NEET Amines — Set 1

Question 1

Which amine forms a yellow oily liquid with $ \ce{HNO2} $?

Accepted Answer

Secondary amines like $ \ce{CH3CH2NHCH3} $ form nitrosamines (yellow oily liquids) with $ \ce{HNO2} $, as per the PDF’s chemical properties.

Question 2

Which preparation method uses $ \ce{KOH} $ and $ \ce{Br2} $ to convert an amide to an amine?

Accepted Answer

The Hoffmann bromamide reaction uses $ \ce{Br2} $ and $ \ce{KOH} $ to convert an amide to an amine with one less carbon, as per the PDF’s preparation methods.

Question 3

Which amine has the highest solubility in water per carbon atom?

Accepted Answer

Methylamine ($ \ce{CH3NH2} $, 1C) has two N-H bonds for hydrogen bonding relative to one carbon, giving the highest solubility per carbon, as per the PDF.

NEET Chemistry: Amines — Practice Set 1

Q1. Which amine forms a yellow oily liquid with \( \ce{HNO2} \)?

Q2. Which preparation method uses \( \ce{KOH} \) and \( \ce{Br2} \) to convert an amide to an amine?

Q3. Which amine has the highest solubility in water per carbon atom?

Q4. Which amine has the lowest basicity in aqueous solution?

Q5. The carbon chain length of the amine from \( \ce{CH3CH2CH2CONH2} \) via Hoffmann bromamide reaction is:

Q6. The IUPAC name of \( \ce{CH3CH2N(CH3)CH2CH3} \) is:

Q7. The product of aniline reacting with excess \( \ce{Br2} \) in water is:

Q8. Which amine shows no reaction with \( \ce{HNO2} \) at 273-278 K?

Q9. The IUPAC name of \( \ce{CH3CH2CH2NHCH3} \) is:

Q10. The product of \( \ce{CH3CH2CH2NO2} \) reduction with \( \ce{Sn/HCl} \) is: