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Chemistry Case study (4 Marks)

Aldehydes, Ketones, and Carboxylic Acids · Bihar Board (BSEB) STD 12 Science (BSEB - English Medium). 10 questions.

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Case study (4 Marks)

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Question 14 Marks
Read the passage given below and answer the following questions:

Fehling's reagent: Fehling's reagent is a mixture of two solutions. Fehllng's solution A is aqueous copper sulphate solution. Fehling's solution Bis alkaline sodium potassium tartarate (Rochelle salt).

$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{CH(OH)COONa}\\\text{CuSo}_{4\text{(aq)}}+|\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{CH(OH)COOK}$

It is a mild oxidising agent. It is weaker than Tollens' reagent. It oxidises only aliphatic aldehydes to carboxylate ions and itself gets reduced to reddish brown precipitate of cuprous oxide. Aromatic aldehydes do not respond to Fehling's test. This reaction is used for the test of aliphatic aldehydes known as Fehling's reagent test.

In these questions (Q. No. l-iv), a statement of assertion followed by a statement ofreason is given. Choose the correct answer out of the following choices.

  1. Assertion and reason both are correct statements and reason is correct explanation for assertion.
  2. Assertion and reason both are correct statements but reason is not correct explanation for assertion.
  3. Assertion is correct statement but reason is wrong statement.
  4. Assertion is wrong statement but reason is correct statement.
  1. Assertion: Fehling's solution can be used to distinguish between acetaldehyde and acetone.

Reason: Fehling's reagent is a mixture of two solutions.

  1. Assertion: Aromatic aldehydes can be distinguished from aliphatic aldehydes by Fehling's solution.

Reason: Aromatic aldehydes reduce Fehling's solution, but aliphatic aldehydes do not.

  1. Assertion: Fehling's solution oxidises acetaldehyde to acetic acid but not benzaldehyde to benzoic acid.

Reason: The C-H bond of -CHO group in benzaldehyde is stronger than in acetaldehyde.

  1. Assertion: CH3CHO and C6H5CH2CHO cannot be distinguished chemically by Fehling's solution.

Reason: CH3CHO and C6H5CH2CHO cannot be distinguished chemically by Fehling's solution.

  1. Assertion: Formaldehyde, when heated with Fehling's reagent produces a reddish brown ppt, of Cu.

Reason: Fehling's reagent oxidises fonnaldehyde to formate ion.

Answer
  1. (b) Assertion and reason both are correct statements but reason is not correct explanation for assertion.

Explanation:

All aliphatic aldehydes give red ppt, with Fehling's solution, but ketones do not reduce Pehling's solution.

  1. (c) Assertion is correct statement but reason is wrong statement.

Explanation:

Aliphatic aldehydes reduce Fehling's solution, but aromatic aldehydes do not.

  1. (a) Assertion and reason both are correct statements and reason is correct explanation for assertion.

Explanation:

Fehling's solution is a mild oxidising agent. It cannot oxidise aromatic aldehydes to corresponding carboxylate ion.

  1. (b) Assertion and reason both are correct statements but reason is not correct explanation for assertion.

Explanation:

CH3CHO and C6H5CH2CHO both are aliphatic aldehydes, hence cannot be distinguished by Fehling's solution. CH3CHO contains CH3CO-group whereas C6H5CH2CHO does not contain any CH3CO-group. Thus, CH3CHO will give yellow ppt. with 12 and NaOH but C6H5CH2CHO will not.

  1. (d) Assertion is wrong statement but reason is correct statement.

Explanation:

Formaldehyde when heated with Fehllng's reagent, undergo oxidation to give formate ion and produce reddish brown ppt. of Cu20.

$\text{HCHO}+2\text{Cu}^{2+}+5\text{OH}^-\rightarrow\text{HCOO}^-+\text{Cu}_2\text{O}+3\text{H}_2\text{O}\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{Reddish brown ppt.}$

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Question 24 Marks
Read the passage given below and answer the following questions:

Aldehydes and ketones undergo nucleophilic addition reactions.

Carbonyl carbon is electron deficient hence acts as an electrophi le. Nucleophile attacks on the electrophili c carbon atom of the carbonyl group from a direction perpendicular to the plane of the molecule.

ln this process, hybridisation of carbon atom changes from sp2 to sp3 and a tetrahedral alkoxide ion is formed as intermediate. This intermediate captures proton from the reaction medium to give the neutral product. Aldehydes are generally more reactive than ketones in nucleophilic addition reactions.

In these questions (Q. No. l-iv), a statement of assertion followed by a statement ofreason is given. Choose the correct answer out of the following choices.

  1. Assertion and reason both are correct statements and reason is correct explanation for assertion.
  2. Assertion and reason both are correct statements but reason is not correct explanation for assertion.
  3. Assertion is correct statement but reason is wrong statement.
  4. Assertion is wrong statement but reason is correct statement.
  1. Assertion: Benzaldehyde is more reactive than ethanal towards nucleophitic attack.

Reason: The overall effect of -I and +R effect of phenyl group decreases the electron density on the carbon atom of >c=o group in benzaldehyde.

  1. Assertion: (CH3)3CCOC(CH3)3 and acetone can be distinguished by the reaction with NaHSO3.

Reason: HSO3 is the nucleophile in bisulphite addition.

  1. Assertion: Ease of nucleophilic addition of the compounds (I). CH3CHO(II) and CH3COCH3(III) is I > II > III.

Reason: Aldehydes and ketones undergo nucleophilic addition reactions.

  1. Assertion: The formation of cyanohydrin from an aldehyde or ketone occurs very slowly with pure HCN. The is reaction is catalysed by a base.

Reason: Base generates CN- ion which is a stronger nucleophile.

  1. Assertion:  is more reactive towards nucleophilic addition reaction than .

Reason: Reactivity of carbonyl group is due to electrophilic nature of carbonyl carbon.

Answer
  1. (a) Assertion and reason both are correct statements and reason is correct explanation for assertion.
  2. (b) Assertion and reason both are correct statements but reason is not correct explanation for assertion.

Explanation:

$\text{HSO}_3^-$ is a bulky nucleophile, hence, cannot attack on sterically hindered ketones.

  1. (d) Assertion is wrong statement but reason is correct statement.

Explanation:

Aromatic aldehydes and ketones are less reactive than the corresponding aliphatic analogues towards nucleophilic addition reactions due to the +R effect of benzene ring. Further, aldehydes are more reactive than ketones due to +I effect and steric effect of alkyl group.

Therefore, the ease of nucleophili c addition will follow the order:

  1. (a) Assertion and reason both are correct statements and reason is correct explanation for assertion.

Explanation:

Formation of cyanohydrin from an aldehyde or ketone occurs very slowly with pure HCN because it is feebly ionised. This reaction is catalysed by a base. Base generates CN- ion which is a stronger nuclephile and readily adds to carbonyl compound.

  1. (b) Assertion and reason both are correct statements but reason is not correct explanation for assertion.

Explanation:

Electron withdrawing group (-NO2) increases the reactivity towards nucleophilic addition reactions, whereas electron donating group (-CH3) decreases the reactivity towards nucleophilic addition reactions.

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Question 34 Marks

Read the passage given below and answer the following questions:

(A), (B) and (C) are three non-cyclic functional isomers of a carbonyl compound with molecular formula C4H8O. Isomers (A) and (C) give positive Tollen's test whereas isomer (B) does not give Tollen's test but gives positive iodoform test. Isomers (A) and (B) on reduction with $\frac{\text{Zn(Hg)}}{\text{conc.}}.$ HCl give the same product (D).

The following questions are multiple choice questions. Choose the most appropriate answer:

  1. Compound A is:

  1. $\text{CH}_3-\text{CH}-\text{CHO}\\\ \ \ \ \ \ \ \ \ \ \ \ \ |\\ \ \ \ \ \ \ \ \ \ \ \ \ \text{CH}_3$

  2. $\text{CH}_3\text{CH}_2\text{CH}_2\text{CHO}$

  3. $\ \ \ \ \ \ \ \ \ \ \ \ \text{O}\\\ \ \ \ \ \ \ \ \ \ \ \ \ ||\\\text{CH}_3-\text{C}-\text{CH}_2-\text{CH}_3$

  4. None of these.
  1. Compound (C) is:
  1. Iso-butyraldehyde
  2. Butyraldehyde
  3. Crotonaldehyde
  4. Acrolein
  1. Compound (B) can be obtained by:

  1. $\text{CH}_3-\text{C}\equiv\text{C}-\text{CH}_2-\text{CH}_3\xrightarrow[333\text{K}]{\text{dil.H}_2\text{SO}_4+\text{HgSO}_4}$

  2. $(\text{CH}_3\text{CH}_2\text{COO})_2\text{Ca}\xrightarrow{\text{Dry distill}}$

  3. $\text{CH}_3-\text{C}\equiv\text{C}-\text{CH}_3\xrightarrow[\frac{\text{H}_2\text{O}_2}{\text{NaOH}}]{\frac{\text{B}_2\text{H}_6}{\text{THF}}}$

  4. $\text{CH}_3-\text{CH}=\text{CH}-\text{CH}_3\xrightarrow[\frac{\text{ZN}}{\text{H}_2\text{O}}]{\text{O}_3}$

  1. Out of (A), (B) and ( C) isomers, which one is least reactive towards addition of HCN?
  1. A
  2. B
  3. C
  4. All are equally reactive.
  1. What will be the product when (B) reacts with ethylene glycol in presence of HCl gas?
  1.  

  1.  

  1.  

  1. None of these.
Answer
  1. (b) $\text{CH}_3\text{CH}_2\text{CH}_2\text{CHO}$

Explanation:

As (A) and (C) gives 'positive Tollens' test thus these two should be aldehydes while (B) should be a ketone (does not give Tollen's test) with $-\text{C}-\text{CH}_3\\\ \ \ ||\\\ \ \ \text{O}$ group (as it gives positive iodofonn test).

Three isomers are,

  1. (d) Acrolein
  2. (c) $\text{CH}_3-\text{C}\equiv\text{C}-\text{CH}_3\xrightarrow[\frac{\text{H}_2\text{O}_2}{\text{NaOH}}]{\frac{\text{B}_2\text{H}_6}{\text{THF}}}$

Explanation:

  1. (b) B

Explanation:

(B) is least reactive among the three isomers towards addition of HCN. Aldehydes are more reactive than ketones towards n ucleophilic addition reactions.

  1. (a) 

Explanation:

When butanone reacts with ethylene glycol in presence of HCl, it forms a ketal.

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Question 44 Marks

Read the passage given below and answer the following questions:

The addition reaction of enol or enolate to the carbonyl functional group of aldehyde or ketone is known as aldol addition. The $\beta-$ hydroxyaldehyde or $\beta-$hydroxyketone so obtained undergo dehydration in second step to produce a conjugated enone. The first part of reaction is an addition reaction and the second part is an elimination reaction. Carbonyl compound having $\propto-$hydrogen undergoes aldol condensation reaction.

The following questions are multiple choice questions. Choose the most appropriate answer:

  1. Condensation reaction is the reverse of which of the following reaction?
  1. Lock and key hypothesis.
  2. Oxidation.
  3. Hydrolysis.
  4. Glycogen formation.
  1. Which of the following compounds would be the main product of an aldol condensation of acetaldehyde and acetone?
  1. CH3CH = CHCHO
  2. CH3CH = CHCOCH3
  3. (CH3)2C = CHCHO
  4. (CH3)2C = CHCOCH3
  1. Which combination of carbonyl compounds gives phenyl vinyl ketone by an aldol condensation?

  1. Acetophenone and Formaldehyde.
  2. Acetophenone and acetaldehyde.
  3. Benzaldehyde and acetaldehyde.
  4. Benzaldehyde and acetone.
  1. Which of the following will undergo aldol condensation?
  1. HCHO
  2. CH3CH2OH
  3. C6H5CHO
  4. CH3CH2CHO
  1. Which of the following does not undergo aldol condensation?
  1. CH3CHO
  2. CH3CH2CHO
  3. CH3COCH3
  4. C3H2CHO
Answer
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Question 54 Marks
Read the passage given below and answer the following questions:

The addition reaction of enol or enolate to the carbonyl functional group of aldehyde or ketone is known as aldol addition. The β-hydroxyaldehyde or ββ-hydroxyketone so obtained undergo dehydration in second step to produce a conjugated enone. The first part of reaction is an addition reaction and the second part is an elimination reaction. Carbonyl compound having αα-hydrogen undergoes aldol condensation reaction.

  1. Condensation reaction is the reverse of which of the following reaction?
  1. Lock and key hypothesis
  2. Oxidation
  3. Hydrolysis
  4. Glycogen formation
  1. Which of the following compounds would be the main product of an aldol condensation of acetaldehyde and acetone?
  1. CH3CH=CHCHO
  2. CH3CH=CHCOCH3
  3. (CH3)2C=CHCHO
  4. (CH3)2C=CHCOCH3
  1. Which combination of carbonyl compounds gives phenyl vinyl ketone by an aldol condensation?

  1.  Acetophenone and Formaldehyde
  2. Acetophenone and acetaldehyde
  3. Benzaldehyde and acetaldehyde
  4. Benzaldehyde and acetone
  5. Which of the following will undergo aldol condensation?
  1. HCHO
  2. CH3CH2OH
  3. C6H5CHO
  4. CH3CH2CHO
Answer
  1. (c) Hydrolysis

Explanation:

Condensation reaction is the reverse of hydrolysis, which splits a chemical entity into two parts through the action of the polar water molecule.

  1. (b) CH3CH=CHCOCH3

Explanation:

  1. (a) Acetophenone and Formaldehyde
  1. (d) CH3CH2CHO
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Question 64 Marks
Read the passage given below and answer the following questions :
When an aldehyde with no a-hydrogen reacts with concentrated aqueous NaOH, half the aldehyde is converted to carboxylic acid salt and other half is converted to an alcohol. In other words, half of the reactant is oxidized
and other half is reduced. This reaction is known as Cannizzaro reaction

The following questions are multiple choice questions. Choose the most appropriate answer:
  1. A mixture of benzaldehyde and formaldehyde on heating with aqueous NaOH solution gives:
  1. Benzyl alcohol and sodium formate.
  2. Sodium benzoate and methyl alcohol.
  3. Sodium benzoate and sodium formate.
  4. Benzyl alcohol and methyl alcohol.
  1. Which of the following compounds will undergo Cannizzaro reaction?
  1. CH3CHO
  2. CH3COCH3
  3. C6H5CHO
  4. C6H5CH2CHO
  1. Trichloroacetaldehyde is subjected to Cannizzaro's reaction by using NaOH. The mixture of the products contains sodium trichloroacetate ion and another compound. The other compounds is:
  1. 2, 2, 2-trichloroethanol
  2. Trichloromethanol
  3. 2, 2, 2-trichloropropanol
  4. Chloroform
  1. Which of the following reaction will not result in the formation of carbon-carbon bonds?
  1. Cannizzaro reaction
  2. Wurtz reaction
  3. Reimer- Tiemann reaction
  4. Friedel - Crafts acylation
Answer
  1. (a) Benzyl alcohol and sodium formate.

Explanation:

It is an example of cross Cannizzaro reaction where aromatic aldehyde gets reduced to alcohol and aliphatic aldehyde gets oxidised to its sodium salt (both aldehydes must not contain any αα-hydrogen).

CBSE 12th Chemistry Aldehydes , Ketones and Carboxylic Acids Case Study Questions With Solution 2021

  1. (c) C6H5CHO
  1. (a) 2, 2, 2-trichloroethanol

Explanation:

The Cannizzaro product of given reaction yields 2, 2, 2-trichloroethanol.

CBSE 12th Chemistry Aldehydes , Ketones and Carboxylic Acids Case Study Questions With Solution 2021

  1. (a) Cannizzaro reaction

Explanation:

C - C bond is not formed in Cannizzaro reaction while other reactions result in the formation of C - C bond.

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Question 74 Marks

Read the passage given below and answer the following questions:

Aldehydes and ketones are reduced to primary and secondary alcohols respectively by NaBH4 or LiAlH4 as well as catalytic hydrogenation. The carbonyl group of aldehydes and ketones is reduced to  group on treatment with Zn-Hg and cone. HCl (Clemmensen reduction) or with hydrazine followed by NaOH or KOH in highly boiling solvent such as ethylene glycol (Wolff-Kishner reduction).Aldehydes differ from ketones in their oxidation reactions. Aldehydes are easily oxidised to carboxylic acids on treatment with HNO3, KMnO4, K2Cr2O7 etc. Even mild oxidising agents mainlyTollens' reagent and Fehling's solution also oxidise aldehydes. Ketones are generally oxidised under vigorous conditions i.e., strong oxidising agents and at elevated temperatures, to give mixture of carboxylic acids having lesser number of C-atoms than the parent ketone.

The following questions are multiple choice questions. Choose the most appropriate answer:

  1. Which of the following cannot be made by reduction of ketone or aldehyde with NaBH4 in methanol?
  1. 1-Butanol
  2. 2-Butanol
  3. 2-Methyl-1-propanol
  4. 2-Methyl-2-propanol
  1. The carbonyl compound producing an optically active product by reaction with LiAlH4 is:
  1. Propanone
  2. Butanone
  3. 3-pentanone
  4. Benzophenone
  1. A substance C4H10O (X) yields on oxidation a compound C4H8O which gives an oxime and a positive iodoform test. The substance X on treatment with cone. H2SO4 gives C4H8. The structure of the compound (X) is:
  1. CH3CH2CH2CH2OH
  2. CH3CH(OH)CH2CH3
  3. (CH3)3COH
  4. CH3CH2 - O - CH2CH3
  1. In the oxidation of  by acidified K2Cr2O7, the products are:

  1. $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{O}\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ||\\\text{CH}_3-\ ^\text{14}\text{C}-\text{OH}$ and $\text{CH}_3\text{CH}_2\text{COOH}$

  2. $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{O}\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ||\\\text{CH}_3(\text{CH}_2)_2\text{COOH}-\text{C}-\text{OH}$ and $ \ \ \ \ \ \ \ 14\\\text{CH}_3\text{CH}_2\text{COOH}$

  3. $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 14\\\text{CH}_3\text{CH}_2\text{CH}_2\text{COOH}+\text{HCOOH}$

  4. None of these.

  1. The appropriate reagent for the following transformation is:

  1. $\text{Na}_2\text{NH}_2,^-\text{OH}$

  2. $\text{NaBH}_4$

  3. $\frac{\text{H}_2}{\text{Ni}}$

  4. $\text{AICl}_3$

Answer
  1. (d) 2-Methyl-2-propanol

Explanation:

Methyl-2-propanol is $\ \ \ \ \ \ \ \ \ \ \ \ \text{OH}\\\ \ \ \ \ \ \ \ \ \ \ \ \ |\\\text{CH}_3-\text{C}-\text{CH}_3.\\\ \ \ \ \ \ \ \ \ \ \ \ \ |\\\ \ \ \ \ \ \ \ \ \ \ \ \text{CH}_3$ It cannot be obtained by reduction of an aldehyde or ketone with NaBH4.

  1. (b) Butanone

Explanation:

  1. (b) CH3CH(OH)CH2CH3

Explanation:

$\ \ \ \ \ \ \ \ \text{OH}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{O}\\ \ \ \ \ \ \ \ \ \ \text{|}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ||\\\text{CH}_3\text{CH}-\text{CH}_2\text{CH}_3\xrightarrow{\text{Oxidation}}\text{CH}_3\text{CCH}_2\text{CH}_3\\\text{2-Butanol(C}_4\text{H}_{10}\text{O})\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{2-Butanone}$

2-Butanone forms oxime on reaction with hydroxylamine (NH2OH) and also gives positive i odofonn test.

$\ \ \ \ \ \ \ \ \text{OH}\\ \ \ \ \ \ \ \ \ \ \text{|}\\\text{CH}_3\text{CH}\text{CH}_2\text{CH}_3\xrightarrow{\text{H}_2\text{SO}_4(\text{conc.})}\text{CH}_3\text{CH}=\text{CH}\text{CH}_3+\text{H}_2\text{O}\\\ \ \ \ (\text{C}_4\text{H}_{10}\text{O})\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{2-Butene}(\text{C}_4\text{H}_8)$

  1. (a) $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{O}\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ||\\\text{CH}_3-\ ^\text{14}\text{C}-\text{OH}$ and $\text{CH}_3\text{CH}_2\text{COOH}$

Explanation:

  1. (a) $\text{Na}_2\text{NH}_2,^-\text{OH}$

Explanation:

This reaction is Wolff-Kishner reduction. The reagents used for this reduction are $\frac{\text{NH}_2\text{NH}_2}{\text{KOH}}.$

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Question 84 Marks

Read the passage given below and answer the following questions:

When an aldehyde with no et-hydrogen reacts with concentrated aqueous NaOH, half the aldehyde is converted to carboxylic acid salt and other half is converted to an alcohol. In other words, half of the reactant is oxidized and other half is reduced. This reaction is known as Cannizzaro reaction.

The following questions are multiple choice questions. Choose the most appropriate answer:

  1. A mixture of benzaldehyde and formaldehyde on heating with aqueous NaOH solution gives:
  1. Benzyl alcohol and sodium formate.
  2. Sodium benzoate and methyl alcohol.
  3. Sodium benzoate and sodium formate.
  4. Benzyl alcohol and methyl alcohol.
  1. Which of the following compounds will undergo Cannizzaro reaction?
  1. CH3CHO
  2. CH3COCH3
  3. C6H5CHO
  4. C6H5CH2CHO
  1. Trichloroacetaldehyde is subjected to Cannizzaro's reaction by using NaOH. The mixture of the products contains sodium trichloroacetate ion and another compound. The other compounds is:
  1. 2, 2, 2-trichloroethanol.
  2. Trichloromethanol.
  3. 2, 2, 2-trichloropropanol.
  4. Chloroform.
  1. In Cannizzaro reaction given below:

$2\text{PhCHO}\xrightarrow{\stackrel{-}{\hbox{ OH}}}\text{PhCH}_2+\text{OH}+\text{PhCO}_2^-$ the slowest step is:

  1. The attack -OH at the carboxyl group.
  2. The transfer of hydride to the carbonyl group.
  3. The abstraction of proton from the carboxylic group.
  4. The deprotonation of PhCH2OH.
  1. Which of the following reaction will not result in the formation of carbon-carbon bonds?
  1. Cannizzaro reaction.
  2. Wurtz reaction.
  3. Reimer-Tiemann reaction.
  4. Friedel-Crafts' acylation.
Answer
  1. (a) Benzyl alcohol and sodium formate.

Explanation:

It is an example of cross Cannizzaro reaction where aromatic aldehyde gets reduced to alcohol and aliphatic aldehyde gets oxidised to its sodium salt (both aldehydes must not contain any $\propto-$hydrogen).

  1. (c) C6H5CHO
  2. (a) 2, 2, 2-trichloroethanol.

​​​​​​​Explanation:

The Cannizzaro product of given reaction yields 2, 2, 2-trichloroethanol.

  1. (b) The transfer of hydride to the carbonyl group.

​​​​​​​​​​​​​​​​​​​​​Explanation:

Hydride transfer is the slowest step.

  1. (a) Cannizzaro reaction.

​​​​​​​​​​​​​​​​​​​​​​​​​​​​Explanation:

C-C bond is not formed in Cannizzaro reaction while other reactions result in the formation of C-C bond.

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Question 94 Marks
Read the passage given below and answer the following questions:

Carboxylic acids having an $\alpha$-hydrogen atom when treated with chlorine or bromine in the presence of small amount of red phosphorus gives $\alpha$-halocarboxytic acids. The reaction is known as Hell-Volhard-Zelinsky reaction.

$\text{R}-\text{CH}_2-\text{COOH}+\text{X}_2\xrightarrow{\text{red p}}\text{R}-\text{CH}-\text{COOH}\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{X}\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (\text{X = Cl, Br)}$

When sodium salt of carboxylic acid is heated with soda lime it loses carbon dioxide and gives hydrocarbon with less number of C-atoms.

$\text{R}-\text{COOH}\xrightarrow{\text{NaOH}}\text{R}-\text{COONa}\xrightarrow[\Delta]{\text{NaOH}+\text{CaO}}\text{R}-\text{H}+\text{Na}_2\text{CO}_3\\\text{Carboxylic}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{Sod.}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{Alkane}\\\ \ \ \ \ \text{acid}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{carboxylate}$

In these questions (Q. No. l-iv), a statement of assertion followed by a statement ofreason is given. Choose the correct answer out of the following choices.

  1. Assertion and reason both are correct statements and reason is correct explanation for assertion.
  2. Assertion and reason both are correct statements but reason is not correct explanation for assertion.
  3. Assertion is correct statement but reason is wrong statement.
  4. Assertion is wrong statement but reason is correct statement.
  1. Assertion: (CH3)3CCOOH does not give H.V.Z reaction.

Reason: (CH3)3CCOOH does not have $\alpha$-hydrogen atom.

  1. Assertion: H.V.Z. reaction involves the treatment of carboxylic acids having $\alpha$-hydrogens with Cl2 or Br2 in presence of small amount of red phosphorus.

Reason: Phosphorus reacts with halogens to form phosphorus trihalides.

  1. Assertion: Propionic acid with $\frac{\text{Br}_2}{\text{P}}$ yields CH2Br - CHBr - COOH.

Reason: Propionic acid has two $\alpha$-hydrogen atoms.

  1. Assertion: C6H5COCH2COOH undergoes decarboxylation easily than C6H5COCH2COOH.

Reason: C6H5COCH2COOH is $\beta$-keto acid.

  1. Assertion: On heating 3-methylbutanoic acid with soda lime, isobutane is obtained.

Reason: Soda lime is a mixture of NaOH + CaO in the ratio 3 : 1.

Answer
  1. (a) Assertion and reason both are correct statements and reason is correct explanation for assertion.
  2. (c) Assertion is correct statement but reason is wrong statement.

Explanation:

Phosphorus converts a little of the acid into acid chloride which is more reactive than the parent carboxylic acid. Thus, it is the acid chloride, not the acid itself, that undergoes chlorination at the $\alpha$-carbon.

  1. (d) Assertion is wrong statement but reason is correct statement.

Explanation:

Bromination occurs at $\alpha$-positions.

  1. (a) Assertion and reason both are correct statements and reason is correct explanation for assertion.

Explanation:

$\beta$-ketoacids are unstable acids. They readily undergo decarboxylation through a cyclic transition state.

  1. (b) Assertion and reason both are correct statements but reason is not correct explanation for assertion.

Explanation:

$\text{CH}_3-\text{CH}-\text{CH}_2\text{COOH}\xrightarrow{\frac{\text{NaOH}}{\text{CaO}}}\\\ \ \ \ \ \ \ \ \ \ \ \ \ |\\\ \ \ \ \ \ \ \ \ \ \ \ \ \text{CH}_3\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{CH}_3-\text{CH}-\text{CH}_3+\text{Na}_2\text{CO}_3\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{CH}_3$

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Question 104 Marks

Read the passage given below and answer the following questions:

Aldehydes and ketones having acetyl group $\left(\begin{array}{c}\ \ \ \ \ \ \ \ \ \ \ \ \text{O}\\\ \ \ \ \ \ \ \ \ \ \ \ ||\\\ \ \ \text{CH}_3-\text{C}-\end{array}\right)$ are oxidised by sodium hypohalate (NaOX) or halogen and alkali (X2 + OH-) to corresponding sodium salt having one carbon atoms less than the carbonyl compound and give a haloform.

$\ \ \ \ \ \ \ \ \text{O}\\\ \ \ \ \ \ \ \ ||\\\text{R}-\text{C}-\text{CH}_3\xrightarrow[\text{orX}_2+\text{NaOH}]{\text{NaOX}}\\ \ \ \ \ \ \ \ \ \text{O}\\\ \ \ \ \ \ \ \ ||\\\text{R}-\text{C}-\stackrel{-}{\hbox{ O}}\stackrel{+}{\hbox{Na}}+\text{CHX}_3(\text{X = Cl, Br, I})$

Sodium hypoiodite (NaOl) when treated with compounds containing CH3CO-group gives yellow precipitate of iodoform. Haloform reaction does not affect a carbon-carbon double bond present in the compound.

The following questions are multiple choice questions. Choose the most appropriate answer:

  1. Which of the following compounds will give positive iodoform test?
  1. Isopropyl alcohol.
  2. Propionaldehyde.
  3. Ethylphenyl ketone.
  4. Benzyl alcohol.
  1. Which of the following compounds is not formed in iodoform reaction of acetone?
  1. CH3COCH2l
  2. lCH2COCH2l
  3. CH3COCHl2
  4. CH3COCl3
  1. For the given set of reactions,

starting compound A corresponds to:

  1. In the following reaction sequence, the correct structures of E, F and G are:

(* implies 13C labelled carbon)

  1. An organic compound 'A' has the molecular formula C3H6O. It undergoes iodoform test. When saturated with HCl it gives 'B' of molecular formula C9H14O. 'A' and 'B' respectively are:
  1. propanal and mesityl oxide.
  2. Propanone and mesityl oxide.
  3. propanone and 2,6-dimethyl-2,5-hepta-dien-4-one.
  4. propanone and propionaldehyde.
Answer
  1. (a) Isopropyl alcohol.

Explanation:

Iodoform test is given by the organic compounds having $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{O}\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ||\\\ \ \ \text{CH}_3-\text{C}-$ or $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{OH}\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |\\\ \ \ \text{CH}_3-\text{CH}-$ group.

$\text{CH}_3-\text{CH}-\text{OH}\\ \ \ \ \ \ \ \ \ \ \ \ \ \ |\\ \ \ \ \ \ \ \ \ \ \ \ \ \text{CH}_3$: Isopropyl alcohol

CH3CH2CHO: Propionaldehyde

$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{O}\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ||\\\ \ \ \text{C}_2\text{H}_5-\text{C}-\text{C}_6\text{H}_5$: Ethylphenyl ketone

C6H5 - CH2 - OH: Benzyl alcohol

Therefore, isopropyl alcohol will give positive iodofonn test.

  1. (b) lCH2COCH2l

Explanation:

Iodoform reaction of acetone occurs in following steps:

$\ \ \ \ \ \ \ \ \ \ \ \ \text{O}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{O}\\ \ \ \ \ \ \ \ \ \ \ \ \ \ ||\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ||\\\text{CH}_3-\text{C}-\text{CH}_3+\text{NaOl}\rightarrow\text{CH}_3-\text{C}-\text{CH}_2\text{l}+\text{NaOH}$

$\ \ \ \ \ \ \ \ \ \ \ \ \text{O}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{O}\\ \ \ \ \ \ \ \ \ \ \ \ \ \ ||\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ||\\\text{CH}_3-\text{C}-\text{CH}_2\text{l}+\text{NaOl}\rightarrow\text{CH}_3-\text{C}-\text{CHl}_2+\text{NaOH}$

$\ \ \ \ \ \ \ \ \ \ \ \ \text{O}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{O}\\ \ \ \ \ \ \ \ \ \ \ \ \ \ ||\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ||\\\text{CH}_3-\text{C}-\text{CHl}_2+\text{NaOl}\rightarrow\text{CH}_3-\text{C}-\text{Cl}_3+\text{NaOH}$

$\ \ \ \ \ \ \ \ \ \ \ \ \text{O}\\ \ \ \ \ \ \ \ \ \ \ \ \ \ ||\\\text{CH}_3-\text{C}-\text{Cl}_3+\text{NaOH}\rightarrow\text{CH}_3-\text{COONa}+\text{CHl}_3$

  1. (c) 

Explanation:

Given reagents indicate the presence of -COCH3 group in the starting compound A. Further, since the -COOH group introduced in B due to iodofonn reaction is absent in the final product, B should be a $\beta$-keto acid. Hence, A should have structure given in option (c).

  1. (d)

Explanation:

  1. (c) propanone and 2,6-dimethyl-2,5-hepta-dien-4-one.

Explanation:

Since compound A(C3H6O) undergoes iodoform test, it must be CH3COCH3 (propanone). Further, the compound 'B' obtained from 'A' has three times more the number of carbon atoms as in 'A' (propanone), 'B' must be phorone, i.e.,2, 6-dimethyl-2, 5-heptadien-4-one.

$(\text{CH}_3)\text{C=O}+\text{H}_3\text{CCOCH}_3+{\text{O=C(CH}_3)}_2\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ _{\text{A,}\ \text{propanone(3 molecules)}}$

$\xrightarrow{\text{HCl}}(\text{CH}_3)\text{C}=\text{CHCOCH}={\text{C(CH}_3)}_2\\ \ \ \ \ \ \ _{2,6-\text{dimethyl -2,5-heptadien -4-one}}$

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