Answer:(b) $CH_3 CH_2 CO CH_3$

Explanation:

But-1-yne on reaction with water in presence of $ \mathrm{Hg}^{2+}$ ions as a catalyst produces butan-2-one.

Now, consider the incorrect options:

(a) This option suggests the formation of butan-1-one. However, the hydration of but-1-yne in the presence of $ \mathrm{Hg}^{2+}$ ions follows Markovnikov’s rule, leading to the formation of a more stable ketone, which is butan-2-one, not butan-1-one.

(c) This option suggests the formation of butan-1-ol. The hydration of alkynes in the presence of $ \mathrm{Hg}^{2+}$ ions does not produce alcohols directly. Instead, it forms ketones or aldehydes depending on the structure of the alkyne.

(d) This option suggests the formation of butan-2-ol. Similar to option (c), the hydration of alkynes in the presence of $ \mathrm{Hg}^{2+}$ ions does not produce alcohols directly. The reaction mechanism leads to the formation of a ketone, specifically butan-2-one in this case.

Answer:(a) $CH_3 CHO$

Explanation:

Reactivity of carbonyl compounds can be decided by two factors

(i) Steric factor Lesser the steric factor greater will be its reactivity.

(ii) Electronic factor Greater the number of alkyl group lesser will be its electrophilicity.

Hence, $ \mathrm{CH}_{3}-\mathrm{CHO}$ is most reactive towards nucleophilic addition reaction.

Now, consider the incorrect options:

(b) The presence of two alkyl groups (ethyl groups) increases the steric hindrance around the carbonyl carbon, making it less accessible to nucleophiles. Additionally, the electron-donating effect of the alkyl groups reduces the electrophilicity of the carbonyl carbon, making it less reactive towards nucleophilic addition.

(c) The presence of a phenyl group (benzene ring) attached to the carbonyl carbon significantly reduces its reactivity towards nucleophilic addition. The phenyl group is bulky, increasing steric hindrance, and it also has a resonance effect that delocalizes the positive charge on the carbonyl carbon, reducing its electrophilicity.

(d) The presence of a benzene ring and an additional methyl group in acetophenone, leading to increased steric hindrance and decreased reactivity.

Answer:(c) ethanol $<$ phenol $<$ acetic acid $<$ chloroacetic acid

Explanation:

Phenol is more stable than alcohol due to formation of more stable conjugate base after removal of $ \mathrm{H}^{+}$ from phenol.

On the other hand, carboxylic acid is more acidic than phenol due to formation of more stable conjugate base after removal of $ \mathrm{H}^{+}$ as compared to phenol.

Chloroacetic acid is more acidic than acetic acid due to the presence of electron withdrawing chlorine group attached to $\alpha$-carbon of carboxylic acid.

Hence, correct choice is (c).

Now, consider the incorrect options:

(a) Phenol is more acidic than ethanol because the phenoxide ion formed after deprotonation of phenol is resonance-stabilized, whereas the ethoxide ion formed from ethanol is not. Additionally, chloroacetic acid is more acidic than acetic acid due to the electron-withdrawing effect of the chlorine atom, which stabilizes the conjugate base. Therefore, the order given in option (a) is incorrect.

(b) Acetic acid is less acidic than chloroacetic acid because the electron-withdrawing chlorine atom in chloroacetic acid stabilizes the conjugate base more effectively than the hydrogen atom in acetic acid. Therefore, the order given in option (b) is incorrect.

(d) Ethanol is the least acidic among the given compounds because its conjugate base (ethoxide ion) is not stabilized by resonance or electron-withdrawing groups. Phenol is more acidic than ethanol due to resonance stabilization of the phenoxide ion. Acetic acid is more acidic than phenol because the carboxylate ion is stabilized by resonance. Chloroacetic acid is the most acidic due to the electron-withdrawing chlorine atom. Therefore, the order given in option (d) is incorrect.

Answer:(b) phenol and benzoyl chloride in the presence of pyridine

Explanation:

Compound $ \mathrm{Ph}-\mathrm{COO} - \mathrm{Ph}$ can be prepared by the reduction of

This is an example of Schotten-Baumann reaction.

Now, consider the incorrect options:

(a) Phenol and benzoic acid in the presence of $ \mathrm{NaOH}$: This reaction would not produce the desired compound because benzoic acid and phenol in the presence of NaOH would lead to the formation of the phenoxide ion and the benzoate ion, but not the ester $ \mathrm{Ph-O-\stackrel{\substack{\mathrm{O}\\ ||}}{C}-Ph}$. Esterification typically requires an acid catalyst, not a base like NaOH.

(c) Phenol and benzoyl chloride in the presence of $ \mathrm{ZnCl}_{2}$: While benzoyl chloride and phenol are the correct reactants, but $ \mathrm{ZnCl}{2}$ is not a suitable catalyst for this reaction. The Schotten-Baumann reaction specifically requires a base like pyridine to neutralize the HCl produced during the reaction, which $ \mathrm{ZnCl}{2}$ cannot do.

(d) Phenol and benzaldehyde in the presence of palladium: Benzaldehyde is not the correct reactant for forming the ester $ \mathrm{Ph-O-\stackrel{\substack{\mathrm{O}\\ ||}}{C}-Ph}$ . Benzaldehyde would more likely participate in reactions such as the Cannizzaro reaction or aldol condensation, but not in the formation of an ester with phenol.

Answer:(c) Fehling’s solution

Explanation:

The reagent which does not reacted with both, acetone and benzaldehyde is Fehling’s soltuion. Fehling’s soltuion is a mild oxidizing agent. It does not oxidize aromatic aldehydes such as acetaldehyde and ketones such as acetone.

Now, consider the incorrect options:

(a) Sodium hydrogen sulphite: Sodium hydrogen sulphite reacts with both acetone and benzaldehyde. It forms bisulfite addition compounds with carbonyl compounds, including both ketones and aldehydes.

(b) Phenyl hydrazine: Phenyl hydrazine reacts with both acetone and benzaldehyde to form hydrazones. It reacts with the carbonyl group present in both compounds.

(d) Grignard reagent: Grignard reagents react with both acetone and benzaldehyde. They add to the carbonyl group of ketones and aldehydes to form alcohols after hydrolysis.

Answer:(d) $ \mathrm{CH}_{3} \mathrm{CHO}$

Explanation:

Necessary condition for Cannizzaro reaction is absence of $\alpha$-hydrogen atom. So, $ \mathrm{CH}_{3} \mathrm{CHO}$ will not give Cannizzaro reaction while other three compounds have no $\alpha$-hydrogen. Hence, they will give Cannizzaro reaction.

Now, consider the incorrect options:

This compound does not have any $\alpha$-hydrogen atoms, making it suitable for the Cannizzaro reaction.

(b) $ \mathrm{C_6H_5CHO}$: Benzaldehyde also lacks $\alpha$-hydrogen atoms, which makes it capable of undergoing the Cannizzaro reaction.

(c) $ \mathrm{HCHO}$: Formaldehyde does not have an $\alpha$-hydrogen atom, thus it can participate in the Cannizzaro reaction.

Answer:(b)

Explanation:

Benzaldehyde is an aromatic aldehyde having no $\alpha$ hydrogen. So, on reaction with aqueous ${KOH}$ solution it undergo cannizzaro reaction to produce benzyl alcohol and potassium benzoate.

Answer:(d) Prop-1-en-2-ol, tautomerism

Explanation:

Chemical reaction can be shown as

$[A]$ is prop-1-en-2-ol, which undergo tautomerism to form acetone.

Now, consider the incorrect options:

(a) Prop-1-en-2-ol, metamerism: Metamerism is a type of isomerism where compounds have the same molecular formula but different alkyl groups on either side of a functional group. In this reaction, the isomerism involved is tautomerism, not metamerism.

(b) Prop-1-en-1-ol, tautomerism: The structure of ‘A’ is not Prop-1-en-1-ol. The correct structure is Prop-1-en-2-ol. Therefore, this option is incorrect because it specifies the wrong structure.

(c) Prop-2-en-2-ol, geometrical isomerism: The structure of ‘A’ is not Prop-2-en-2-ol. Additionally, the type of isomerism involved in the reaction is tautomerism, not geometrical isomerism.

Answer:(b) positional isomers

Explanation:

Chemical reaction can be shown as

Thus, $ \mathrm{CH} _{3} -CH(OH) -CH_3$ and $ \mathrm{CH} _{3}-\mathrm{CH} _{2}-\mathrm{CH} _{2} \mathrm{OH}$ are positional isomers.

Hence, option (b) is correct.

Now, consider the incorrect options:

(a) identical: Compounds A and C are not identical because they have different structures. Compound A is 2-propanol $ \mathrm{CH} _{3} -CH(OH) -CH_3$, while compound C is 1-propanol $ \mathrm{CH} _{3}-\mathrm{CH} _{2}-\mathrm{CH} _{2} \mathrm{OH}$. Identical compounds must have the same molecular structure and connectivity, which is not the case here.

(c) functional isomers: Functional isomers have the same molecular formula but different functional groups. In this case, both compounds A and C are alcohols (they both contain the -OH group), so they are not functional isomers. They differ in the position of the -OH group, making them positional isomers instead.

(d) optical isomers: Optical isomers (enantiomers) are non-superimposable mirror images of each other and typically occur in compounds with chiral centers. Neither compound A (2-propanol) nor compound C (1-propanol) has a chiral center, so they cannot be optical isomers.

Answer:

Explanation:

(c) lodoform test is used to test presence of $-\mathrm{COCH}_{3}$ group which is converted into $-\mathrm{COOH}$ group.

The reaction is shown as

Now, consider the incorrect options:

Tollen’s reagent: Tollen’s reagent is used to test for the presence of aldehydes. It oxidizes aldehydes to carboxylic acids but does not react with ketones or the $-\mathrm{COCH}_{3}$ group(present in the provided reactant), making it unsuitable for this conversion.

Benzoyl peroxide: Benzoyl peroxide is a radical initiator used in polymerization reactions and as an oxidizing agent. It does not specifically react with the $-\mathrm{COCH}_{3}$ group to convert it into a $-\mathrm{COOH}$ group, making it unsuitable for this conversion.

$ \mathrm{Sn}$ and $ \mathrm{NaOH}$ solution: Tin (Sn) and sodium hydroxide (NaOH) are typically used in reduction reactions, such as the reduction of nitro groups to amines. They do not facilitate the conversion of the $-\mathrm{COCH}_{3}$ group to a $-\mathrm{COOH}$ group, making them unsuitable for this conversion.

Answer:(b) Butan-2-ol

Explanation:

(b) Butan-2-ol on oxidation with alkaline $ \mathrm{KMnO}_{4}$ solution produces butanone as follows

Now, consider the incorrect options:

Butan-1-ol: On oxidation with alkaline $KMnO_4$, butan-1-ol will not produce butanone. Instead, it will undergo oxidation to form butanoic acid. This is because primary alcohols are typically oxidized to carboxylic acids under these conditions.

Both (a) and (b): This option is incorrect because, as mentioned, butan-1-ol does not produce butanone upon oxidation with alkaline $KMnO_4$. Only butan-2-ol produces butanone.

None of these: This option is incorrect because butan-2-ol does indeed produce butanone upon oxidation with alkaline $KMnO_4$, as correctly identified in the answer.

Answer:(a) zinc amalgam $+\mathrm{HCl}$

Explanation:

Clemmensen reduction is a chemical reaction that involves the reduction of carbonyl compounds (aldehydes and ketones) to alkanes. This reaction is particularly useful for converting carbonyl groups into methylene groups.

The key reagents in the Clemmensen reduction are zinc amalgam and hydrochloric acid $( HCl )$ . Zinc amalgam is a mixture of zinc and mercury, which serves as the reducing agent.

In the presence of zinc amalgam and HCl , the carbonyl compound undergoes reduction. The carbonyl group $(\mathrm{C}=\mathrm{O})$ is converted into a methylene group $(\mathrm{C}-\mathrm{H}_2)$, effectively removing the oxygen and adding hydrogen.

For example : If we take acetone $(\mathrm{CH}_3 \mathrm{C}(=\mathrm{O}) \mathrm{CH}_3)$ as a carbonyl compound, the Clemmensen reduction will convert it into propane $(\mathrm{CH}_3 \mathrm{CH}2 \mathrm{CH} 3)$ . During this process, water $(\mathrm{H}_2 \mathrm{O})$ is also produced as a byproduct.

Therefore, in the Clemmensen reduction, the carbonyl compound is treated with zinc amalgam and hydrochloric acid $( HCl )$ .

Now, consider the incorrect options:

(b) Sodium amalgam $ + $ $ \mathrm{HCl}$: Sodium amalgam is not used in Clemmensen reduction. The specific reagent required is zinc amalgam, which has different chemical properties and reactivity compared to sodium amalgam.

(c) Zinc amalgam $ + $ nitric acid: Nitric acid is an oxidizing agent and would not facilitate the reduction of a carbonyl group to a $ \mathrm{CH}_{2}$ group. Instead, it would likely oxidize the compound further.

(d) Sodium amalgam $+$ $ \mathrm{HNO}_{3}$ : Similar to option (c), nitric acid is an oxidizing agent and would not be suitable for the reduction process. Additionally, sodium amalgam is not the correct reagent for Clemmensen reduction.

Answer:(b, d)

Explanation:

Necessary condition for aldol condensation is the presence of atleast one $\alpha$ $ \mathrm{H}$-atom. Hence, $ \mathrm{C} _{6} \mathrm{H} _{5} \mathrm{CHO}$ and $\left(\mathrm{CH} _{3}\right) _{3} \mathrm{CCHO}$ do not undergo aldol condensation as the both do not have any $\alpha$-hydrogen atom.

Now, consider the incorrect options:

(a) Compound (a) undergoes aldol condensation because it has at least one $\alpha$-hydrogen atom.

(c) Compound (c) undergoes aldol condensation because it has at least one $\alpha$-hydrogen atom.

Answer: (b, c)

Explanation:

Treatment of compound with $NaOH$ yields sodium phenoxide and sodium by means of nucleophilic substitution reaction as follows:

Now, consider the incorrect options:

Phenol: Phenol is not formed directly in this reaction because the nucleophilic substitution reaction with NaOH results in the formation of sodium phenoxide, not free phenol. Phenol would only be formed if the sodium phenoxide were subsequently acidified.

Benzophenone: Benzophenone is not formed in this reaction because the reaction mechanism involves the cleavage of the ester bond, leading to the formation of sodium phenoxide and sodium benzoate. There is no pathway in this reaction that would result in the formation of benzophenone.

Answer:(b, d)

Explanation:

Clemmensen reduction is used to convert cyclohexanone into cyclohexane and benzophenone into diphenyl methane as follows

Reagent used in Clemmensen reduction is zinc amalgam in hydrochloric acid, i.e., $ \mathrm{Zn}(\mathrm{Hg})$ in $ \mathrm{HCl}$.

Now, consider the incorrect options:

(a) Benzaldehyde into benzyl alcohol: Clemmensen reduction is not suitable for reducing aldehydes to alcohols. It specifically reduces carbonyl groups (C=O) in ketones and aldehydes to methylene groups $(CH_2)$, but it does not convert aldehydes to alcohols.

(c) Benzoyl chloride into benzaldehyde: Clemmensen reduction does not convert acyl chlorides (like benzoyl chloride) to aldehydes. It reduces carbonyl compounds to hydrocarbons, but it does not perform partial reductions to form aldehydes from acyl chlorides.

Answer:(a, c)

Explanation:

Grignard reaction and aldol condensation is used to increase the number of carbon atom in the chain as follows

Grignard reaction

Aldol condensation

While other two reactions Cannizzaro reaction and HVZ reaction don’t lead to increase in number of carbon atom.

Now, consider the incorrect options:

Cannizzaro’s reaction: This reaction involves the disproportionation of non-enolizable aldehydes in the presence of a strong base, resulting in the formation of an alcohol and a carboxylic acid. It does not lead to an increase in the number of carbon atoms in the chain.

HVZ reaction: The Hell-Volhard-Zelinsky (HVZ) reaction is used for the halogenation of carboxylic acids at the alpha position. This reaction does not increase the number of carbon atoms in the chain.

Answer: ( a, b )

Explanation:

(a) Benzophenone can be obtained by Friedel-Craft acylation reaction.

(b) Benzophenone can also be obtained by the reaction between benzoyl chloride and diphenyl cadmium.

Now, consider the incorrect options:

(c) benzoyl chloride + phenyl magnesium chloride: This reaction would produce a tertiary alcohol (triphenylmethanol) rather than benzophenone. The Grignard reagent (phenyl magnesium chloride) reacts with benzoyl chloride to form an intermediate which, upon hydrolysis, yields the tertiary alcohol.

(d) benzene + carbon monoxide + $ZnCl_2$: This reaction is known as the Gattermann-Koch reaction, which typically produces benzaldehyde, not benzophenone. The reaction involves the formylation of benzene to introduce a formyl group (-CHO) rather than a ketone group (C=O) bonded to two phenyl groups.

Answer: (a, b)

Explanation:

Since, carbonyl compound is a planar molecule hence two orientation of molecule regarding attack of nucleophile is possible as follows

Since, the product contains a chiral carbon, therefore, attack of nucleophile can occur either from front side attack or rear side attack giving enantiomeric products. Hence, (a) and (b) are the correct choices.

Now, consider the incorrect options:

(c) is incorrect because it does not represent the intermediate formed during the nucleophilic addition reaction. The intermediate should have a tetrahedral geometry around the carbonyl carbon, but option (c) does not show this structure.

(d) is incorrect because it also does not represent the correct intermediate. The intermediate should have a negatively charged oxygen atom (alkoxide ion) and a tetrahedral carbon center, which is not depicted in option (d).

Answer:

Butan-1-ol contains an $ O-H $ group and hence undergoes intermolecular H -bonding.

Therefore, its b.p. is much higher than that of butanal $\left(\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CHO}\right)$ which does not contain an $ \mathrm{O}-\mathrm{H}$ group and hence does not undergo H -bonding. Instead, it has only weak dipole-dipole interactions.

Answer:

Pentan-2-one is a methyl ketone (i.e., contains the grouping $ \mathrm{CH}_3 \mathrm{CO}$ ) and hence undergoes Iodoform test on treatment with $ \mathrm{I}_2 / \mathrm{NaOH}$ to give yellow ppt. of iodoform. In contrast, pentan-3-one is not a methyl ketone and hence does not give yellow ppt. of $ \mathrm{CHI}_3$ on treatment with $ \mathrm{I}_2 / \mathrm{NaOH}$.

Answer:

The IUPAC names of the compounds are :

(a)

3-phenylprop-2-en-1-al

(b)

Cyclohexanecarbaldehyde

(c)

$\underset{\text { }}{\mathrm{CH} _{3}}-\mathrm{CH} _{2}-\underset{\mathrm{O}}{\underset{||}{C}}-\mathrm{CH} _{2}-\mathrm{CHO}$

3-oxo-pentan-1-al

(d)

$ \mathrm{CH} _{3}-\mathrm{CH}=\mathrm{CH}-\mathrm{CHO}$

But-2-en-1-al

Answer:

(a) 4-nitropropiophenone

(b) 2-hydroxycyclopentanecarbaldehyde

(c) Phenyl acetaldehyde

Answer:

The IUPAC names of the following structures are:

Ethane-1, 2-dial

Benzene-1, 4- dicarbaldehyde

Answer:

It is the commercial method for preparing benzaldehyde. Benzal chloride can be obtained by photochlorination of toluene i.e., chlorination of toluene in presence of sunlight. Then, benzal chloride on heating with boiling water produces benzaldehyde .

Answer:

Benzene, on reaction with benzoyl chloride undergo formation of benzophenone through intermediate benzoylinium cation.

The name of the reaction is Friedel-Craft acylation reaction.

Answer:

Being an unsymmetrical ketone, oxidation occurs on either side of the $ \mathrm{C}=\mathrm{O}$ group giving a mixture of 2methylpropanoic acid, 3-methylbutanoic acid and propan-2-one. Propan-2-one on further oxidation gives a mixture of ethanoic acid and methanoic acid.

Further oxidation of methanoic acid gives $ \mathrm{CO}_2$ and $ \mathrm{H}_2 \mathrm{O}$.

Since $-I$-effect of F is much stronger than that of Cl , therefore, $ \mathrm{FCH}_2 \mathrm{COOH}$ is a stronger acid than $ \mathrm{ClCH}_2 \mathrm{COOH}$. Further, $ \mathrm{C}_6 \mathrm{H}_5$ group has a weak -I -effect though much weaker than those of F and Cl , therefore, $ \mathrm{C}_6 \mathrm{H}_5 \mathrm{CH}_2 \mathrm{COOH}$ is a stronger acid than $ \mathrm{CH}_3 \mathrm{COOH}$ but weaker than $ \mathrm{FCH}_2 \mathrm{COOH}$ and $ \mathrm{ClCH}_2 \mathrm{COOH}$.

Now $ \mathrm{CH}_3 \mathrm{COOH}$ is a much stronger acid than $ \mathrm{CH}_3 \mathrm{CH}_2 \mathrm{OH}$ because $ \mathrm{CH}_3 \mathrm{COO}^{-}$ ion left after the loss of a proton is stabilized by resonance but no such stabilization is possible for $ \mathrm{CH}_3 \mathrm{CH}_2 \mathrm{O}^{-}$ ion (obtained after loss of a proton from $ \mathrm{CH}_3 \mathrm{CH}_2 \mathrm{OH}$ ).

Thus, the overall, acidic strength decreases in the order :

$ \mathrm{FCH}_2 \mathrm{COOH}>\mathrm{ClCH}_2 \mathrm{COOH}>\mathrm{C}_6 \mathrm{H}_5 \mathrm{CH}_2 \mathrm{COOH}>\mathrm{CH}_3 \mathrm{COOH}>\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{OH}$.

Answer:

Propanal and 2-methylpropanal both have $\alpha$ -hydrogens and hence can undergo cross aldol condensation in which each one of them can act either an electrophile or a nucleophile thereby giving two cross aldol products. Thus,

Besides, cross aldol condensation each aldehyde can also undergo self or simple aldol condensation giving two self or simple aldol products.

Answer:

(i) Since compound ’ $A$ ’ is obtained by oxidation of compound ’ $B$ ’ and reduction of compound ’ $B$ ’ with $ \mathrm{LiAlH}_4$ gives back compound ’ $A$ ‘, therefore, both compounds ’ $A$ ’ and ’ $B$ ’ have the same number of carbon atoms.

(ii) Since compound ’ $A$ ’ when heated with compound ’ $B$ ’ in presence of $ \mathrm{H}_2 \mathrm{SO}_4$, producess fruity smell of compound ’ $C$ ‘, therefore, ’ $C$ ’ must be an ester, ’ $A$ ’ must be a carboxylic acid and ’ $B$ ’ must be an alcohol.

$\underset{(B)\text{Alcohol}}{RCH_2OH}\quad \underset{[O]}{\xrightarrow{\text{Alkaline} {KMnO_4}}}\quad \underset{ (A)\text {Carboxylic acid} } {RCOOH}$

$\underset{(A)\text{Carboxylic acid}}{RCOOH}\quad \xrightarrow{ {LiAlH_4}}\quad \underset{ (B) \text {Alcohol} } {RCH_2OH}$

$\underset{\text{Acid}}{ {RCOOH}} \quad + \quad \underset{\text {Alcohol} } {RCH_2OH}\quad \underset{\Delta} {\xleftrightharpoons{H_2SO_4}} \quad \underset{\text {Ester (Fruity smell )} } {RCOOCH_2R\quad} +\quad H_2O $

Thus, ’ $A$ ’ is a carboxylic acid, ’ $B$ ’ is an alcohol and ’ $C$ ’ is an ester.

Answer:

Due to the presence of + ve charge on the nitrogen atom of $ \mathrm{NO}_2$ group, the - I-effect of $-\mathrm{NO}_2$ group is much stronger than that of F. Therefore, $ \mathrm{O}_2 \mathrm{NCH}_2 \mathrm{COOH}$ is a stronger acid than $ \mathrm{FCH}_2 \mathrm{COOH}$.

$ \mathrm{C}_6 \mathrm{H}_5$ group, on the other hand, has a weak -I-effect and hence $ \mathrm{C}_6 \mathrm{H}_5 \mathrm{COOH}$ is a weaker acid than both $ \mathrm{FCH}_2 \mathrm{COOH}$ and $ \mathrm{NO}_2 \mathrm{CH}_2 \mathrm{COOH}$.

Thus, the overall acidic strength decreases in the order :

$ \mathrm{O}_2 \mathrm{NCH}_2 \mathrm{COOH}>\mathrm{FCH}_2 \mathrm{COOH}>\mathrm{C}_6 \mathrm{H}_5 \mathrm{COOH} . $

Answer:

The $>\mathrm{C}=\mathrm{C}<$ undergoes electrophilic addition reactions while $>\mathrm{C}=\mathrm{O}$ shows nucleophilic addition reactions. This difference in behaviour is due to the different shapes of their $\pi$-electron clouds.

The $\pi$ electron cloud of $>\mathrm{C}=\mathrm{C}<$ bond due to similar electronegatives of the two carbon atoms is almost symmetrical and surrounds both the carbon atoms equally.

Consequently, if a reagent is to attack one of the carbon atoms of the $>\mathrm{C}=\mathrm{C}<$ bond, it has to pass through this electron cloud. Since the $\pi$-electron cloud consists of loosely held electrons, therefore, it can readily allow electrophiles to react.

As a result, the typical reactions of $>\mathrm{C}=\mathrm{C}<$ are electrophilic addition reactions.

In contrast, the $\pi$-electron cloud of the $>\mathrm{C}=\mathrm{O}$ is unsymmetrical i.e., shifts towards oxygen due to greater electronegativity of O than C . As a result, the C -atom of the $>\mathrm{C}=\mathrm{O}$ bond acquires a partial +ve charge and hence is readily attacked by nucleophiles. Thus, the typical reactions of $>\mathrm{C}=\mathrm{O}$ are nucleophilic addition reactions.

(a) Unsymmetrical $\pi$-electron cloud of $>\mathrm{C}=\mathrm{O}$ bond

(b) Symmetrical $\pi$-electron cloud of $>\mathrm{C}=\mathrm{C}<$ bond

Answer:

Due to the presence of lone pairs of electrons on the oxygen atom of the OH group, the carboxylic acids may be regarded as a resonance hybrid of structures (I and II) :

Similarly, a carbonyl group of aldehydes and ketones may be regarded as a resonance hybrid of structures (III and IV):

Due to contribution of structure (IV), the carbonyl carbon in aldehydes and ketones is electrophilic. However, due to contribution of structure (II), the electrophilic character of carboxyl carbon is reduced.

In other words, carbonyl carbon of carboxyl group is less electrophilic than carbonyl carbon in aldehydes and ketones and hence nucleophilic addition reactions of aldehydes and ketones do to take place with carboxylic acids.

Answer:

Complete chemical conversion can be done as:

$ \underset{\text{Bromoethane}}{\mathrm{CH_3-Br}} \xrightarrow{\text{Mg/ether}} \underset{\substack{\text{[A]} \\ \text{Methyl magnesium} \\ \text{bromide}}}{\mathrm{CH_3MgBr}} \xrightarrow[{\text{(ii) Water}}]{\text{(i) } \mathrm{CO_2}} \underset{\substack{\text{[B]} \\ \text{Ethanoic aicd} }}{\mathrm{CH_3COOH}} \xrightarrow[\text { (Esterification) }]{\mathrm{CH}_3 \mathrm{OH} / \mathrm{H}^{+}} \underset{\substack{\text{[C]} \\ \text{Methyl ethanoate}}}{\mathrm{CH_3COOCH_3}} $

Hence,

$A=\mathrm{CH} _{3} \mathrm{MgBr}$,

$B=\mathrm{CH} _{3} \mathrm{COOH}$

$C=CH_3$ $COOCH_3$

Answer:

The aliphatic carboxylic acids are stronger acids than alcohols and phenols . The difference in the relative strength can be understood if we compare the resonance hybrids of carboxylate ion, alkoxide ion and phenoxide ion .

The electron charge on the carboxylate ion is more dispersed in comparison to the phenoxide ion since there are two electronegative oxygen atoms in carboxylate ion as compared to only one oxygen atom in phenoxide ion. In other words the carboxylate ion is relatively more stable as compared to phenoxide ion.

Thus release of $H^{+}$ from carboxylic acid is comparatively easier or it behaves as a stronger acid than phenol.

Answer:

The complete chemical transformation can be shown as

Answer:

Preparation of ethyl benzene from acylation of benzene and reduction can be shown as:

The direct alkylation can not be performed because there is polysubstitution product is formed. Due to disadvantage of polysubstitution that Friedel-Craft’s alkylation reaction is not used for preparation of alkylbenzenes. Instead of Friedel-Craft’s acylation is used.

Answer:

Yes. Gattermann-Koch reaction can be considered similar to Friedel-Crafts acylation. The reason being that in Friedel-Crafts acylation reactions, benzene (or any other arene) is treated with an acid chloride in presence of anhyd. $ \mathrm{AlCl}_3$. Since HCOCl (formyl chloride) is not stable, therefore, in Gattermann-Koch reaction, it is prepared in situ by reacting CO with HCl gas in presence of anhydrous $ \mathrm{AlCl}_3$.

Thus, Gattermann-Koch reaction is similar to Friedel-Crafts acylation reaction.

Answer:

A. $\rightarrow(4) $

B. $\rightarrow(5) $

C. $\rightarrow(1) $

D. $\rightarrow(2) $

E. $\rightarrow(3) $

Answer:

A. $\rightarrow(2)$

B. $\rightarrow(5)$

C. $\rightarrow(4)$

D. $\rightarrow (1)$

E. $\rightarrow (3)$

Answer:

A. $\rightarrow(3)$

B. $\rightarrow(4) \quad$

C. $\rightarrow(1)$

D. $\rightarrow(2)$

Answer:

A. $\rightarrow (5) $

B. $\rightarrow(4) $

C. $\rightarrow(1) $

D. $\rightarrow(2)$

E. $\rightarrow (6)$

F. $\rightarrow (3)$

Answer:(a) Assertion and reason both are correct and reason is correct explanation of assertion.

Formaldehyde is planar molecule due to $s p^{2}$ hybridised carbon atom.

Answer:(e) Assertion and reason both are correct statements but reason is not correct explanation of assertion.

Compounds containing - $ \mathrm{CHO}$ group are easily oxidised to corresponding carboxylic acids. And the reason is due to electron withdrawing nature of $ \mathrm{C}=\mathrm{O}$ group, $ \mathrm{C}-\mathrm{H}$ bond in aldehydes is weak and easily oxidised to the corresponding carboxylic acids even with mild oxidising agent like Fehling’s solution and Tollen’s reagents.

Answer:(d) Assertion is wrong statement but reason is correct statement.

Correct assertion is the $\alpha$-hydrogen atom in carbonyl compounds is acidic in nature due to presence of electron withdrawing carbonyl group. The anion formed after loss of $\alpha$-hydrogen atom is resonance stabilised.

Answer:(c) Assertion is correct statement but reason is wrong statement.

Aromatic aldehyde and formaldehyde undergo Cannizzaro reaction due to absence of $\alpha-\mathrm{H}$ - atom lead to formation of carboxylic acid and alcohols of corresponding aldehyde.

Answer:(d) Assertion is wrong statement but reason is correct statement.

Aldehydes but not ketones react with Tollen’s reagent to form silver mirror. Reason is correct statement as aldehyde and ketone both contain carbonyl group.

Answer:

(i) Since compound ’ B ’ gives Fehling’s test, therefore, it must be aldehyde. Further since aldehyde ’ B ’ gives iodoform on treatment with $ \mathrm{I}_2$ and NaOH , therefore, ’ B ’ must be acetaldehyde $\left(\mathrm{CH}_3 \mathrm{CHO}\right)$.

(ii) Since alkene ‘A’ $({MF} C { }5 \mathrm{H}{10}$ ) contains five carbon atoms and one of the products of ozonolysis is ’ B ’ $\left(\mathrm{CH}_3 \mathrm{CHO}\right)$ which contains two carbon atoms, therefore, the other product of ozonolysis, i.e., ’ C ’ must contain three carbon atoms.

(iii) Since compound ’ C ’ does not give Fehling’s test, it must be a ketone. Further since ketone ’ C ’ contains three carbon atoms and gives iodoform on treatment with $ \mathrm{I}_2$ and NaOH , therefore, ketone ’ C ’ must be acetone $\left(\mathrm{CH}_3 \mathrm{COCH}_3\right)$.

(iv) Write the products of ozonolysis, i.e., ‘B’ $\left(\mathrm{CH}_3 \mathrm{CHO}\right)$ and ’ C’ $\left(\mathrm{CH}_3 \mathrm{COCH}_3\right)$ side by side with their $ \mathrm{C}=\mathrm{O}$ groups facing each other. Remove the oxygen atoms and join the remaining fragments by a double bond, the structure of alkene ’ A ’ is 2-methylbut-2-ene.

(v) Formation of iodoform from ’ B ’ and ’ C ’ may be explained as follows :

$\underset{\substack{\text { Acetaldehyde }(B)}}{\mathrm{CH}_3 \mathrm{CHO}}+3 \mathrm{I}_2+4 \mathrm{NaOH} \xrightarrow{\Delta} \underset{\text { Iodoform }}{\mathrm{CHI}_3}+\underset{\text { Sod. formate }}{\mathrm{HCOONa}}+3 \mathrm{NaI}+3 \mathrm{H}_2 \mathrm{O}$

$\underset{\text { Acetone }(\mathrm{C})}{\mathrm{CH}_3 \mathrm{COCH}_3}+3 \mathrm{I}_2+4 \mathrm{NaOH} \xrightarrow{\Delta} \underset{\text { Iodoform }}{\mathrm{CHI}_3}+\underset{\text { Sod. acetate }}{\mathrm{CH}_3 \mathrm{COONa}}+3 \mathrm{NaI}+3 \mathrm{H}_2 \mathrm{O}$

Answer:

(i) Since aromatic compound ’ A ’ $(\left.\mathrm{MF}_8 \mathrm{H}_8 \mathrm{O}\right)$ gives positive 2, 4-DNP test, it must be an aldehyde or a ketone.

(ii) Since compound ’ A ’ does not give Tollens’ test or Fehling’s test, therefore, ‘A’ must be a ketone.

(iii) Since compound ’ A ’ on treatment with $ \mathrm{I}_2 / \mathrm{NaOH}$, gives yellow ppt. of compound ’ B ‘, therefore, compound ’ $B$ ’ must be iodoform and the ketone ’ $A$ ’ must be a methyl ketone.

(iv) Since methyl ketone ‘A’ on drastic oxidation with $ \mathrm{KMnO}_4$ gives a carboxylic acid ‘C’ $\left(\mathrm{MF} \mathrm{C}_7 \mathrm{H}_6 \mathrm{O}_2\right)$, therefore, ’ C ’ must be benzoic acid and compound ’ A ’ must be acetophenone $\left(\mathrm{C}_6 \mathrm{H}_5 \mathrm{COCH}_3\right)$.

(v) If ’ A ’ is acetophenone, then all the reactions described above may be explained as follows

Answer:

The functional isomers of the carbonyl compound with molecular formula $ \mathrm{C}_3 \mathrm{H}_6 \mathrm{O}$ are :

$ \underset{\text { Propanal (I) }}{\mathrm{CH} _{3} \mathrm{CH} _{2} \mathrm{CHO}} \quad \quad \text { and } \quad \quad \underset{\text { Propanone (II) }}{\mathrm{CH} _{3} \mathrm{COCH} _{3}} $

Due to less +I -effect of one $ \mathrm{CH}_3 \mathrm{CH}_2$ group as compared to two $ \mathrm{CH}_3$ groups in $ \mathrm{CH}_3 \mathrm{COCH}_3$ and less steric hindrance, compound I, $\left(\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CHO}\right)$ will react faster with HCN than compound II $\left(\mathrm{CH}_3 \mathrm{COCH}_3\right)$.

The reaction will not proceed to completion since it is a reversible reaction. Addition of strong acid inhibits the reaction because the formation of $ \mathrm{CN}^{-}$ from HCN is prevented.

Answer:

(i) Since liquid ‘A’ forms a white crystalline solid on treatment with $ \mathrm{NaHSO}_3$, it may be an aldehyde or a methyl ketone.

Further since liquid ’ A ’ forms a bright silver mirror on treatment with a freshly prepared ammoniacal solution of $ \mathrm{AgNO}_3$, therefore, liquid ’ A ’ is an aldehyde.

(ii) Since liquid ‘B’ forms a white crystalline solid on treatment with $ \mathrm{NaHSO}_3$, it may be an aldehyde or a methyl ketone.

Further since liquid ’ B ’ does not give test with ammoniacal $ \mathrm{AgNO}_3$ solution, therefore, liquid ’ B ’ must be a methyl ketone.

Chemical equations for the reactions discussed are :