Hydrocarbons Test

Result

Hydrocarbons Test - 21

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

Direction (Q. Nos. 1 - 7) This section contains 7 multiple choice questions. Each question has four choices (a), (b), (c) and (d), out of which ONLY ONE option is correct.
Q. What is the major addition product in the following reactions?

A

B

C

D
Question 2
1 / -0

What is the major product in the following reaction?

A

B

C

D
Question 3
1 / -0

Consider the following reaction.
Molecular formula of both P and Q are C6H₁₃l
tetra substituted alkene isom er of A. Hence, P is most likely

A
2, 2-dimethyl-3-iodobutane

B
2-iodo-2, 3-dimethyl butane

C
1-iodo-3, 3-dimethyl butane

D
2, 3-diiodo-2, 3-dimethyl butan

Solution

Molecular formula of P and Q is C6H13I and this 3 degree reaction mechanism is SN1 mechanism so by that I molecules from HI attack as nucleophile to compound and when the reaction with KOH and C2H5OH rearrangement of I take place and the molecule formed is 2 iodo-2, 3- dimethyl butane.
Question 4
1 / -0

If trans-2-pentene is treated with Br₂in CCI₄ in presence of Lewis acid FeBr₃

A
a pair of enantiomers of 2, 3-dibromopentane, but in unequal amounts are formed

B
a racemic mixture of 2, 3-dibromopentane is formed

C
a pure enantiomer of 2, 3-dibromopentane is formed

D
a pair of diastereomers of 2, 3 dibromopentane is formed

Solution

Trans-2-pentene on reaction with Br₂ in CCl4 produces mainly (±) 2,3dibromopentene. A molecule of bromine is added across C=C double bond. A racemic mixture is obtained as the product.
Question 5
1 / -0

3-methyl-1-pentene has a chiral carbon. If a pure enantiomer of 3-methyl-1-pentene is treated with HBr, the correct statement regarding major bromoalkane formed is

A
a pure enantiomer is formed

B
racemic mixture is formed

C
optically inactive product is formed

D
meso-product is formed

Solution

The correct answer is Option A.
3−Methyl−pent−2−ene in reaction with HBr forms an additional product.
A molecule of HBr is added to C=C double bond.
The addition follows anti-Markovnikov's rule. Br is added to less substituted C atoms.
The number of possible stereoisomers for the products is four. The product has 2 chiral C atoms and is unsymmetrical.

An enantiomer will rotate the plane of polarization the equal angle but opposite direction. A racemic mixture will not rotate the plane of polarization.
Question 6
1 / -0

Consider the following reaction.
Which of the following would not be formed in the above reaction?

A
2, 3-dibromo-2-methyl butane

B
2-chloro-3-bromo-2-methyl butane

C
3-bromo-2-methyl-2-butanol

D
2, 3-dichloro-2-methyi butane
Question 7
1 / -0

HBr reacts with CH₂ = CH – OCH₃ under anhydrous conditions at room temperature to give
[AlEEE 2006]

A
BrCH₂— CHO and CH₃OH

B
Br— CH₂— CH₂— OCH₃

C
H₃C – CHBr – OCH₃

D
CH₃CHO + CH₃Br

Solution

Methyl vinyl ether is a very reactive gas. It is hydrolysed rapidly by dilute acids at room temperature to give methanol and aldehyde. However, under anhydrous conditions at room temperature, it undergoes many addition reactions at the double bond.
Electrophilic addition reaction more favourable.
Question 8
1 / -0

Choices for the correct combination of elements from Column I and Column lI are given as options (a), (b), (c) and (d), out of which one is correct.
Consider the reactions in Column I and match with the properties of products from Column II.

A
a

B
b

C
c

D
d