Answer

(i) Cyanohydrin:

Cyanohydrins are organic compounds having the formula $\mathrm{RR}^{{ }^{\prime 2}} \mathrm{C}(\mathrm{OH}) \mathrm{CN}$, where $\mathrm{R}$ and $\mathrm{R}^{\text {“2 }}$ can be alkyl or aryl groups.

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

$\mathrm{\underset{\text{Ketone}}{RR^\prime C=O} + HCN \xrightarrow{NaCN} \underset{\text{Cyanohydrin}}{RR^\prime C(OH)CN}}$

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.

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

(iii) Semicarbarbazone:

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

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

(iv) Aldol:

$\beta-$ 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.

(v) Hemiacetal:

Hemiacetals are $\alpha$ - alkoxyalcohols

Aldehyde reacts with one molecule of a monohydric alcohol in the presence of dry $\mathrm{HCl}$ gas.

(vi) Oxime:

Oximes are a class of organic compounds having the general formula $\mathrm{RR}^{*} \mathrm{CNOH}$, where $\mathrm{R}$ is an organic side chain and $\mathrm{R}^{\text {“2 }}$ is either hydrogen or an organic side chain. If $\mathrm{R}^{2}$ is $\mathrm{H}$, then it is known as aldoxime and if $\mathrm{R}^{\text {“2 }}$ is an organic side chain, it is known as ketoxime.

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

(vii) Ketal:

Ketals are gem - dialkoxyalkanes in which two alkoxy 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.

General structure of a ketal

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

(viii) Imine:

Imines are chemical compounds containing a carbon nitrogen double bond.

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

(ix) 2, 4 - DNP - derivative:

2, 4 - dinitrophenylhydragones are 2,4 - DNP - derivatives, which are produced when aldehydes or ketones react with 2, 4 - dinitrophenylhydrazine in a weakly acidic medium.

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

(x) Schiff’s base:

Schiff’s base (or azomethine) is a 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 $\mathrm{R_1} \mathrm{R_2} \mathrm{C}=\mathrm{NR_3}$. Hence, it is an imine.

It is named after a scientist, Hugo Schiff.

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

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

Answer

(i)

(ii)

(iii)

(iv)

(v)

(vi)

(vii)

(viii)

Answer

(i) $\mathrm{CH_3} \mathrm{CO}\left(\mathrm{CH_2}\right)_{4} \mathrm{CH_3}$

IUPAC name: Heptan-2-one

Common name: Methyl n-propyl ketone

(ii) $\mathrm{CH_3} \mathrm{CH_2} \mathrm{CHBrCH_2} \mathrm{CH}\left(\mathrm{CH_3}\right) \mathrm{CHO}$

IUPAC name: 4-Bromo-2-methylhaxanal

Common name: ($\gamma$-Bromo-$\alpha$-methyl-caproaldehyde)

(iii) $\mathrm{CH_3}\left(\mathrm{CH_2}\right)_{5} \mathrm{CHO}$

IUPAC name: Heptanal

(iv) $\mathrm{Ph}-\mathrm{CH}=\mathrm{CH}-\mathrm{CHO}$

IUPAC name: 3-phenylprop-2-enal

Common name: $\beta-$Pheynolacrolein (v)

IUPAC name: Cyclopentanecarbaldehyde

(vi) $\mathrm{PhCOPh}$

IUPAC name: Diphenylmethanone

Common name: Benzophenone

Answer

(i)

(ii)

(iii)

(iv)

(v)

(vi)

Answer

(i)

(ii)

(iii)

(iv)

(v)

Answer

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

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

Compound (iv) Benzophenone is a ketone having no $\alpha$-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)

(v)

(vi)

(vii)

Cannizzaro reaction

(i)

(iii)

(ix)

Answer

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

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

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

Answer

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

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

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

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

Answer

It is given that the compound (with molecular formula $\mathrm{C_9} \mathrm{H_10} \mathrm{O}$ ) 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 - $\mathrm{CHO}$ group is directly attached to a benzene ring and this benzaldehyde is ortho-substituted. Hence, the compound is 2-ethylbenzaldehyde.

2-Ethylbenzaldehyde

The given reactions can be explained by the following equations.

Answer

An organic compound $A$ with molecular formula $\mathrm{C_8} \mathrm{H_{16}} \mathrm{O_2}$ 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, $\mathrm{B}$ and $\mathrm{C}$ 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, $\mathrm{C}$ 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 $\mathrm{C_8} \mathrm{H_{16}} \mathrm{O_2}$ is butylbutanoate.

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

Answer

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

Hence, the given compounds can be arranged according to their increasing reactivities toward $\mathrm{HCN}$ as:

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

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

$ \mathrm{R}-\mathrm{COOH} \longrightarrow \mathrm{R}-\mathrm{COO}^{-}+\mathrm{H}^{+} $

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, $-\mathrm{CH_3}$ group has $+\mathrm{I}$ effect and $\mathrm{Br}$ group has - I effect. Thus, acids containing $\mathrm{Br}$ are stronger.

Now, the $+l$ effect of isopropyl group is more than that of $n$-propyl group. Hence, $\left(\mathrm{CH_3}\right)_{2} \mathrm{CHCOOH}$ is a weaker acid than $\mathrm{CH_3} \mathrm{CH_2} \mathrm{CH_2} \mathrm{COOH}$.

Also, the - I effect grows weaker as distance increases. Hence, $\mathrm{CH_3} \mathrm{CH}(\mathrm{Br}) \mathrm{CH_2} \mathrm{COOH}$ is a weaker acid than $\mathrm{CH_3} \mathrm{CH_2} \mathrm{CH}(\mathrm{Br}) \mathrm{COOH}$.

Hence, the strengths of the given acids increase as:

$\left(\mathrm{CH_3}\right)_{2} \mathrm{CHCOOH}<\mathrm{CH_3} \mathrm{CH_2} \mathrm{CH_2} \mathrm{COOH}<\mathrm{CH_3} \mathrm{CH}(\mathrm{Br}) \mathrm{CH_2} \mathrm{COOH}<\mathrm{CH_3} \mathrm{CH_2} \mathrm{CH}(\mathrm{Br}) \mathrm{COOH}$

(iii) As we have seen in the previous case, electron-donating groups decrease the strengths of acids, while electronwithdrawing groups increase the strengths of acids. As methoxy group is an electron-donating group, 4methoxybenzoic 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

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.

$\underset{\text{Propanal}}{\mathrm{CH_3} \mathrm{CH_2} \mathrm{CHO}}+ \underset{\text{Tollen’s reagent}}{2\left[\mathrm{Ag}\left(\mathrm{NH_3}\right)_{2}\right]^{+}+3 \mathrm{OH}} \longrightarrow \underset{\text{Propanoate ion}}{\mathrm{CH_3} \mathrm{CH_2} \mathrm{COO}^{-}}+\underset{\text{Silver mirror}}{\mathrm{Ag} \downarrow}+4 \mathrm{NH_3}+2 \mathrm{H_2} \mathrm{O}$

(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 $\mathrm{Cu_2} \mathrm{O}$, but propanone being a ketone does not.

$\mathrm{\underset{\text{Propanal}}{CH_3CH_2CHO}+2CU^{2+} +5OH^-} \longrightarrow \mathrm{\underset{\text{Propanoate ion}}{CH_3CH_2COO^-}+ \underset{\substack{\text{Cuprous oxide }\\ \text{(Red -brown ppt)}}}{Cu_2O}+ 3H_2O}$

(c) lodoform 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.

$\mathrm{\underset{\text{Propane}}{CH_3COCH_3}+ 3 NaOI}\longrightarrow \mathrm{\underset{\text{Sodium acetate}}{CH_3COONa}+\underset{\substack{\text{Iodoform}\\ \text{(Yellow ppt)}}}{{CHI_3}}+ 2NaOH}$

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

lodoform 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.

$\mathrm{\underset{\text{Acetophenone}}{C_6H_5COH_3}+\underset{\substack{\text{Sodium}\\ \text{hypoiodite}}}{3NaOI}} \longrightarrow \mathrm{\underset{\substack{\text{Sodium}\\ \text{benzoate}}}{C_6H_5COONa} + \underset{\substack{\text{Iodoform}\\ \text{(yellow ppt)}}}{CHI_3} + 2NaOH}$

$\mathrm{\underset{\text{Benzophenone}}{C_6H_5COC_6H_5}+ NaOI}\longrightarrow \text{No yellow ppt of } \mathrm{CHI_3}$

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

Ferric chloride test:

Phenol reacts with neutral $\mathrm{FeCl_3}$ to form an iron-phenol complex giving violet colouration.

$ \underset{\text { Phenol }}{6 \mathrm{C_6} \mathrm{H_5} \mathrm{OH}+} \mathrm{FeCl_3} \longrightarrow \underset{\substack{\text { Iron-phenol complex } \\ \text { (Violet colour) }}}{\left[\mathrm{Fe}\left(\mathrm{OC_6} \mathrm{H_5}\right)_{6}\right]^{3-}}+3 \mathrm{H}^{+}+3 \mathrm{Cl}^{-} $

But benzoic acid reacts with neutral $\mathrm{FeCl_3}$ to give a buff coloured ppt. of ferric benzoate.

$ \underset{\text { Benzoic acid }}{3 \mathrm{C_6H_5}OH}+\mathrm{FeCl_3} \longrightarrow \underset{\substack{\text{Ferric benzoate} \\ \text{(Buff coloured ppt)}}}{\mathrm{(C_6H_5COO)_3Fe}} + \mathrm{3HCl} $

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

Sodium bicarbonate test:

Acids react with $\mathrm{NaHCO_3}$ to produce brisk effervescence due to the evolution of $\mathrm{CO_2}$ gas.

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

$\mathrm{\underset{\text{Benzoic acid}}{C_6H_5COOH}+ NaHCO_3} \longrightarrow \mathrm{\underset{\text{Sodium benzoate}}{C_6H_5COONa} + CO_2 \uparrow + H_2O}$

$\mathrm{C_6H_5COOC_2H_5 + NaHCO_3 \longrightarrow } \text{No effervescence due to evolution of } \mathrm{CO_2} \text{ gas}$

(v) Pentan-2-one and pentan-3-one can be distinguished by iodoform test.

lodoform test:

Pentan-2-one is a methyl ketone. Thus, it responds to this test. But pentan-3-one not being a methyl ketone does not respond to this test.

(vi) Benzaldehyde and acetophenone can be distinguished by the following tests.

(a) Tollen’s Test

Aldehydes respond to Tollen’s test. Benzaldehyde being an aldehyde reduces Tollen’s reagent to give a red-brown precipitate of $\mathrm{Cu_2} \mathrm{O}$, but acetophenone being a ketone does not.

$\underset{\text{Benzaldehyde}}{\mathrm{C_6H_5CHO}} + \underset{\text{Tollen’s reagen}}{\mathrm{2[Ag(NH_3)_2]^+}} + \mathrm{3OH^-} \longrightarrow \underset{\text{Benzoate ion}}{\mathrm{C_6H_5COO^-}} + \underset{\text{Silver mirror}}{\mathrm{Ag \downarrow}}+ \mathrm{4NH_3 + 2H_2O}$

Benzaldehyde Tollen’s reagent Benzoate ion Silver mirror

(b) lodoform test

Acetophenone being a methyl ketone undergoes oxidation by sodium hypoiodite ( $\mathrm{NaOI}$ ) to give a yellow ppt. of iodoform. But benzaldehyde does not respond to this test.

$\underset{\text{Acetophenone}}{\mathrm{C_6} \mathrm{H_5} \mathrm{COCH_3}}+3 \mathrm{NaOI} \longrightarrow \underset{\text{Sodium benzoate}}{\mathrm{C_6} \mathrm{H_5} \mathrm{COONa}}+ \underset{\substack{\text{Iodoform}\\ \text{(Yellow ppt)}}}{\mathrm{CHI_3}}+2 \mathrm{NaOH}$

(vii) Ethanal and propanal can be distinguished by iodoform test.

lodoform test

Aldehydes and ketones having at least one methyl group linked to the carbonyl carbon atom responds to the iodoform test. Ethanal having one methyl group linked to the carbonyl carbon atom responds to this test. But propanal does not have a methyl group linked to the carbonyl carbon atom and thus, it does not respond to this state.

$ \underset{\text{Ethanal}}{\mathrm{CH_3CHO}} + \mathrm{3NaOI} \longrightarrow \underset{\substack{\text{Sodium}\\ \text{methanoate}}}{\mathrm{HCOONa}} + \underset{\substack{\text{Iodoform}\\ \text{(yellow ppt)}}}{\mathrm{CHI_3}} + \mathrm{2NaOH} $

Answer

(i)

(ii)

(iii)

(iv)

(v)

Answer

(i)

(ii)

(iii)

(iv)

(v)

(vi)

(vii)

(viii)

(ix)

Answer

(i) 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, dirnethylaniline, etc. This process involves the substitution of an acetyl group for an active hydrogen atom. Acetyl chloride and acetic anhydride are commonly used as acetylating agents.

For example, acetylation of ethanol produces ethyl acetate.

$\underset{\text{Ethanol}}{\mathrm{CH_3} \mathrm{CH_2} \mathrm{OH}}+\underset{\substack{\text{Acetyl}\\ \text{chloride}}}{\mathrm{CH_3} \mathrm{COCl}} \xrightarrow{\text { Pyridine }} \underset{\text{Ethylacetate}}{\mathrm{CH_3} \mathrm{COOC_2} \mathrm{H_5}}+\mathrm{HCl}$

(ii) Cannizzaro reaction:

The self oxidation-reduction (disproportionation) reaction of aldehydes having no $\alpha$-hydrogens on treatment with concentrated alkalis is known as the Cannizzaro reaction. In this reaction, two molecules of aldehydes participate where one is reduced to alcohol and the other is oxidized to carboxylic acid.

For example, when ethanol is treated with concentrated potassium hydroxide, ethanol and potassium ethanoate are produced.

(iii) Cross-aldol condensation:

When aldol condensation is carried out between two 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 $\alpha$-hydrogens, four compounds are obtained as products.

For example, ethanal and propanal react to give four products.

(iv) 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.

Decarboxylation also takes place when aqueous solutions of alkali metal salts of carboxylic acids are electrolyzed. This electrolytic process is known as Kolbe’s electrolysis.