In presence of conc. NaOH and heat, 2-formyl benzoic acid and glyoxalic acid undergoes decarboxylation and form benzaldehyde and formaldehyde respectively, which undergoes Cannizaro reaction in basic medium and form benzal alcohol and sodium formate.

Reaction: When ester reacts with the sulphuric acid. After that, the oxygen will take the proton. After that, water will attack on the electrophilic carbon. After that, there is proton transfer. After that, the benzoic acid and phenol are formed as a product. Mechanism:

C 6 H 12 O 2 X ⟶ H 3 O + C 2 H 5 CO 2 H + C 3 H 7 OH , X is.

C 6 H 12 O 2 X ⟶ H 3 O + C 2 H 5 CO 2 H + C 3 H 7 OH The compound X is an ester group. When ester reacts with the acidic medium. After that, the oxygen will take the proton. After that, water will attack the electrophilic carbon. After that, there is proton transfer. After that, the propanoic acid and propanol are formed as a product. Reaction:

In the given reaction the product is: CH 3 - CH 2 - C | | O - OCH 3 + CH 3 - C | | O - OCH 3 → ( 2 ) CH 3 OH ( 1 ) CH 2 O ⊝ Na + → ∆ H 2 O Product

CH 3 - CH 2 - C | | O - CH 2 - C | | O - OCH 3

CH 3 - C | | O - C | CH 3 H - COOCH

H - C | | O - CH 2 - CH - COOCH 3

The correct statement about the synthesis of erythritol C C H 2 O H 4 used in the preparation of PETN is

The synthesis requires three aldol condensations and one Cannizzaro reaction.

Alpha hydrogen of ethanol and methanol are involved in this reaction.

The synthesis requires two aldol condensation and two Cannizzaro reactions.

The synthesis requires four aldol condensations between methanol and ethanol.

E M B I B E

Alpha Question Bank for Engineering: Chemistry>Chapter 27 - Aldehydes, Ketones and Carboxylic Acids>Exercise-4>Q 63

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IMPORTANT

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When semicarbazide reacts with a ketone (or aldehyde) to form semicarbazone. Only one nitrogen atom of semicarbazide acts as a nucleophile and attacks the carbonyl carbon of the ketone. The product of the reaction, consequently, is $R_{2} C = N - NH - {CONH}_{2}$ rather than $R_{2} C = NCONH - {NH}_{2}$ . What factor account for the fact that two nitrogen atoms of semicarbazide are relatively non-nucleophilic?

Important Questions on Aldehydes, Ketones and Carboxylic Acids

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Alpha Question Bank for Engineering: Chemistry>Chapter 27 - Aldehydes, Ketones and Carboxylic Acids>Exercise-4>Q 64

Arrange the following compounds in decreasing orders of $K_{eq}$ for hydrate formation.

(a) Question Image

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Alpha Question Bank for Engineering: Chemistry>Chapter 27 - Aldehydes, Ketones and Carboxylic Acids>Exercise-4>Q 65

Which acetone when treated with excess of benzaldehyde in the presence of base results in the crossed condensation which adds two equivalents of benzaldehyde and expels two equivalents of water and forms

[X] .

Identify the structures of

[X]

when

[X]

reacts with

{NH}_{2} OH .

How many stereo isomers are formed?

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Alpha Question Bank for Engineering: Chemistry>Chapter 27 - Aldehydes, Ketones and Carboxylic Acids>Exercise-4>Q 68

Question Image

Product is:

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IMPORTANT

Alpha Question Bank for Engineering: Chemistry>Chapter 27 - Aldehydes, Ketones and Carboxylic Acids>Exercise-4>Q 69

An organic compound

(A), C_{8} H_{4} O_{3}

, in dry benzene in the presence of anhydrous

{AlCl}_{3}

gives compound

(B)

. The compound

(B)

on treatment with

{PCl}_{5}

, followed by reaction with

H_{2} / Pd ({BaSO}_{4})

gives compound

(C)

, which on reaction with hydrazine gives a cyclised compound

(D) (C_{14} H_{10} N_{2})

Identify

(A), (B), (C)

and

(D)

. Explain the formation of

(D)

from

(C)

.

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Alpha Question Bank for Engineering: Chemistry>Chapter 27 - Aldehydes, Ketones and Carboxylic Acids>Exercise-4>Q 70

Question Image find $A$ and $B$ .

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IMPORTANT

Alpha Question Bank for Engineering: Chemistry>Chapter 27 - Aldehydes, Ketones and Carboxylic Acids>Exercise-4>Q 71

C_{6} H_{12} O_{2} (X) \overset{H_{3} O^{+}}{⟶} C_{2} H_{5} {CO}_{2} H + C_{3} H_{7} OH, X

is.

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IMPORTANT

Alpha Question Bank for Engineering: Chemistry>Chapter 27 - Aldehydes, Ketones and Carboxylic Acids>Exercise-4>Q 72

In the given reaction the product is:

${CH}_{3} - {CH}_{2} - \underset{O}{\underset{| |}{C}} - {OCH}_{3} + {CH}_{3} - \underset{O}{\underset{| |}{C}} - {OCH}_{3} \to_{(2) {CH}_{3} OH}^{(1) {CH}_{2} O^{⊝} {Na}^{+}} \to_{∆}^{H_{2} O} Product$