318048
In the presence of peroxide, \(\mathrm{HCl}\) and \(\mathrm{HI}\) do not give anti-Markovnikoff's addition of alkenes because
1 One of the steps is endothermic in \(\mathrm{HCl}\) and \(\mathrm{HI}\)
2 Both \(\mathrm{HCl}\) and \(\mathrm{HI}\) are strong acids
3 \(\mathrm{HCl}\) is oxidising and the \(\mathrm{HI}\) is reducing
4 All the steps are exothermic is \(\mathrm{HCl}\) and \(\mathrm{HI}\)
Explanation:
Anti-Markovnikov's addition is possible only in case of \(\mathrm{HBr}\) and not in \(\mathrm{HCl}\) and \(\mathrm{HI}\). In \(\mathrm{HBr}\) both the chain initiation and propagation steps are exothermic, while in \(\mathrm{HCl}\), first step is exothermic, and second step is endothermic and in \(\mathrm{HI}\), no step is exothermic. Hence \(\mathrm{HCl}\) and \(\mathrm{HI}\) do not undergo anti-Markovnikov's addition.
JEE - 2014
CHXI13:HYDROCARBONS
318049
In the presence of peroxide, hydrogen chloride and hydrogen iodide do not give anti-Markovnikov's addition to alkenes because
1 Both are highly ionic
2 One is oxidising and the other is reducing
3 One of the steps is endothermic in both cases
4 All steps are exothermic in both cases.
Explanation:
The reaction of \(\mathrm{HCl}\) with carbon radical in case of \(\mathrm{HCl}\), and addition of iodine radical to double bond in case of HI are endothermic steps.
CHXI13:HYDROCARBONS
318050
Which of the following reaction depicts incorrect product?
4 \(\mathrm{B}\) is less stable and formed with faster rate.
Explanation:
\(2^{\circ}\) carbocation is more stable than \(1^{\circ}\) carbocation.
JEE - 2021
CHXI13:HYDROCARBONS
318052
3-Methyl-pent-2-ene on reaction with \(\mathrm{HBr}\) in presence of peroxide forms an addition product. The number of possible stereoisomers for the product is
1 Six
2 Zero
3 Two
4 Four
Explanation:
If two chirality centers are created as a result of an addition reaction four stereoisomers can be obtained as products.
318048
In the presence of peroxide, \(\mathrm{HCl}\) and \(\mathrm{HI}\) do not give anti-Markovnikoff's addition of alkenes because
1 One of the steps is endothermic in \(\mathrm{HCl}\) and \(\mathrm{HI}\)
2 Both \(\mathrm{HCl}\) and \(\mathrm{HI}\) are strong acids
3 \(\mathrm{HCl}\) is oxidising and the \(\mathrm{HI}\) is reducing
4 All the steps are exothermic is \(\mathrm{HCl}\) and \(\mathrm{HI}\)
Explanation:
Anti-Markovnikov's addition is possible only in case of \(\mathrm{HBr}\) and not in \(\mathrm{HCl}\) and \(\mathrm{HI}\). In \(\mathrm{HBr}\) both the chain initiation and propagation steps are exothermic, while in \(\mathrm{HCl}\), first step is exothermic, and second step is endothermic and in \(\mathrm{HI}\), no step is exothermic. Hence \(\mathrm{HCl}\) and \(\mathrm{HI}\) do not undergo anti-Markovnikov's addition.
JEE - 2014
CHXI13:HYDROCARBONS
318049
In the presence of peroxide, hydrogen chloride and hydrogen iodide do not give anti-Markovnikov's addition to alkenes because
1 Both are highly ionic
2 One is oxidising and the other is reducing
3 One of the steps is endothermic in both cases
4 All steps are exothermic in both cases.
Explanation:
The reaction of \(\mathrm{HCl}\) with carbon radical in case of \(\mathrm{HCl}\), and addition of iodine radical to double bond in case of HI are endothermic steps.
CHXI13:HYDROCARBONS
318050
Which of the following reaction depicts incorrect product?
4 \(\mathrm{B}\) is less stable and formed with faster rate.
Explanation:
\(2^{\circ}\) carbocation is more stable than \(1^{\circ}\) carbocation.
JEE - 2021
CHXI13:HYDROCARBONS
318052
3-Methyl-pent-2-ene on reaction with \(\mathrm{HBr}\) in presence of peroxide forms an addition product. The number of possible stereoisomers for the product is
1 Six
2 Zero
3 Two
4 Four
Explanation:
If two chirality centers are created as a result of an addition reaction four stereoisomers can be obtained as products.
318048
In the presence of peroxide, \(\mathrm{HCl}\) and \(\mathrm{HI}\) do not give anti-Markovnikoff's addition of alkenes because
1 One of the steps is endothermic in \(\mathrm{HCl}\) and \(\mathrm{HI}\)
2 Both \(\mathrm{HCl}\) and \(\mathrm{HI}\) are strong acids
3 \(\mathrm{HCl}\) is oxidising and the \(\mathrm{HI}\) is reducing
4 All the steps are exothermic is \(\mathrm{HCl}\) and \(\mathrm{HI}\)
Explanation:
Anti-Markovnikov's addition is possible only in case of \(\mathrm{HBr}\) and not in \(\mathrm{HCl}\) and \(\mathrm{HI}\). In \(\mathrm{HBr}\) both the chain initiation and propagation steps are exothermic, while in \(\mathrm{HCl}\), first step is exothermic, and second step is endothermic and in \(\mathrm{HI}\), no step is exothermic. Hence \(\mathrm{HCl}\) and \(\mathrm{HI}\) do not undergo anti-Markovnikov's addition.
JEE - 2014
CHXI13:HYDROCARBONS
318049
In the presence of peroxide, hydrogen chloride and hydrogen iodide do not give anti-Markovnikov's addition to alkenes because
1 Both are highly ionic
2 One is oxidising and the other is reducing
3 One of the steps is endothermic in both cases
4 All steps are exothermic in both cases.
Explanation:
The reaction of \(\mathrm{HCl}\) with carbon radical in case of \(\mathrm{HCl}\), and addition of iodine radical to double bond in case of HI are endothermic steps.
CHXI13:HYDROCARBONS
318050
Which of the following reaction depicts incorrect product?
4 \(\mathrm{B}\) is less stable and formed with faster rate.
Explanation:
\(2^{\circ}\) carbocation is more stable than \(1^{\circ}\) carbocation.
JEE - 2021
CHXI13:HYDROCARBONS
318052
3-Methyl-pent-2-ene on reaction with \(\mathrm{HBr}\) in presence of peroxide forms an addition product. The number of possible stereoisomers for the product is
1 Six
2 Zero
3 Two
4 Four
Explanation:
If two chirality centers are created as a result of an addition reaction four stereoisomers can be obtained as products.
318048
In the presence of peroxide, \(\mathrm{HCl}\) and \(\mathrm{HI}\) do not give anti-Markovnikoff's addition of alkenes because
1 One of the steps is endothermic in \(\mathrm{HCl}\) and \(\mathrm{HI}\)
2 Both \(\mathrm{HCl}\) and \(\mathrm{HI}\) are strong acids
3 \(\mathrm{HCl}\) is oxidising and the \(\mathrm{HI}\) is reducing
4 All the steps are exothermic is \(\mathrm{HCl}\) and \(\mathrm{HI}\)
Explanation:
Anti-Markovnikov's addition is possible only in case of \(\mathrm{HBr}\) and not in \(\mathrm{HCl}\) and \(\mathrm{HI}\). In \(\mathrm{HBr}\) both the chain initiation and propagation steps are exothermic, while in \(\mathrm{HCl}\), first step is exothermic, and second step is endothermic and in \(\mathrm{HI}\), no step is exothermic. Hence \(\mathrm{HCl}\) and \(\mathrm{HI}\) do not undergo anti-Markovnikov's addition.
JEE - 2014
CHXI13:HYDROCARBONS
318049
In the presence of peroxide, hydrogen chloride and hydrogen iodide do not give anti-Markovnikov's addition to alkenes because
1 Both are highly ionic
2 One is oxidising and the other is reducing
3 One of the steps is endothermic in both cases
4 All steps are exothermic in both cases.
Explanation:
The reaction of \(\mathrm{HCl}\) with carbon radical in case of \(\mathrm{HCl}\), and addition of iodine radical to double bond in case of HI are endothermic steps.
CHXI13:HYDROCARBONS
318050
Which of the following reaction depicts incorrect product?
4 \(\mathrm{B}\) is less stable and formed with faster rate.
Explanation:
\(2^{\circ}\) carbocation is more stable than \(1^{\circ}\) carbocation.
JEE - 2021
CHXI13:HYDROCARBONS
318052
3-Methyl-pent-2-ene on reaction with \(\mathrm{HBr}\) in presence of peroxide forms an addition product. The number of possible stereoisomers for the product is
1 Six
2 Zero
3 Two
4 Four
Explanation:
If two chirality centers are created as a result of an addition reaction four stereoisomers can be obtained as products.
318048
In the presence of peroxide, \(\mathrm{HCl}\) and \(\mathrm{HI}\) do not give anti-Markovnikoff's addition of alkenes because
1 One of the steps is endothermic in \(\mathrm{HCl}\) and \(\mathrm{HI}\)
2 Both \(\mathrm{HCl}\) and \(\mathrm{HI}\) are strong acids
3 \(\mathrm{HCl}\) is oxidising and the \(\mathrm{HI}\) is reducing
4 All the steps are exothermic is \(\mathrm{HCl}\) and \(\mathrm{HI}\)
Explanation:
Anti-Markovnikov's addition is possible only in case of \(\mathrm{HBr}\) and not in \(\mathrm{HCl}\) and \(\mathrm{HI}\). In \(\mathrm{HBr}\) both the chain initiation and propagation steps are exothermic, while in \(\mathrm{HCl}\), first step is exothermic, and second step is endothermic and in \(\mathrm{HI}\), no step is exothermic. Hence \(\mathrm{HCl}\) and \(\mathrm{HI}\) do not undergo anti-Markovnikov's addition.
JEE - 2014
CHXI13:HYDROCARBONS
318049
In the presence of peroxide, hydrogen chloride and hydrogen iodide do not give anti-Markovnikov's addition to alkenes because
1 Both are highly ionic
2 One is oxidising and the other is reducing
3 One of the steps is endothermic in both cases
4 All steps are exothermic in both cases.
Explanation:
The reaction of \(\mathrm{HCl}\) with carbon radical in case of \(\mathrm{HCl}\), and addition of iodine radical to double bond in case of HI are endothermic steps.
CHXI13:HYDROCARBONS
318050
Which of the following reaction depicts incorrect product?
4 \(\mathrm{B}\) is less stable and formed with faster rate.
Explanation:
\(2^{\circ}\) carbocation is more stable than \(1^{\circ}\) carbocation.
JEE - 2021
CHXI13:HYDROCARBONS
318052
3-Methyl-pent-2-ene on reaction with \(\mathrm{HBr}\) in presence of peroxide forms an addition product. The number of possible stereoisomers for the product is
1 Six
2 Zero
3 Two
4 Four
Explanation:
If two chirality centers are created as a result of an addition reaction four stereoisomers can be obtained as products.
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