Aldehydes & Ketones LEVEL I. Phenol (enol form) Phenol is aromatic, so equiulibrium is shifted to the right hand side. b) O

Transcription

1 Subjective Problems Aldehydes & Ketones LEVEL I 1. a) 2,4cyclohexadiene-1-one (keto form) Phenol (enol form) Phenol is aromatic, so equiulibrium is shifted to the right hand side. b) Base This ketone is more acidic because the resulting enolate ion obey's uckel's rule and is thus more stable. 2. a) i) l 2 2 l 2 2 (Et) 2 Et l K Et 2 (Et) 2 KMn 4 2 = (Et) 2 Na 3 2 This is called protection of carbonyl group. ii) Proceed via same procedure b) = J = K = 3. a) fast 3 2 slow 3 3 = 2 Br Br Br Br 3 NANEMIS

2 In the r.d.s. bromine molecule is not used np. ence the rate law does not contain the concentration of Br 2 i.e. the rate law is zero order with respect to bromine. b) i) nly double ( = ) bond is affected A = ii) arbonyl group as well as double bond is reduced B = iii) = iv) D = v) D D E = a) Ph Ph Ph 1. MgI 2. 3 D 3 Ph() 3 Br2 Ph Ph K r Br 2 Br Ph 3 P Ph 3 P 2 Ph All3 3 3 Na2 Ph 3 P = Ph Ph Ph = Ph Ph 3 P b) A: 6 5 B: Ph()Ph : 6 5 = innamic acid D: 6 5Br Br 5. This reaction is carried out in a solution of Na containing deuterium oxide (D 2 ). If the solvent supplies the hydrogen benzyl alcohol produced will contain some deuterium. If no deuterium is present the hydrogen must have come from a molecule of benzaldehyde. 6. a) A contains one side chain only, now = NANEMIS

3 Also the D.B.E. of A is 8 1 6/2 = 6 Three double bonds and one ring account for benzene. ence the side chain contains two double bonds = 6 5 A on treatment with Na and then acid gives B, The side chain is now Since acidification was necessary, this suggests that a Na salt was formed first i.e. side chain now contain a carboxyl group. Also the D.B.E. of B is 8 1 8/2 = 5. ence the side chain now contains one double bond (4 accounted for by benzene). A structure of B which fits the fact is = Since keto-group has been reduced and aldehyde group oxidises then is a case of internal cannizaro reaction. P Ph Ph = shift Proton exchange 2 Ph 2 2 [ Ph 2 Ph ] 3 b) The salt of the ketone is resonance stabilized etc. 7. a) N N having R N 2 having R I will be more readily protonated than (II). Alternatively protonated (I) is more stabilised by resonance than is protonated (II) N (Ia) (IIa) N NANEMIS

4 In (Ia) there is extended conjugation and only one charge is involved. In IIa there is not this extended conjugation and the relative close proximity of two positive charges is a destabilising factor. ence Ia is more stable than IIa. 8. a) Ph 2 Sol 2 Phl 2 N PhN 2 Li(Et) 2 Al 2 Ph 9. a) I does not undergo haloform reaction whereas II does. b) A: 3 > 3 3 > > B: F 3 > 3 > 2 = 10. Ph Ph Ph a) N Syn phenyl Ph N antiphenyl Ph N Ph Ph N PhNPh Ph Ph NPh ere anti migration takes place b) i) The bond in >=N prevents free rotation and threfore geometric isomerism exists if the groups on the carbonyl carbon are different ii) The order of electropositivity is Mg Zn d. The order of ionic character is Mg Zn - d in M R s. The order of nucleophilicity of R is MgR ZnR dr. Therefore, BrMg 2 Et reacts with = of the ester Br 2 Et. BrZn 2 Et can only react with RR = to be the Reformatsky reaction. v) In presence of Lewis acid, due to formation of anilinium ion, unexpected m.substituted derivatives are formed. 11. = N = ()N 1, 2 addition is favoured because of high reactivity of aldehyde group = = Because of conjugation in acraldehyde the charges and the distances between them are both greater than the corresponding values in propionadehyde. ence the former has greater = chanism NANEMIS

5 = = = = 14. Since ozonolysis involves fission at double or triple bonds, the formation of 3 products indicates the presence of two sites of attack. ence an alkyne is excluded. The positions of the two double bonds may be deduced as follows. 2 = therefore occurs at end of chain. 2 2 = therefore occurs at end of chain 2 = therefore occurs in the chain 15. i) NaB 4, NaB 4 can reduce ii) LiB 4, NaB 4 same reason but not 2 Et NANEMIS

6 LEVEL II = P(Ph) = a) 3 ( 2 ) ( 2 ) 3 ( 2 ) b) (i) Ph 2 Ph (ii) (c) 2 Base (B) B Ph = Ph Ph Ph Ph 2 Ph NANEMIS

7 4. a) 2-methyl 1,3-cyclohexane dione is more acidic because its enolate ion is stabilised by an additional resonance structure B. of A b) In the case of the acid the double bond is activated due to the presence of a group 2 2 N N 2 = N 2 N a) (A) Ph() 2 (B) Ph 2 b) Analogous to Pinacol - Pinacolone rearrangement. ere in first step diazotisation occurs. 6. It is an example of Knoevenagel reaction. This type condensation is favoured when 2 ( 2 Et) 2 is present in excess and pipesidine is base. N Et 22(2Et)2 Et[(2Et)2]2 7. ( ) 3 = 2 NaB ( ) 3 = ( ) 3 = 2 ( ) 3 = ( ) 3 = 2 2. NaB 4 reduces group and the resulting alcohol undergoes the allylic rearrangement to give more stable alcohol (in which there is increased conjugation). 8. a) We are given that A reduction B 2S 4 B forms mono-ozonide, D / Zn 2 D 3 NANEMIS

8 The compound A gives a haloform reaction; it must contain 3 group. The compound contains a double bond as it forms mono-ozonide D. Since, the compound D on hydrolysis gives only 3, the structure of would be 3 = () 2-butenal The compound is obtained by dehydration of B, thus the latter should be (B) 2-butanol Finally, B is obtained by the reduction of A. ence, the compound A should be butanone (A) The equations involved are as follows: [ ] (A) (B) Zn/ S = 3 () 9. keto acids undergo decarboxylation through the keto form via a cyclic transition state which results in the formation of enol form of ketone product. The enol then changes to more stable keto form. If decarboxylation of the given keto acids could follow this route then 2 2 (I) (II) Structure (I) is very unstable because bridgehead carbon cannot chance its hybridization from sp 3 to sp a) (A) chanism: Mg Mg 2 2e 2 2 (B) NANEMIS

9 2 2e 2 2, 2 1,2 3 shift b) 2 = Mg 2 = 2 Pinacolone 3 Mg Mg / = (Pinacol pinacolone rearr.) 11. a) The corresponding hydrate is very stable due to intra molecular hydrogen bonding 2 Ph Ph Ph b) A: B: Ph B: 12. A is ketone, as it forms oxime but does not reduce Tollen s reagent. D is aldehyde but if has no 3 group E is ketone containing 3 fragment NANEMIS

10 Let consider D & E are R 2 & RR respectively. R R R 2 = R 2 R R R ] ] [ 3 R 2 = [ R 2 2 As A is 6 12 so R = R = R = 3 3 A is 3 2 is 3 2 = 3 R R R B is Et 3 D is 3 2 E is N,, N 2 2 N 2, 2 2 () Erythrose Br Br 2, 2 2 ()Threose Note: The erythro isomer is the one that is convertible (in principle at least) into a meso structure, whereas the threo isomer is convertible into a racemic modification. The names of these compounds are the basis for designations erythro and threo acid to specify certain configurations of compounds containing two chiral carbons. 14. a) aqk ( 2) b) aq K 3 2 ( 2 ) 3 2 = 3 NANEMIS

11 c) aq. (Self aldol) d) Sodium dimethyl amide or Lithium diisopropyl amide (only one product) Totally convert any of the aldehydes to the carbanion by using strong base, followed by reaction with the other aldehyde. This is done to avoid multiple products. 15. A: 3 B: 3 3 : D: E:.N 2 NANEMIS

12 LEVEL III 1. i) ii) (A) (A) 2. Erythrose is an aldehyde and contains three - 's cleavage by 3I 4 show that it has the following structure, containing two chiral carbons: = * * = Erythrose xidation of () erythrose yield a dicarboxylic acid, 2 * () * () 2. Since the acid is optically inactive, it must have the meso configuration, showing that () erythrose is structure (I) or (II) N 3 2 or N Same meso (I) meso-acid (II) acid () Threose must be a diastereomer of erythrose, either structure (III) or (IV) 2 (III) N 3 Active acid 2 (IV) N 3 Active acid (X) Br2/Na Vigorous oxidation 2 (Y) 2 an form anhydride NANEMIS

13 3 2 (Z) Na/a 3 3-tyl anisole 5. a) b) = N 2 c) 3 = 3 3 d) e) 7. A) B) 3 2 ) D) 3 E) 3 2 F) i) NANEMIS

14 ii) 3 iii) 2Et l SN2 reaction 2Et 9. i) 6 5 () ii) 6 5 = 6 5 = = 6 5 = = Tautomerisation Since ozonolysis of A gives two aldehydes, the compound A contains the carboncarbon double bond. In fact, the molecule of A contains two double bonds as it is successively oznolyzed products will remain same as in the compound A. ence, it may be concluded that the ozonolysis products include two molecules of D( 3 ) and one molecule of E( ). From this, we derive the structure of A as shown in the following. 3-3 (D) (E) (D) The structures of B and are as follows. 3 3 = = 3 2, 4-hexadiene (A) NANEMIS

15 3 = = = 3 2-hexadiene (B) 2 The structure of F is as follows. 322 = a) N hexane () KMn Butanoic acid (F) b) S S c) 3 2 = N d) N In large quantities All 3 complexes with = group to give m - bromoacetophenone. In small amounts,. owever it acts as a catalyst to form phenacyl bromide. All 3 3 All 3 3 Br ---BrAll 3 Br All 3 3 Br 3 Br NANEMIS

16 13. i) The carbonyl group in aldehydes and ketones add on N resulting in the formation of an anion where the negative charge resides on oxygen. owever if a nucleophile adds on to N an alkene the negative charge resides on carbon. Since carbon is much less strongly electron attracting than oxygen this species is less stable and hence not readily formed. ii) In alkenes the double bond joins two carbon atoms and there is not resultant polarity. In carbonyl compounds, the carbonyl group is highly polar and the high partial positive charge on the atom makes it subsceptible to nucleophilic attack. iii) The positive inductive effect of the second alkyl radical reinforces that of the first one decreasing still further the partial positive charge on the carbonyl carbon atom. This reduces the attraction of the atom for nucleophilic reagents. ence ketones are less electrophilic. iv) The > = group in aldehydes activates the atom attached to the carbonyl group. This is due to the relaying of the I effect of the oxygen atom to the bond so that partial positive charge is created on the atom. The result of this activation is that the atom of the group can be oxidised readily to a () group. Thus aldehydes are reducers. v) Br is strongly polar and is hence readily added to the polarised > = group. The > addition product Br is however unstable and decomposes to give the original carbonyl compound and Br. 14. (a) (b) (c) 3 = A: 3 = 2 3 : B: Br D: NANEMIS