Aldehydes, Ketones and Carboxylic Acids

Correct order of acidic strength of major product formed in the given reaction is

$ \text { (B) } \mathrm{CH}_3 \mathrm{CH}_2-\mathrm{CHO} \xrightarrow[\Delta]{\mathrm{Ag}\left(\mathrm{NH}_3\right)_2 \mathrm{OH}} \mathrm{CH}_3-\mathrm{CH}_2-\mathrm{COOH} $

$ \text { (C) } \mathrm{CH}_4+\mathrm{O}_2 \xrightarrow[\mathrm{Na}_2 \mathrm{Cr}_2 \mathrm{O}_7 / \mathrm{H}^{+}]{\mathrm{MO}_2 \mathrm{O}_3} \mathrm{HCOOH} $

$ \text { (D) } \mathrm{Ph}-\mathrm{CH}_2 \mathrm{MgBr}+\mathrm{CO}_2 \xrightarrow[\mathrm{H}_3 \mathrm{O}^{\oplus}]{\text { Dryether }} \mathrm{Ph}-\mathrm{CH}_2-\mathrm{COOH} $

C > A > D > B

$ \text { Major product of reaction (1) will be } $

$ \text { As } 3^{\circ} \text { radical more stable } $

(A) $Q$, $R$ and $S$ give the $2,4$-DNP test because each of them contains a carbonyl group $(>\!C=O)$ of aldehyde/ketone. In this test, aldehydes and ketones form an orange/yellow precipitate of the $2,4$-dinitrophenylhydrazone.

(C) $Q$ and $R$ give a sooty flame because they contain an aromatic ring. Aromatic compounds have a higher carbon content, so they burn with incomplete combustion and produce a smoky (sooty) flame.

(E) Only $Q$ gives Tollens’ test because $Q$ is an aldehyde. Aldehydes reduce Tollens’ reagent (ammoniacal silver nitrate) to metallic silver, so a silver deposit/mirror is obtained. Ketones ($R$ and $S$) do not give this test.

'P' and 'Q' can be differentiated by Fehling's test.

P gives positive Fehling test

Q gives negative Fehling test

Statement I: False.

A crossed aldol between two different aldehydes gives four products only in the uncontrolled case when both aldehydes have α‑H (2 self‑aldols + 2 crossed). It is not “always”: e.g., benzaldehyde (no α‑H) cannot self‑aldol, so fewer products result; and under controlled conditions you can also avoid mixtures.

Statement II: False.

Semicarbazide forms semicarbazones with both aldehydes and ketones, and aldehydes (benzaldehyde) generally react faster than ketones (acetophenone) under the usual “optimum pH” conditions. So it will not give a condensation product with acetophenone only.

Correct option: B — Both Statement I and Statement II are false.

$ \mathrm{x}=\mathrm{H}_2(\text { gas }), \mathrm{y}=\mathrm{CO}_2(\text { gas }) $

Now find the molar mass of each gas:

The molar mass of $\mathrm{H}_2$ is $2\ \text{g mol}^{-1}$.

The molar mass of $\mathrm{CO}_2$ is $12+16\times 2=44\ \text{g mol}^{-1}$.

Sum of molar mass $=2+44=46$

(A) This statement is true. Sodium hypochlorite oxidises iodide ion to hypoiodite. So, the reaction is:

$\mathrm{NaOCl}+\mathrm{KI} \rightarrow \mathrm{NaCl}+\mathrm{KOI}$

(B) This statement is false. $\mathrm{KOI}$ (hypoiodite) behaves as an oxidising agent in haloform (iodoform) conditions, not as a reducing agent.

(C) This statement is true. In iodoform reaction, any compound having a $\mathrm{-COCH_3}$ (methyl ketone) group gives iodoform test. An $\alpha,\beta$-unsaturated methyl ketone still contains the $\mathrm{-COCH_3}$ group, so it gives iodoform reaction.

gives iodoform reaction.

(D) This statement is false. Isopropyl alcohol $\left(\mathrm{CH_3CHOHCH_3}\right)$ gives iodoform test because it is oxidised (in presence of $\mathrm{I_2/NaOH}$) to acetone $\left(\mathrm{CH_3COCH_3}\right)$, which is a methyl ketone.

(E) This statement is false. Methanoic acid $\left(\mathrm{HCOOH}\right)$ does not have the required $\mathrm{-COCH_3}$ group (or the $\mathrm{CH_3CH(OH)-}$ group) for iodoform test, so it does not give a positive iodoform test.

The Iodoform Test

The iodoform test is used to detect the presence of a specific structural group in a molecule. A compound gives a positive iodoform test if it contains either:

1. A methyl ketone group ($CH_3-C(=O)-$ group)

2. An alcohol group that can be oxidized to a methyl ketone. This means the alcohol must have a $CH_3-CH(OH)-$ group.

A positive test is indicated by the formation of a yellow precipitate of iodoform ($CHI_3$) when the compound is warmed with iodine ($I_2$) and an aqueous solution of sodium hydroxide ($NaOH$).

The general reaction is:

$ R-CO-CH_3 \xrightarrow{I_2, NaOH} CHI_3 \downarrow + R-COONa $

(Yellow ppt)

Now, let's analyze each pair of compounds one by one.

A. Methanol ($CH_3OH$) and Ethanol ($CH_3CH_2OH$)

• Methanol ($CH_3OH$): This is a primary alcohol. It does not have the required $CH_3-CH(OH)-$ group. Therefore, Methanol gives a negative iodoform test.

• Ethanol ($CH_3CH_2OH$): This is the only primary alcohol that gives a positive iodoform test. Under the reaction conditions, ethanol is first oxidized to acetaldehyde ($CH_3CHO$). Acetaldehyde has a methyl ketone group ($CH_3-C(=O)-$H). Thus, Ethanol gives a positive iodoform test.

Since one gives a positive test and the other gives a negative test, they can be differentiated.

B. Acetic acid ($CH_3COOH$) and Propanoic acid ($CH_3CH_2COOH$)

• Carboxylic acids do not give the iodoform test. The acidic proton of the $-COOH$ group reacts with the base ($NaOH$), but the resulting carboxylate ion does not undergo iodination like ketones do.

• Therefore, both Acetic acid and Propanoic acid give a negative iodoform test.

Since both give a negative test, they cannot be differentiated.

C. Cyclohexene and Cyclohexanone

• Cyclohexene: This is an alkene. It does not have a methyl ketone group or an alcohol group. It gives a negative iodoform test.

• Cyclohexanone: This is a ketone, but it does not have a methyl ($CH_3$) group attached to the carbonyl ($C=O$) group. The structure is $-CH_2-C(=O)-CH_2-$. Therefore, it gives a negative iodoform test.

Since both give a negative test, they cannot be differentiated.

D. Diethyl ether ($CH_3CH_2-O-CH_2CH_3$) and Pentan-3-one ($CH_3CH_2-C(=O)-CH_2CH_3$)

• Diethyl ether: Ethers do not have the required structural groups for this test. It gives a negative iodoform test.

• Pentan-3-one: This is a ketone, but the carbonyl group is bonded to two ethyl ($CH_3CH_2-$) groups, not a methyl group. Therefore, it gives a negative iodoform test.

Since both give a negative test, they cannot be differentiated.

E. Anisole ($C_6H_5-O-CH_3$) and Acetone ($CH_3-C(=O)-CH_3$)

• Anisole: This is an ether. It does not have the required structural group. It gives a negative iodoform test.

• Acetone (Propanone): This is a methyl ketone. It has the $CH_3-C(=O)-$ group. Therefore, it gives a positive iodoform test.

Since one gives a positive test and the other gives a negative test, they can be differentiated.

Conclusion

Based on our analysis, the iodoform test can be used to differentiate the pairs in A and E.

Therefore, the correct choice is Option C: A & E Only.

In compounds given in options (B) and (C), the special carbon (anomeric carbon) is attached to two –OR groups.

Such structures are called acetals. In acetals, there is no anomeric –OH group present (they are not in the hemiacetal form), so the ring cannot open to give a free aldehyde group.

Since Tollen's reagent reacts only with a free aldehyde group, compounds (B) and (C) do not give Tollen's test.

(On cross aldol reaction, a single structure is obtained but it can show geometrical isomerism).

A reagent is $NH_2NH_2,\ KOH$ (hydrazine in strong base). This is used to reduce aldehydes/ketones to hydrocarbons by Wolff–Kishner reduction.

So, A $\rightarrow$ III.

B reagent is $Ag(NH_3)_2OH$ (ammoniacal silver nitrate). This gives silver mirror with aldehydes: Tollen’s test.

So, B $\rightarrow$ I.

C reagent is aqueous $CuSO_4$ + sodium potassium tartrate (Rochelle salt) + $KOH$; together they form Fehling’s solution, used in Fehling’s test for aldehydes (red $Cu_2O$ ppt).

So, C $\rightarrow$ IV.

D reagent is $Zn$–$Hg,\ HCl$ which reduces carbonyl group to $-CH_2-$ in acidic medium: Clemmensen reduction.

So, D $\rightarrow$ II.

Hence the correct matching is:

$A-III,\ B-I,\ C-IV,\ D-II$

Correct option: B.

A. $H_2$, Pd–BaSO$_4$ is used to reduce acyl chlorides to aldehydes (poisoned catalyst) $\Rightarrow$ Rosenmund Reduction (II).

B. SnCl$_2$, HCl reduces nitriles to iminium salt which on hydrolysis gives aldehyde $\Rightarrow$ Stephen Reaction (IV).

C. CrO$_2$Cl$_2$ in CS$_2$ oxidises benzylic methyl group $(\text{-CH}_3)$ to aldehyde $(\text{-CHO})$ $\Rightarrow$ Etard Reaction (I).

D. CO, HCl, anhyd. AlCl$_3$ formylates aromatic ring to give an aldehyde $\Rightarrow$ Gattermann–Koch Reaction (III).

So the correct matching is:

$A\!-\!II,\; B\!-\!IV,\; C\!-\!I,\; D\!-\!III$

Correct option: D.

Two chiral centre and due to presence of -COOH compound dissolves in $\mathrm{NaHCO}_3$.

Aldol condensation reaction

The molecules listed below do not undergo the iodoform reaction:

Butanal

2-Pentanone

Pentanal

3-Pentanol

(A) $\mathrm{R}-\mathrm{C} \equiv \mathrm{N} \xrightarrow[\text { mild condition }]{\text { (i) } \mathrm{H}^{+} / \mathrm{H}_2 \mathrm{O}} \mathrm{R}-\mathrm{CONH}_2$

Under mild condition amide is formed because this reaction is typically slow if further more heat will supplied then it gets convert in to $-$COOH .

(B) $\mathrm{R}-\mathrm{MgX} \xrightarrow[\text { (ii) } \mathrm{H}_3 \mathrm{O}^{+}]{\text {(i) } \mathrm{CO}_2} \mathrm{R}-\mathrm{COOH}$

(C) $\mathrm{R}-\mathrm{C} \equiv \mathrm{N} \xrightarrow[\text { (ii) } \mathrm{H}_3 \mathrm{O}^{+}]{\text {(i) } \mathrm{SnCl}_2 / \mathrm{HCl}} \mathrm{R}-\mathrm{CHO}$

(D) $\mathrm{R} \cdot \mathrm{CH}_2 \cdot \mathrm{OH} \xrightarrow{\mathrm{PCC}} \mathrm{R}-\mathrm{CHO}$

Vanillin does not give self-aldol reaction due to lack of acidic H for condensation.

Compounds having at least $1 ~\alpha-\mathrm{H}$ will react with $\mathrm{KMnO}_4$ and give benzoic acid.

Total 5 compounds.

Ans. (2) A, C and E only.

$\mathrm{A}, \mathrm{C}, \mathrm{D}$ produce $\mathrm{CO}_2$ with aqueous $\mathrm{NaHCO}_3$ solution.

$\mathrm{A}, \mathrm{C}, \mathrm{D}$ acids are stronger acid than $\mathrm{H}_2 \mathrm{CO}_3$ (Carbonic acid)

The rate of nucleophilic addition decreased due to steric crowding around carbonyl carbon \& increased by electron withdrawing group if the steric crowding is same hence the reactivity towards nucleophilic addition will be

$\mathrm{k}_{\mathrm{eq}}=2280$ is for HCHO

$\mathrm{k}_{\mathrm{eq}}=2000$ is for chloral

Both data is given in clayden and warren book.

$\mathrm{k}_{\mathrm{eq}}>1$ because HCHO and chloral are more electrophilic.

This is an example of Tollen's reaction i.e. multiple cross aldol followed by cross Cannizaro reaction

Due to NGP effect of phenyl ring Nucleophilic substitution of Br will occurs.

The reaction involved here is the Tollen's test, which detects aldehydes and alpha-hydroxy ketones, as they are able to reduce the ammoniacal silver nitrate solution to metallic silver, producing a mirror-like coating on the glassware. This test specifically distinguishes these reducing sugars from non-reducing sugars and is often used to identify the presence of an aldehyde functional group.

Let's analyze each option:

A. Formic acid - Formic acid ($HCOOH$) is a carboxylic acid but importantly, it can also reduce silver ions to metallic silver because it can act as a reducing agent just like aldehydes. Thus, it will give a positive Tollen's test.

B. Formaldehyde ($HCHO$) is the simplest aldehyde, and aldehydes are known to give positive results with the Tollen's test since they can be oxidized to carboxylic acids, reducing the silver ions in the process to metallic silver.

C. Benzaldehyde ($C_6H_5CHO$) is an aromatic aldehyde and, similar to formaldehyde, it can be oxidized to an acid, thereby reducing the silver ions to silver in the Tollen's test, giving a positive result.

D. Acetone ($CH_3COCH_3$) is a ketone. Generally, ketones do not undergo oxidation easily and do not reduce ammoniacal silver nitrate, and thus, do not give a silver mirror with the Tollen's reagent.

Therefore, the compounds that will give a silver mirror with ammoniacal silver nitrate (positive Tollen's test) are Formic acid, Formaldehyde, and Benzaldehyde, but not Acetone.

The correct answer from the options given is therefore:

Option C: A, B, and C only.

is known as Gattermann-Koch reaction.

Given reaction is HVZ reaction.

The major product (P) formed in the given reaction is 2-hydroxy propanoic acid.

Formaldehyde is more reactive due to least hindrance.