NEET Chemistry Aldehydes, Ketones and Carboxylic Acids Questions Solved


Aldehydes and ketones will not form crystalline derivatives with                            

(a) sodium bisulphite

(b) phenyl hydrazine

(c) semicarbazide hydrochloride

(d) dihydrogen sodium phosphate

(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.

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Acetone reacts with iodine (I2) to form iodoform in the presence of

(a) CaCO3          (b) NaOH          (c) KOH            (d) MgCO3

(b) 3I2 +4NaOHCHI3 + CH3COONa +3NaI +3H2O

                                                   Iodoform 

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

(a) 2-methyl-2-butene

(b) 3-methyl-1-butene

(c) Cyclopentane

(d) 2-methyl-1-butene

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

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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

(c) Acetic acid and phenol

(d) Phenol

(a) CH3COOH + NaHCO3 CH3COONa + H2O + CO2

  Acetic acid       Sodium 

                       Carbonate

Difficulty Level:

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The reagent used for the preparation of acetaldehyde from acetophenone is

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

 

(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:

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Iodoform test is not given by

(a) 2-pentanone                  (b) ethanol

(c) ethanal                          (d) 3-pentanone

(d) The compounds which contain either CH3-CO-group or  group give positive iodoform test. In 2-pentanone, (CH3CH2CH2COCH3)CH3CHO and C2H5OH, these groups are present, thus they give iodoform as follows

CH3-COCH2-CH2CH3 +3I2 +4NaOHCHI3+ CH3CH2CH2COONa +3NaI +3H2O

                                                            Iodoform(yellow ppt.)

CH3CHO +3I2 +4NaOHCHI3 +HCOONa +3NaI +3H2O

                                     Iodoform(yellow ppt.)

C2H5OHI2CH3CHO

CH3CHO +3I2 +4NaOHCHI3 +HCOONa +3NaI +3H2O#

                                      Iodoform

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

 +I2 +NaOHNo reaction.

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

(a) I2/NaOH only               (b) NaHSO3 only

(c) NaCN/HCl                     (d) both of (a) and (b)   

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

Difficulty Level:

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Products of the following reaction                                   

  CH3CC.CH2CH3 (ii) hydrolysis(i) O3 ... are

(a) CH3CHO + CH3CH2CHO

(b) CH3COOH + CH3COCH3

(c) CH3COOH + HOOC.CH2CH3

(d) CH3COOH + CO2

(c) 

 

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Which alkene on ozonolysis gives CH3CH2CHO and CH3COCH3?                                          

(a) CH3CH2CH=C<CH3CH3

(b) CH3CH2CH=CHCH2CH3

(c) CH3CH2CH=CHCH3

(d) CH3-C|=CHCH3           CH3

(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

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The major organic product formed from the following reaction

 (ii) LiAlH4  (iii) H2O(i) CH3NH2 ... is

(a)                    (b) 

(c)                  (d) 

 

(b) CH2NH2(ii) LiAlH4  (iii) H2OHor 

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