NEET Amines — Set 20

Question 1

The amide that produces propanamine via Hoffmann bromamide reaction is:

Accepted Answer

Hoffmann bromamide reaction converts an amide ($ \ce{RCONH2} $) to an amine with one less carbon. For propanamine ($ \ce{CH3CH2CH2NH2} $, 3 carbons), the amide must have 4 carbons: butanamide ($ \ce{CH3CH2CH2CONH2} $).

Question 2

The product of $ \ce{C6H5NO2} $ reduction with $ \ce{Sn/HCl} $ is:

Accepted Answer

Reduction of nitrobenzene ($ \ce{C6H5NO2} $) with $ \ce{Sn/HCl} $ yields aniline ($ \ce{C6H5NH2} $), as per the PDF’s preparation methods.

Question 3

Which method reduces $ \ce{RCONH2} $ to $ \ce{RNH2} $ with one less carbon?

Accepted Answer

The Hoffmann bromamide reaction converts an amide ($ \ce{RCONH2} $) to an amine ($ \ce{RNH2} $) with one less carbon, as per the PDF.

NEET Chemistry: Amines — Practice Set 20

Q1. The amide that produces propanamine via Hoffmann bromamide reaction is:

Q2. The product of \( \ce{C6H5NO2} \) reduction with \( \ce{Sn/HCl} \) is:

Q3. Which method reduces \( \ce{RCONH2} \) to \( \ce{RNH2} \) with one less carbon?

Q4. Which of the following amines can form hydrogen bonds with water?

Q5. Which reagent converts aniline to acetanilide?

Q6. The product of \( \ce{CH3CH2NH2} \) reacting with \( \ce{HNO2} \) at 273-278 K is:

Q7. The product of \( \ce{CH3NH2} \) with \( \ce{CHCl3} \) and alcoholic KOH is:

Q8. Which amine reacts with \( \ce{CHCl3} \) and alcoholic KOH to produce a compound with a characteristic odor?

Q9. Which amine shows the least solubility in water?

Q10. Which amine produces a yellow oily compound when treated with \( \ce{HNO2} \)?