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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.
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
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.
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.
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.
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.
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
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.
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.
In which of the following reactions, ketone is formed as the major organic product?- a)
- b)
- c)
- d)
Correct answer is option 'A,C'. Can you explain this answer?
In which of the following reactions, ketone is formed as the major organic product?
a)
b)
c)
d)
|
Infinity Academy answered |
If acid derivatives like nitrile, acid chlorid e or ester is taken in excess in Grignard synthesis, second addition of Grignard’s reagent on carbonyl product does not succeed and carbonyls are obtained as major products.
In option (b), carboxylic acids and in option (d), an aldehyde is formed.
In option (b), carboxylic acids and in option (d), an aldehyde is formed.
All of the following reaction gives atleast one ketone as a significant organic product except- a)
- b)
- c)
- d)
Correct answer is option 'B'. Can you explain this answer?
All of the following reaction gives atleast one ketone as a significant organic product except
a)
b)
c)
d)
|
|
Vivek Rana answered |
It gives aldehydes as major product.
What compound is produced when cyclohexene is treated with concentrated KMnO4?- a)adipic acid
- b)succinic acid
- c)hexanoic acid
- d)cyclohexanecarboxylic acid
Correct answer is option 'A'. Can you explain this answer?
What compound is produced when cyclohexene is treated with concentrated KMnO4?
a)
adipic acid
b)
succinic acid
c)
hexanoic acid
d)
cyclohexanecarboxylic acid
|
Dishani Kulkarni answered |
Conc KMnO4 will cause oxidation and ring opening forming adipic acid.
Which one of the following compounds will be most readily dehydrated? [2010]- a)
- b)
- c)
- d)
Correct answer is option 'D'. Can you explain this answer?
Which one of the following compounds will be most readily dehydrated? [2010]
a)
b)
c)
d)
|
Raghav Khanna answered |
The intermediate is carbocation which is
destabilised by C = O group (present on ��-
carbon to the –OH group) in the first three
cases. In (d), α–hydrogen is more acidic which
can be removed as water. Moreover, the
positive charge on the intermediate carbocation
is relatively away from the C = O group.
destabilised by C = O group (present on ��-
carbon to the –OH group) in the first three
cases. In (d), α–hydrogen is more acidic which
can be removed as water. Moreover, the
positive charge on the intermediate carbocation
is relatively away from the C = O group.
∴ Correct choice : (d)
Clemmensen reduction of a ketone is carried outin the presence of which of the following ? [2011]- a)Glycol with KOH
- b)Zn-Hg with HCl
- c)Li Al H4
- d)H2 and Pt as catalyst
Correct answer is option 'B'. Can you explain this answer?
Clemmensen reduction of a ketone is carried outin the presence of which of the following ? [2011]
a)
Glycol with KOH
b)
Zn-Hg with HCl
c)
Li Al H4
d)
H2 and Pt as catalyst
|
Sankar Banerjee answered |
Clemmensen Reduction and Its Mechanism:
The Clemmensen reduction is a chemical reaction that involves the conversion of a carbonyl group in a ketone to a methylene group using zinc amalgam and hydrochloric acid. This process is particularly effective for reducing ketones to alkanes.
Role of Zn-Hg with HCl:
- In the Clemmensen reduction, zinc amalgam (Zn-Hg) serves as the reducing agent while hydrochloric acid (HCl) acts as a catalyst.
- The reaction occurs at high temperatures, typically refluxing conditions, to facilitate the reduction of the carbonyl group.
- The zinc amalgam reacts with the carbonyl group in the ketone, leading to the formation of an alkyl zinc intermediate.
- The alkyl zinc intermediate is unstable and undergoes protonation by HCl to form the corresponding alkane.
Comparison with Other Reagents:
- Glycol with KOH is commonly used for cleaving ethers, not for the Clemmensen reduction of ketones.
- LiAlH4 is a powerful reducing agent that can reduce ketones to alcohols, but it is not typically used in the Clemmensen reduction.
- H2 and Pt as a catalyst are commonly used for hydrogenation reactions, not for the Clemmensen reduction.
In conclusion, the Clemmensen reduction of a ketone is carried out in the presence of zinc amalgam and hydrochloric acid. These reagents work together to reduce the carbonyl group in the ketone to a methylene group, producing the desired alkane product.
Which of the following reactions will not resultin the formation of carbon-carbon bonds?- a)Reimer-Tieman reaction [2010]
- b)Cannizaro reaction
- c)Wurtz reaction
- d)Friedel-Crafts acylation
Correct answer is option 'B'. Can you explain this answer?
Which of the following reactions will not resultin the formation of carbon-carbon bonds?
a)
Reimer-Tieman reaction [2010]
b)
Cannizaro reaction
c)
Wurtz reaction
d)
Friedel-Crafts acylation
|
Aashna Mukherjee answered |
Note that new C–C bond is formed is a, c and d.
The common name for pentanedioic acid is:- a)succinic acid
- b)oxalic acid
- c)glutaric acid
- d)pimelic acid
Correct answer is option 'C'. Can you explain this answer?
The common name for pentanedioic acid is:
a)
succinic acid
b)
oxalic acid
c)
glutaric acid
d)
pimelic acid
|
|
Aryan Sen answered |
Pentanedioic acid is glutaric acid.
The compound obtained when acetaldehyde reacts with dilute aqueous sodium hydroxide exhibits:- a)geometrical isomerism
- b)both optical and geometrical isomerism
- c)optical isomerism.
- d)neither optical nor geometrical isomerism
Correct answer is option 'A'. Can you explain this answer?
The compound obtained when acetaldehyde reacts with dilute aqueous sodium hydroxide exhibits:
a)
geometrical isomerism
b)
both optical and geometrical isomerism
c)
optical isomerism.
d)
neither optical nor geometrical isomerism
|
Amrutha Pillai answered |
CH3CH=CHCHO is the product which is suitably substituted so will show geometrical isomer.
The relative reactivities of acyl compounds towardsnucleophilic substitution are in the order of : [2008]- a)Acyl chloride > Acid anhydride > Ester >Amide
- b)Ester > Acyl chloride > Amide > Acidanhydride
- c)Acid anhydride > Amide > Ester > Acylchloride
- d)Acyl chloride > Ester > Acid anhydride >Amide
Correct answer is option 'A'. Can you explain this answer?
The relative reactivities of acyl compounds towardsnucleophilic substitution are in the order of : [2008]
a)
Acyl chloride > Acid anhydride > Ester >Amide
b)
Ester > Acyl chloride > Amide > Acidanhydride
c)
Acid anhydride > Amide > Ester > Acylchloride
d)
Acyl chloride > Ester > Acid anhydride >Amide
|
Diya Datta answered |
The relative reactivities of acyl compounds
towards nucleophilic substitution follow
the order Acyl halides > Acid anhydride >
Ester > Amide. Thus the correct answer is
(a).
towards nucleophilic substitution follow
the order Acyl halides > Acid anhydride >
Ester > Amide. Thus the correct answer is
(a).
An organic compound ‘A’ on treatment with NH3 gives ‘B’ which on heating gives ‘C’, ‘C’ whentreated with Br2 in the presence of KOH produces ethylamine. Compound ‘A’ is:[2011 M]- a)CH3COOH
- b)CH3 CH2 CH2 COOH
- c)
- d)CH3CH2COOH
Correct answer is option 'D'. Can you explain this answer?
An organic compound ‘A’ on treatment with NH3 gives ‘B’ which on heating gives ‘C’, ‘C’ whentreated with Br2 in the presence of KOH produces ethylamine. Compound ‘A’ is:[2011 M]
a)
CH3COOH
b)
CH3 CH2 CH2 COOH
c)
d)
CH3CH2COOH
|
Surbhi Das answered |
Since, C when heated with Br2 in presence
of KOH produces ethylamine, hence it must
be propanamide and hence the organic
compound (A) will be propanoic acid. The
reactions follows.
of KOH produces ethylamine, hence it must
be propanamide and hence the organic
compound (A) will be propanoic acid. The
reactions follows.
Many naturally occurring aldehydes and ketones are used in the blending of perfumes and flavouring agents. But the preferred ones are- a)higher ketones
- b)lower aldehydes
- c)lower ketones
- d)higher aldehydes
Correct answer is option 'D'. Can you explain this answer?
Many naturally occurring aldehydes and ketones are used in the blending of perfumes and flavouring agents. But the preferred ones are
a)
higher ketones
b)
lower aldehydes
c)
lower ketones
d)
higher aldehydes
|
Pritam Malik answered |
Higher aldehydes are preferred.
Which of the following is correct?
- a)On reduction any aldehyde gives secondary alcohol
- b)Reaction of vegetable oil with H2SO4 gives glycerine
- c)Alcoholic iodine with NaOH gives iodoform
- d)Sucrose on reaction with NaCl gives invert sugar
Correct answer is option 'C'. Can you explain this answer?
Which of the following is correct?
a)
On reduction any aldehyde gives secondary alcohol
b)
Reaction of vegetable oil with H2SO4 gives glycerine
c)
Alcoholic iodine with NaOH gives iodoform
d)
Sucrose on reaction with NaCl gives invert sugar
|
Pooja Pillai answered |
Which of the following represents the correctorder of the acidity in the given compounds? [2007]- a)FCH2COOH > CH3COOH > BrCH2COOH >ClCH2COOH
- b)BrCH2COOH > ClCH2COOH > FCH2COOH> CH3COOH
- c)FCH2COOH > ClCH2COOH > BrCH2COOH> CH3COOH
- d)CH3 COOH > BrCH2COOH > ClCH2COOH >FCH2COOH
Correct answer is option 'C'. Can you explain this answer?
Which of the following represents the correctorder of the acidity in the given compounds? [2007]
a)
FCH2COOH > CH3COOH > BrCH2COOH >ClCH2COOH
b)
BrCH2COOH > ClCH2COOH > FCH2COOH> CH3COOH
c)
FCH2COOH > ClCH2COOH > BrCH2COOH> CH3COOH
d)
CH3 COOH > BrCH2COOH > ClCH2COOH >FCH2COOH
|
Sneha Basak answered |
Electron withdrawing substituent (like
halogen, —NO2, C6H5 etc.) would disperse
the negative charge and hence stabilise the
carboxylate ion and thus increase acidity
of the parent acid. On the other hand,
electron-releasing substituents would
intensify the negative charge, destabilise
the carboxylate ion and thus decrease
acidity of the parent acid.
Electronegativity decreases in order
F > Cl > Br
and hence –I effect also decreases in the
same order, therefore the correct option is
halogen, —NO2, C6H5 etc.) would disperse
the negative charge and hence stabilise the
carboxylate ion and thus increase acidity
of the parent acid. On the other hand,
electron-releasing substituents would
intensify the negative charge, destabilise
the carboxylate ion and thus decrease
acidity of the parent acid.
Electronegativity decreases in order
F > Cl > Br
and hence –I effect also decreases in the
same order, therefore the correct option is
FCH2COOH > ClCH2COOH > BrCH2COOH> CH3COOH
Only One Option Correct TypeDirection (Q. Nos. 1-12) This section contains 12 multiple choice questions. Each question has four choices (a), (b), (c) and (d), out of which ONLY ONE is correct.Q. An optically active organic compound has molecular formula C5H12O(X). X on oxidation with CrO3/H2SO4 gives an achiral C5H10O. Hence, X could be- a)
- b)
- c)
- d)
Correct answer is option 'C'. Can you explain this answer?
Only One Option Correct Type
Direction (Q. Nos. 1-12) This section contains 12 multiple choice questions. Each question has four choices (a), (b), (c) and (d), out of which ONLY ONE is correct.
Q.
An optically active organic compound has molecular formula C5H12O(X). X on oxidation with CrO3/H2SO4 gives an achiral C5H10O. Hence, X could be
a)
b)
c)
d)
|
|
Rishika Patel answered |
Oxidation of X giving ketone as well as X is chiral, it must be a secondary alcohol with a-chiral carbon.
Which reagent below cannot reduce an acid chloride to an aldehyde?- a)Pd/BaSO4
- b)Li[{(CH3)3CO}3AIH] at -80°C
- c)Na/C2H5OH
- d)[HAI {CH2CH(CH3)2}2] at - 80°C
Correct answer is option 'C'. Can you explain this answer?
Which reagent below cannot reduce an acid chloride to an aldehyde?
a)
Pd/BaSO4
b)
Li[{(CH3)3CO}3AIH] at -80°C
c)
Na/C2H5OH
d)
[HAI {CH2CH(CH3)2}2] at - 80°C
|
|
Raghavendra Rane answered |
Na /C2 H5OH further reduces aldehydes to alcohols.
Which of the following reaction will not produce an aldehyde?- a)
- b)
- c)
- d)
Correct answer is option 'C'. Can you explain this answer?
Which of the following reaction will not produce an aldehyde?
a)
b)
c)
d)
|
Preeti Iyer answered |
Anti vicinal-diols do not undergo oxidative cleavage with HIO4.
Aldehydes are produced on reduction of the following by DIBAL – H- a)esters
- b)nitriles
- c)both nitriles and esters
- d)none of these
Correct answer is option 'C'. Can you explain this answer?
Aldehydes are produced on reduction of the following by DIBAL – H
a)
esters
b)
nitriles
c)
both nitriles and esters
d)
none of these
|
Ashwin Yadav answered |
Reduction of Esters and Nitriles by DIBAL H
Aldehydes can be produced by the reduction of both esters and nitriles using DIBAL H.
Reduction of Esters:
- When esters are treated with DIBAL H, they are reduced to aldehydes.
- The reaction proceeds via a two-step process where the ester is first converted to an aldehyde intermediate, which is then further reduced to the final aldehyde product.
- This reduction reaction is useful in organic synthesis to selectively produce aldehydes from esters.
Reduction of Nitriles:
- Similarly, nitriles can also be reduced to aldehydes using DIBAL H.
- Nitriles are first converted to imines, which are then reduced to aldehydes.
- This reaction provides a method for the conversion of nitriles to aldehydes in organic synthesis.
Conclusion:
- Therefore, the reduction of both esters and nitriles by DIBAL H can lead to the formation of aldehydes.
- This versatile reagent allows for the selective synthesis of aldehydes from these functional groups, providing a valuable tool for organic chemists.
The oxidation of toluene to benzoic acid can be stopped at the aldehyde stage, to give benzadehyde, The reagent used for the purpose is one of the following.- a)Chromyl chloride
- b)Carbon monoxide and hydrogen chloride
- c)Chlorine
- d)Chromic oxide
Correct answer is option 'A'. Can you explain this answer?
The oxidation of toluene to benzoic acid can be stopped at the aldehyde stage, to give benzadehyde, The reagent used for the purpose is one of the following.
a)
Chromyl chloride
b)
Carbon monoxide and hydrogen chloride
c)
Chlorine
d)
Chromic oxide
|
Preethi Bose answered |
This is Etard Reaction.
Which of the following compounds will give ayellow precipitate with iodine and alkali ?[2012 M]- a)Acetophenone
- b)Acetamide
- c)Methyl acetate
- d)2-Hydroxypropane
Correct answer is option 'A,D'. Can you explain this answer?
Which of the following compounds will give ayellow precipitate with iodine and alkali ?[2012 M]
a)
Acetophenone
b)
Acetamide
c)
Methyl acetate
d)
2-Hydroxypropane
|
Nayanika Reddy answered |
It is iodoform reaction. Acetophenone
and 2-Hydroxypropane 
both give a yellowprecipitate of CHI3 (iodoform) with iodine
and alkali.
Consider the following reaction,
All of the following reagents can bring about the above transformation except- a)MnO2
- b)[(CH3)3CO]3AI/Acetone
- c)CrO3/Pyridine/HC
- d)KMnO4/OH-
Correct answer is option 'D'. Can you explain this answer?
Consider the following reaction,
All of the following reagents can bring about the above transformation except
a)
MnO2
b)
[(CH3)3CO]3AI/Acetone
c)
CrO3/Pyridine/HC
d)
KMnO4/OH-
|
Gauri Datta answered |
Alkaline permanganate (Baeyer’s reagent) also oxidises olefinic bonds to syn vicinal-diols.
If all the ketone isomers of C6H10O are reduced independently with NaBH4 , how many of them will produce racemic mixture of alcohols?
Correct answer is '5'. Can you explain this answer?
If all the ketone isomers of C6H10O are reduced independently with NaBH4 , how many of them will produce racemic mixture of alcohols?
|
Anuj Iyer answered |
All the above shown ketones on reduction with NaBH4 will produce racemic mixtures of alcohols. The isomer below already has a chiral carbon, show enantiomerism. A pure enantiomer of this ketone, on reduction with NaBH4, will produce pair of diastereomers.
In Hell – Volhard Zelinsky reaction, halogen reacts with- a)ketones
- b)carboxylic acids
- c)ethers
- d)aldehydes
Correct answer is option 'B'. Can you explain this answer?
In Hell – Volhard Zelinsky reaction, halogen reacts with
a)
ketones
b)
carboxylic acids
c)
ethers
d)
aldehydes
|
Ashwini Chakraborty answered |
Alpha Hydrogen containing carboxylic acids undergo HVZ reaction.
Chapter doubts & questions for Aldehydes, Ketones and Carboxylic Acids - 4 Months Preparation for NEET 2025 is part of NEET exam preparation. The chapters have been prepared according to the NEET exam syllabus. The Chapter doubts & questions, notes, tests & MCQs are made for NEET 2025 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests here.
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