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All questions of Aldehydes, Ketones and Carboxylic Acids for NEET Exam
The IUPAC name of CH3CHO is:- a)Acetaldehyde
- b)Ethanal
- c)Formaldehyde
- d)Methanal
Correct answer is option 'A'. Can you explain this answer?
The IUPAC name of CH3CHO is:
a)
Acetaldehyde
b)
Ethanal
c)
Formaldehyde
d)
Methanal
|
Ambition Institute answered |
- The functional group is an aldehyde; −CHO and the given compound has two carbon atoms.
- Thus, the IUPAC name of the compound is ethanal.
Write the IUPAC name of (CH3)2CHCHO?- a)2,2-Dimethylpropanal
- b)3-Hydroxypropanal
- c)But-3-en-2-one
- d)None of these
Correct answer is option 'D'. Can you explain this answer?
Write the IUPAC name of (CH3)2CHCHO?
a)
2,2-Dimethylpropanal
b)
3-Hydroxypropanal
c)
But-3-en-2-one
d)
None of these
|
Neha Sharma answered |
The IUPAC name of (CH3)2CHCHO is 2-methylpropanal and its chemical name is Isobutyraldehyde.
Reaction
Primary amine + CHCl3 + KOH → product, here product will be - [AIEEE-2002]- a)Cyanide
- b)Isocyanide
- c)Amine
- d)Alcohol
Correct answer is option 'B'. Can you explain this answer?
Reaction
Primary amine + CHCl3 + KOH → product, here product will be - [AIEEE-2002]
Primary amine + CHCl3 + KOH → product, here product will be - [AIEEE-2002]
a)
Cyanide
b)
Isocyanide
c)
Amine
d)
Alcohol
|
Krishna Iyer answered |
The correct answer is Option B.
When primary amine reacts with CHCl3 in alcoholic KOH, the product is isocyanide which is foul smelling.
R−NH2+CHCl3+3KOH -ΔR−NC+3KCl+3H2O
This reaction is known as carbylamine test / Isocyande test. This test is not given by secondary or tertiary amines.
When primary amine reacts with CHCl3 in alcoholic KOH, the product is isocyanide which is foul smelling.
R−NH2+CHCl3+3KOH -ΔR−NC+3KCl+3H2O
This reaction is known as carbylamine test / Isocyande test. This test is not given by secondary or tertiary amines.
Which of the following statements are correct in case of the carbonyl bond between carbon and oxygen?- a)Carbon is the nucleophilic centre and Oxygen is the electrophilic centre.
- b)Oxygen is the nucleophilic centre and Carbon is the electrophilic centre.
- c)Carbon and Oxygen double bond is polarised.
- d)Both ‘b’ and ‘c’ are correct
Correct answer is option 'C'. Can you explain this answer?
Which of the following statements are correct in case of the carbonyl bond between carbon and oxygen?
a)
Carbon is the nucleophilic centre and Oxygen is the electrophilic centre.
b)
Oxygen is the nucleophilic centre and Carbon is the electrophilic centre.
c)
Carbon and Oxygen double bond is polarised.
d)
Both ‘b’ and ‘c’ are correct
|
|
Nandini Patel answered |
The double bonds in alkenes and double bonds in carbonyl groups are VERY different in terms of reactivity. The C=C is less reactive due to C=O electronegativity attributed to the oxygen and its two lone pairs of electrons. One pair of the oxygen lone pairs are located in 2s while the other pair are in 2p orbital where its axis is directed perpendicular to the direction of the pi orbitals. The Carbonyl groups properties are directly tied to its electronic structure as well as geometric positioning. For example, the electronegativity of oxygen also polarizes the pi bond allowing the single bonded substituent connected to become electron withdrawing.
Which of the following statement about C=O and C=C is correct?- a)Both consist of a sigma and pi bond
- b)C=O is polar but C=C is non-polar
- c)Both a and b are correct
- d)Both C=O and C=C undergo nucleophilic addition reactions
Correct answer is option 'C'. Can you explain this answer?
Which of the following statement about C=O and C=C is correct?
a)
Both consist of a sigma and pi bond
b)
C=O is polar but C=C is non-polar
c)
Both a and b are correct
d)
Both C=O and C=C undergo nucleophilic addition reactions
|
Knowledge Hub answered |
- The first bond formed is a sigma bind and the second one is a pi bond.
- O has a higher electronegativity than C and hence the electron cloud will be shifted towards the O atom, making the compound polar.
- This is not possible in C=C.
Among the following functional groups, which of these is not a carbonyl compound?- a)alcohols
- b)aldehydes
- c)Carboxylic acid
- d)ketones
Correct answer is option 'A'. Can you explain this answer?
Among the following functional groups, which of these is not a carbonyl compound?
a)
alcohols
b)
aldehydes
c)
Carboxylic acid
d)
ketones
|
Pari answered |
In carbonyl compound C=O is present bt in alchol OH is present. So alchol is not a carbonyl compound
Correct IUPAC name of the following compound is :
- a)3-(Hepta-2,4,6-trienyl)-4 bromo cyclopenta-2, 4, -dien-1-ol
- b)7-(2-Bromo-4-hydroxy cyclopenta-1,4-dienyl)hepta-1,3,5-triene
- c)7-(5-Bromo-3-hydroxycyclopenta-1,4-dienyl)hepta-1,3,5-triene
- d)3-Bromo-4-(hepta-2,4,6-trienyl)cyclopenta-2,4-dien-1-oll
Correct answer is option 'D'. Can you explain this answer?
Correct IUPAC name of the following compound is :
a)
3-(Hepta-2,4,6-trienyl)-4 bromo cyclopenta-2, 4, -dien-1-ol
b)
7-(2-Bromo-4-hydroxy cyclopenta-1,4-dienyl)hepta-1,3,5-triene
c)
7-(5-Bromo-3-hydroxycyclopenta-1,4-dienyl)hepta-1,3,5-triene
d)
3-Bromo-4-(hepta-2,4,6-trienyl)cyclopenta-2,4-dien-1-oll
|
|
Krishna Iyer answered |
The correct answer is option D
The IUPAC name for the given compound is:
3-Bromo-4-(hepta-2,4,6-trienyl)cyclopenta-2,4-dien-1-oll
according to the rule of nearly attached halogen count first.so we count the the chain where bromo is near
The IUPAC name for the given compound is:
3-Bromo-4-(hepta-2,4,6-trienyl)cyclopenta-2,4-dien-1-oll
according to the rule of nearly attached halogen count first.so we count the the chain where bromo is near
The correct IUPAC name of the folllowing compound is
- a)5,6-Diethyl-8-methyl dec-6-ene
- b)5,7-Diethyl-3-methyl dec-4-ene
- c)5,6-Diethyl-3-methyl dec-4-ene
- d)2,4,5-Triethylnon-3-ene
Correct answer is option 'C'. Can you explain this answer?
The correct IUPAC name of the folllowing compound is
a)
5,6-Diethyl-8-methyl dec-6-ene
b)
5,7-Diethyl-3-methyl dec-4-ene
c)
5,6-Diethyl-3-methyl dec-4-ene
d)
2,4,5-Triethylnon-3-ene
|
Gunjan Lakhani answered |
The IUPAC name of the given compound is 5, 6 - diethyl -2- methyldec - 4 - ene.
The parent hydrocarbon contains 10 carbon atoms and a double bond.
It is called dec-4-ene.
One methyl group is present on second carbon atom and two ethyl groups are present on fifth and sixth carbon atoms.
The parent hydrocarbon contains 10 carbon atoms and a double bond.
It is called dec-4-ene.
One methyl group is present on second carbon atom and two ethyl groups are present on fifth and sixth carbon atoms.
Acetone is isomeric to:- a)n-propyl alcohol
- b)propanal
- c)ethyl methyl ether
- d)isopropyl alcohol
Correct answer is option 'B'. Can you explain this answer?
Acetone is isomeric to:
a)
n-propyl alcohol
b)
propanal
c)
ethyl methyl ether
d)
isopropyl alcohol
|
Lavanya Menon answered |
Acetone (CH3COCH3) and Propanal (CH3CH2CHO) are functional isomers.
- Acetone:
- Propanal:
Propanone and prop-2-en-1-ol are examples of which type of isomerism?
- a)Functional isomers
- b)Chain isomers
- c)Tautomers
- d)Position isomers
Correct answer is option 'A'. Can you explain this answer?
Propanone and prop-2-en-1-ol are examples of which type of isomerism?
a)
Functional isomers
b)
Chain isomers
c)
Tautomers
d)
Position isomers
|
Gayatri Banerjee answered |
Functional isomers
Explanation: Propanone (CH3COCH3) and prop-2-en-1-ol (CH2=CHCH2OH) are examples of functional isomers because they have the same molecular formula (C3H6O) but different functional groups. Propanone has a carbonyl group (C=O) while prop-2-en-1-ol has an alcohol group (OH) and a carbon-carbon double bond (C=C).
Explanation: Propanone (CH3COCH3) and prop-2-en-1-ol (CH2=CHCH2OH) are examples of functional isomers because they have the same molecular formula (C3H6O) but different functional groups. Propanone has a carbonyl group (C=O) while prop-2-en-1-ol has an alcohol group (OH) and a carbon-carbon double bond (C=C).
How many structural isomers can compound with molecular formula ‘C3H6O’ have?- a)6
- b)4
- c)5
- d)11
Correct answer is option 'D'. Can you explain this answer?
How many structural isomers can compound with molecular formula ‘C3H6O’ have?
a)
6
b)
4
c)
5
d)
11
|
Ritu Singh answered |
The Isomers of C3H6O include:
Hence we can see that a total of 11 isomers of C3H6O are possible.
What is the common name of 2-methyl-propanal?- a)formaldehyde
- b)Isobutyraldehyde
- c)carbaldehyde
- d)acetaldehyde
Correct answer is 'B'. Can you explain this answer?
What is the common name of 2-methyl-propanal?
a)
formaldehyde
b)
Isobutyraldehyde
c)
carbaldehyde
d)
acetaldehyde
|
|
Nikita Singh answered |
- Isobutyraldehyde is the chemical compound with the formula (CH₃)₂CHCHO.
- It is an aldehyde, isomeric with n-butyraldehyde.
- Isobutyraldehyde is manufactured, often as a side-product, by the hydroformylation of propene. Its odour is described as that of wet cereal or straw.
CH3CHO and C6H5CH2CHO can be distinguished chemically by: [2012]
- a)Benedict test
- b)Iodoform test
- c)Tollen’s reagent test
- d)Fehling solution test
Correct answer is option 'B'. Can you explain this answer?
CH3CHO and C6H5CH2CHO can be distinguished chemically by: [2012]
a)
Benedict test
b)
Iodoform test
c)
Tollen’s reagent test
d)
Fehling solution test
|
Ayush Choudhury answered |
CH3CHO and C6H5CH2CHO both are aldehydes so they can give the test of Tollen’s reagent, Fehling's solution and Benedict’s solution.
The carbonyl compounds with the structure R-CO-CH3 can only give the Iodoform test.
CH3CHO is the only aldehyde which reacts with NaOH and I2 to give yellow crystals of Iodoform while C6H5CH2CHO doesn’t react with it.
So, the iodoform test is used to distinguish between CH3CHO and C6H5CH2CHO compounds.
CH3CHO + 3I2 + 4NaOH ⟶ CHI3 + HCOONa + 3NaI +3H2O
C6H5CH2CHO + I2 + 4NaOH⟶ No reaction.
The general formula CnH2nO2 could be for open chain : [AIEEE-2003]- a)Diols
- b)Dialdehydes
- c)Diketones
- d)Carboxylic acid
Correct answer is option 'D'. Can you explain this answer?
The general formula CnH2nO2 could be for open chain : [AIEEE-2003]
a)
Diols
b)
Dialdehydes
c)
Diketones
d)
Carboxylic acid
|
Hansa Sharma answered |
The correct answer is option D
The general formula CnH2nO2 could be for open chain carboxylic acids. Thus, butanoic acid CH3−CH2−CH2−COOH has formula C4H8O2. And, decanoic acid CH3−(CH2)8−COOH has formula C10H20O2..
The general formula CnH2nO2 could be for open chain carboxylic acids. Thus, butanoic acid CH3−CH2−CH2−COOH has formula C4H8O2. And, decanoic acid CH3−(CH2)8−COOH has formula C10H20O2..
In the chemical reaction, CH3CH2NH2 + CHCl3 + 3KOH → (A) + (B) + 3H2O, the compounds (A) and (B)
are respectively - [AIEEE-2007]- a)C2H5CN and 3KCl
- b)CH3CH2CONH2 and 3KCl
- c)C2H5NC and K2CO3
- d)C2H5NC and 3KCl
Correct answer is option 'D'. Can you explain this answer?
In the chemical reaction, CH3CH2NH2 + CHCl3 + 3KOH → (A) + (B) + 3H2O, the compounds (A) and (B)
are respectively - [AIEEE-2007]
are respectively - [AIEEE-2007]
a)
C2H5CN and 3KCl
b)
CH3CH2CONH2 and 3KCl
c)
C2H5NC and K2CO3
d)
C2H5NC and 3KCl
|
|
Lavanya Menon answered |
The correct answer is option D
In the chemical reaction,
CH3CH2NH2+CHCl3+3KOH→(A)+(B)+3H2O
The compounds(A) and (B) are C2H5NC and 3KCl respectively. The given reaction is carbylamine test/Isocyanide test in which aliphatic or aromatic primary amines on heating with chloroform and alcoholic potassium hydroxide give foul smelling alkyl isocyanides or carbylamines. Secondary or tertiary amines do not give this test.
In the chemical reaction,
CH3CH2NH2+CHCl3+3KOH→(A)+(B)+3H2O
The compounds(A) and (B) are C2H5NC and 3KCl respectively. The given reaction is carbylamine test/Isocyanide test in which aliphatic or aromatic primary amines on heating with chloroform and alcoholic potassium hydroxide give foul smelling alkyl isocyanides or carbylamines. Secondary or tertiary amines do not give this test.
Identify the pair of enantiomers amongst the given pairs:- a)
- b)
- c)
- d)
Correct answer is option 'B'. Can you explain this answer?
Identify the pair of enantiomers amongst the given pairs:
a)
b)
c)
d)
|
|
Nandini Iyer answered |
The correct answer is Option B.
The 1st option has R configuration and 2nd has S configuration hence they are non super imposable mirror images of each other and are enantiomers.
The 1st option has R configuration and 2nd has S configuration hence they are non super imposable mirror images of each other and are enantiomers.
Propionic acid with Br2|P yields a dibromo product. Its structure would be: [2009]- a)
- b)CH2Br – CH2 – COBr
- c)
- d)CH2 Br – CHBr – COOH
Correct answer is option 'C'. Can you explain this answer?
Propionic acid with Br2|P yields a dibromo product. Its structure would be: [2009]
a)
b)
CH2Br – CH2 – COBr
c)
d)
CH2 Br – CHBr – COOH
|
Anand Jain answered |
This reaction is an example of Hell - Volhard
Zelinsky reaction. In this reaction acids
containing α– H on treatment with X2 /P
give di-halo substituted acid.
Zelinsky reaction. In this reaction acids
containing α– H on treatment with X2 /P
give di-halo substituted acid.
Self condensation of two moles of ethyl acetatein presence of sodium ethoxide yields [2006]- a)acetoacetic ester
- b)methyl acetoacetate
- c)ethyl propionate
- d)ethyl butyrate
Correct answer is option 'A'. Can you explain this answer?
Self condensation of two moles of ethyl acetatein presence of sodium ethoxide yields [2006]
a)
acetoacetic ester
b)
methyl acetoacetate
c)
ethyl propionate
d)
ethyl butyrate
|
Ishani Nambiar answered |
It is an example of Claisen condensation. The
product is acetoacetic ester.
product is acetoacetic ester.
CH3CHO and C6H5CH2CHO can be distinguished chemically by:- a)Benedict test
- b)Iodoform test
- c)Tollen's reagent test
- d)None of these
Correct answer is option 'B'. Can you explain this answer?
CH3CHO and C6H5CH2CHO can be distinguished chemically by:
a)
Benedict test
b)
Iodoform test
c)
Tollen's reagent test
d)
None of these
|
Samridhi Bajaj answered |
CH3CHO will give iodoform test and C6H5CH2CHO will not give iodoform test.
Rate of the reaction :
is fastest when Z is : [AIEEE-2004]- a)OCOCH3
- b)NH2
- c)OC2H5
- d)Cl
Correct answer is option 'D'. Can you explain this answer?
Rate of the reaction :
is fastest when Z is : [AIEEE-2004]
a)
OCOCH3
b)
NH2
c)
OC2H5
d)
Cl
|
Preeti Iyer answered |
The correct answer is Option D.
Cl− is the best leaving group being the weakest nucleophile out of NH2− , Cl− , OC2H5 and CH3COO−.
So, when Z is Cl− , the ion leaves faster thus making the reaction the fastest when Z is Cl− .
Cl− is the best leaving group being the weakest nucleophile out of NH2− , Cl− , OC2H5 and CH3COO−.
So, when Z is Cl− , the ion leaves faster thus making the reaction the fastest when Z is Cl− .
One or More than One Options Correct TypeDirection (Q. Nos. 9-14) This section contains 6 multiple choice questions. Each question has four choices (a), (b), (c) and (d), out of which ONE or MORE THAN ONE are correct.Q. Which of the following form(s) yellow precipitate with NaOH/I2 solution?- a)CH3CHO
- b)
- c)
- d)CH3CH2OH
Correct answer is option 'A,B,C,D'. Can you explain this answer?
One or More than One Options Correct Type
Direction (Q. Nos. 9-14) This section contains 6 multiple choice questions. Each question has four choices (a), (b), (c) and (d), out of which ONE or MORE THAN ONE are correct.
Q.
Which of the following form(s) yellow precipitate with NaOH/I2 solution?
a)
CH3CHO
b)
c)
d)
CH3CH2OH
|
Pioneer Academy answered |
Carbonyls with 
group or an alcohol with a methyl group and a “H” at α-carbon gives iodoform test.
group or an alcohol with a methyl group and a “H” at α-carbon gives iodoform test. Only One Option Correct TypeDirection (Q. Nos. 1-8) This section contains 8 multiple choice questions. Each question has four choices (a), (b), (c) and (d), out of which ONLY ONE is correct.Q. Tollen’s reagent used for the distinction of aldehydes with ketones is- a)Pb(NO3),-NH3(aq)
- b)Cu(NO3)2-NH3(aq)
- c)AgNO3-NH3(aq)
- d)Cu(ll) citrate-NH3(aq)
Correct answer is option 'C'. Can you explain this answer?
Only One Option Correct Type
Direction (Q. Nos. 1-8) This section contains 8 multiple choice questions. Each question has four choices (a), (b), (c) and (d), out of which ONLY ONE is correct.
Q.
Tollen’s reagent used for the distinction of aldehydes with ketones is
a)
Pb(NO3),-NH3(aq)
b)
Cu(NO3)2-NH3(aq)
c)
AgNO3-NH3(aq)
d)
Cu(ll) citrate-NH3(aq)
|
Geetika Shah answered |
Tollen’s reagent is ammoniacal solution of silver nitrate in which silver remains as [Ag(NH3)2]+.
A hydrocarbon A (C10H18) is capable of showing both enantiomerism as well as diastereomerism. Treatment of A either with HgSO4 / H2SO4 or B2H6 / H2O2 -NaOH results in the same carbonyl compound B. Also,
C can also be obtained as one of the product in the following reaction.
Consider the reaction given below,
Q. How many different alcohols are expected?- a)1
- b)2
- c)3
- d)4
Correct answer is option 'D'. Can you explain this answer?
A hydrocarbon A (C10H18) is capable of showing both enantiomerism as well as diastereomerism. Treatment of A either with HgSO4 / H2SO4 or B2H6 / H2O2 -NaOH results in the same carbonyl compound B. Also,
C can also be obtained as one of the product in the following reaction.Consider the reaction given below,
Q.
How many different alcohols are expected?
a)
1
b)
2
c)
3
d)
4
|
Yoganand Pendem answered |
C can be pentanal or 2- Me butanal .Both on rxn with CH3MgBr followed by hydrolysis yield enantiomeric pair of secondary alcohols.So a total of 2+2=4 different alcohols are formed.
When CH2 = CHCOOH is reduced with LiAlH4, the compound obtained is: [AIEEE-2003]
a) CH3CH2CH2OH
b) CH3CH2CHO
c) CH2=CH-CH2OH
d) CH3CH2COOH
Correct answer is option 'C'. Can you explain this answer?
|
|
Hansa Sharma answered |
The correct answer is option C
LiAlH4 can reduce the carboxylic acid group without affecting the double bond because alkene is an electron-rich species.
LiAlH4 can reduce the carboxylic acid group without affecting the double bond because alkene is an electron-rich species.
Only One Option Correct TypeDirection (Q, Nos. 1-9) This section contains 9 multiple choice questions. Each question has four choices (a), (b), (c) and (d), out of which ONLY ONE is correct.Q. Which of the following will give a racemic mixture on reduction with NaBH4 followed by acid work-up?- a)
- b)
- c)
- d)
Correct answer is option 'B'. Can you explain this answer?
Only One Option Correct Type
Direction (Q, Nos. 1-9) This section contains 9 multiple choice questions. Each question has four choices (a), (b), (c) and (d), out of which ONLY ONE is correct.
Q.
Which of the following will give a racemic mixture on reduction with NaBH4 followed by acid work-up?
a)
b)
c)
d)
|
Amisha answered |
The ans is B . coz after the action of NaBH4 ,chiral carbon is present only in the product of B.And hence it can give both R and S configuration.
Which one of the following methods is neither meant for the synthesis nor for separation of amines ?
[AIEEE-2005]- a)Hofmann method
- b)Hinsberg method
- c)Curtius reaction
- d)Wurtz reaction
Correct answer is option 'D'. Can you explain this answer?
Which one of the following methods is neither meant for the synthesis nor for separation of amines ?
[AIEEE-2005]
[AIEEE-2005]
a)
Hofmann method
b)
Hinsberg method
c)
Curtius reaction
d)
Wurtz reaction
|
|
Nikita Singh answered |
The correct answer is Option D.
Wurtz reaction is used to produce alkanes from alkyl halides.
2RX+2Na (in ether)→R−R+2NaX
Example :
2CH3Cl+2Na (in ether)→C2H6+2NaCl
So, the Wurtz reaction has nothing to do with amines.
2RX+2Na (in ether)→R−R+2NaX
Example :
2CH3Cl+2Na (in ether)→C2H6+2NaCl
So, the Wurtz reaction has nothing to do with amines.
In the anion HCOO- the two carbon-oxygen bonds are found to be of equal length. What is the reason for it ? [AIEEE-2003]- a)The anion HCOO- has two resonating structures
- b)The anion is obtained by removal of a proton from the acid molecule
- c)Electronic orbitals of carbon atom are hybridised
- d)The C=O bond is weaker than the C - O bond
Correct answer is option 'A'. Can you explain this answer?
In the anion HCOO- the two carbon-oxygen bonds are found to be of equal length. What is the reason for it ? [AIEEE-2003]
a)
The anion HCOO- has two resonating structures
b)
The anion is obtained by removal of a proton from the acid molecule
c)
Electronic orbitals of carbon atom are hybridised
d)
The C=O bond is weaker than the C - O bond
|
|
Rohit Shah answered |
InHCOO-iontwo carbon-oxygen bonds are found to be of equal length because this ion is stabilized by resonance and in resonance both C-O bonds have partial double bond character.
Which of the following could result as a product in the aldol condensation reaction?- a)4-methyl-3-penten-2-on
- b)4-methyl-4-penten-2-one
- c)4-methyl-5-hexen-2-one
- d)3-methyl-4-hexen-2-one
Correct answer is option 'A'. Can you explain this answer?
Which of the following could result as a product in the aldol condensation reaction?
a)
4-methyl-3-penten-2-on
b)
4-methyl-4-penten-2-one
c)
4-methyl-5-hexen-2-one
d)
3-methyl-4-hexen-2-one
|
Maulik Verma answered |
It is an α, β-unsaturated ketone which can be formed in an aldol condensation followed by dehydration.
The incorrect statement regarding oxo process for synthesis of an aldehyde is- a)Mixture of CO and H2 is allowed to react with an alkene
- b)Co2(CO)8 may be used as a catalyst
- c)[Co(CO)4H] may act as a catalyst
- d)This process can also be used in the same manner for the synthesis of ketone
Correct answer is option 'D'. Can you explain this answer?
The incorrect statement regarding oxo process for synthesis of an aldehyde is
a)
Mixture of CO and H2 is allowed to react with an alkene
b)
Co2(CO)8 may be used as a catalyst
c)
[Co(CO)4H] may act as a catalyst
d)
This process can also be used in the same manner for the synthesis of ketone
|
|
Yoganand Pendem answered |
Oxo process involves formulation of alkene to form aldehyde only not ketone
The oxidation of toluene to benzoic acid can be stopped at the aldehyde stage. The reaction is called?- a)Etard reaction
- b)Stephen reaction
- c)Friedel – crafts reaction
- d)Gatterman – Koch reaction
Correct answer is option 'A'. Can you explain this answer?
The oxidation of toluene to benzoic acid can be stopped at the aldehyde stage. The reaction is called?
a)
Etard reaction
b)
Stephen reaction
c)
Friedel – crafts reaction
d)
Gatterman – Koch reaction
|
Aryan Sen answered |
Etard reaction
The oxidation of toluene to benzoic acid can be stopped at the aldehyde stage by using the Etard reaction. The Etard reaction is a chemical reaction that involves the oxidation of aromatic hydrocarbons using chromyl chloride (CrO2Cl2) as the oxidizing agent. This reaction is commonly used to convert alkylbenzenes, such as toluene, into their corresponding aldehydes.
Reaction Mechanism
The reaction proceeds through several steps:
1. Formation of the chromyl chloride complex: Chromyl chloride is formed by the reaction of chromic acid (H2CrO4) with sodium chloride (NaCl). The chromyl chloride complex acts as the oxidizing agent in the Etard reaction.
2. Activation of the aromatic ring: The aromatic ring of toluene is activated through the coordination of the chromyl chloride complex with the electron-rich aromatic system. This coordination weakens the carbon-hydrogen bonds adjacent to the aromatic ring, making them more susceptible to oxidation.
3. Oxidation of toluene to benzaldehyde: The activated toluene undergoes oxidation by the chromyl chloride complex, resulting in the formation of benzaldehyde. The chromyl chloride complex acts as a strong electrophile, attacking the activated carbon-hydrogen bond and forming a carbocation intermediate. This intermediate is then hydrolyzed to yield benzaldehyde.
4. Further oxidation to benzoic acid: If the reaction is allowed to continue, benzaldehyde can undergo further oxidation to form benzoic acid. However, to stop the reaction at the aldehyde stage, the reaction conditions can be carefully controlled, such as by adjusting the reaction temperature or using a milder oxidizing agent.
Applications
The Etard reaction has several applications in organic synthesis. It is commonly used for the selective oxidation of alkylbenzenes to their corresponding aldehydes, as demonstrated in the oxidation of toluene to benzaldehyde. This reaction is useful for the synthesis of various aromatic aldehydes, which are important intermediates in the preparation of pharmaceuticals, fragrances, and other organic compounds.
Conclusion
The oxidation of toluene to benzoic acid can be stopped at the aldehyde stage by using the Etard reaction. This reaction involves the oxidation of toluene to benzaldehyde using chromyl chloride as the oxidizing agent. By carefully controlling the reaction conditions, it is possible to halt the reaction at the aldehyde stage and prevent further oxidation to benzoic acid. The Etard reaction has important applications in organic synthesis, particularly in the preparation of aromatic aldehydes.
The oxidation of toluene to benzoic acid can be stopped at the aldehyde stage by using the Etard reaction. The Etard reaction is a chemical reaction that involves the oxidation of aromatic hydrocarbons using chromyl chloride (CrO2Cl2) as the oxidizing agent. This reaction is commonly used to convert alkylbenzenes, such as toluene, into their corresponding aldehydes.
Reaction Mechanism
The reaction proceeds through several steps:
1. Formation of the chromyl chloride complex: Chromyl chloride is formed by the reaction of chromic acid (H2CrO4) with sodium chloride (NaCl). The chromyl chloride complex acts as the oxidizing agent in the Etard reaction.
2. Activation of the aromatic ring: The aromatic ring of toluene is activated through the coordination of the chromyl chloride complex with the electron-rich aromatic system. This coordination weakens the carbon-hydrogen bonds adjacent to the aromatic ring, making them more susceptible to oxidation.
3. Oxidation of toluene to benzaldehyde: The activated toluene undergoes oxidation by the chromyl chloride complex, resulting in the formation of benzaldehyde. The chromyl chloride complex acts as a strong electrophile, attacking the activated carbon-hydrogen bond and forming a carbocation intermediate. This intermediate is then hydrolyzed to yield benzaldehyde.
4. Further oxidation to benzoic acid: If the reaction is allowed to continue, benzaldehyde can undergo further oxidation to form benzoic acid. However, to stop the reaction at the aldehyde stage, the reaction conditions can be carefully controlled, such as by adjusting the reaction temperature or using a milder oxidizing agent.
Applications
The Etard reaction has several applications in organic synthesis. It is commonly used for the selective oxidation of alkylbenzenes to their corresponding aldehydes, as demonstrated in the oxidation of toluene to benzaldehyde. This reaction is useful for the synthesis of various aromatic aldehydes, which are important intermediates in the preparation of pharmaceuticals, fragrances, and other organic compounds.
Conclusion
The oxidation of toluene to benzoic acid can be stopped at the aldehyde stage by using the Etard reaction. This reaction involves the oxidation of toluene to benzaldehyde using chromyl chloride as the oxidizing agent. By carefully controlling the reaction conditions, it is possible to halt the reaction at the aldehyde stage and prevent further oxidation to benzoic acid. The Etard reaction has important applications in organic synthesis, particularly in the preparation of aromatic aldehydes.
In the propanoate ion- a)both the carbon – oxygen bonds are the same length.
- b)the carbon atom bears a – 1 charge.
- c)one of the oxygen atoms bears a – 1 charge
- d)the carbon – oxygen double bond is shorter.
Correct answer is option 'A'. Can you explain this answer?
In the propanoate ion
a)
both the carbon – oxygen bonds are the same length.
b)
the carbon atom bears a – 1 charge.
c)
one of the oxygen atoms bears a – 1 charge
d)
the carbon – oxygen double bond is shorter.
|
Vivek Rana answered |
This is because of Resonance.
Which is the most suitable reagent for the following transformation?
- a)Zn(Hg)-HCI
- b)N2H4/NaOH/Heat
- c)LiAIH4
- d)NaBH4
Correct answer is option 'A'. Can you explain this answer?
Which is the most suitable reagent for the following transformation?
a)
Zn(Hg)-HCI
b)
N2H4/NaOH/Heat
c)
LiAIH4
d)
NaBH4
|
|
Geetika Shah answered |
Clemmensen reduction is suitable for reductio n of carbonyls containing additional acidic functional group.
A liquid was mixed with ethanol and a drop of concentrated H2SO4 was added. A compound with a fruity
smell was formed. The liquid was : [AIEEE-2009]- a)CH3OH
- b)HCHO
- c)CH3COCH3
- d)CH3COOH
Correct answer is option 'D'. Can you explain this answer?
A liquid was mixed with ethanol and a drop of concentrated H2SO4 was added. A compound with a fruity
smell was formed. The liquid was : [AIEEE-2009]
smell was formed. The liquid was : [AIEEE-2009]
a)
CH3OH
b)
HCHO
c)
CH3COCH3
d)
CH3COOH
|
|
Swati Verma answered |
The correct answer is Option D.
The fruity smell is evolved due to the formation of ester when ethanol and acid are treated in the presence of concentrated H2SO4.
The reaction is
CH3COOH + CH3CH2OH → CH3COOC2H5
The reaction is
CH3COOH + CH3CH2OH → CH3COOC2H5
Arrange the following compounds in decreasing order of their acid strength: i) trichloroacetic acid ii) trifluoroacetic acid iii) acetic acid and iv) formic acid- a)trifluoroacetic acid, trichloroacetic acid, formic acid and acetic acid
- b)formic acid., trifluoroacetic acid, trichloroacetic acid, and acetic acid
- c)trichloroacetic acid, trifluoroacetic acid, acetic acid and formic acid.
- d)trifluoroacetic acid, formic acid acetic acid and Propan – 1 – ol, 4 – methylphenol
Correct answer is option 'A'. Can you explain this answer?
Arrange the following compounds in decreasing order of their acid strength: i) trichloroacetic acid ii) trifluoroacetic acid iii) acetic acid and iv) formic acid
a)
trifluoroacetic acid, trichloroacetic acid, formic acid and acetic acid
b)
formic acid., trifluoroacetic acid, trichloroacetic acid, and acetic acid
c)
trichloroacetic acid, trifluoroacetic acid, acetic acid and formic acid.
d)
trifluoroacetic acid, formic acid acetic acid and Propan – 1 – ol, 4 – methylphenol
|
Tanuja Kapoor answered |
Acidic strength of carboxylic acid -
– More acidic than phenols or alcohols.
– Acidity increase with the presence of a group with -I effect in the alkyl group.Whereas it decreases with the presence of +I group.
– Acidity increases with increase in the number of halogen atoms on 
- position.
- position.– It decreases with increasing distance of halogen from 
– It increases with increase in the electronegativity of halogen.
- CF3COOH > CCl3COOH > HCOOH > CH3 COOH
The compound formed in the positive test for nitrogen with the Lassaigne solution of an organic compound is - [AIEEE-2004]- a)Fe4[Fe(CN)6]3
- b)Na3[Fe(CN)6]
- c)Fe(CN)3
- d)Na4[Fe(CN)5NOS]
Correct answer is option 'A'. Can you explain this answer?
The compound formed in the positive test for nitrogen with the Lassaigne solution of an organic compound is - [AIEEE-2004]
a)
Fe4[Fe(CN)6]3
b)
Na3[Fe(CN)6]
c)
Fe(CN)3
d)
Na4[Fe(CN)5NOS]
|
|
Neha Sharma answered |
The correct answer is option A
It is ferri ferrocyanide
6NaCN+FeSO4→Na4[Fe(CN)6]+Na2SO4
3Na4[Fe(CN)6]+2Fe2(SO4)3→Fe4[Fe(CN)6]3+6Na2SO4
Prussian blue
It is ferri ferrocyanide
6NaCN+FeSO4→Na4[Fe(CN)6]+Na2SO4
3Na4[Fe(CN)6]+2Fe2(SO4)3→Fe4[Fe(CN)6]3+6Na2SO4
Prussian blue
Iodoform test is not given by [1999]- a)2-Pentanone
- b)Ethanol
- c)Ethanal
- d)3-Pentanone
Correct answer is option 'D'. Can you explain this answer?
Iodoform test is not given by [1999]
a)
2-Pentanone
b)
Ethanol
c)
Ethanal
d)
3-Pentanone
|
Priyanka Iyer answered |
Iodoform test is exhibited by ethyl alcohol acetaldehyde, acetone methyl ketones and those alcohols which possess CH3CH(OH)- group. As 3-pentanone does not contain
CH3CO-group as therefore it does not give iodoform test.
CH3CO-group as therefore it does not give iodoform test.
Which of the following is incorrect? [2001]- a)NaHSO3 is used in detection of carbonylcompound
- b)FeCl3 is used in detection of phenolic group
- c)Tollen reagent is used in detection ofunsaturation
- d)Fehling solution is used in detection ofglucose
Correct answer is option 'C'. Can you explain this answer?
Which of the following is incorrect? [2001]
a)
NaHSO3 is used in detection of carbonylcompound
b)
FeCl3 is used in detection of phenolic group
c)
Tollen reagent is used in detection ofunsaturation
d)
Fehling solution is used in detection ofglucose
|
|
Priyanka Iyer answered |
Tollen's reagent is used to detect of
aldehydes. Tollen's reagent is an ammonical
solution of silver nitrate. When aldehyde is
added to Tollen's reagent, silver oxide is
reduced to metallic silver which deposits as
mirror.
aldehydes. Tollen's reagent is an ammonical
solution of silver nitrate. When aldehyde is
added to Tollen's reagent, silver oxide is
reduced to metallic silver which deposits as
mirror.
One mole of a symmetrical alkane on ozonolysis gives two moles of an aldehyde having molecular mass of 44u. The alkene is: - a)Ethene
- b)Propene
- c)1-butene
- d)2-butene
Correct answer is option 'D'. Can you explain this answer?
One mole of a symmetrical alkane on ozonolysis gives two moles of an aldehyde having molecular mass of 44u. The alkene is:
a)
Ethene
b)
Propene
c)
1-butene
d)
2-butene
|
|
Priya Patel answered |
One mole of a symmetrical alkene on ozonolysis gives two moles of an aldehyde having a molecular weight of mass of 44 u. The alkene is
CH3CH=CHCH3 +O3 ------2CH3CHO
molecular mass of CH3CHO is 12*2+4*1+16 =44
Thus the alkene is 2 butene and the aldehyde formed is ethanal(acetaldehyde)
The compound formed as a result of oxidation of ethyl benzene by KMnO4 is – [AIEEE-2007]- a)benzophenone
- b)acetophenone
- c)benzoic acid
- d)benzyl alcohol
Correct answer is option 'C'. Can you explain this answer?
The compound formed as a result of oxidation of ethyl benzene by KMnO4 is – [AIEEE-2007]
a)
benzophenone
b)
acetophenone
c)
benzoic acid
d)
benzyl alcohol
|
Rishu Verma answered |
Any benzene derivative ( benzene alkyl ) which have at least one alpha hydrogen gives benzoic acid on oxidation with kmno4 ........what ever the lenght or no of carbon atom on benzene if it has alpha hydrogen it will oxidise to carboxylic grp
Consider the following reaction,
(A pure enantiomer)Q. The incorrect statement regarding X is- a)It is an aldehyde
- b)Product has inverted configuration at the α-carbon
- c)X is a racemic mixture
- d)Product can be either laevorotatory or dextrorotatory isomer
Correct answer is option 'C'. Can you explain this answer?
Consider the following reaction,
(A pure enantiomer)
Q.
The incorrect statement regarding X is
a)
It is an aldehyde
b)
Product has inverted configuration at the α-carbon
c)
X is a racemic mixture
d)
Product can be either laevorotatory or dextrorotatory isomer
|
Om Desai answered |
X will be a pure enantiomer of aldehyde.
An organic compound A (C7H16O) shows both enantiomerism and diastereomerism. Treatment of a pure enantiomer of A with Na2CrO4 /Dil. H2SO4 gives B (C7H14O) - Also A on dehydration with concentrated H2SO4 gives a single alkene C (C7H14). Ozonolysis of C followed by work-up with Zn-H2O gives D (C5H10O) as one of the product which gives racemic mixture on reduction with NaBH4.Q. If B is reduced with LiAIH4 followed by acid hydrolysis will give- a)a pure enantiomer
- b)a racemic mixture
- c)a pair of diastereomers
- d)an achiral alcohol
Correct answer is option 'C'. Can you explain this answer?
An organic compound A (C7H16O) shows both enantiomerism and diastereomerism. Treatment of a pure enantiomer of A with Na2CrO4 /Dil. H2SO4 gives B (C7H14O) - Also A on dehydration with concentrated H2SO4 gives a single alkene C (C7H14). Ozonolysis of C followed by work-up with Zn-H2O gives D (C5H10O) as one of the product which gives racemic mixture on reduction with NaBH4.
Q.
If B is reduced with LiAIH4 followed by acid hydrolysis will give
a)
a pure enantiomer
b)
a racemic mixture
c)
a pair of diastereomers
d)
an achiral alcohol
|
|
Kashish Garg answered |
Since, A is completely saturated, it must contain more than one chiral carbon atoms in order to show both enantiomerism and diastereomerism. Also, C on ozonolysis gives D (C5H10O) as one product, other product must be CH3SHO. Hence, C must be 
B is enantiomeric. If a pure enantiomer of B is reduced with LiAIH4, pair of diastereomers would be formed.
B is enantiomeric. If a pure enantiomer of B is reduced with LiAIH4, pair of diastereomers would be formed.
A strong base can abstract an α – hydrogen from- a)Alkane.
- b)Amine
- c)Ketone
- d)Alkene
Correct answer is option 'C'. Can you explain this answer?
A strong base can abstract an α – hydrogen from
a)
Alkane.
b)
Amine
c)
Ketone
d)
Alkene
|
|
Vijay Bansal answered |
After deprotonation the negative charge will be in conjugation with the pi orbital of carbonyl..so base will prefer to abstract alpha hydrogen from ketone.
Which one of the following is reduced with zinc and hydrochloric acid to give the corresponding
hydrocarbon : [AIEEE-2004]- a)Butan-2-one
- b)Acetic acid
- c)Acetamide
- d)Ethyl acetate
Correct answer is option 'A'. Can you explain this answer?
Which one of the following is reduced with zinc and hydrochloric acid to give the corresponding
hydrocarbon : [AIEEE-2004]
hydrocarbon : [AIEEE-2004]
a)
Butan-2-one
b)
Acetic acid
c)
Acetamide
d)
Ethyl acetate
|
|
Priya Patel answered |
Butan-2-one will get reduced into butane when treated with zinc and hydrochloric acid following Clemmensen reduction reaction whereas zn/HCl do not reduce ester,acid and amide.
Only One Option Correct TypeDirection (Q. Nos. 1-10) This section contains 10 multiple choice questions. Each question has four choices (a), (b), (c) and (d), out of which ONLY ONE is correct.Q. Arrange the following in the increasing order of reactivity with NH3.I. CH2O
II. CH3CHO
III. CH3—CO—CH3
- a)I < II < III < IV
- b)IV < III < II < I
- c)III < IV < I < II
- d)II < I < IV < III
Correct answer is option 'B'. Can you explain this answer?
Only One Option Correct Type
Direction (Q. Nos. 1-10) This section contains 10 multiple choice questions. Each question has four choices (a), (b), (c) and (d), out of which ONLY ONE is correct.
Q.
Arrange the following in the increasing order of reactivity with NH3.
I. CH2O
II. CH3CHO
III. CH3—CO—CH3
II. CH3CHO
III. CH3—CO—CH3
a)
I < II < III < IV
b)
IV < III < II < I
c)
III < IV < I < II
d)
II < I < IV < III
|
Harsharma answered |
I think this is because of steric hindrance.... because NH3 attack carbonyl carbon as nucleophile (Sn2)...and then steric hindrance is main factor
IUPAC name of compound
- a)3-Bromo7-chloro-7ethyl-5-(1,1-dimethyethyl)-5-(2-methylpropyl)-3-methylnonane
- b)3-Bromo7-chloro-5-(1,1-dimethyethyl)-7-ethyl-3methyl-5-(2-methylpropyl)nonane
- c)3-Bromo7-chloro-7ethyl-3-,methyl-5-(1,1-dimethyethyl)-5-(2-methylpropyl)nonane
- d)3-Bromo-5-(1,1-dimethyethyl)-5-(2-methylpropyl)-7-chloro-7ethyl-5--3-methylnonane
Correct answer is option 'B'. Can you explain this answer?
IUPAC name of compound
a)
3-Bromo7-chloro-7ethyl-5-(1,1-dimethyethyl)-5-(2-methylpropyl)-3-methylnonane
b)
3-Bromo7-chloro-5-(1,1-dimethyethyl)-7-ethyl-3methyl-5-(2-methylpropyl)nonane
c)
3-Bromo7-chloro-7ethyl-3-,methyl-5-(1,1-dimethyethyl)-5-(2-methylpropyl)nonane
d)
3-Bromo-5-(1,1-dimethyethyl)-5-(2-methylpropyl)-7-chloro-7ethyl-5--3-methylnonane
|
Infinity Academy answered |
The correct answer is option B
First of all, we should number the carbon atoms from right to left for the above molecule. According to IUPAC rule, we have to name the functional groups alphabetically. Hence, bromine which is attached to 3rd carbon is named as 3-bromo, chlorine which is attached to 7th carbon is named as 7-chloro and ethyl group which is also attached to 7th carbon is named as 7-ethyl. 5-(1,1-dimethyethyl) indicates that, 2 methyl groups are attached to the 1st carbon of ethyl group which is attached to the 5th carbon of the given molecule. 5-(2-methylpropyl) indicates that, a methyl group is attached to the 2nd carbon of propyl group which is attached to the 5th carbon of the given molecule. Finally, since the molecule has nine atoms and it is an alkane, the word nonane is added to the IUPAC name.
First of all, we should number the carbon atoms from right to left for the above molecule. According to IUPAC rule, we have to name the functional groups alphabetically. Hence, bromine which is attached to 3rd carbon is named as 3-bromo, chlorine which is attached to 7th carbon is named as 7-chloro and ethyl group which is also attached to 7th carbon is named as 7-ethyl. 5-(1,1-dimethyethyl) indicates that, 2 methyl groups are attached to the 1st carbon of ethyl group which is attached to the 5th carbon of the given molecule. 5-(2-methylpropyl) indicates that, a methyl group is attached to the 2nd carbon of propyl group which is attached to the 5th carbon of the given molecule. Finally, since the molecule has nine atoms and it is an alkane, the word nonane is added to the IUPAC name.
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