CH_2CH_2OH, [B] = CH_3CH_2CH_2Cl, [D] = CH_3CH_2CH_2CH_2NH_2$.

Aniline ($C_6H_5NH_2$) on oxidation with $Na_2Cr_2O_7$ and $H_2SO_4$ forms p-benzoquinone ($C_6H_4O_2$). The reaction involves the oxidation of the amino group to a quinone structure. Hence, the correct answer is p-benzoquinone.

Hinsberg's reagent is benzene sulphonyl chloride ($C_6H_5SO_2Cl$). It is used to distinguish between primary, secondary, and tertiary amines. Primary amines form a soluble sulfonamide, secondary amines form an insoluble sulfonamide, and tertiary amines do not react. Hence, the correct answer is benzene sulphonyl chloride.