Haloalkanes and Haloarenes Test

Result

Haloalkanes and Haloarenes Test - 51

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Weekly Quiz Competition

Question 1
1 / -0

Reagent which can convert an alkyl amine into alkyl chloride.

A
Tilden reagent (NOCI)

B
Lucas reagent

C
Hisnberg's reagent

D
None

Solution

Alkylamine + Tilden's Reagent $$\rightarrow $$ Alkyl chloride.
$$R-{ NH }_{ 2 }+NoCl\rightarrow R-Cl+NO-{ NH }_{ 2 }$$
$$\hookrightarrow $$ Tiden's Regent.
Question 2
1 / -0

Which of the following has the lowest barrier to rotation about the indicated bond ?

A

B

C
Both A and B

D

Solution

lowest barsier to rotation indicates
easier to rotate ( activation energy is less)
So option $$B$$ is corrcet
Question 3
1 / -0

$${ CH }_{ 2 }={ CH }_{ 2 }+{ Br }_{ 2 }\xrightarrow { NaCl } \quad Products$$. Which of the following product is not formed in the above reaction?

A
$$Br{ CH }_{ 2 }{ CH }_{ 2 }Br$$

B
$$Cl{ CH }_{ 2 }{ CH }_{ 2 }Cl$$

C
$$Br{ CH }_{ 2 }{ CH }_{ 2 }Cl$$

D
All are formed

Solution

Solution:- (B) $$Cl-C{H}_{2}-C{H}_{2}-Cl$$
$$C{H}_{2} = C{H}_{2} + {Br}_{2} \xrightarrow{NaCl} {C{H}_{2}}^{+}-C{H}_{2}-Br$$
The $$\pi$$-electrons will attack an electrophile forming the carbocation.
Now the carbocation will be attacked by a nucleophile.
Hence the product will be a mixture of $$Br-C{H}_{2}-C{H}_{2}-Br$$ and $$Br-C{H}_{2}-C{H}_{2}-Cl$$.
Question 4
1 / -0

Which of the following statements regarding the $$S_N1$$ reaction shown by alkyl halide is NOT CORRECT?

A
The added nucleophile plays no kinetic role in $$S_N1$$ reaction.

B
The $$S_N1$$ reaction involves the inversion of configuration of the optically active substrate

C
The $$S_N1$$ reaction on the chiral starting material ends up with racemization of the product.

D
The more stable the carbocation intermediate the faster the $$S_N1$$ reaction

Solution
Question 5
1 / -0

Alkane can be iodinated in the presence of:

A
$$HI$$

B
$$I_2$$ and $$P$$

C
$$HIO_3$$

D
$$PI_3$$

Solution

Iodination of an alkane is carried out in the presence of oxidizing agent because one of the products of this reaction is hydrogen iodide and this is a strong reducing agent and converts alkyl iodide back to an alkane.

Therefore to counter this difficulty, the iodination of alkanes are carried out in the presence of a strong oxidizing agent like iodic acid $$(HIO_3)$$ which oxidizes the hydrogen iodide formed.
Question 6
1 / -0

The racemic mixture of Alanine ($$CH_8-CH-COOH$$) can be resolved by
$$NH_2$$
1) (+)-2-Butanol
2) (l)-2-Chlorobutanoic acid
3) ()-2-Butanol
4) $$(dil$$ mix)-2-Chlorobutanoic acid

A
1&2 only

B
1&3 only

C
2&4only

D
3&4only

Solution
Question 7
1 / -0

Alkyl halides can be obtained by all methods excepts?

A
$$CH_{3}-CH_{2}-+HCl\overset{znci_{2}}{\rightarrow}$$

B
$$CH_{2}=CH_CH_{3}-+HBr\rightarrow $$

C
both a and b

D
none of these

Solution
Question 8
1 / -0

$$CH_3-CH_3+Cl_2\xrightarrow{hv}$$?

A
$$C_2H_5Cl$$

B
$$Cl_2$$

C
$$CH_2(Cl)-CH_2(Cl)$$

D
All of the above

Solution

The reaction given in the question is free radical halogenation reaction of alkanes in the presence of sunlight. The free radical halogenation in the given conditions gives the major product as $${C_2}{H_5}Cl$$.
The whole reaction will be:
$$C{H_3} - C{H_3}(g) + C{l_2} \to {C_2}{H_5}Cl(g) + HCl(g)$$
Hence, the correct option will be option A
Question 9
1 / -0

$${CH}_{3}-I+2Na+I-{CH}_{3}\xrightarrow [ ]{dry \; ether } $$ _______

A
$${C}_{2}{H}_{5}-{C}_{2}{H}_{5}$$

B
$${CH}_{4}$$

C
$${C}_{3}{H}_{8}$$

D
$${CH}_{3}-{CH}_{3}$$

Solution

When methyl Iodide is treated with sodium in the presence of dry ether, Ethane is formed .It is
called Wurtz reaction.
Dry ether
$$CH_3−I + 2Na + I−CH_3 \rightarrow CH_3−CH_3 + 2NaI$$
So option $$D$$ is correct
Question 10
1 / -0

Above reaction is an example of 1-4 -elimination .Predict the product.

A
$$ C{ H }_{ 3 }-CH=C=C{ H }_{ 2 }$$

B
$$C{ H }_{ 3 }-C=C=C{ H }_{ 3 }$$

C
$$ C{ H }_{ 3 }-C{ H }_{ 2 }=C=C{ H } $$

D
$${ H }_{ 2 }C=CH-CH=C{ H }_{ 2 }$$