NEET Chemistry Aldehydes, Ketones and Carboxylic Acids Questions Solved

Aldehydes and ketones will not form crystalline derivatives with

(a) sodium bisulphite

(b) phenyl hydrazine

(d) dihydrogen sodium phosphate

#Preparation & properties of Aldehydes and Ketones

(d) Dihydrogen sodium phosphate (NaH₂PO₄) does not react with aldehydes and ketones because NaH₂PO₄ does not have any lone pair of electron on phosphorus atom, so it cannot act as a nucleophile.

Difficulty Level:

(a)
25%
(b)
21%
(c)
20%
(d)
35%

Acetone reacts with iodine $(I_{2})$ to form iodoform in the presence of

(a) Ca ${CO}_{3}$ (b) NaOH (c) KOH (d) Mg ${CO}_{3}$

#Preparation & properties of Aldehydes and Ketones

(b) + $3 I_{2}$ +4NaOH $\to {CHI}_{3}$ + ${CH}_{3}$ COONa +3NaI +3 $H_{2} O$

Iodoform

Difficulty Level:

(a)
9%
(b)
64%
(c)
23%
(d)
5%

Which of the following compounds with molecular formula, C₅H₁₀ yields acetone on ozonolysis?

(a) 2-methyl-2-butene

(b) 3-methyl-1-butene

(d) 2-methyl-1-butene

#Preparation & properties of Aldehydes and Ketones

(a) 2-methyl-2-butene (molecular formula C₅H₁₀) yields acetone on ozonolysis.

Difficulty Level:

(a)
58%
(b)
18%
(c)
16%
(d)
10%

Among acetic acid, phenol and n-hexanol which one of the following compound will react with NaHCO₃ solution to give sodium salt and CO₂?

(a) Acetic acid

(b) n-hexanol

(d) Phenol

#Preparation & properties of Aldehydes and Ketones

(a) CH₃COOH + NaHCO₃ $\to$ CH₃COONa + H₂O + CO₂

Acetic acid Sodium

Carbonate

Difficulty Level:

(a)
53%
(b)
14%
(c)
25%
(d)
10%

The reagent used for the preparation of acetaldehyde from acetophenone is

(1) NaHSO₃ (2) C₆H₅NHNH₂ (3) NH₂OH (4) NaOH and I₂

#Preparation & properties of Aldehydes and Ketones

(1) NaHSO₃ gives the addition reaction with aldehyde and only aliphatic ketone. Acetophenone is the aromatic ketone so it does not give the addition product with NaHSO₃ aldehyde from the addition product with NaHSO₃ which on treatment with acid or base give again aldehyde.

Difficulty Level:

(a)
25%
(b)
29%
(c)
15%
(d)
33%

Iodoform test is not given by

(a) 2-pentanone (b) ethanol

#Preparation & properties of Aldehydes and Ketones

(d) The compounds which contain either ${CH}_{3} - CO -$ group or group give positive iodoform test. In 2-pentanone, $({CH}_{3} {CH}_{2} {CH}_{2} {COCH}_{3})$ , ${CH}_{3} CHO$ and $C_{2} H_{5} OH$ , these groups are present, thus they give iodoform as follows

${CH}_{3} - {COCH}_{2} - {CH}_{2} {CH}_{3} + 3 I_{2} + 4 NaOH \to {CHI}_{3} ↓ + {CH}_{3} {CH}_{2} {CH}_{2} COONa + 3 NaI + 3 H_{2} O$

Iodoform(yellow ppt.)

${CH}_{3} CHO + 3 I_{2} + 4 NaOH \to {CHI}_{3} ↓ + HCOONa + 3 NaI + 3 H_{2} O$

Iodoform(yellow ppt.)

$C_{2} H_{5} OH \overset{I_{2}}{\to}$ ${CH}_{3} CHO$

${CH}_{3} CHO + 3 I_{2} + 4 NaOH \to {CHI}_{3} ↓ + HCOONa + 3 NaI + 3 H_{2} O$ #

Iodoform

but due to absence of or group in 3-pentanone, it does not give iodoform.

+ $I_{2}$ +NaOH $\to$ No reaction.

Difficulty Level:

(a)
12%
(b)
25%
(c)
22%
(d)
43%

Two isomeric ketones 3-Pentanones and 2-pentanone can be distinguished by

(a) I₂/NaOH only (b) NaHSO₃ only

(d) Both Iodoform reaction and reaction with NaHSO₃ can distinguish between 3-pentanone and 2-pentanone.

Difficulty Level:

(a)
53%
(b)
12%
(c)
5%
(d)
33%

Products of the following reaction

CH₃C $\equiv$ C.CH₂CH₃ $\to_{(ii) hydrolysis}^{(i) O_{3}}$ ... are

(a) CH₃CHO + CH₃CH₂CHO

(b) CH₃COOH + CH₃COCH₃

(d) CH₃COOH + CO₂

#Preparation & properties of Aldehydes and Ketones

(c)

Difficulty Level:

(a)
38%
(b)
20%
(c)
41%
(d)
3%

Which alkene on ozonolysis gives CH₃CH₂CHO and CH₃COCH₃?

(a) CH₃CH₂CH=C $<_{{CH}_{3}}^{{CH}_{3}}$

(b) CH₃CH₂CH=CHCH₂CH₃

(d) ${CH}_{3} - \underset{|}{C} = {CHCH}_{3} {CH}_{3}$

#Preparation & properties of Aldehydes and Ketones

(a) When O₃ reacts with alkene, it forms ozonide, which on reaction with Zn and acid or H2/Ni gives aldehydes and/or ketones. These products helps in locating the position of a double bond as