Bihar Board - Class 12 Chemistry - Chapter 4: Aldehydes, Ketones and Carboxylic Acids Long Answer Question

Question. 1. Explain the following and give an example:
(i) Cyanohydrin
(ii) Acetal
(iii) Semicarbazone
(iv) Aldol
(v) Hemiacetal
(vil) Oxime
(vii) Ketal
(viii) Amine
(ix) 2 , 4-DNP-derivative
(x) Schiff’s base.
Answer⇒ (i) Cyanohydrin : Aldehydes and ketones react with hydrogen cyanide. HCN to form the additional products called cyanohydrins.
(ii) Acetal : It is di alkoxy compound formed by the reaction of aldehyde and two moles of alcohols, e.g.
(iii) Semicarbazone : Aldehydes and ketones react with semicarbazide to form semicarbazones.
(iv) Aldol : The compounds having alcoholic group and aldehyde or ketone group are called aldols, e.g.,
(v) Hemi-acetal : It is formed by the reaction of aldehyde with one mole of alcohol. It contains both alcohol and alkoxy group e.g.,
(vi) Oxime : When aldehydes and ketones react with NH2OH (hydroxylamine) oxides are formed.
.C=O+H₂N C-NOH+H₂O
.                                                        oxime
(vii) Ketal : Ketones react with dihydroxy compounds to form cyeli ketals.
(viii) Amine : It is an organic derivative of ammonia in which one or more hydrogen atoms are replaced with alkyl or aryl groups.
NH₃           HR-NH₂      R-NH-R  R- -R
Ammonia primary          Secondary    Tertiary
.amine        amine               amine            amine
(ix) 2 , 4-DNP derivative : Its structure is It is short name 2 , 4-Dinitrophenyl hydrazine. 2 , 4- DNP derivatives are yellow, orange or red solids, useful for characterisation of aldehydes and ketones.
(x) Schiff’s base : They are the class of compounds having general formula C6H5-N=CR2, where ‘R’ is alkyl or aryl group or hydrogen atoms. It is formed by reaction of primary aromatic amine with aldehydes or ketones. It turns aldehyde pink.

Question. 2. Write equations of the following reactions :
(i) Friedel crafts reaction-alkylation of anisole.
(ii) Nitration of anisole.
(iii) Bromination of anisole in ethanoic acid medium.
(iv) Friedel craft’s acetylation of anisole.
Answer ⇒

Question. 3. Explain phenol is more acidic than aliphatic alcohol, Why?
Answer ⇒ The strength of acid depends on the stability of its conjugate base. When phenol releases Ht and forms phenoxide ion (C₆H₅O-) which is a conjugate base of phenol which under resonance stabilizes it and hence phenol is more acidic.
But in case of Aliphatic alcohol (ROH) it also forms alkoxide ion  after loosing H+ but it is not stabilized due to inability to form resonance.

I, II and III is the resonating structure of phenoxide ions showing resonance.
   R -OH                R-O^-+H⁺  
Alkyl Alcohol                        Alkoxide
But in case of Aliphatic alcohol it does not shows resonance and the reaction becomes reversible.

Question . 4. What is Lucas's test of turbidity of 1 alcohol, 2 alcohol and 3 Alcohol ?
Answer⇒ Lucas reagent is anhydrous ZnCl₂+Conc. HCl .
(R₃C-OH) ₃Alcohol (Tertiary Alcohol) becomes turbid immediately by the formation of chloride with lucas reagent.
2 Alcohol (Secondary alcohol R2C-H) becomes Turbid after 5 minute by the formation of chloride with lucas reagent.
1 Alcohol (Primary Alcohol) becomes turbid at last by the formation of chloride with lucas reagent.
The amount of turbidity formation depends on the stability of carbocation which is as follows :
3 Carbocation > 2 Carbocation > 1Carbocation R₃C+ > R₂C+ H > R+ CH₂

Q. 5. How will you obtain aldehydes from primary alcohols ?
Answer ⇒ Preparation of aldehydes from primary alcohols :
(i) By the oxidation of alcohols : An aldehyde is prepared when a primary alcohol is oxidized by acidified potassium dichromate under controlled conditions.
(ii) By catalytic dehydrogenation of alcohol : By passing vapors of a primary alcohol over hot reduced copper at 573 K, we get an aldehyde.

Question. 6. How will you prepare ketones from alcohol ?
Answer⇒ Preparation of ketones from alcohols : Ketones can be prepared from secondary alcohols either ‘ by oxidation or by catalytic dehydrogenation.
(i) Oxidation of secondary alcohol:
(ii) Catalytic dehydrogenation of secondary alcohol :

Question . 7. Discuss the preparation of aldehydes and ketones from alkenes.
Answer ⇒ (i) By oxidation of alkenes with ozone followed by the reduction of the ozonide by zinc and ethanoic
acid or hydrogen on palladium, aldehydes and ketones are formed.
(ii) If out of the two doubly-bonded carbon atoms in alkene only one carbon contains a hydrogen atom, a mixture of aldehyde and ketone will be obtained.

Question. 8. Predict the product formed when cyclohexane carbaldehyde reacts with following agents :
(i) PhMgBr then H₃O
(ii) Tollen’s reagent
(iii) Semicarbazide and weak acid
(iv) Excess ethanol and acid
(v) Zinc amalgam and dilute hydrochloric acid.
Answer:  

Question. 9. Describe the following:
(i) Acetylation
(ii) Cannizzaro reaction
(iii) Cross aldol condensation
(iv) Decarboxylation.
Answer ⇒ (i) Acetylation : Acetylation is the replacement of H atom of alcohols, amines and other such compounds with acetyl group, e.g.,
C₂H₅OH + CICOCH₃  CH₃COOC₂H₅ + HCl

(ii) Cannizzaro’s reaction : In this reaction two molecules of aldehydes (without the a-hydrogen atom) react in presence of concentrated alkali by mutual oxidation-reaction forming one molecule of each of the corresponding alcohol and acid. Thus
Methanol when heated with alkali gives methanol and methanoic acid.

(iii) Cross-aldol condensation : Aldol condensation of a mixture of two different aldehydes or/and ketones each containing a -hydrogen gives a mixture of four products. Two products are made up of two molecules of same carbonyl compounds and are the same as in case of simple aldol condensation. The other two products arise from the reaction between one molecule each of two different carbonyl compounds. These are called cross aldol condensation products and the reaction leading to formation of these products is known as cross-aldol condensation. Example :
(simple aldol condensation products)
(cross aldol condensation products)
(iv) Decarboxylation : Carboxylic acids lose carbon dioxide, when their sodium salts are heated with soda lime (NaOH +CaO). The reaction is known as Decarboxylation.

Question. 10. (a) Write IUPAC name of the compound
(b) Account for the following:
(i) Aldehydes are more reactive than Ketones towards nucleophilic addition reaction.
(ii) Chloroacetic acid is stronger acid than acetic acid
Answer ⇒
(a)3-oxopentanoic acid
(b) (i) The reactivity of the carbonyl group towards the nucleophilic addition reaction depends upon the magnitude of positive charge on the carbonyl carbon atom. Introduction of alkyl group (+I effect) decreases the reactivity ketone has two alkyl groups while the aldehydes have only one. So aldehydes are more reactive towards nucleophilic addition reactions.
(ii) In presence of electron withdrawing group (-I effect) strength of carboxylic acids increases. Hence chloroacetic acid is stronger than acetic acid.

Question. 11. Why is benzoic acid a stronger acid than phenol ?
Answer⇒ Benzoic acid is a stronger acid than phenol because the benzoate ion is stabilized by two equivalent resonance structures in which the negative charge is present at the more electronegative oxygen atom.
The conjugate base of phenol, a phenoxide ion, has non-equivalent resonance structure in which the negative charge is at the less electronegative carbon atom. Thus, the benzoate ion is more stable than phenoxide ion. Hence benzoic acid is a stronger acid than phenol.


Question. 12. (a) What happens when :
(i) Aldehyde is treated with methyl magnesium bromide and then hydrolysed.
(ii) Acetyl chloride is heated with hydrogen in presence of boiling Xylene Pd-supported by BaSO₄
(iii) Propanone is treated with I2 and NaOH.
(d) Identify the products A and B in the following reactions.
(i) CH₃CHBr [A][B]
(ii) (CH₃COO)₂Ca [A][B]
Answer ⇒ (a) (i) Isopropyl alcohol (2) is obtained
(ii) Acetaldehyde is obtained
(iii) Iodoform is obtained
(b) (i) A = CH₃CH₃OH Ethyl alcohol
B= CH₃COOH Acetic acid
(ii) A = CH₃COH₃ Acetone
B = CH₃CH₂CH₃ Propane

Question 13. Why are ketones less reactive than aldehydes?
Benzaldehyde is less reactive than Acetaldehyde. Why?
Answer:
1. Ketones are less reactive than aldehydes because in ketones there are two alky group attached with carbonyl group, due to the positive inductive effect (+I) of both the alkyl group the positive charge on the carbon atom decreases. Hence, the sensitivity of ketones to the nucleophilic reagents decreases. In aldehydes, they have only one alkyl group so they are more reactive than ketones.
2. -CHO  The group of benzaldehyde becomes stable due to resonance with the benzene ring whereas resonance is not found in acetaldehyde. Benzaldehyde is aromatic and aldehyde is aliphatic.

Question 14. Among formaldehyde, acetaldehyde and acetone which is more reactive and why? Explain. Answer: Among HCHO , CH₃CHO and CH₃COCH₃ , HCHO is more reactive. This can be explained on the basis of: 1. Electron releasing effect: Alkyl groups are electron releasing in nature due to which the magnitude of positive charge on carbonyl carbon decreases and hence it becomes less susceptible to nucleophilic attack. 2. Steric effect: The bulkier groups in ketones hinder the approach of the nucleophile to the carbonyl carbon. This is known as the steric effect. Thus, HCHO with negligible electron releasing effect as well as steric effect is more reactive.

Question 15. Write down the difference between compounds containing aldehydic group and the ketonic group. Answer: 

Question 16. Describe the following:  Acetylation Cannizzaro reaction Cross – aldol condensation Decarboxylation.

Answer :  1. Acetylation: The introduction of an acetyl functional group into an organic compound is known as acetylation. It is usually carried out in the presence of a base such as pyridine, dimethylaniline, etc. This process involves the substitution of an acetyl group for an active hydrogen atom. Acetyl chloride and acetic anhydride are commonly used as acetyl – lating agents. For example, acetylation of ethanol produces ethyl acetate. CH₂CH₂OH +CH₃COCl  CH₃COOC₂H₅ + HCl

2. Cannizzaro reaction: Aldehydes which do not contain -hydrogen like HCHO , C₆H₅CHO react with a cone. NaOH solution to form methyl alcohol and formic acid. This reaction is called Cannizzaro reaction. 2HCHO HCOOH + CH₃OH                                Formic acid    Methyl alcohol C₆H₅CH0 +C₆H₅CHC₆H₅CH₂OH +C₆H5COO-Na+

3. Cross – aldol condensation: When aldol condensation is carried out between different aldehydes or two different ketones or an aldehyde and a ketone, then the reaction is called a Cross – aldol condensation. If both the reactants contain - hydrogens, four compounds are obtained as products.

4. Decarboxylation: Decarboxylation refers to the reaction in which carboxylic acids lose carbon dioxide to form hydrocarbons when their sodium salts are heated with soda – lime. R-COONa + NaOH RH + Na₂CO₃ CH3COONa + NaOH CH₄+Na₂CO₃ Sodium Acetate Methane Decarboxylation also takes place when aqueous solutions of alkali metal salts of carboxylic acids are electrolysed. This electrolytic process is known as Kolbe’s electrolysis.

Question 17. Describe the laboratory method of preparation of acetone. Draw labelled diagrams and write down chemical equations.

Answer :  In the laboratory, acetone is prepared by dry distillation of anhydrous calcium acetate. (CH₃COO)₂CaCH₃COCH₃+CaCO₃ Method 30-40 gm calcium acetate mixed with an equal amount of sodium acetate is heated in a glass retort fitted with a condenser and receiver. Acetone is collected in the receiver. The acetone obtained is not pure. To purify this, it is shaken with a saturated solution of sodium bisulphite then crystals of acetone sodium bisulfite salt separate out. The crystal is washed and heated with sodium carbonate and then dried over anhydrous and CaCl2 then distilled at 56C to get pure acetone. (CH₃)₂C=O+NaHSO3 (CH3)2C(OH)SO₃Na₂(CH₃)₂C(OH)SO₃Na+ Na₂CO_3 2(CH₃)₂C=O +₂Na₂SO₃+H₂O +CO₂

Question 18. Write down the following reaction giving example and equation :

Iodoform reaction
Tischenko reaction Gattermann –
Koch synthese Rosenmund’s reaction.

Answer :  1. Iodoform (Haloform) reaction: Acetaldehyde or methyl ketone reacts with iodine in presence of alkali to form yellow coloured iodoform. This reaction is known as the Iodoform test. 2NaOH +I₂  NaI+NaOI +H₂O  R-CO-CH₃+3NaOI R-COONa +CHI₃+2NaOH                                                                   Iodoform

2. Tishchenko reaction: Two molecules of benzaldehyde are coupled together in presence of aluminum ethoxide or isopropoxide then benzyl benzoate (ester) is formed. 2C₆H₅CHO C₆H₅COOCH₂C₆H₅

3. Gattermann – Koch synthese: Mixture of CO and HCl bubbles through a solution of aromatic hydrocarbon in either solution in the presence of anhydrous AlCl3, then benzaldehyde is formed. C₆H₆+CO+HClC₆H₅CHO+HCl                                                        Benzaldehyde

4. Rosenmund’s reaction: Aldehydes are obtained by the reduction of acid chloride with hydrogen in boiling xylene in presence of a catalyst Pd suspended in BaSO4. RCOCl +H2 RCHO +HCl CH3COCl +H2 CH3CHO +HCl C6H5COCl +H2C6H5CHO + HCl This reaction is called the Rosenmund reaction.

Question 19. Give a quick vinegar method of preparation of acetic acid. Give its reaction with phosphorus pentoxide and phosphorus pentachloride and write its two uses.

Answer :  In this process, a dilute aqueous solution of ethyl alcohol is oxidized in presence of enzyme Mycoderma aceti.

CH3CH2OHCH3COOH + H2O

In this process, a wooden vat is fitted with two wooden plates having holes. Between these plates is filled by beech wood savings, moistened with old vinegar solution which is the chief source of Mycoderma aceti. A 10 % Aqueous solution of ethyl alcohol is dropped slowly from the top of the vat and air is passed at a controlled rate through the holes near the bottom of the vat. Ethyl alcohol is oxidized to acetic acid. This process is called the quick vinegar process because vinegar is formed very quickly.

(a) Reaction with P2O5: 2CH3COOH(CH3CO)2O+H2O (b) Reaction with PCl5: CH3CHOOH+PCl5CH3COCl +POCl3+HCl Uses: As a reagent and solvent in the lab. As vinegar in the preparation of pickles, chutney etc. In the preparation of methyl acetate, ethyl acetate and other esters.

Question 20. Write a brief note on : Claisen condensation Benzoin condensation.

Answer :  1. Claisen condensation: When aromatic aldehyde reacts with aliphatic aldehyde or ketone with - hydrogen, in presence of weak base, ,- unsaturated aldehyde or ketone is formed. This type of condensation is called Claisen condensation. C6H5CHO +H2CHCHO C6H5CH=CH CH O+H2O                                                         Cinnamaldehyde C6H5CHO +H2CHCOCH3 C6H5CH=CH COCH3+H2O                                                   Benzylidene acetone

2. Benzoin condensation: Two molecules of benzaldehyde in presence of alcoholic KCN or NaCN condenses to form benzoin.

Question 21. How will you convert ethanal into the following compounds :  (i) Butan -1 ,3- diol (ii) But -2- enal (iii) But -2-  enoic acid.

Answer : 

Question 22.What happens when, (only equation):

On reacting acetone with Grignard reagent?
Reaction of acetone with chloroform in presence of KOH?
Benzaldehyde reacts with aniline? On heating sodium salt of carboxylic acid with soda lime?
Benzene reacts with acetyl chloride in presence of anhydrous AlCl3?

Answer :  1. Reaction of acetone with Grignard reagent.
2. Reaction of acetone with chloroform.
3. Reaction of benzaldehyde with aniline
4. Reaction of sodium salt of carboxylic acid with soda lime. 5. Reaction of benzene with CH3COCl.

Question 23. How will you obtain the following from acetic acid (Give only equations):

Acetamide
Ethyl acetate
Acetic Anhydride
Trichloro acetic acid.

Answer :  1. Acetamide: CH3COOH+NH3CH3COONH4CH3CONH2
2. Ethyl acetate: CH3COOH+C2H5OHCH3COOC2H5+H2O
3. Acetic anhydride:
4. Trichloro acetic acid: CH3-COOH+3Cl2  CCl3-COOH +3HCl Question.

24. Give equation of the following reactions :
(i) Oxidation of propane-1-ol with alkaline KMnO4 solution.
(ii) Bromine in CS2 with phenol.
(iii) Dilute HNO3 with phenol.
(iv) Treating phenol with chloroform in presence of aqueous NaOH.

Answer :  (i) CH3CH2CH2OH  CH3CH3CH2COOH 1-propanol     Propanoic acid
(ii) A mixture of o-bromo phenol and p-bromo is formed.
(iii) A mixture of o-nitrophenol and  p-nitrophenol is formed.
(iv) Salicylaldehyde formed.

Question.25. Which acid of each pair shown here would you expect to be stronger?
(i) CH3CO2H or FCH2COH 
(ii) FCH2CO2H or ClCH2CO2H
(iii) FCH2CH2CH2CO2H or CH3CHFCH2CO2H

Answer: CH3 group with +I effect increases the electron density on O-atom of O-Hbond in the carboxyl group and cleavage of bond becomes difficult. Therefore, it decreases the acid strength. F has very strong -I effect i.e., electron withdrawing effect. It decreases the electron density on the O Atom and cleavage of the bond becomes easy. Therefore acid strength increases. It is further related to the (i) number of Atoms present in the molecule and (ii) relative position of the Fatom in the carbon atom chain. In the light of given discussion: 

(i) FCH2COH is a stronger acid. 
(ii) FCH2CO2H is a stronger acid. 
(iii) CH3CHFCH2CO2H is a stronger acid. 

(iv)  is a stronger acid.

हिंदी के सभी अध्याय के महत्वपूर्ण प्रशन उत्तर के लिए अभी Download करें Vidyakul App - Free Download Click Here