NEET Amines — Set 14

Question 1

Which of the following amines has the highest basic strength in aqueous solution?

Accepted Answer

Basic strength depends on electron availability on nitrogen. $ \ce{(C2H5)2NH} $ (diethylamine) has two electron-donating ethyl groups and benefits from solvation effects, making it the strongest base among the options.

Question 2

Which amine, when treated with $ \ce{CH3COCl} $, forms a compound with two nitrogen atoms?

Accepted Answer

None of the listed amines form a compound with two nitrogens with $ \ce{CH3COCl} $; however, if misinterpreted, $ \ce{CH3NH2} $ forms $ \ce{CH3NHCOCH3} $ (one N). Assuming a typo or contextual error, the intent might be a single N product, but all options yield one N. Correcting intent: primary amines like $ \ce{CH3NH2} $ react typically. Here, $ \ce{CH3NH2} $ is chosen for standard reaction.

Question 3

Which amine would require the least amount of $ \ce{HCl} $ to form a salt per mole?

Accepted Answer

Tertiary amines like $ \ce{(CH3CH2)3N} $ have one lone pair and need only one mole of $ \ce{HCl} $ to form a salt, unlike primary or secondary amines with additional N-H bonds, as per the PDF’s basicity.

NEET Chemistry: Amines — Practice Set 14

Q1. Which of the following amines has the highest basic strength in aqueous solution?

Q2. Which amine, when treated with \( \ce{CH3COCl} \), forms a compound with two nitrogen atoms?

Q3. Which amine would require the least amount of \( \ce{HCl} \) to form a salt per mole?

Q4. Which of the following is a secondary amine?

Q5. The correct order of basic strength in aqueous solution is:

Q6. An amine \( X \) reacts with \( \ce{H2SO4} \) at 453-473 K to form \( \ce{H2NC6H4SO3H} \). \( X \) is:

Q7. Which of the following is a product of the reduction of nitrobenzene using \( \ce{Sn/HCl} \)?

Q8. The reaction of aniline with \( \ce{H2SO4} \) at 453-473 K produces:

Q9. Which of the following is insoluble in alkali when reacted with Hinsberg’s reagent?

Q10. A compound \( X \) on treatment with \( \ce{Br2} \) and \( \ce{KOH} \) yields \( \ce{CH3NH2} \). \( X \) is: