Name Reaction flash cards

1. 1.Aldol condensation — reagents and product?

   An aldehyde/ketone with an $\alpha$-H, treated with dilute base (dil. NaOH/KOH) or acid, gives a $\beta$-hydroxy aldehyde/ketone (aldol). On heating it dehydrates to an $\alpha,\beta$-unsaturated carbonyl. e.g. $2\,CH_3CHO \to CH_3CH(OH)CH_2CHO \to CH_3CH{=}CHCHO$ (crotonaldehyde).

   Hint: Needs $\alpha$-H; self-condensation of ethanal.

2. 2.Cannizzaro reaction — condition and outcome?

   An aldehyde with no $\alpha$-H, on treatment with conc. NaOH/KOH, undergoes disproportionation: one molecule is oxidised to the carboxylate salt and another reduced to the alcohol. e.g. $2\,HCHO + NaOH \to CH_3OH + HCOONa$.

   Hint: No $\alpha$-H; self redox.

3. 3.Cross-Cannizzaro reaction — what happens?

   Two different aldehydes lacking $\alpha$-H with conc. alkali; usually HCHO is the reductant (best H-donor) so it is oxidised to $HCOO^-$ while the other aldehyde is reduced to its alcohol. e.g. $C_6H_5CHO + HCHO \xrightarrow{NaOH} C_6H_5CH_2OH + HCOONa$.

   Hint: HCHO sacrifices itself; benzaldehyde → benzyl alcohol.

4. 4.Perkin reaction — reagents and product?

   An aromatic aldehyde + aliphatic acid anhydride, with the sodium (or potassium) salt of the acid as base, gives an $\alpha,\beta$-unsaturated aromatic acid. e.g. $C_6H_5CHO + (CH_3CO)_2O \xrightarrow{CH_3COONa} C_6H_5CH{=}CHCOOH$ (cinnamic acid).

   Hint: Benzaldehyde + acetic anhydride → cinnamic acid.

5. 5.Claisen condensation — reagents and product?

   Two ester molecules (with $\alpha$-H) with a strong base like sodium ethoxide give a $\beta$-keto ester. e.g. $2\,CH_3COOC_2H_5 \xrightarrow{NaOC_2H_5} CH_3COCH_2COOC_2H_5$ (ethyl acetoacetate).

   Hint: Ester + ester; product is acetoacetic ester.

6. 6.Claisen-Schmidt / crossed aldol: benzaldehyde + acetophenone with base gives?

   An $\alpha,\beta$-unsaturated ketone (chalcone): $C_6H_5CHO + CH_3COC_6H_5 \xrightarrow{OH^-} C_6H_5CH{=}CHCOC_6H_5$.

   Hint: Aromatic aldehyde (no $\alpha$-H) + methyl ketone.

7. 7.Wittig reaction — reagents and product?

   An aldehyde/ketone reacts with a phosphorus ylide (from $R_3P$ = triphenylphosphine + alkyl halide, then base) to give an alkene, converting $C{=}O$ into $C{=}C$. e.g. $R_2C{=}O + Ph_3P{=}CR'_2 \to R_2C{=}CR'_2 + Ph_3P{=}O$.

   Hint: Ylide; $C{=}O \to C{=}C$; byproduct triphenylphosphine oxide.

8. 8.Reformatsky reaction — reagents and product?

   An aldehyde/ketone + an $\alpha$-halo ester (e.g. $BrCH_2COOC_2H_5$) + Zn (in dry ether) gives a $\beta$-hydroxy ester after workup. Zinc forms an organozinc enolate that adds to the carbonyl.

   Hint: Zn + $\alpha$-bromoester; $\beta$-hydroxy ester.

9. 9.Reimer-Tiemann reaction — reagents and product?

   Phenol + $CHCl_3$ + aq. NaOH (then acid hydrolysis) introduces a $-CHO$ group ortho to $-OH$, giving salicylaldehyde. Intermediate is dichlorocarbene $:CCl_2$.

   Hint: Phenol → salicylaldehyde; carbene.

10. 10.Reimer-Tiemann with $CCl_4$ instead of $CHCl_3$ gives?

    Phenol + $CCl_4$ + NaOH gives salicylic acid (a $-COOH$ group ortho to $-OH$) instead of the aldehyde.

    Hint: $CCl_4$ → acid, not aldehyde.

11. 11.Kolbe's reaction (Kolbe-Schmitt) — reagents and product?

    Sodium phenoxide + $CO_2$ under pressure ($\sim 4{-}7$ atm, 400 K), then acidification, gives salicylic acid (2-hydroxybenzoic acid). Electrophilic carboxylation ortho to $O^-$.

    Hint: Sodium phenoxide + $CO_2$ → salicylic acid → aspirin.

12. 12.Kolbe electrolysis — reagents and product?

    Electrolysis of an aqueous sodium/potassium salt of a carboxylic acid gives an alkane (with one fewer... actually double the alkyl) at the anode: $2\,RCOO^- \to R{-}R + 2CO_2 + 2e^-$. e.g. $CH_3COONa \to CH_3CH_3$ (ethane).

    Hint: Anodic decarboxylative dimerisation; acetate → ethane.

13. 13.Wurtz reaction — reagents and product?

    Two alkyl halides + Na metal in dry ether give a symmetrical alkane with double the carbons: $2\,RX + 2Na \to R{-}R + 2NaX$. e.g. $2\,CH_3Br \to CH_3CH_3$.

    Hint: Alkyl halide + Na/dry ether; symmetrical alkane.

14. 14.Wurtz-Fittig reaction — reagents and product?

    An aryl halide + an alkyl halide + Na in dry ether gives an alkyl-substituted arene (alkylbenzene). e.g. $C_6H_5Br + CH_3Br + 2Na \to C_6H_5CH_3 + 2NaBr$.

    Hint: Aryl X + alkyl X + Na → alkylbenzene (toluene).

15. 15.Fittig reaction — reagents and product?

    Two aryl halides + Na in dry ether give a biaryl (diaryl). e.g. $2\,C_6H_5Br + 2Na \to C_6H_5{-}C_6H_5$ (biphenyl) $+ 2NaBr$.

    Hint: Two aryl halides + Na → biphenyl.

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