Haloalkanes and Haloarenes Test

Result

Haloalkanes and Haloarenes Test - 84

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Weekly Quiz Competition

Question 1
1 / -0

The decreasing order of the acidic character of the following is :

A
(I) > (II) > (III) > (IV)

B
(I) > (III) > (II) > (IV)

C
(IV) > (III) > (II) > (I)

D
(IV) > (II) > (III) > (I)

Solution

Cyclo alkanes are more acidic than the corresponding alkanes .
Question 2
1 / -0

$$C_3H_7Cl \xrightarrow {KOH(alc.)} (A) \xrightarrow {770\ K}{Cl_2 (g)} (X)$$
(X) can be:

A
Vinyl chloride

B
Allyl chloride

C
Ethyl chloride

D
Ethyl iodide

Solution

$$CH_3-CH_2-CH_2-Cl\xrightarrow {alc.KOH}(A) \Rightarrow CH_3-CH=CH_2$$

$$(X) \Rightarrow$$ Allylic chlorination

$$Cl-CH_2-CH=CH_2$$

Allyl chloride

Option B is correct.
Question 3
1 / -0

Which method is feasible for the preparation of compound (X) only ?

A

B

C

D

Solution

So, the method is method A
Hence, Option "A" is the correct answer.
Question 4
1 / -0

The chemical reaction
$${ { (CH }_{ 3 }) }_{ 3 }CBr\xrightarrow [ { -H }_{ 2 }O,-KBr ]{ alc.KOH } { { (CH }_{ 3 }) }_{ 2 }C={ CH }_{ 2 }$$
is an example of:

A
Nucleophilic subsitution

B
Electrophilic subsitution

C
Free radical subsitution

D
$$\beta$$-Elimination

Solution

$$E1$$ and $$E2$$ are both called $$\beta$$-elimination because $$\beta-H$$ (most acidic) is removed by base.

But $$E2$$ is also called anti-or trans-elimination.

Option D is correct.
Question 5
1 / -0

Which statement is correct about the above reaction?

A
Both products (B) and (C) are optically active.

B
Both products (B) and (C) are optically inactive.

C
Product (B) is optically inactive but product (C) is optically active.

D
Product (B) is formed by anti-addition but product (C) is formed by syn-addition of $$H_2$$.
Solution
Both reagents convert alkyne to alkene
P-2catalyst adds by $$syn-addition$$, while Birch reduction adds by $$anti-addition$$.
hence option (C) is correct.
Question 6
1 / -0

The compound (A) is:

A

B

C

D
None of these

Solution
Question 7
1 / -0

Arrange the compounds in increasing order of rate of reaction towards nucleophilic substitution.

A
$$ (a)< (b)< (c) $$

B
$$ (a)< (c)< (b) $$

C
$$ (c)< (b)< (a) $$

D
$$ (b)< (c)< (a) $$
Question 8
1 / -0

In the question arrange the compounds in increasing order of rate of reaction towards nucleophilic substitution.

A
(I) < (II) < (III)

B
(I) < (III) < (II)

C
(III) < (II) < (I)

D
(II) < (III) < (I)

Solution

Presence of electron donating group (-CH$$_3$$) increases the reactivity towards nucleophilic substitution and the effect of o-position is greater than at m-position because of hyperconjugation and inductive effects.

So, the order becomes (II) < (III) < (I)
Hence, Option "D" is the correct answer.
Question 9
1 / -0

Ethyl alcohol is obtained when ethyl chloride is boiled with

A
alcoholic KOH

B
aqueous KOH

C
water

D
aqueous $$KMnO_4$$

Solution

Ethyl alcohol is obtained when ethyl chloride is boiled with aqueous KOH.

Reaction:
$$C_2H_5Cl + KOH_{(aq)} \rightarrow C_2H_5OH + KCl$$

Hence, Option "B" is the correct answer.
Question 10
1 / -0

Which of the following will be the major and minor product?

A
(A) is major product and (B) is minor products.

B
(A) is minor product and (B) is major product

C
Both (A) and (B) are major products.

D
Only (B) is formed and (A) is not formed.

Solution

According to Saytzeff's rule, the more substituted products is more stable and is formed as major product. Hence (A) is major product (80%) while (B) is the minor (20%) and less stable product.

Hence, Option "A" is the correct answer.