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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.
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
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.
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
What is the IUPAC name of Acrolein?- a)pent-2-enal
- b)but-1-enal
- c)prop-2-enal
- d)but-2-enal
Correct answer is option 'C'. Can you explain this answer?
What is the IUPAC name of Acrolein?
a)
pent-2-enal
b)
but-1-enal
c)
prop-2-enal
d)
but-2-enal
|
|
Nandini Iyer answered |
IUPAC name of acrolein is prop-2-enal, a name derived from that of acrylic acid, the parent carboxylic acid.
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
|
Abhijeet Menon answered |
Functional isomers:
Functional isomers are compounds that have the same molecular formula but differ in their functional groups. In other words, functional isomers have different chemical properties because they have different functional groups.
Propanone (CH3COCH3) and prop-2-en-1-ol (CH3CH=CHCH2OH) are examples of functional isomers because they have different functional groups. Propanone has a ketone functional group whereas prop-2-en-1-ol has an alcohol functional group.
Ketones and alcohols have different chemical properties and reactivities because of the differences in their functional groups. Therefore, propanone and prop-2-en-1-ol are functional isomers.
Functional isomers are compounds that have the same molecular formula but differ in their functional groups. In other words, functional isomers have different chemical properties because they have different functional groups.
Propanone (CH3COCH3) and prop-2-en-1-ol (CH3CH=CHCH2OH) are examples of functional isomers because they have different functional groups. Propanone has a ketone functional group whereas prop-2-en-1-ol has an alcohol functional group.
Ketones and alcohols have different chemical properties and reactivities because of the differences in their functional groups. Therefore, propanone and prop-2-en-1-ol are functional isomers.
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:
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.
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.
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..
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.
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.
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. 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.
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.
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]+.
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.
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.
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
|
Riya Banerjee answered |
The correct answer is option C
The compound formed as a result of oxidation of ethyl benzene by alkaline KMnO4 is benzoic acid.
Other reagents that can be used are acidified potassium dichromate or dilute nitric acid.
The compound formed as a result of oxidation of ethyl benzene by alkaline KMnO4 is benzoic acid.
Other reagents that can be used are acidified potassium dichromate or dilute nitric acid.
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.
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)
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)
|
Om Desai 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.
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.
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.
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
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.
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.
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.
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.
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.
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.
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
Consider the acidity of the carboxylic acids :
(1) PhCOOH(2) 0-NO2C6H4COOH(3) p-NO2C6H4COOH(4) m-NO2C6H4COOH
Which of the following order is correct ? [AIEEE-2004]- a)2 > 3 > 4 > 1
- b)2 > 4 > 3 > 1
- c)2 > 4 > 1 > 3
- d)1 > 2 > 3 > 4
Correct answer is option 'A'. Can you explain this answer?
Consider the acidity of the carboxylic acids :
(1) PhCOOH
(1) PhCOOH
(2) 0-NO2C6H4COOH
(3) p-NO2C6H4COOH
(4) m-NO2C6H4COOH
Which of the following order is correct ? [AIEEE-2004]
Which of the following order is correct ? [AIEEE-2004]
a)
2 > 3 > 4 > 1
b)
2 > 4 > 3 > 1
c)
2 > 4 > 1 > 3
d)
1 > 2 > 3 > 4
|
|
Vivek answered |
Ortho derivative of Benzoic acid is most acidic regardless of the nature of the group at that position. Para derivative is on the 2nd position due to electron withdrawing effect due to -R and -I effect. Meta derivative comes 3rd due to -I effect only (as meta position is not involved in resonance). At last comes Benzoic acid which is least acidic due to absence of any electron withdrawing group.
On mixing ethyl acetate with aqueous sodium chloride, the composition of the resultant solution is :
[AIEEE-2004]- a)CH3Cl + C2H5COONa
- b)CH3COONa + C2H5OH
- c)CH3COCl + C2H5OH + NaOH
- d)CH3COOC2H5 + NaCl
Correct answer is option 'D'. Can you explain this answer?
On mixing ethyl acetate with aqueous sodium chloride, the composition of the resultant solution is :
[AIEEE-2004]
[AIEEE-2004]
a)
CH3Cl + C2H5COONa
b)
CH3COONa + C2H5OH
c)
CH3COCl + C2H5OH + NaOH
d)
CH3COOC2H5 + NaCl
|
|
Preeti Iyer answered |
aqueous NaCl is a neutral compound. Thus no reaction takes place between ethyl acetate and aqueous NaCl.
Thus resultant solution will be CH3COOC2H5 + NaCl
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 statements is not correct?- a)A compound whose molecule has D configuration will always be dextrorotatory
- b)A compound whose molecule has D configuration may be dextrorotatory or levorotatory
- c)A compound whose molecule has R configuration may be dexrotatory or levorotatory
- d)A compound whose molecule has L configuration may be dextrorotatory or levorotatory
Correct answer is option 'A'. Can you explain this answer?
Which of the following statements is not correct?
a)
A compound whose molecule has D configuration will always be dextrorotatory
b)
A compound whose molecule has D configuration may be dextrorotatory or levorotatory
c)
A compound whose molecule has R configuration may be dexrotatory or levorotatory
d)
A compound whose molecule has L configuration may be dextrorotatory or levorotatory
|
Avantika Saha answered |
The configuration in a compound is independent of its physical properties (optical activity)
How manyh assymmetric carbon atoms are present in
(i) 2-Dimethyl cyclohexane
(ii) 3-Methyl cyclopentene
(iii) 3-Methylcyclohexene- a)2,1,1
- b)1,1,1
- c)2,0.2
- d)2,0,1
Correct answer is option 'A'. Can you explain this answer?
How manyh assymmetric carbon atoms are present in
(i) 2-Dimethyl cyclohexane
(ii) 3-Methyl cyclopentene
(iii) 3-Methylcyclohexene
(i) 2-Dimethyl cyclohexane
(ii) 3-Methyl cyclopentene
(iii) 3-Methylcyclohexene
a)
2,1,1
b)
1,1,1
c)
2,0.2
d)
2,0,1
|
|
Aarya Khanna answered |
Asymmetric carbon atoms are those carbon atoms which are attached to four different groups or atoms. These carbon atoms are also known as chiral centers.
(i) 2-Dimethyl cyclohexane:
- Cyclohexane has no asymmetric carbon atoms.
- When two methyl groups are attached to cyclohexane, it becomes 2,3-dimethyl cyclohexane.
- The carbon at position 2 is attached to two methyl groups and two hydrogen atoms. Therefore, it is not an asymmetric carbon atom.
- The carbon at position 3 is attached to one methyl group, one hydrogen atom, one methyl group, and one cyclohexane ring. Therefore, it is an asymmetric carbon atom.
- Hence, there is one asymmetric carbon atom in 2-dimethyl cyclohexane.
(ii) 3-Methyl cyclopentene:
- Cyclopentene has one asymmetric carbon atom.
- When a methyl group is attached to cyclopentene at position 3, it becomes 3-methyl cyclopentene.
- The carbon at position 3 is attached to one methyl group, one hydrogen atom, one double bond with carbon, and one cyclopentene ring. Therefore, it is an asymmetric carbon atom.
- Hence, there is one asymmetric carbon atom in 3-methyl cyclopentene.
(iii) 3-Methylcyclohexene:
- Cyclohexene has one asymmetric carbon atom.
- When a methyl group is attached to cyclohexene at position 3, it becomes 3-methylcyclohexene.
- The carbon at position 3 is attached to one methyl group, one hydrogen atom, one double bond with carbon, and one cyclohexene ring. Therefore, it is an asymmetric carbon atom.
- Hence, there is one asymmetric carbon atom in 3-methylcyclohexene.
Therefore, the correct answer is option 'A' (2,1,1) as there are 2 asymmetric carbon atoms in total, with 1 in 2-dimethyl cyclohexane, 1 in 3-methyl cyclopentene, and 1 in 3-methylcyclohexene.
(i) 2-Dimethyl cyclohexane:
- Cyclohexane has no asymmetric carbon atoms.
- When two methyl groups are attached to cyclohexane, it becomes 2,3-dimethyl cyclohexane.
- The carbon at position 2 is attached to two methyl groups and two hydrogen atoms. Therefore, it is not an asymmetric carbon atom.
- The carbon at position 3 is attached to one methyl group, one hydrogen atom, one methyl group, and one cyclohexane ring. Therefore, it is an asymmetric carbon atom.
- Hence, there is one asymmetric carbon atom in 2-dimethyl cyclohexane.
(ii) 3-Methyl cyclopentene:
- Cyclopentene has one asymmetric carbon atom.
- When a methyl group is attached to cyclopentene at position 3, it becomes 3-methyl cyclopentene.
- The carbon at position 3 is attached to one methyl group, one hydrogen atom, one double bond with carbon, and one cyclopentene ring. Therefore, it is an asymmetric carbon atom.
- Hence, there is one asymmetric carbon atom in 3-methyl cyclopentene.
(iii) 3-Methylcyclohexene:
- Cyclohexene has one asymmetric carbon atom.
- When a methyl group is attached to cyclohexene at position 3, it becomes 3-methylcyclohexene.
- The carbon at position 3 is attached to one methyl group, one hydrogen atom, one double bond with carbon, and one cyclohexene ring. Therefore, it is an asymmetric carbon atom.
- Hence, there is one asymmetric carbon atom in 3-methylcyclohexene.
Therefore, the correct answer is option 'A' (2,1,1) as there are 2 asymmetric carbon atoms in total, with 1 in 2-dimethyl cyclohexane, 1 in 3-methyl cyclopentene, and 1 in 3-methylcyclohexene.
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.
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
Chapter doubts & questions for Aldehydes, Ketones and Carboxylic Acids - Chapter-wise Tests for JEE Main & Advanced 2025 is part of JEE exam preparation. The chapters have been prepared according to the JEE exam syllabus. The Chapter doubts & questions, notes, tests & MCQs are made for JEE 2025 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests here.
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