Carboxylic Acid AND Amines flash cards

1. 1.What is the general functional group of a carboxylic acid, and why is it named so?

   The carboxyl group $-COOH$, a combination of carbonyl ($C=O$) and hydroxyl ($-OH$) on the same carbon. General formula $R-COOH$ (or $C_nH_{2n}O_2$ for saturated monocarboxylic acids).

   Hint: Carbonyl + hydroxyl on one carbon.

2. 2.Describe the structure of the carboxyl carbon (hybridization and geometry).

   The carboxyl carbon is $sp^2$ hybridized with trigonal planar geometry (bond angles $\approx 120^\circ$). The $C=O$ and $C-O(H)$ bonds lie in one plane, allowing resonance/conjugation.

   Hint: Trigonal planar, one carbon.

3. 3.Why are the two C–O bond lengths in the carboxylate ion ($RCOO^-$) equal?

   Because of resonance delocalization: the negative charge is spread equally over both oxygens, giving two equivalent bonds of intermediate length (between $C-O$ single $\approx 143$ pm and $C=O$ double $\approx 121$ pm, both $\approx 127$ pm).

   Hint: Symmetric resonance spreads the charge.

4. 4.Why do carboxylic acids have abnormally high boiling points compared to alcohols of similar mass?

   They form stable intermolecular hydrogen-bonded dimers (cyclic, two H-bonds), effectively doubling the molecular mass in the liquid/vapour and raising the boiling point above comparable alcohols.

   Hint: Cyclic dimer with two H-bonds.

5. 5.Why are lower carboxylic acids (formic to butanoic) completely miscible with water?

   The $-COOH$ group forms hydrogen bonds with water (both donor and acceptor). As the hydrocarbon chain lengthens, hydrophobic character dominates and solubility falls sharply beyond $\sim C_5$.

   Hint: H-bonding vs chain length.

6. 6.Why is a carboxylic acid a stronger acid than the corresponding alcohol?

   The carboxylate anion $RCOO^-$ is stabilized by resonance (charge on two O), whereas the alkoxide $RO^-$ has no such delocalization. A more stable conjugate base means a stronger acid.

   Hint: Compare stability of $RCOO^-$ vs $RO^-$.

7. 7.Why is a carboxylic acid a stronger acid than phenol?

   Carboxylate resonance places negative charge on two electronegative oxygens (equivalent structures), whereas phenoxide resonance puts charge on carbon atoms of the ring. Carboxylate is far more stabilized, so $RCOOH$ ($pK_a\approx 4$–5) is stronger than phenol ($pK_a\approx 10$).

   Hint: Charge on O vs on ring carbons.

8. 8.How does an electron-withdrawing group (EWG) near the $-COOH$ affect acidity?

   EWGs (like $-Cl$, $-NO_2$, $-CN$) stabilize the carboxylate anion by dispersing negative charge, increasing acidity. Effect grows with number of EWGs and decreases with distance from $-COOH$.

   Hint: Stabilize the anion, stronger acid.

9. 9.How does an electron-donating group (EDG) affect carboxylic acid acidity?

   EDGs (like $-CH_3$, alkyl) intensify the negative charge on the carboxylate, destabilizing it and decreasing acidity. Hence acetic acid is weaker than formic acid.

   Hint: Push electrons in, weaker acid.

10. 10.QUESTION: Arrange in increasing acid strength: HCOOH, $CH_3COOH$, $ClCH_2COOH$, $Cl_2CHCOOH$.

    $CH_3COOH < HCOOH < ClCH_2COOH < Cl_2CHCOOH$. Methyl (EDG) makes acetic weakest; $H$ vs $CH_3$ makes formic stronger; each Cl (EWG, $-I$) further increases acidity, two Cl > one Cl.

    Hint: EDG weakens, each Cl strengthens.

11. 11.QUESTION: Which is more acidic: chloroacetic acid $ClCH_2COOH$ or fluoroacetic acid $FCH_2COOH$? Why?

    Fluoroacetic acid is more acidic. F is more electronegative than Cl, so its $-I$ (inductive electron-withdrawing) effect is stronger, better stabilizing the carboxylate.

    Hint: More electronegative halogen, stronger $-I$.

12. 12.QUESTION: Order the acidity of the isomeric chlorobutanoic acids: 2-, 3-, and 4-chlorobutanoic acid.

    $4$-chloro $< 3$-chloro $< 2$-chloro butanoic acid. The inductive $-I$ effect of Cl weakens with distance from $-COOH$, so the closer Cl (2-position) gives the strongest acid.

    Hint: Inductive effect falls with distance.

13. 13.Why are dicarboxylic acids' first ionization ($K_{a1}$) stronger than acetic acid?

    The second $-COOH$ acts as an EWG stabilizing the mono-anion. Also, in the ionized form, the remaining $-COOH$ withdraws electrons. Hence oxalic, malonic etc. have $K_{a1}$ larger than acetic.

    Hint: Second COOH is electron-withdrawing.

14. 14.Why is $K_{a2} \ll K_{a1}$ for a dicarboxylic acid?

    Removing a second proton from an already negative species ($^-OOC-R-COO^-$ formation) is opposed by electrostatic repulsion between the two negative charges, making the second ionization much harder.

    Hint: Pulling H off an anion is hard.

15. 15.Give three general laboratory methods to prepare carboxylic acids.

    (1) Oxidation of primary alcohols/aldehydes/alkylbenzenes. (2) Hydrolysis of nitriles, esters, amides, acid halides/anhydrides. (3) Carbonation of Grignard reagents ($RMgX + CO_2$).

    Hint: Oxidation, hydrolysis, Grignard + CO2.

16. 16.How is a carboxylic acid prepared from a Grignard reagent?

    $RMgX$ adds to dry ice/$CO_2$ giving $RCOO^-MgX^+$, which on acidic workup ($H_3O^+$) yields $RCOOH$. Product has one more carbon than the halide precursor.

    Hint: Grignard + $CO_2$, then $H_3O^+$; chain +1.

17. 17.How are carboxylic acids prepared from nitriles? How does the carbon count change?

    Acidic or basic hydrolysis of $R-C\equiv N$ gives $R-COOH$ (via amide intermediate). Starting from $R-X \to R-CN$ (with $KCN$), the acid has one more carbon than the alkyl halide.

    Hint: $RX \to RCN \to RCOOH$, chain +1.

18. 18.What product forms when a primary alcohol is oxidized by strong oxidants like $KMnO_4/H^+$ or $K_2Cr_2O_7/H^+$?

    A carboxylic acid $RCOOH$ (via aldehyde intermediate). To stop at the aldehyde one must use mild reagents (e.g., PCC) or distill it off.

    Hint: Strong oxidant takes 1° alcohol all the way.

19. 19.What is the product of vigorous oxidation of toluene (or any alkylbenzene with a benzylic H) with hot $KMnO_4$?

    Benzoic acid ($C_6H_5COOH$). Regardless of the alkyl chain length, any alkyl group with a benzylic hydrogen is oxidized down to $-COOH$; tert-butylbenzene (no benzylic H) is not oxidized.

    Hint: Side chain becomes COOH; needs benzylic H.

20. 20.Write the reaction and mechanism type of Fischer esterification.

    $RCOOH + R'OH \underset{}{\overset{H^+}{\rightleftharpoons}} RCOOR' + H_2O$. Acid-catalyzed, reversible, nucleophilic acyl substitution (addition–elimination). Excess alcohol or removal of water drives it forward.

    Hint: Acid-catalyzed, reversible; alcohol + acid $\to$ ester.

21. 21.In Fischer esterification, isotopic labeling shows which bond breaks?

    The C–OH bond of the acid breaks (acyl–oxygen cleavage). Using $^{18}O$-labeled alcohol shows the label appears in the ester, confirming the acid loses its $-OH$ as part of water.

    Hint: Acid loses OH; alcohol keeps its O.

22. 22.Why does esterification of a carboxylic acid slow down with bulky (branched) alcohols or acids?

    Steric hindrance at the tetrahedral intermediate. Rate order: $CH_3OH > 1^\circ > 2^\circ > 3^\circ$ alcohol; similarly branching near $-COOH$ slows the reaction.

    Hint: Crowding the tetrahedral intermediate.

23. 23.How can a carboxylic acid be reduced to a primary alcohol? Why not with $NaBH_4$?

    Use $LiAlH_4$ (or $B_2H_6$): $RCOOH \to RCH_2OH$. $NaBH_4$ is too mild to reduce the resonance-stabilized, relatively unreactive $-COOH$ group.

    Hint: $LiAlH_4$ / diborane; $NaBH_4$ too weak.

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