Amines — Questions & Answers

Sample questions from Amines

1. Q1. Which explanation correctly accounts for why treatment of an ortho-substituted aryl diazonium salt (bearing a bulky group adjacent to the diazonium position) can show altered reactivity (sometimes reduced yield) in Sandmeyer-type reactions compared to the unsubstituted or para-substituted analogue?

   • A.Diazonium salts show identical reactivity regardless of neighbouring substituents
   • B.Ortho substituents always increase the yield of Sandmeyer-type reactions
   • C.Steric effects are irrelevant to radical substitution mechanisms
   • D.Steric hindrance from the adjacent bulky ortho substituent can impede the copper-mediated approach and radical substitution mechanism at the diazonium-bearing carbon, reducing reaction efficiency compared to positions with less steric encumbrance✓
   Solution

   Even though the Sandmeyer mechanism proceeds via a radical pathway rather than a classical backside SN2 attack, steric crowding from a bulky adjacent ortho substituent can still impede the necessary approach of the copper catalyst/radical intermediate to the reacting carbon, this can measurably reduce yields or alter the efficiency of Sandmeyer-type substitutions at sterically encumbered diazonium positions compared to less hindered ones.

2. Q2. A quaternary ammonium salt such as tetraethylammonium bromide is useful as a phase-transfer catalyst because

   • A.it is strongly basic
   • B.the large lipophilic cation carries anions from the aqueous phase into the organic phase where they react as naked, highly reactive nucleophiles✓
   • C.it decomposes to give a tertiary amine in situ
   • D.it dissolves only in water
   Solution

   The bulky alkyl groups make the cation soluble in organic solvents, so the salt ferries anions such as CN- or OH- across the interface; in the organic phase these anions are poorly solvated and therefore unusually reactive.

3. Q3. Given pKb(ammonia) = 4.75, pKb(methylamine) = 3.36 and pKb(aniline) = 9.38, which statement is quantitatively correct?

   • A.methylamine is about 1.4 times more basic than ammonia
   • B.aniline is more basic than ammonia by a factor of 10^4.63
   • C.all three have basicities within one order of magnitude
   • D.methylamine is about 25 times more basic than ammonia, and ammonia is about 4 x 10^4 times more basic than aniline✓
   Solution

   A difference of 1.39 in pKb corresponds to a Kb ratio of 10^1.39 = 24.5, so methylamine is roughly 25 times the stronger base; the gap of 4.63 pKb units between ammonia and aniline corresponds to 10^4.63 = 4.3 x 10^4, with ammonia the stronger of the two because aniline has the larger pKb.

4. Q4. Why is the sulfonamide formed by a primary amine in the Hinsberg reaction soluble in aqueous KOH?

   • A.It has no nitrogen atom left
   • B.The remaining N-H hydrogen is rendered acidic by the strongly electron-withdrawing sulfonyl group, allowing deprotonation to a soluble potassium salt✓
   • C.It reacts with KOH to release ammonia gas
   • D.It is insoluble in all aqueous media
   Solution

   The sulfonyl group withdraws electron density strongly, making the one remaining N-H hydrogen on the primary amine's sulfonamide product sufficiently acidic to be removed by KOH, forming a water-soluble potassium salt, which reprecipitates as the free sulfonamide on acidification.

5. Q5. Why does treatment of aniline directly with concentrated nitric and sulfuric acids (without protection) tend to give a substantial proportion of meta-nitroaniline as a by-product, contrary to the expected ortho/para-directing nature of the amino group?

   • A.Meta-nitroaniline cannot form under any nitration conditions
   • B.Aniline always gives exclusively meta product under any nitration conditions
   • C.Protonation of aniline has no effect on the directing behaviour of the nitrogen substituent
   • D.Under the strongly acidic conditions, a significant fraction of aniline is protonated to the anilinium ion, the positively charged -NH3+ group is a meta-directing, deactivating substituent (similar to other positively charged groups), so nitration of the protonated fraction gives meta product, while any unprotonated aniline still gives ortho/para product✓
   Solution

   Because the reaction medium is strongly acidic, a substantial fraction of the aniline present exists as the protonated anilinium ion, this positively charged nitrogen group behaves as a deactivating, meta-directing substituent (its lone pair is tied up in the N-H bond to the extra proton and cannot donate to the ring), the observed nitration product mixture is therefore a composite of ortho/para product from any free aniline and meta product from the protonated fraction, explaining the unexpectedly significant meta-nitroaniline by-product.

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