3CHO< CH3CH2OH
Question - 4 : - Arrange the following compounds in increasing order oftheir reactivity in nucleophilic addition reactions.
Answer - 4 : -
(i)Ethanal, Propanal, Propanone, Butanone.
(ii)Benzaldehyde, p-Tolualdehyde, p-Nitrobenzaldehyde, Acetophenone.
Hint:Consider steric effect and electronic effect.
Answer
(i)
The +I effect of the alkyl group increases in the order:
Ethanal < Propanal < Propanone < Butanone
The electron density at the carbonyl carbon increases withthe increase in the +I effect. As a result, the chances of attack by anucleophile decrease. Hence, the increasing order of the reactivities of thegiven carbonyl compounds in nucleophilic addition reactions is:
Butanone < Propanone < Propanal < Ethanal
(ii)
The +I effect is more in ketone than inaldehyde. Hence, acetophenone is the least reactive in nucleophilic addition reactions.Among aldehydes, the +I effect is the highest in p-tolualdehydebecause of the presence of the electron-donating −CH3 groupand the lowest in p-nitrobezaldehyde because of the presence of theelectron-withdrawing −NO2 group. Hence, the increasing order of thereactivities of the given compounds is:
Acetophenone < p-tolualdehyde< Benzaldehyde < p-Nitrobenzaldehyde
Question - 5 : - Predict the products of the followingreactions:
Answer - 5 : -
Question - 6 : - Show how each of the following compounds can be converted to benzoic acid.
Answer - 6 : -
(i) Ethylbenzene (ii) Acetophenone
(iii) Bromobenzene (iv) Phenylethene(Styrene)
Answer
(i)
(ii)
(iii)
(iv)
Question - 7 : - Which acid of each pair shown here would you expect to bestronger?
Answer - 7 : -
(i) CH3CO2H orCH2FCO2H
(ii)CH2FCO2H orCH2ClCO2H
(iii) CH2FCH2CH2CO2H or CH3CHFCH2CO2H
(iv)
Answer
(i)
The +I effect of −CH3 groupincreases the electron density on the O−H bond. Therefore, release of protonbecomes difficult. On the other hand, the −I effect of F decreases the electrondensity on the O−H bond. Therefore, proton can be released easily. Hence, CH2FCO2H isa stronger acid than CH3CO2H.
(ii)
F has stronger −I effect than Cl. Therefore,CH2FCO2H can release proton more easily than CH2ClCO2H.Hence, CH2FCO2H is stronger acid than CH2ClCO2H.
(iii)
Inductive effect decreases with increase indistance. Hence, the +I effect of F in CH3CHFCH2CO2H ismore than it is in CH2FCH2CH2CO2H. Hence, CH3CHFCH2CO2H isstronger acid than CH2FCH2CH2CO2H.
(iv)
Due to the −I effect of F, it is easier torelease proton in the case of compound (A). However, in the case of compound(B), release of proton is difficult due to the +I effect of −CH3 group.Hence, (A) is a stronger acid than (B).
Question - 8 : - What is meant by the following terms? Give an example ofthe reaction in each case.
Answer - 8 : -
(i) Cyanohydrin (ii) Acetal
(iii) Semicarbazone (iv) Aldol
(v) Hemiacetal (vi) Oxime
(vii) Ketal (vii) Imine
(ix) 2,4-DNP-derivative (x) Schiff’sbase
Answer
(i) Cyanohydrin:
Cyanohydrins are organic compounds having the formulaRR′C(OH)CN, where R and R′ can be alkyl or aryl groups.
Aldehydes and ketones react with hydrogen cyanide (HCN) inthe presence of excess sodium cyanide (NaCN) as a catalyst to fieldcyanohydrin. These reactions are known as cyanohydrin reactions.
Cyanohydrins are useful synthetic intermediates.
(ii) Acetal:
Acetals are gem−dialkoxy alkanes in which two alkoxygroups are present on the terminal carbon atom. One bond is connected to analkyl group while the other is connected to a hydrogen atom.
When aldehydes are treated with two equivalents of amonohydric alcohol in the presence of dry HCl gas, hemiacetals are producedthat further react with one more molecule of alcohol to yield acetal.
(iii) Semicarbarbazone:
Semicarbazones are derivatives of aldehydes and ketonesproduced by the condensation reaction between a ketone or aldehyde andsemicarbazide.
Semicarbazones are useful for identification andcharacterization of aldehydes and ketones.
(iv) Aldol:
A β-hydroxy aldehyde or ketoneis known as an aldol. It is produced by the condensation reaction of twomolecules of the same or one molecule each of two different aldehydes orketones in the presence of a base.
(v) Hemiacetal:
Hemiacetals are α−alkoxyalcohols
General structure of a hemiacetal
Aldehyde reacts with one molecule of a monohydric alcoholin the presence of dry HCl gas.
(vi) Oxime:
Oximes are a class of organic compounds having the generalformula RR′CNOH, where R is an organic side chain and R′ is either hydrogen oran organic side chain. If R′ is H, then it is known as aldoxime and if R′ is anorganic side chain, it is known as ketoxime.
On treatment with hydroxylamine in a weakly acidic medium,aldehydes or ketones form oximes.
(vii) Ketal:
Ketals are gem−dialkoxyalkanes in which two alkoxy groupsare present on the same carbon atom within the chain. The other two bonds ofthe carbon atom are connected to two alkyl groups.
Ketones react with ethylene glycol in the presence of dryHCl gas to give a cyclic product known as ethylene glycol ketals.
(viii) Imine:
Imines are chemical compounds containing a carbon nitrogendouble bond.
Imines are produced when aldehydes and ketones react withammonia and its derivatives.
(ix) 2, 4−DNP−derivative:
2, 4−dinitrophenylhydragones are 2, 4−DNP−derivatives,which are produced when aldehydes or ketones react with 2,4−dinitrophenylhydrazine in a weakly acidic medium.
To identify and characterize aldehydes and ketones, 2,4−DNP derivatives are used.
(x) Schiff’s base:
Schiff’s base (or azomethine) is a chemicalcompound containing a carbon-nitrogen double bond with the nitrogen atomconnected to an aryl or alkyl group-but not hydrogen. They have the generalformula R1R2C = NR3. Hence, it is an imine.
It is named after a scientist, Hugo Schiff.
Aldehydes and ketones on treatment with primary aliphaticor aromatic amines in the presence of trace of an acid yields a Schiff’s base.
Question - 9 : - Name the following compounds according to IUPAC system ofnomenclature:
Answer - 9 : -
(i) CH3CH(CH3)CH2CH2CHO
(ii) CH3CH2COCH(C2H5)CH2CH2Cl
(iii) CH3CH=CHCHO
(iv) CH3COCH2COCH3
(v) CH3CH(CH3)CH2C(CH3)2COCH3
(vi) (CH3)3CCH2COOH
(vii) OHCC6H4CHO-p
Answer
(i) 4-methylpentanal
(ii) 6-Chloro-4-ethylhexan-3-one
(iii) But-2-en-1-al
(iv) Pentane-2,4-dione
(v) 3,3,5-Trimethylhexan-2-one
(vi) 3,3-Dimethylbutanoicacid
(vii) Benzene-1,4-dicarbaldehyde
Question - 10 : - Draw the structures of the following compounds.
Answer - 10 : -
(i) 3-Methylbutanal (ii) p-Nitropropiophenone
(iii) p-Methylbenzaldehyde (iv) 4-Methylpent-3-en-2-one
(v) 4-Chloropentan-2-one (vi) 3-Bromo-4-phenylpentanoic acid
(vii) p,p’-Dihydroxybenzophenone (viii) Hex-2-en-4-ynoic acid
Answer
(i)
(iv)
(v)
(vi)
