12- ALDEHYDES, KETONES AND CARBOXYLIC ACIDS


Here we are going through the ncert solutions for class 12 chemistry chapter 12- aldehydes ketones and carboxylic acids. so before going through the ncert solutions make sure to go through the textbook that helps you to understand the solutions more easily

 ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​​​ Chapter​​ 12​​ - Aldehydes, Ketones And Carboxylic Acids

 

 ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​​​  ​​ ​​ ​​​​  ​​ ​​​​ INTEXT ANSWERS

 

Question 12.1:

Write the structures of the following compounds.

(i) α-Methoxypropionaldehyde

(ii) 3-Hydroxybutanal

(iii) 2-Hydroxycyclopentane carbaldehyde

(iv) 4-Oxopentanal

(v) Di-sec-butyl ketone

(vi) 4-Fluoroacetophenone

ANSWER:

(i)

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/271/5804/NCERT(INTEXT)_27-11-08_Utpal_12_Chemistry_12_8_html_m2f44bb19.jpg

(ii)

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/271/5804/NCERT(INTEXT)_27-11-08_Utpal_12_Chemistry_12_8_html_m76191634.jpg

(iii)

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/271/5804/NCERT(INTEXT)_27-11-08_Utpal_12_Chemistry_12_8_html_34c05f41.jpg

(iv)

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/271/5804/NCERT(INTEXT)_27-11-08_Utpal_12_Chemistry_12_8_html_1cc84da8.jpg

(v)

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/271/5804/NCERT(INTEXT)_27-11-08_Utpal_12_Chemistry_12_8_html_7aa9d942.jpg

(vi)

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/271/5804/NCERT(INTEXT)_27-11-08_Utpal_12_Chemistry_12_8_html_4a9688af.jpg



Question 12.2:

Write the structures of products of the following reactions;

(i)

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/271/5805/NCERT(INTEXT)_27-11-08_Utpal_12_Chemistry_12_8_html_m20f04397.jpg

(ii)

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/271/5805/NCERT(INTEXT)_27-11-08_Utpal_12_Chemistry_12_8_html_m6c665586.jpg

(iii)

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/271/5805/NCERT(INTEXT)_27-11-08_Utpal_12_Chemistry_12_8_html_m75764933.jpg

(iv)

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/271/5805/NCERT(INTEXT)_27-11-08_Utpal_12_Chemistry_12_8_html_975c3e2.jpg

ANSWER:

  •  

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/271/5805/NCERT(INTEXT)_27-11-08_Utpal_12_Chemistry_12_8_html_53c3a6e4.jpg

  •  

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/271/5805/NCERT(INTEXT)_27-11-08_Utpal_12_Chemistry_12_8_html_m754e5a1.jpg

  •  

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/271/5805/NCERT(INTEXT)_27-11-08_Utpal_12_Chemistry_12_8_html_3382db64.jpg

(iv)

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/271/5805/NCERT(INTEXT)_27-11-08_Utpal_12_Chemistry_12_8_html_m5d88adcf.jpg



Question 12.3:

Arrange the following compounds in increasing order of their boiling points.

CH3CHO, CH3CH2OH, CH3OCH3, CH3CH2CH3

ANSWER:

The molecular masses of the given compounds are in the range 44 to 46. CH3CH2OH undergoes extensive intermolecular H-bonding, resulting in the association of molecules. Therefore, it has the highest boiling point. CH3CHO is more polar than CH3OCH3 and so CH3CHO has stronger intermolecular dipole − dipole attraction than CH3OCH3 CH3CH2CH3 has only weak van der Waals force. Thus, the arrangement of the given compounds in the increasing order of their boiling points is given by:

CH3CH2CH3 < CH3OCH3 < CH3CHO < CH3CH2OH



Question 12.4:

Arrange the following compounds in increasing order of their reactivity in nucleophilic addition reactions.

(i)Ethanal, Propanal, Propanone, Butanone.

(ii)Benzaldehyde, p-Tolualdehyde, p-Nitrobenzaldehyde, Acetophenone.

Hint:Consider steric effect and electronic effect.

ANSWER:

(i)

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/271/5807/NCERT(INTEXT)_27-11-08_Utpal_12_Chemistry_12_8_html_m681a0c91.jpg

The +I effect of the alkyl group increases in the order:

Ethanal < Propanal < Propanone < Butanone

The electron density at the carbonyl carbon increases with the increase in the +I effect. As a result, the chances of attack by a nucleophile decrease. Hence, the increasing order of the reactivities of the given carbonyl compounds in nucleophilic addition reactions is:

Butanone < Propanone < Propanal < Ethanal

(ii)

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/271/5807/NCERT(INTEXT)_27-11-08_Utpal_12_Chemistry_12_8_html_m7e4c2a56.jpg

The +I effect is more in ketone than in aldehyde. Hence, acetophenone is the least reactive in nucleophilic addition reactions. Among aldehydes, the +I effect is the highest in p-tolualdehyde because of the presence of the electron-donating −CH3 group and the lowest in p-nitrobezaldehyde because of the presence of the electron-withdrawing −NO2 group. Hence, the increasing order of the reactivities of the given compounds is:

Acetophenone < p-tolualdehyde < Benzaldehyde
p-Nitrobenzaldehyde

 

Question 12.5:

Predict the products of the following reactions:

(i)

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/271/5808/NCERT(INTEXT)_27-11-08_Utpal_12_Chemistry_12_8_html_m64fb7212.jpg

(ii)

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/271/5808/NCERT(INTEXT)_27-11-08_Utpal_12_Chemistry_12_8_html_m4127564c.jpg

(iii)

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/271/5808/NCERT(INTEXT)_27-11-08_Utpal_12_Chemistry_12_8_html_63382d2e.jpg

(iv)

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/271/5808/NCERT(INTEXT)_27-11-08_Utpal_12_Chemistry_12_8_html_3ac259e4.jpg

ANSWER:

(i)

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/271/5808/NCERT(INTEXT)_27-11-08_Utpal_12_Chemistry_12_8_html_201740dc.jpg

(ii)

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/271/5808/NCERT(INTEXT)_27-11-08_Utpal_12_Chemistry_12_8_html_6a3b7966.jpg

(iii)

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/271/5808/NCERT(INTEXT)_27-11-08_Utpal_12_Chemistry_12_8_html_m37ffe1ff.jpg

(iv)

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/271/5808/NCERT(INTEXT)_27-11-08_Utpal_12_Chemistry_12_8_html_13788c2d.jpg



Question 12.6:

Give the IUPAC names of the following compounds:

(i) PhCH2CH2COOH (ii) (CH3)2C=CHCOOH

(iii) https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/271/5809/NCERT(INTEXT)_27-11-08_Utpal_12_Chemistry_12_8_html_m17f7df4f.jpg (iv) https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/271/5809/NCERT(INTEXT)_27-11-08_Utpal_12_Chemistry_12_8_html_m39224e94.jpg

ANSWER:

(i) 3-Phenylpropanoic acid

(ii) 3-Methylbut-2-enoic acid

(iii) 2-Methylcyclopentanecarboxylic acid

(iv)2,4,6-Trinitrobenzoic acid



​​ Question 12.7:

Show how each of the following compounds can be converted to benzoic acid.

(i) Ethylbenzene (ii) Acetophenone

(iii) Bromobenzene (iv) Phenylethene (Styrene)

ANSWER:

(i)

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/271/5810/NCERT(INTEXT)_27-11-08_Utpal_12_Chemistry_12_8_html_578e26c1.jpg

(ii)

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/271/5810/NCERT(INTEXT)_27-11-08_Utpal_12_Chemistry_12_8_html_5246c72.jpg

(iii)

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/271/5810/NCERT(INTEXT)_27-11-08_Utpal_12_Chemistry_12_8_html_m188ee7a6.jpg

(iv)

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/271/5810/NCERT(INTEXT)_27-11-08_Utpal_12_Chemistry_12_8_html_m3c42a4e.jpg



Question 12.8:

Which acid of each pair shown here would you expect to be stronger?

(i) CH3CO2H or CH2FCO2H

(ii)CH2FCO2H or CH2ClCO2H

(iii) CH2FCH2CH2CO2H or CH3CHFCH2CO2H

(iv)

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/271/5811/NCERT(INTEXT)_27-11-08_Utpal_12_Chemistry_12_8_html_m7d7a6d1a.jpg

ANSWER:

(i)

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/271/5811/NCERT(INTEXT)_27-11-08_Utpal_12_Chemistry_12_8_html_5ae35121.jpg

The +I effect of −CH3 group increases the electron density on the O−H bond. Therefore, release of proton becomes difficult. On the other hand, the −I effect of F decreases the electron density on the O−H bond. Therefore, proton can be released easily. Hence, CH2FCO2H is a stronger acid than CH3CO2H.

(ii)

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/271/5811/NCERT(INTEXT)_27-11-08_Utpal_12_Chemistry_12_8_html_796b09db.jpg

F has stronger −I effect than Cl. Therefore, CH2FCO2H can release proton more easily than CH2ClCO2H. Hence, CH2FCO2H is stronger acid than CH2ClCO2H.

(iii)

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/271/5811/NCERT(INTEXT)_27-11-08_Utpal_12_Chemistry_12_8_html_m241e0508.jpg

Inductive effect decreases with increase in distance. Hence, the +I effect of F in CH3CHFCH2CO2H is more than it is in CH2FCH2CH2CO2H. Hence, CH3CHFCH2CO2H is stronger acid than CH2FCH2CH2CO2H.

(iv)

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/271/5811/NCERT(INTEXT)_27-11-08_Utpal_12_Chemistry_12_8_html_m3f0466aa.jpg

Due to the −I effect of F, it is easier to release 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).



EXERCISE ANSWERS

 

Question 12.1:

What is meant by the following terms? Give an example of the reaction in each case.

(i) Cyanohydrin (ii) Acetal

(iii) Semicarbazone (iv) Aldol

(v) Hemiacetal (vi) Oxime

(vii) Ketal (vii) Imine

(ix) 2,4-DNP-derivative (x) Schiff’s base

ANSWER:

(i) Cyanohydrin:

Cyanohydrins are organic compounds having the formula RR′C(OH)CN, where R and R′ can be alkyl or aryl groups.

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/271/5812/NCERT_15-12-08_Uptal_Chemistry_12_12_20_html_615f56b3.jpg

Aldehydes and ketones react with hydrogen cyanide (HCN) in the presence of excess sodium cyanide (NaCN) as a catalyst to field cyanohydrin. These reactions are known as cyanohydrin reactions.

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/271/5812/NCERT_15-12-08_Uptal_Chemistry_12_12_20_html_m4433a2a5.gif

Cyanohydrins are useful synthetic intermediates.

(ii) Acetal:

Acetals are gem−dialkoxy alkanes in which two alkoxy groups are present on the terminal carbon atom. One bond is connected to an alkyl group while the other is connected to a hydrogen atom.

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/271/5812/NCERT_15-12-08_Uptal_Chemistry_12_12_20_html_m321215d4.jpg

When aldehydes are treated with two equivalents of a monohydric alcohol in the presence of dry HCl gas, hemiacetals are produced that further react with one more molecule of alcohol to yield acetal.

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/271/5812/NCERT_15-12-08_Uptal_Chemistry_12_12_20_html_m27d0fc2e.jpg

(iii) Semicarbarbazone:

Semicarbazones are derivatives of aldehydes and ketones produced by the condensation reaction between a ketone or aldehyde and semicarbazide.

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/271/5812/NCERT_15-12-08_Uptal_Chemistry_12_12_20_html_5d2ef7ba.jpg

Semicarbazones are useful for identification and characterization of aldehydes and ketones.

(iv) Aldol:

β-hydroxy aldehyde or ketone is known as an aldol. It is produced by the condensation reaction of two molecules of the same or one molecule each of two different aldehydes or ketones in the presence of a base.

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/271/5812/NCERT_15-12-08_Uptal_Chemistry_12_12_20_html_m36270e66.jpg

(v) Hemiacetal:

Hemiacetals are α−alkoxyalcohols

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/271/5812/NCERT_15-12-08_Uptal_Chemistry_12_12_20_html_m456e6442.jpg

General structure of a hemiacetal

Aldehyde reacts with one molecule of a monohydric alcohol in the presence of dry HCl gas.

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/271/5812/NCERT_15-12-08_Uptal_Chemistry_12_12_20_html_2992a23e.jpg

(vi) Oxime:

Oximes are a class of organic compounds having the general formula RR′CNOH, where R is an organic side chain and R′ is either hydrogen or an organic side chain. If R′ is H, then it is known as aldoxime and if R′ is an organic side chain, it is known as ketoxime.

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/271/5812/NCERT_15-12-08_Uptal_Chemistry_12_12_20_html_3f42772c.jpg

On treatment with hydroxylamine in a weakly acidic medium, aldehydes or ketones form oximes.

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/271/5812/NCERT_15-12-08_Uptal_Chemistry_12_12_20_html_m5eb66c8f.jpg

(vii) Ketal:

Ketals are gem−dialkoxyalkanes in which two alkoxy groups are present on the same carbon atom within the chain. The other two bonds of the carbon atom are connected to two alkyl groups.

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/271/5812/NCERT_15-12-08_Uptal_Chemistry_12_12_20_html_68e8fe87.jpg

Ketones react with ethylene glycol in the presence of dry HCl gas to give a cyclic product known as ethylene glycol ketals.

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/271/5812/NCERT_15-12-08_Uptal_Chemistry_12_12_20_html_m48138f13.jpg

(viii) Imine:

Imines are chemical compounds containing a carbon​​ ­nitrogen double bond.

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/271/5812/NCERT_15-12-08_Uptal_Chemistry_12_12_20_html_m5f60d4a6.jpg

Imines are produced when aldehydes and ketones react with ammonia and its derivatives.

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/271/5812/NCERT_15-12-08_Uptal_Chemistry_12_12_20_html_55e14875.jpg

(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.

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/271/5812/NCERT_15-12-08_Uptal_Chemistry_12_12_20_html_m7c88589e.jpg

To identify and characterize aldehydes and ketones, 2, 4−DNP derivatives are used.

(x) Schiff’s base:

Schiff’s base (or azomethine) is a chemical compound containing a carbon-nitrogen double bond with the nitrogen atom connected to an aryl or alkyl group-but not hydrogen. They have the general formula R1R2C = NR3. Hence, it is an imine.

It is named after a scientist, Hugo Schiff.

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/271/5812/NCERT_15-12-08_Uptal_Chemistry_12_12_20_html_m44907053.jpg

Aldehydes and ketones on treatment with primary aliphatic or aromatic amines in the presence of trace of an acid yields a Schiff’s base.

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/271/5812/NCERT_15-12-08_Uptal_Chemistry_12_12_20_html_56eed273.jpg

 

Question 12.2:

Name the following compounds according to IUPAC system of nomenclature:

(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-Dimethylbutanoic acid

(vii) Benzene-1,4-dicarbaldehyde

 

Question 12.3:

Draw the structures of the following compounds.

(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)

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/271/5814/NCERT_15-12-08_Uptal_Chemistry_12_12_20_html_75dad4a6.jpg

(ii)

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/271/5814/NCERT_15-12-08_Uptal_Chemistry_12_12_20_html_1b5acbeb.jpg

(iii)

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/271/5814/NCERT_15-12-08_Uptal_Chemistry_12_12_20_html_7cbd7566.jpg

(iv)

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/271/5814/NCERT_15-12-08_Uptal_Chemistry_12_12_20_html_m746d5dba.jpg

(v)

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/271/5814/NCERT_15-12-08_Uptal_Chemistry_12_12_20_html_798d9b76.jpg

(vi)

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/271/5814/NCERT_15-12-08_Uptal_Chemistry_12_12_20_html_m3f12c262.jpg

(vii)

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/271/5814/NCERT_15-12-08_Uptal_Chemistry_12_12_20_html_m60620871.jpg

(viii)

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/271/5814/NCERT_15-12-08_Uptal_Chemistry_12_12_20_html_m74ab30ff.jpg

 

Question 12.4:

Write the IUPAC names of the following ketones and aldehydes. Wherever possible, give also common names.

(i) CH3CO(CH2)4CH3 (ii) CH3CH2CHBrCH2CH(CH3)CHO

(iii) CH3(CH2)5CHO (iv) Ph-CH=CH-CHO

(v) https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/271/5815/NCERT_15-12-08_Uptal_Chemistry_12_12_20_html_1e01f4a0.jpg (vi) PhCOPh

ANSWER:

(i) CH3CO(CH2)4CH3

IUPAC name: Heptan-2-one

Common name: Methyl n-propyl ketone

(ii) CH3CH2CHBrCH2CH(CH3)CHO

IUPAC name: 4-Bromo-2-methylhaxanal

Common name: (γ-Bromo-α-methyl-caproaldehyde)

(iii) CH3(CH2)5CHO

IUPAC name: Heptanal

(iv) Ph-CH=CH-CHO

IUPAC name: 3-phenylprop-2-enal

Common name: β-Pheynolacrolein

(v)

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/271/5815/NCERT_15-12-08_Uptal_Chemistry_12_12_20_html_1e01f4a0.jpg

IUPAC name: Cyclopentanecarbaldehyde

(vi)PhCOPh

IUPAC name: Diphenylmethanone

Common name: Benzophenone

 

Question 12.5:

Draw structures of the following derivatives.

(i) The 2,4-dinitrophenylhydrazone of benzaldehyde

(ii) Cyclopropanone oxime

(iii) Acetaldehydedimethylacetal

(iv) The semicarbazone of cyclobutanone

(v) The ethylene ketal of hexan-3-one

(vi) The methyl hemiacetal of formaldehyde

ANSWER:

(i)

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/271/5816/NCERT_15-12-08_Uptal_Chemistry_12_12_20_html_m5341e5ce.jpg

(ii)

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/271/5816/NCERT_15-12-08_Uptal_Chemistry_12_12_20_html_m76f76ee5.jpg

(iii)

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/271/5816/NCERT_15-12-08_Uptal_Chemistry_12_12_20_html_m4011e930.jpg

(iv)

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/271/5816/NCERT_15-12-08_Uptal_Chemistry_12_12_20_html_41241471.jpg

(v)

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/271/5816/NCERT_15-12-08_Uptal_Chemistry_12_12_20_html_4ecf8766.jpg

(vi)

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/271/5816/NCERT_15-12-08_Uptal_Chemistry_12_12_20_html_m57099d32.jpg



Question 12.6:

Predict the products formed when cyclohexanecarbaldehyde reacts with following reagents.

(i) PhMgBr and then H3O+

(ii)Tollens’ reagent

(iii) Semicarbazide and weak acid

(iv)Excess ethanol and acid

(v) Zinc amalgam and dilute hydrochloric acid

ANSWER:

(i)

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/271/5817/NCERT_15-12-08_Uptal_Chemistry_12_12_20_html_m208138bf.jpg

(ii)

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/271/5817/NCERT_15-12-08_Uptal_Chemistry_12_12_20_html_m4454198d.jpg

(iii)

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/271/5817/NCERT_15-12-08_Uptal_Chemistry_12_12_20_html_m1fb94dd9.jpg

(iv)

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/271/5817/NCERT_15-12-08_Uptal_Chemistry_12_12_20_html_m16754dda.jpg

(v)

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/271/5817/NCERT_15-12-08_Uptal_Chemistry_12_12_20_html_431ca2ee.jpg

 

Question 12.7:

Which of the following compounds would undergo aldol condensation, which the Cannizzaro reaction and which neither? Write the structures of the expected products of aldol condensation and Cannizzaro reaction.

(i) Methanal (ii) 2-Methylpentanal

(iii) Benzaldehyde (iv) Benzophenone

(v) Cyclohexanone (vi) 1-Phenylpropanone

(vii) Phenylacetaldehyde (viii) Butan-1-ol

(ix) 2, 2-Dimethylbutanal

ANSWER:

Aldehydes and ketones having at least one α-hydrogen undergo aldol condensation. The compounds (ii) 2−methylpentanal, (v) cyclohexanone, (vi) 1-phenylpropanone, and (vii) phenylacetaldehyde contain one or more α-hydrogen atoms. Therefore, these undergo aldol condensation.

Aldehydes having no α-hydrogen atoms undergo Cannizzaro reactions. The compounds (i) Methanal, (iii) Benzaldehyde, and (ix) 2, 2-dimethylbutanal do not have any α-hydrogen. Therefore, these undergo cannizzaro reactions.

Compound (iv) Benzophenone is a ketone having no α-hydrogen atom and compound (viii) Butan-1-ol is an alcohol. Hence, these compounds do not undergo either aldol condensation or cannizzaro reactions.

Aldol condensation

(ii)

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/271/5818/NCERT_15-12-08_Uptal_Chemistry_12_12_20_html_m2c3e54d7.jpg

(v)

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/271/5818/NCERT_15-12-08_Uptal_Chemistry_12_12_20_html_56c1fed5.jpg

(vi)

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/271/5818/NCERT_15-12-08_Uptal_Chemistry_12_12_20_html_m579284f5.jpg

(vii)

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/271/5818/NCERT_15-12-08_Uptal_Chemistry_12_12_20_html_41bafdd9.jpg

Cannizzaro reaction

(i)

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/271/5818/NCERT_15-12-08_Uptal_Chemistry_12_12_20_html_m71ec06c8.jpg

(iii)

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/271/5818/NCERT_15-12-08_Uptal_Chemistry_12_12_20_html_m61d6b552.jpg

(ix)

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/271/5818/NCERT_15-12-08_Uptal_Chemistry_12_12_20_html_m1be46cda.jpg

 

Question 12.8:

How will you convert ethanal into the following compounds?

(i) Butane-1, 3-diol (ii) But-2-enal (iii) But-2-enoic acid

ANSWER:

(i) On treatment with dilute alkali, ethanal produces 3-hydroxybutanal gives butane-1, 3-diol on reduction.

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/271/5819/NCERT_15-12-08_Uptal_Chemistry_12_12_20_html_4c189c81.jpg

(ii) On treatment with dilute alkali, ethanal gives 3-hydroxybutanal which on heating produces but-2-enal.

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/271/5819/NCERT_15-12-08_Uptal_Chemistry_12_12_20_html_66fa8fcd.jpg

(iii) When treated with Tollen’s reagent, But-2-enal produced in the above reaction produces but-2-enoic acid .

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/271/5819/NCERT_15-12-08_Uptal_Chemistry_12_12_20_html_74f29828.jpg

 

Question 12.9:

Write structural formulas and names of four possible aldol condensation products from propanal and butanal. In each case, indicate which aldehyde acts as nucleophile and which as electrophile.

ANSWER:

(i) Taking two molecules of propanal, one which acts as a nucleophile and the other as an electrophile.

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/271/5820/NCERT_15-12-08_Uptal_Chemistry_12_12_20_html_ma02af1b.jpg

(ii) Taking two molecules of butanal, one which acts as a nucleophile and the other as an electrophile.

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/271/5820/NCERT_15-12-08_Uptal_Chemistry_12_12_20_html_5280cb79.jpg

(iii) Taking one molecule each of propanal and butanal in which propanal acts as a nucleophile and butanal acts as an electrophile.

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/271/5820/NCERT_15-12-08_Uptal_Chemistry_12_12_20_html_m5f29d4bd.jpg

(iv) Taking one molecule each of propanal and butanal in which propanal acts as an electrophile and butanal acts as a nucleophile.

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/271/5820/NCERT_15-12-08_Uptal_Chemistry_12_12_20_html_765990ea.jpg

 

Question 12.10:

An organic compound with the molecular formula C9H10O forms 2, 4-DNP derivative, reduces Tollens’ reagent and undergoes Cannizzaro reaction. On vigorous oxidation, it gives 1, 2-benzenedicarboxylic acid. Identify the compound.

ANSWER:

It is given that the compound (with molecular formula C9H10O) forms 2, 4-DNP derivative and reduces Tollen’s reagent. Therefore, the given compound must be an aldehyde.

Again, the compound undergoes cannizzaro reaction and on oxidation gives 1, 2-benzenedicarboxylic acid. Therefore, the −CHO group is directly attached to a benzene ring and this benzaldehyde is ortho-substituted. Hence, the compound is 2-ethylbenzaldehyde.

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/271/5821/NCERT_15-12-08_Uptal_Chemistry_12_12_20_html_6c33b1aa.jpg

The given reactions can be explained by the following equations.

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/271/5821/NCERT_15-12-08_Uptal_Chemistry_12_12_20_html_c71eddd.jpg

 

Question 12.11:

An organic compound (A) (molecular formula C8H16O2) was hydrolysed with dilute sulphuric acid to give a carboxylic acid (B) and an alcohol (C). Oxidation of (C) with chromic acid produced (B). (C) on dehydration gives but-1-ene.Write equations for the reactions involved.

ANSWER:

An organic compound A with molecular formula C8H16O2 gives a carboxylic acid (B) and an alcohol (C) on hydrolysis with dilute sulphuric acid. Thus, compound A must be an ester. Further, alcohol C gives acid B on oxidation with chromic acid. Thus, B and C must contain equal number of carbon atoms.

Since compound A contains a total of 8 carbon atoms, each of B and C contain 4 carbon atoms.

Again, on dehydration, alcohol C gives but-1-ene. Therefore, C is of straight chain and hence, it is butan-1-ol.

On oxidation, Butan-1-ol gives butanoic acid. Hence, acid B is butanoic acid.

Hence, the ester with molecular formula C8H16O2 is butylbutanoate.

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/271/5822/NCERT_15-12-08_Uptal_Chemistry_12_12_20_html_2e1f716c.jpg

All the given reactions can be explained by the following equations.

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/271/5822/NCERT_15-12-08_Uptal_Chemistry_12_12_20_html_66a88b35.jpg

 

Question 12.12:

Arrange the following compounds in increasing order of their property as indicated:

(i) Acetaldehyde, Acetone, Di-tert-butyl ketone, Methyl tert-butyl ketone (reactivity towards HCN)

(ii) CH3CH2CH(Br)COOH, CH3CH(Br)CH2COOH, (CH3)2CHCOOH, CH3CH2CH2COOH (acid strength)

(iii) Benzoic acid, 4-Nitrobenzoic acid, 3,4-Dinitrobenzoic acid, 4-Methoxybenzoic acid (acid strength)

ANSWER:

(i) When HCN reacts with a compound, the attacking species is a nucleophile, CN. Therefore, as the negative charge on the compound increases, its reactivity with HCN decreases. In the given compounds, the +I effect increases as shown below. It can be observed that steric hindrance also increases in the same

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/271/5823/NCERT_15-12-08_Uptal_Chemistry_12_12_20_html_3b41cfeb.jpg

Hence, the given compounds can be arranged according to their increasing reactivities toward HCN as:

Di-tert-butyl ketone < Methyl tert-butyl ketone < Acetone < Acetaldehyde

(ii) After losing a proton, carboxylic acids gain a negative charge as shown:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/271/5823/NCERT_15-12-08_Uptal_Chemistry_12_12_20_html_4cc111f.gif

Now, any group that will help stabilise the negative charge will increase the stability of the carboxyl ion and as a result, will increase the strength of the acid. Thus, groups having +I effect will decrease the strength of the acids and groups having −I effect will increase the strength of the acids. In the given compounds, −CH3 group has +I effect and Br group has −I effect. Thus, acids containing Br are stronger.

Now, the +I effect of isopropyl group is more than that of n-propyl group. Hence, (CH3)2CHCOOH is a weaker acid than CH3CH2CH2COOH.

Also, the −I effect grows weaker as distance increases. Hence, CH3CH(Br)CH2COOH is a weaker acid than CH3CH2CH(Br)COOH.

Hence, the strengths of the given acids increase as:

(CH3)2CHCOOH < CH3CH2CH2COOH < CH3CH(Br)CH2COOH < CH3CH2CH(Br)COOH

(iii) As we have seen in the previous case, electron-donating groups decrease the strengths of acids, while electron-withdrawing groups increase the strengths of acids. As methoxy group is an electron-donating group, 4-methoxybenzoic acid is a weaker acid than benzoic acid. Nitro group is an electron-withdrawing group and will increase the strengths of acids. As 3,4-dinitrobenzoic acid contains two nitro groups, it is a slightly stronger acid than 4-nitrobenzoic acid. Hence, the strengths of the given acids increase as:

4-Methoxybenzoic acid < Benzoic acid < 4-Nitrobenzoic acid
< 3,4-Dinitrobenzoic acid

 

Question 12.13:

Give simple chemical tests to distinguish between the following pairs of compounds.

(i) Propanal and Propanone

(ii) Acetophenone and Benzophenone

(iii) Phenol and Benzoic acid

(iv) Benzoic acid and Ethyl benzoate

(v) Pentan-2-one and Pentan-3-one

(vi) Benzaldehyde and Acetophenone

(vii) Ethanal and Propanal

ANSWER:

(i) Propanal and propanone can be distinguished by the following tests.

(a) Tollen’s test

Propanal is an aldehyde. Thus, it reduces Tollen’s reagent. But, propanone being a ketone does not reduce Tollen’s reagent.

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/271/5824/NCERT_15-12-08_Uptal_Chemistry_12_12_20_html_10733b1f.jpg

(b) Fehling’s test

Aldehydes respond to Fehling’s test, but ketones do not.

Propanal being an aldehyde reduces Fehling’s solution to a red-brown precipitate of Cu2O, but propanone being a ketone does not.

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/271/5824/NCERT_15-12-08_Uptal_Chemistry_12_12_20_html_m6dbc22de.jpg

(c) Iodoform test:

Aldehydes and ketones having at least one methyl group linked to the carbonyl carbon atom respond to iodoform test. They are oxidized by sodium hypoiodite (NaOI) to give iodoforms. Propanone being a methyl ketone responds to this test, but propanal does not.

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/271/5824/NCERT_15-12-08_Uptal_Chemistry_12_12_20_html_m69c3a9bc.jpg

(ii) Acetophenone and Benzophenone can be distinguished using the iodoform test.

Iodoform test:

Methyl ketones are oxidized by sodium hypoiodite to give yellow ppt. of iodoform. Acetophenone being a methyl ketone responds to this test, but benzophenone does not.

https://img-nm.mnimgs.com/img/study_content/content_ck_images/images/Q12_13.png

(iii) Phenol and benzoic acid can be distinguished by ferric chloride test.

Ferric chloride test:

Phenol reacts with neutral FeCl3 to form an iron-phenol complex giving violet colouration.

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/271/5824/NCERT_15-12-08_Uptal_Chemistry_12_12_20_html_m775702a5.jpg

But benzoic acid reacts with neutral FeClto give a buff coloured ppt. of ferric benzoate.

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/271/5824/NCERT_15-12-08_Uptal_Chemistry_12_12_20_html_ma02b7c.jpg

(iv) Benzoic acid and Ethyl benzoate can be distinguished by sodium bicarbonate test.

Sodium bicarbonate test:

Acids react with NaHCO3 to produce brisk effervescence due to the evolution of CO2 gas.

Benzoic acid being an acid responds to this test, but ethylbenzoate does not.