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The functional group given below is characteristic of organic _____ . - a)ketones
- b)acids
- c) aldehydes
- d)esters
Correct answer is 'A'. Can you explain this answer?
The functional group given below is characteristic of organic _____ .
a)
ketones
b)
acids
c)
aldehydes
d)
esters
|
Preeti Iyer answered |
Alkenes do not undergo mercuration, indeed they undergo oxymercuration , a process in which an alkene is converted into an alcohol.
How many different isomeric alkynes on catalytic hydrogenation gives the same 3-ethyl hexane?- a)2
- b)3
- c)4
- d)6
Correct answer is option 'C'. Can you explain this answer?
How many different isomeric alkynes on catalytic hydrogenation gives the same 3-ethyl hexane?
a)
2
b)
3
c)
4
d)
6
|
Divya Menon answered |
Isomeric Alkynes and Catalytic Hydrogenation
Concept: Isomers are molecules that have the same molecular formula but different structural arrangements. Alkynes are hydrocarbons having at least one triple bond between two carbon atoms. Catalytic hydrogenation is the addition of hydrogen to an unsaturated organic compound in the presence of a catalyst. Catalytic hydrogenation of alkynes involves the addition of two hydrogen atoms across the triple bond, resulting in an alkane.
Explanation:
The given compound is 3-ethylhexane. The molecular formula of 3-ethylhexane is C8H18. The possible isomeric alkynes of C8H18 are:
1. Octyne-1
2. Octyne-2
3. Octyne-3
4. Octyne-4
5. Octyne-5
6. Octyne-6
Catalytic hydrogenation of octyne-1 and octyne-5 would result in the formation of the same product, i.e., 3-ethylhexane. Similarly, catalytic hydrogenation of octyne-2 and octyne-4 would also result in the formation of 3-ethylhexane. However, catalytic hydrogenation of octyne-3 and octyne-6 would result in the formation of different products.
Therefore, the total number of different isomeric alkynes that on catalytic hydrogenation gives the same 3-ethylhexane is 4.
Answer: The correct option is (C) 4.
Concept: Isomers are molecules that have the same molecular formula but different structural arrangements. Alkynes are hydrocarbons having at least one triple bond between two carbon atoms. Catalytic hydrogenation is the addition of hydrogen to an unsaturated organic compound in the presence of a catalyst. Catalytic hydrogenation of alkynes involves the addition of two hydrogen atoms across the triple bond, resulting in an alkane.
Explanation:
The given compound is 3-ethylhexane. The molecular formula of 3-ethylhexane is C8H18. The possible isomeric alkynes of C8H18 are:
1. Octyne-1
2. Octyne-2
3. Octyne-3
4. Octyne-4
5. Octyne-5
6. Octyne-6
Catalytic hydrogenation of octyne-1 and octyne-5 would result in the formation of the same product, i.e., 3-ethylhexane. Similarly, catalytic hydrogenation of octyne-2 and octyne-4 would also result in the formation of 3-ethylhexane. However, catalytic hydrogenation of octyne-3 and octyne-6 would result in the formation of different products.
Therefore, the total number of different isomeric alkynes that on catalytic hydrogenation gives the same 3-ethylhexane is 4.
Answer: The correct option is (C) 4.
Which of the following statements correctly describes the general reactivity of alkynes?- a)Alkynes are more reactive than alkenes
- b)Alkynes reacts as an electrophile
- c)Unlike alkenes, alkynes do not undergo electrophilic addition reaction
- d)An alkene is electron rich molecule, therefore react as a nucleophile
Correct answer is option 'D'. Can you explain this answer?
Which of the following statements correctly describes the general reactivity of alkynes?
a)
Alkynes are more reactive than alkenes
b)
Alkynes reacts as an electrophile
c)
Unlike alkenes, alkynes do not undergo electrophilic addition reaction
d)
An alkene is electron rich molecule, therefore react as a nucleophile
|
|
Preeti Khanna answered |
The correct answer is option D
The clouds of electrons surrounding the sigma bond makes an alkyne an electron-rich molecule. They are therefore nucleophiles that react with electrophiles. Thus alkynes, like alkenes, undergo electrophilic addition reactions because of their weak pi bonds. When a proton adds to an alkyne, a vinylic cation is formed.
The clouds of electrons surrounding the sigma bond makes an alkyne an electron-rich molecule. They are therefore nucleophiles that react with electrophiles. Thus alkynes, like alkenes, undergo electrophilic addition reactions because of their weak pi bonds. When a proton adds to an alkyne, a vinylic cation is formed.
How many dichlorocyclohexane would be produced upon free radical chlorination of chlorocyclohexane?- a)4
- b)6
- c)8
- d)9
Correct answer is option 'D'. Can you explain this answer?
How many dichlorocyclohexane would be produced upon free radical chlorination of chlorocyclohexane?
a)
4
b)
6
c)
8
d)
9
|
Rajesh Gupta answered |
Product 1-Does not have any chiral carbon so only one product
Product 2-Two chiral centres so RR, RS and SR possible
Product 3-Two chiral centres so RR, RS and SR possible
Product 4-Two chiral centres but plane of symmetry exists, so only two isomers.
Product 2-Two chiral centres so RR, RS and SR possible
Product 3-Two chiral centres so RR, RS and SR possible
Product 4-Two chiral centres but plane of symmetry exists, so only two isomers.
What is the major bromination product in the following reaction? 
- a)
- b)
- c)
- d)
Correct answer is option 'C'. Can you explain this answer?
What is the major bromination product in the following reaction?
a)
b)
c)
d)
|
Suresh Reddy answered |
The correct answer is option c
Thus, a racemic mixture is obtained. A racemic mixture is one that has an equal amount of left and right handed enantiomers of a chiral molecule.
Thus, a racemic mixture is obtained. A racemic mixture is one that has an equal amount of left and right handed enantiomers of a chiral molecule.
If 3-methyl-2-pentanol is heated with concentrated H2SO4 , in principle how many different alkenes result?
Correct answer is '5'. Can you explain this answer?
If 3-methyl-2-pentanol is heated with concentrated H2SO4 , in principle how many different alkenes result?
|
Krishna Iyer answered |
- methyl 1 pentene
- 3 methyl 2 butene
- 2 ethyl 1 butene
Hence 3 is the correct answer.
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 option is correct.Q. Arrange the following compounds in increasing order of polarity
- a)I < II < III
- b)III < II < I
- c)II < I < III
- d)III < I < II
Correct answer is option 'C'. Can you explain this answer?
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 option is correct.
Q. Arrange the following compounds in increasing order of polarity
a)
I < II < III
b)
III < II < I
c)
II < I < III
d)
III < I < II
|
Om Desai answered |
In case 1, the bond is broken in oxygen’s favor and it will attain its octet. Also, carbon becomes sp2 hybridized, so there is a chance of polarity.
In case 2, if the bond is broken in favor of oxygen, then the ring will become anti-aromatic which is highly unstable and the bond won’t be broken in that way. If the bond is broken in favor of carbon in the ring, then although the ring becomes aromatic but oxygen will bear +ve charge which is very unstable. So, there is no chance to break the bond.
In case 3, if the double bond is broken in favor of oxygen, then oxygen will acquire a negative charge and the ring will become aromatic. So, it is a highly favorable case of double bond breaking.
Therefore, the order of polarity: - III>I>II
In case 2, if the bond is broken in favor of oxygen, then the ring will become anti-aromatic which is highly unstable and the bond won’t be broken in that way. If the bond is broken in favor of carbon in the ring, then although the ring becomes aromatic but oxygen will bear +ve charge which is very unstable. So, there is no chance to break the bond.
In case 3, if the double bond is broken in favor of oxygen, then oxygen will acquire a negative charge and the ring will become aromatic. So, it is a highly favorable case of double bond breaking.
Therefore, the order of polarity: - III>I>II
Propene on ozonolysis forms:- a)Acetaldehyde
- b)Formaldehyde
- c)Both acetaldehyde and formaldehyde
- d)Acetone
Correct answer is option 'C'. Can you explain this answer?
Propene on ozonolysis forms:
a)
Acetaldehyde
b)
Formaldehyde
c)
Both acetaldehyde and formaldehyde
d)
Acetone
|
|
Raghav Bansal answered |
When propene on ozonolysis it yields a new structure called ozonide
and there cleavage takes place and it yields two products namely
1.acetaldehyde
2.formaldehyde
and there cleavage takes place and it yields two products namely
1.acetaldehyde
2.formaldehyde
Direction (Q. Nos. 16 - 18) This section contains a paragraph, describing theory, experiments, data, etc.
Three questions related to the paragraph have been given. Each question has only one correct answer among the four given options (a), (b), (cj and (d).PassageA hydrocarbon with molecular formula C10H18, upon catalytic hydrogenation gives C10H20 (X). X on free radical chlorination gives two monochloro derivatives with their molecular formula C10H19CI that are constitutional isomers.Q. Which of the following satisfy the criteria of X ?- a)
- b)
- c)
- d)
Correct answer is option 'C'. Can you explain this answer?
Direction (Q. Nos. 16 - 18) This section contains a paragraph, describing theory, experiments, data, etc.
Three questions related to the paragraph have been given. Each question has only one correct answer among the four given options (a), (b), (cj and (d).
Three questions related to the paragraph have been given. Each question has only one correct answer among the four given options (a), (b), (cj and (d).
Passage
A hydrocarbon with molecular formula C10H18, upon catalytic hydrogenation gives C10H20 (X). X on free radical chlorination gives two monochloro derivatives with their molecular formula C10H19CI that are constitutional isomers.
Q. Which of the following satisfy the criteria of X ?
a)
b)
c)
d)
|
|
Krishna Iyer answered |
In option c, there are only two different position which can be chlorinated,
So, option c is correct.
So, option c is correct.
When vicinal dibromide is heated with KOH in ethanol (~ 200°C), double dehydrohalogenation takes place giving alkyne. Which of the following fails to give alkyne according to this procedure?- a)
- b)
- c)
- d)
Correct answer is option 'C'. Can you explain this answer?
When vicinal dibromide is heated with KOH in ethanol (~ 200°C), double dehydrohalogenation takes place giving alkyne. Which of the following fails to give alkyne according to this procedure?
a)
b)
c)
d)
|
|
Rajesh Gupta answered |
The correct answer is option C
CH³)²BrC*–CHBr–CH²–CH³
C* has no H attached to it hence the next Bromine cannot perform dehydrohalogenation using this Carbon. Therefore, triple bond formation is not possible in this compound.
CH³)²BrC*–CHBr–CH²–CH³
C* has no H attached to it hence the next Bromine cannot perform dehydrohalogenation using this Carbon. Therefore, triple bond formation is not possible in this compound.
What is true about the compound calicene?
- a)It is highly soluble in water
- b)It exists mainly as dipolar ion with +ve charge in the three membered ring while -ve charge in five membered ring
- c)It exists mainly as dipolar ion with -ve charge in the three membered ring and +ve charge in five membered ring
- d)In solution it shows very high electrical conductivity
Correct answer is option 'A,B,D'. Can you explain this answer?
What is true about the compound calicene?
a)
It is highly soluble in water
b)
It exists mainly as dipolar ion with +ve charge in the three membered ring while -ve charge in five membered ring
c)
It exists mainly as dipolar ion with -ve charge in the three membered ring and +ve charge in five membered ring
d)
In solution it shows very high electrical conductivity
|
|
Raghav Bansal answered |
- In calicene, the electrons move towared the five-membered ring because both rings are aromatic in the resonance contributor that has a negative charge on a carbon of the five-membered ring and a positive charge on a carbon of the three-membered ring
- It is highly soluble in water and in solution it shows very high electrical conductivity
So, Option C is not correct and other Options are Correct.
- a)Fittig reaction
- b)Wurtz’s-Fittig reaction
- c)Ullmann reaction
- d)Wurtz’s reaction
Correct answer is option 'B'. Can you explain this answer?
a)
Fittig reaction
b)
Wurtz’s-Fittig reaction
c)
Ullmann reaction
d)
Wurtz’s reaction
|
|
Om Desai answered |
The correct answer is option B
Wurtz - Fittig reaction:
Aryl halide and alkyl halide couple in presence of sodium metal / dry ether to form alkyl benzene.
For example, bromobenzene reacts with methyl bromide in presence of sodium. dry ether to form toluene.
C6H6 - Br + CH3 - Br + 2Na(dry ether)------> C6H5 - CH3 + 2NaBr
Wurtz - Fittig reaction:
Aryl halide and alkyl halide couple in presence of sodium metal / dry ether to form alkyl benzene.
For example, bromobenzene reacts with methyl bromide in presence of sodium. dry ether to form toluene.
C6H6 - Br + CH3 - Br + 2Na(dry ether)------> C6H5 - CH3 + 2NaBr
How many different heptenes result by partial hydrogenation of all possible, unbranched heptynes?
Correct answer is '5'. Can you explain this answer?
How many different heptenes result by partial hydrogenation of all possible, unbranched heptynes?
|
Neha Joshi answered |
The correct answer is 5.
C=C-C-C-C-C-C , no geometrical isomerism so 1
C-C=C-C-C-C-C 2 , geometrical cis trans
C-C-C=C-C-C-C , cis trans again so 2
∴ 2+2+1=5
C=C-C-C-C-C-C , no geometrical isomerism so 1
C-C=C-C-C-C-C 2 , geometrical cis trans
C-C-C=C-C-C-C , cis trans again so 2
∴ 2+2+1=5
Out of the following compounds , which will be have a zero dipole moment.- a)1,1 - dichloroethylene
- b)cis-1,2-dichlorethylene
- c)trans-1,2-dichlorethylene
- d)none of these.
Correct answer is option 'C'. Can you explain this answer?
Out of the following compounds , which will be have a zero dipole moment.
a)
1,1 - dichloroethylene
b)
cis-1,2-dichlorethylene
c)
trans-1,2-dichlorethylene
d)
none of these.
|
Hansa Sharma answered |
In case of trans-1,2-dichloroethylene
The net dipole moment is zero but in case of cis -1,2 -dicholorethylene.
There is some resultant dipole moment.
The net dipole moment is zero but in case of cis -1,2 -dicholorethylene.
There is some resultant dipole moment.
Organic compounds can be classified even based upon the function groups. Identify the one which is not a functional group
- a)Isocyanide
- b)Isocyano
- c)Carboxyl
- d)Carbonyl
Correct answer is option 'A'. Can you explain this answer?
Organic compounds can be classified even based upon the function groups. Identify the one which is not a functional group
a)
Isocyanide
b)
Isocyano
c)
Carboxyl
d)
Carbonyl
|
|
Krishna Iyer answered |
Isocyanide is a compound and it is not a functional group.
Consider the following bromides :
- a)Both decomposes to form
- b)Both decomposes to forms Br+
- c)I decomposes to give and II gives Br+
- d)I decomposes to give Br+ while II gives
Correct answer is option 'D'. Can you explain this answer?
Consider the following bromides :
a)
Both decomposes to form
b)
Both decomposes to forms Br+
c)
I decomposes to give and II gives Br+
and II gives Br+d)
I decomposes to give Br+ while II gives
|
Geetika Shah answered |
In the compound I, Br will dispatch as Br+ so that a -ve charge appears on carbon which will give us 6π electrons. So ring will become aromatic.(4n+2 π electron is needed for aromaticity)
In compound II, Br will dispatch as Br-. So that carbon has +ve charge and all the double bond will circulate in the ring. This will maintain 4n+2 π electron and the molecule will remain as aromatic.
In compound II, Br will dispatch as Br-. So that carbon has +ve charge and all the double bond will circulate in the ring. This will maintain 4n+2 π electron and the molecule will remain as aromatic.
If 4-methyl cyclopentene is treated with OsO4 followed by work-up with NaHSO3/H2O, how many different diols would result?
Correct answer is '2'. Can you explain this answer?
If 4-methyl cyclopentene is treated with OsO4 followed by work-up with NaHSO3/H2O, how many different diols would result?
|
|
Preeti Iyer answered |
Since, OsO4 + NaHSO4/H2O syn hydroxylation, so two products can be obtained.
Direction (Q, Nos. 13 - 16) This section is based on Statement I and Statement II. Select the correct answer from the codes given below.Q.Statement I : 2-butyne on reduction with Pd/CaCO3 gives c/s-2-butene.
Statement II : Hydrogenation proceed through adsorption mechanism.- a)Both Statement I and Statement II are correct and Statement II is the correct explanation of Statement I
- b)Both Statement I and Statement II are correct and Statement II is not the correct explanation of Statement I
- c)Statement I is correct but Statement II is incorrect
- d)Statement II is correct but Statement I is incorrect
Correct answer is option 'A'. Can you explain this answer?
Direction (Q, Nos. 13 - 16) This section is based on Statement I and Statement II. Select the correct answer from the codes given below.
Q.
Statement I : 2-butyne on reduction with Pd/CaCO3 gives c/s-2-butene.
Statement II : Hydrogenation proceed through adsorption mechanism.
Statement II : Hydrogenation proceed through adsorption mechanism.
a)
Both Statement I and Statement II are correct and Statement II is the correct explanation of Statement I
b)
Both Statement I and Statement II are correct and Statement II is not the correct explanation of Statement I
c)
Statement I is correct but Statement II is incorrect
d)
Statement II is correct but Statement I is incorrect
|
|
Raghav Bansal answered |
A Lindlar catalyst is a heterogeneous catalyst that consists of palladium deposited on calcium carbonate or barium sulphate which is then poisoned with various forms of lead or sulphur. It is used for the hydrogenation of alkynes to cis alkenes (i.e. without further reduction into alkanes) When the triple bond is (2-butyne) hydrogenated over the Lindlar’s catalyst i.e.Pd/CaCO3
it gives predominantly cis alkene(2-butene).
it gives predominantly cis alkene(2-butene).
Which of the following is not a possible termination step in the free radical chlorination of methane?- a)
- b)
- c)
- d)
Correct answer is option 'A'. Can you explain this answer?
Which of the following is not a possible termination step in the free radical chlorination of methane?
a)
b)
c)
d)
|
|
Preeti Iyer answered |
Termination is the last step. So there shouldn't be any free radical atom remaining. In first option there is Cl• remaining it can't be termination step.The steps in free radical halogenation are as
Select the species which is not aromatic.- a)
- b)
- c)
- d)
Correct answer is option 'C'. Can you explain this answer?
Select the species which is not aromatic.
a)
b)
c)
d)
|
|
Gaurav Kumar answered |
Here, we can see that Nitrogen gave its lone pair to make the system aromatic. The same case happens with option b and d. But with option c, Boron is not having any lone pair to donate. So option c is correct answer
Arrange the following in decreasing order of their boiling points
A) n-butane
B) 2-methylbutane
C) n-pentane
D) 2,2-dimethylpropane- a)A>B>C>D
- b)B>C>D>A
- c)D>C>B>A
- d)C>B>D>A
Correct answer is option 'D'. Can you explain this answer?
Arrange the following in decreasing order of their boiling points
A) n-butane
B) 2-methylbutane
C) n-pentane
D) 2,2-dimethylpropane
a)
A>B>C>D
b)
B>C>D>A
c)
D>C>B>A
d)
C>B>D>A
|
|
Geetika Shah answered |
D is the correct answer.
The boiling point of alkanes increases with increase in molecular mass and for the same alkane, the boiling point decreases with branching. Thus, the decreasing order of their boiling points is:
C B D
n−Pentane>2−Methylbutane>2,2−Dimethylpropane
> n−Butane
A
The boiling point of alkanes increases with increase in molecular mass and for the same alkane, the boiling point decreases with branching. Thus, the decreasing order of their boiling points is:
C B D
n−Pentane>2−Methylbutane>2,2−Dimethylpropane
> n−Butane
A
Arrange the halogens F2, Cl2, Br2, I2 in order of their increasing reactivity with alkanes.- a)I2 < Br2 < Cl2 < F2
- b)Br2 < Cl2 < F2 < I2
- c)F2 < Cl2 < Br2 < I2
- d)Br2 < I2 < Cl2 < F2
Correct answer is option 'A'. Can you explain this answer?
Arrange the halogens F2, Cl2, Br2, I2 in order of their increasing reactivity with alkanes.
a)
I2 < Br2 < Cl2 < F2
b)
Br2 < Cl2 < F2 < I2
c)
F2 < Cl2 < Br2 < I2
d)
Br2 < I2 < Cl2 < F2
|
|
Gaurav Kumar answered |
The correct answer is Option A.
Since reactivity decreases down the group as the electronegativity of the halogen decreases down the group. Thus, rate of reaction of alkanes with halogens is
I2 < Br2 < Cl2 <F2
Since reactivity decreases down the group as the electronegativity of the halogen decreases down the group. Thus, rate of reaction of alkanes with halogens is
I2 < Br2 < Cl2 <F2
Addition of halogens to alkenes is an example of:- a)Nucleophilic addition reaction
- b)Electrophilic addition reaction
- c)Electrophilic substitution reaction
- d)Nucleophilic substitution reaction
Correct answer is option 'B'. Can you explain this answer?
Addition of halogens to alkenes is an example of:
a)
Nucleophilic addition reaction
b)
Electrophilic addition reaction
c)
Electrophilic substitution reaction
d)
Nucleophilic substitution reaction
|
Avantika Chakraborty answered |
Halogens can act as electrophiles to attack a double bond in alkene. Double bond represents a region of electron density and therefore functions as a nucleophile.
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 option is correct.Q. How many distinct alkynes exist for C6H10 which gives effervescence on heating with Na?- a)5
- b)2
- c)3
- d)4
Correct answer is option 'D'. Can you explain this answer?
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 option is correct.
Q. How many distinct alkynes exist for C6H10 which gives effervescence on heating with Na?
a)
5
b)
2
c)
3
d)
4
|
|
Upasana Bose answered |
Hence D
Direction (Q. Nos. 1 - 11) This section contains 11 multiple choice questions. Each question has four choices (a), (b), (c) and (d), out of which ONLY ONE option is correct.Q. When light is shined on a mixture of chlorine and ethane, chloroethane is formed besides dichloroethane, trichloroethane and several other products. What reaction condition can optimise the yield of chloroethane?- a)Higher reaction temperature
- b)High concentration of chlorine gas
- c)Excess of ethane reactan
- d)Low reaction temperature
Correct answer is option 'C'. Can you explain this answer?
Direction (Q. Nos. 1 - 11) This section contains 11 multiple choice questions. Each question has four choices (a), (b), (c) and (d), out of which ONLY ONE option is correct.
Q. When light is shined on a mixture of chlorine and ethane, chloroethane is formed besides dichloroethane, trichloroethane and several other products. What reaction condition can optimise the yield of chloroethane?
a)
Higher reaction temperature
b)
High concentration of chlorine gas
c)
Excess of ethane reactan
d)
Low reaction temperature
|
|
Pooja Shah answered |
When chlorine and Ethane are taken with chlorine in excess only then we have more than one product like chloroethane, dichloroethane, trichloroethane etc. To avoid this we should take Ethane in excess because when we will take it then in excess then we will have only single time chlorination and we will get monochloroethane.
Direction (Q. Nos. 17 - 20) This section contains 4 questions. When worked o u t wili result in an integer from 0 to 9 (both inclusive).Q. If 3-bromo-4-methyl hexane is treated with ethanolic KOH solution, how many different alkenes would be formed?
Correct answer is '6'. Can you explain this answer?
Direction (Q. Nos. 17 - 20) This section contains 4 questions. When worked o u t wili result in an integer from 0 to 9 (both inclusive).
Q. If 3-bromo-4-methyl hexane is treated with ethanolic KOH solution, how many different alkenes would be formed?
|
|
Raghav Bansal answered |
2 is the right answer.
Alcoholic KOH causes dehydrohalogenation reaction i.e. elimination reaction.
Direction (Q. Nos, 19 - 22) This section contains 4 questions. When worked out will result in an integer from 0 to 9 (both inclusive).Q. How many different isomers of alkenes {including stereoisomers) exist that all upon catalytic hydrogenation adds one mole of H2 to give the same 2, 2, 3,5-tetramethyl hexane?
Correct answer is '7'. Can you explain this answer?
Direction (Q. Nos, 19 - 22) This section contains 4 questions. When worked out will result in an integer from 0 to 9 (both inclusive).
Q. How many different isomers of alkenes {including stereoisomers) exist that all upon catalytic hydrogenation adds one mole of H2 to give the same 2, 2, 3,5-tetramethyl hexane?
|
Ashish Mishra answered |
7 is correct.
Direction (Q. Nos. 12 -15) This section contains 4 multiple choice questions. Each question has four choices (a), (b), (c) and (d), out of which ONE or MORE THAN ONE are correct.Q. In the following free radical bromination reaction, the im portant product(s) is/are
- a)
- b)
- c)
- d)
Correct answer is option 'B,C'. Can you explain this answer?
Direction (Q. Nos. 12 -15) This section contains 4 multiple choice questions. Each question has four choices (a), (b), (c) and (d), out of which ONE or MORE THAN ONE are correct.
Q. In the following free radical bromination reaction, the im portant product(s) is/are
a)
b)
c)
d)
|
Lavanya Menon answered |
This is the case of allylic substitution. The radical will delocalise itself to two locations. And so, we will get two different positions for radical substitution.
There is also a possibility at C6, but it is not in our option.
There is also a possibility at C6, but it is not in our option.
Which of the following reactions can bring about chlorination of cyclohexane?- a)
- b)
- c)
- d)
Correct answer is option 'A,C'. Can you explain this answer?
Which of the following reactions can bring about chlorination of cyclohexane?
a)
b)
c)
d)
|
|
Suresh Iyer answered |
For SO2Cl2: The reactivity patterns of SO2Cl2 and SOCl2 are quite different. SOCl2 is a good electrophile, and can be thought of as a source of Cl− ions. These ions can go on to react in their typical nucleophilic fashion. SO2Cl2 however is often a Cl2 source, as it readily decomposes giving off sulfur dioxide. Usually, much easier/safer to use this than measuring out (and getting into solution) chlorine gas. The chlorination of simple alkanes by Cl2 gas (or something that makes it in solution) happens by a radical mechanism i.e. Cl⋅ not Cl
For Cl2 and heat/light:
For Cl2 and heat/light:
For Cl with AlCl3: It is used for chlorination of compounds like benzene
For HCl: It is used for halogenations of a double bond.
For HCl: It is used for halogenations of a double bond.
Ethene and ethyne can be distinguished by:- a)Bromine water
- b)KMnO4 solution
- c)Ammoniacal Cuprous chloride solution
- d)Any of the above
Correct answer is option 'C'. Can you explain this answer?
Ethene and ethyne can be distinguished by:
a)
Bromine water
b)
KMnO4 solution
c)
Ammoniacal Cuprous chloride solution
d)
Any of the above
|
|
Nandini Patel answered |
The two hydrocarbons can be easily distinguished by simple chemical tests, as ethyne molecule is supposed to have acidic hydrogen.
1. When ethyne is bubbled through ammoniacal silver nitrate solution , a yellow-white precipitate of silver acetylide would be formed.
C2H2 + 2AgNO3 = Ag2C2 + 2HNO3
2. Similarly, ethyne forms a red precipitate of copper acetylide (Cu2C2) when it is passed through ammoniacal cuprous chloride solution.
Ethene does not react with AgNO3 or Cu2Cl2 solution.
How many monobromo derivatives exists for anthracene?
- a)2
- b)3
- c)4
- d)5
Correct answer is option 'B'. Can you explain this answer?
How many monobromo derivatives exists for anthracene?
a)
2
b)
3
c)
4
d)
5
|
|
Suresh Iyer answered |
The correct answer is Option B.
There are 3 monobromo derivatives exists for anthracene:
1-Chloroanthracene
2-Chloroanthracene
and 9-Chloroanthracene
1-Chloroanthracene
2-Chloroanthracene
and 9-Chloroanthracene
A hydrocarbon with molecular formula C10H18, upon catalytic hydrogenation gives C10H20 (X). X on free radical chlorination gives two monochloro derivatives with their molecular formula C10H19CI that are constitutional isomers.Q. How many different alkenes on hydrogenation, can gives X ?- a)1
- b)2
- c)3
- d)4
Correct answer is option 'A'. Can you explain this answer?
A hydrocarbon with molecular formula C10H18, upon catalytic hydrogenation gives C10H20 (X). X on free radical chlorination gives two monochloro derivatives with their molecular formula C10H19CI that are constitutional isomers.
Q. How many different alkenes on hydrogenation, can gives X ?
a)
1
b)
2
c)
3
d)
4
|
|
Gaurav Kumar answered |
The correct statement(s) regarding 1,2-butadiene(l) and 1,3-butadiene, (II) is/are- a)I has perpendicular pi planes while II has parallel pi planes
- b)Both gives the same product on adding 1.0 equivalent at HBr of -80° C
- c)I gives 2, 2-dibromobutane while II gives 2, 3-dibromobutane, when reacted with excess of HBr at very low temperature
- d)Both have same length of C = C double bonds
Correct answer is option 'A,C'. Can you explain this answer?
The correct statement(s) regarding 1,2-butadiene(l) and 1,3-butadiene, (II) is/are
a)
I has perpendicular pi planes while II has parallel pi planes
b)
Both gives the same product on adding 1.0 equivalent at HBr of -80° C
c)
I gives 2, 2-dibromobutane while II gives 2, 3-dibromobutane, when reacted with excess of HBr at very low temperature
d)
Both have same length of C = C double bonds
|
|
Krishna Iyer answered |
The correct answers are options A & C
Adjacent hybrid orbitals are perpendicular to each other irrespective of the type of the hybridization. In 1,2-butadiene, adjacent 2× bonds are perpendicular to each other while in 1,3-butadiene, 2× bond and
1× bonds are adjacent to each other and the compound is a straight chain, so, the single bone and a double bond are perpendicular to each other and this is true for both the pairs of single and double bonds. Thus , both the pi orbitals in 1,3-butadiene are parallel to each other.
Adjacent hybrid orbitals are perpendicular to each other irrespective of the type of the hybridization. In 1,2-butadiene, adjacent 2× bonds are perpendicular to each other while in 1,3-butadiene, 2× bond and
1× bonds are adjacent to each other and the compound is a straight chain, so, the single bone and a double bond are perpendicular to each other and this is true for both the pairs of single and double bonds. Thus , both the pi orbitals in 1,3-butadiene are parallel to each other.
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 option is correct.
Q. How many different alkenes are formed when 2 -chlorobutane is treated with ethanolic solution of KOH?
- a)1
- b)3
- c)2
- d)4
Correct answer is option 'B'. Can you explain this answer?
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 option is correct.
Q. How many different alkenes are formed when 2 -chlorobutane is treated with ethanolic solution of KOH?
a)
1
b)
3
c)
2
d)
4
|
|
Om Desai answered |
With ethanolic KOH, we have E2 elimination.
So, we have 3 products in all
So, we have 3 products in all
Ethylene reacts with HBr to give:- a)Acetylene
- b)Ethyl alcohol
- c)Acetaldehyde
- d)Ethyl bromide
Correct answer is option 'D'. Can you explain this answer?
Ethylene reacts with HBr to give:
a)
Acetylene
b)
Ethyl alcohol
c)
Acetaldehyde
d)
Ethyl bromide
|
|
Preeti Iyer answered |
Ethylene reacts with HBr to form Ethyl bromide. The reaction propagates as follow:-
H2C=CH2 + HBr → H2C+-CH3 →H2BrC-CH3
Since π cloud is electron rich, so HBr dissociates into H+ and Br-. H+ attacks on alkene to give a carbocation and then Br- attacks to get ethyl bromide.
H2C=CH2 + HBr → H2C+-CH3 →H2BrC-CH3
Since π cloud is electron rich, so HBr dissociates into H+ and Br-. H+ attacks on alkene to give a carbocation and then Br- attacks to get ethyl bromide.
Consider the following reaction sequence.
The correct statement(s) regarding diois formed in the final step is/are - a)Three different diois are formed
- b)One pair of diastereomers is formed
- c)Two enantiomers are formed
- d)Product is optically inactive
Correct answer is option 'A,C'. Can you explain this answer?
Consider the following reaction sequence.
The correct statement(s) regarding diois formed in the final step is/are
a)
Three different diois are formed
b)
One pair of diastereomers is formed
c)
Two enantiomers are formed
d)
Product is optically inactive
|
|
Om Desai answered |
The correct answers are Options A and C.
Cis isomer have:- a)High boiling point than trans isomer
- b)Lower boiling point than trans isomer
- c)Same boiling point
- d)None of the above
Correct answer is option 'A'. Can you explain this answer?
Cis isomer have:
a)
High boiling point than trans isomer
b)
Lower boiling point than trans isomer
c)
Same boiling point
d)
None of the above
|
|
Rajat Kapoor answered |
The cis isomer in this case has a boiling point of 60.3 degC, while the trans isomer has a boiling point of 47.5 degC. In the cis isomer the two polar C-Cl bond dipole moments combine to give an overall molecular dipole, so that there are intermolecular dipole–dipole forces (or Keesom forces), which add to the London.
The compound that is more reactive towards electrophilic nitration:- a)Toluene
- b)Benzene
- c)Nitrobenzene
- d)Benzoic Acid
Correct answer is option 'A'. Can you explain this answer?
The compound that is more reactive towards electrophilic nitration:
a)
Toluene
b)
Benzene
c)
Nitrobenzene
d)
Benzoic Acid
|
|
Rajesh Gupta answered |
From the given example, Toluene is the compound that is more reactive towards electrophilic Nitration. Toluene has a CH3 group on benzene ring and due to plus hyperconjugation it activates the presence of electrons inside the ring. Why nitrobenzene and benzoic acid have - M effect,they will draw electron cloud from the benzene ring and have deactivated the benzene ring from electrophilic Nitration.
Which starting material should be used to produce the compound shown below?
- a)
- b)
- c)
- d)
Correct answer is option 'B'. Can you explain this answer?
Which starting material should be used to produce the compound shown below?
a)
b)
c)
d)
|
|
Suresh Iyer answered |
The correct answer is option B
most stableconfomer of product will be :- a)
- b)
- c)
- d)
Correct answer is option 'C'. Can you explain this answer?
most stableconfomer of product will be :a)
b)
c)
d)
|
|
Om Desai answered |
Most stable conformer of n - propane
What is the correct order of increasing acidic strength of the following?
- a)Ill < I < IV < II
- b)Ill < IV < II < I
- c) Ill < II < IV < I
- d)III<IV<I<II
Correct answer is option 'B'. Can you explain this answer?
What is the correct order of increasing acidic strength of the following?
a)
Ill < I < IV < II
b)
Ill < IV < II < I
c)
Ill < II < IV < I
d)
III<IV<I<II
|
|
Lavanya Menon answered |
Compound I is having the highest acidic strength due to the -I effect of five CF3 substituents.
Compound II is having less acidic strength than I but more than the rest due to the extremely stable conjugate anion formed after deprrotonation.
So, Option B is correct.
Which of the following compounds react most readily with Br2(g)?
- a) C2H2
- b) C3H6
- c)C2H4
- d)C4H10
Correct answer is option 'B'. Can you explain this answer?
Which of the following compounds react most readily with Br2(g)?
a)
C2H2
b)
C3H6
c)
C2H4
d)
C4H10
|
|
Siddharth Iyer answered |
The compound that reacts most readily with Br(g) is C3H6. Here's why:
Explanation:
When a compound reacts with Br(g), it undergoes a substitution reaction called bromination. In this reaction, a Br atom replaces a hydrogen atom in the compound. The reactivity of a compound towards bromination depends on its structure and the stability of the resulting product.
Comparing the compounds:
Let's compare the given compounds and analyze their structures to determine which one is most reactive towards bromination.
a) C2H2:
C2H2 is an alkyne with a triple bond between two carbon atoms. This triple bond is very strong and stable, making it difficult for Br(g) to break it and substitute a hydrogen atom. Therefore, C2H2 is less reactive towards bromination.
b) C3H6:
C3H6 is an alkene with a double bond between two carbon atoms. The double bond is weaker and less stable than a triple bond. Therefore, it is easier for Br(g) to break the double bond and substitute a hydrogen atom. This makes C3H6 more reactive towards bromination compared to C2H2.
c) C2H4:
C2H4 is also an alkene with a double bond between two carbon atoms, similar to C3H6. It has the same structure as C3H6, but it has fewer carbon atoms. Since the number of carbon atoms does not significantly affect the reactivity towards bromination, C2H4 is also reactive towards bromination, but less reactive compared to C3H6.
d) C4H10:
C4H10 is an alkane with only single bonds between carbon atoms. Alkanes are generally less reactive towards bromination because the single bonds are strong and stable. Breaking a single bond to substitute a hydrogen atom is more difficult for Br(g) compared to breaking a double or triple bond. Therefore, C4H10 is the least reactive towards bromination among the given compounds.
Conclusion:
Based on the structural analysis and the stability of the bonds, C3H6 is the most reactive compound towards bromination among the given options.
Explanation:
When a compound reacts with Br(g), it undergoes a substitution reaction called bromination. In this reaction, a Br atom replaces a hydrogen atom in the compound. The reactivity of a compound towards bromination depends on its structure and the stability of the resulting product.
Comparing the compounds:
Let's compare the given compounds and analyze their structures to determine which one is most reactive towards bromination.
a) C2H2:
C2H2 is an alkyne with a triple bond between two carbon atoms. This triple bond is very strong and stable, making it difficult for Br(g) to break it and substitute a hydrogen atom. Therefore, C2H2 is less reactive towards bromination.
b) C3H6:
C3H6 is an alkene with a double bond between two carbon atoms. The double bond is weaker and less stable than a triple bond. Therefore, it is easier for Br(g) to break the double bond and substitute a hydrogen atom. This makes C3H6 more reactive towards bromination compared to C2H2.
c) C2H4:
C2H4 is also an alkene with a double bond between two carbon atoms, similar to C3H6. It has the same structure as C3H6, but it has fewer carbon atoms. Since the number of carbon atoms does not significantly affect the reactivity towards bromination, C2H4 is also reactive towards bromination, but less reactive compared to C3H6.
d) C4H10:
C4H10 is an alkane with only single bonds between carbon atoms. Alkanes are generally less reactive towards bromination because the single bonds are strong and stable. Breaking a single bond to substitute a hydrogen atom is more difficult for Br(g) compared to breaking a double or triple bond. Therefore, C4H10 is the least reactive towards bromination among the given compounds.
Conclusion:
Based on the structural analysis and the stability of the bonds, C3H6 is the most reactive compound towards bromination among the given options.
The compound below has four phenyl rings, but very less stable due to an opposing factor of stability. Therefore, this compound absorb bromine in dark.
How many bromine molecules, when added to this molecule, would make it stable and prevent further bromine addition?
[IIT JEE 2005]
Correct answer is '3'. Can you explain this answer?
The compound below has four phenyl rings, but very less stable due to an opposing factor of stability. Therefore, this compound absorb bromine in dark.
How many bromine molecules, when added to this molecule, would make it stable and prevent further bromine addition?
How many bromine molecules, when added to this molecule, would make it stable and prevent further bromine addition?
[IIT JEE 2005]
|
Vedant Mishra answered |
. what is different between oxidation and reduction
When potassium metal is added to 1, 3, 5, 7-cyclooctatetraene, a highly conducting salt is formed without evolution of H2 gas because- a)reduction of cyclooctatetraene into anionic
- b)reduction of cyclooctatetraene into anionic
- c)reduction of cyclooctatetraene into
- d)oxidation of cyclooctatetraene into
Correct answer is option 'B'. Can you explain this answer?
When potassium metal is added to 1, 3, 5, 7-cyclooctatetraene, a highly conducting salt is formed without evolution of H2 gas because
a)
reduction of cyclooctatetraene into anionic
b)
reduction of cyclooctatetraene into anionic
c)
reduction of cyclooctatetraene into
d)
oxidation of cyclooctatetraene into
|
|
Jyoti Aiims Aspirant answered |
Here 2 potassium ion loses electron to reduce the 1,3,5,7cyclooctatetraene to form anion c8h8^-2
A hydrocarbon X (C14H22)o n treatm ent with H2/Pt gives C14H26. Also X on treatm ent with alkaline KMnO4 followed by hydrolysis of products yields C7H12O2 which on further heating with soda lime gives cyclohexane. Hence, X is- a)
- b)
- c)
- d)
Correct answer is option 'B'. Can you explain this answer?
A hydrocarbon X (C14H22)o n treatm ent with H2/Pt gives C14H26. Also X on treatm ent with alkaline KMnO4 followed by hydrolysis of products yields C7H12O2 which on further heating with soda lime gives cyclohexane. Hence, X is
a)
b)
c)
d)
|
|
Neha Joshi answered |
The correct answers are option B.
As only 4 hydrogen atoms are increased after hydrogenation, there should be only 1 3× bond. So option B is correct.
As only 4 hydrogen atoms are increased after hydrogenation, there should be only 1 3× bond. So option B is correct.
What is the major formed in the reaction?
- a)
- b)
- c)
- d)
Correct answer is option 'C'. Can you explain this answer?
What is the major formed in the reaction?
a)
b)
c)
d)
|
|
Pranavi Chopra answered |
For option a
12 = [Ne]3s2 , 20 = [Ar]4s2 , 4 = [He]2s2 , 88 = [Ra]7s2.
For option b
8 = [He]2s22p4 , 16 = [Ne]3s23p4 , 34 = [Ar]4s23d104p4 , 2 = 1s2
We can see that only in option a, the last electrons enter in similar group. So, option a is correct.
Similarly we can check for option c and d as done above. However they won't fall in the same group.
12 = [Ne]3s2 , 20 = [Ar]4s2 , 4 = [He]2s2 , 88 = [Ra]7s2.
For option b
8 = [He]2s22p4 , 16 = [Ne]3s23p4 , 34 = [Ar]4s23d104p4 , 2 = 1s2
We can see that only in option a, the last electrons enter in similar group. So, option a is correct.
Similarly we can check for option c and d as done above. However they won't fall in the same group.
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