Haloalkanes and Haloarenes Test

Result

Haloalkanes and Haloarenes Test - 67

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Weekly Quiz Competition

Question 1
1 / -0

Iso-propyl bromide on Wurtz reaction gives:

A
Hexane

B
Propane

C
2, 3-dimethylbutane

D
Neo-hexane

Solution

Hence, the correct option is $$\text{C}$$
Question 2
1 / -0

The major product obtained from the monobromination of the above compound with $$Br_2/FeBr_3$$ is:

A

B

C

D

Solution

In the given reactant, the two groups which are present on the benzene rings ($$OCH_3$$ and $$CH_3$$) are electron donating and ortho- para- directing groups.
The ortho- para- directing nature of the groups are in the following order:
$$NH_2>NR_2>OH>OMe>NHAc>Me>X$$
Thus they both will direct $$Br$$ group on ortho- and para- position.
But due to the steric hindrance, the compound given in option B will be a major product.
Question 3
1 / -0

Consider the following reaction, expected to be occuring by $$ ' E-i cb' $$ mechanism
$$PH-CH_2 - \underset { \overset { | }{ Br } }{ CH } -CH_2+C_2H_5OK\xrightarrow [ ]{ C_2H_2OD } $$
Which is least expected compound in the reaction mixture:

A
$$ PhCH = CHCH_3 $$

B
$$ \underset { \overset { | }{ D } }{ PhCH } -\underset { \overset { | }{ Br } }{ CH } -CH_3 $$

C
$$ Ph-CH2-\overset { \underset { | }{ Br } }{ \underset { \overset { | }{ D } }{ C } } -CH3 $$

D
$$ Ph-\underset { \overset { | }{ D } }{ C } =CH-CH_3 $$
Question 4
1 / -0

The antiaromatic compounds are:

A
$$x,w$$

B
$$y,z$$

C
$$x,z$$

D
$$y,w$$

Solution

$$ \begin{array}{l} x=2 \pi e^{-}(\text {aromatic }) \\ y=4 \pi e^{-}(\text {anti-aromatic }) \\ z=10 \pi e^{-}(\text {aromatic }) \\ \omega=12 \pi e^{-}(\operatorname{anti}-\text { aromatic }) \\ \text { Hence }(D) y, w \text { is correct option. } \end{array} $$
Question 5
1 / -0

Deduce structure $$(A)$$.

$$(A)\ { C }_{ 8 }{ H }_{ 10 }\xrightarrow [ ]{ KMn{ O }_{ 4 } } (B)\quad { C }_{ 8 }{ H }_{ 6 }{ O }_{ 4 }\xrightarrow [ Fe ]{ { Br }_{ 2 } } (C)\quad { C }_{ 8 }{ H }_{ 5 }{ BrO }_{ 4 }$$ (one product only)

A

B

C

D
Question 6
1 / -0

$$\xrightarrow [ 2Na ]{

dry\ ether }$$

A

B

C

D

Solution
Question 7
1 / -0

Wurtz reactions have low yield because of ___________

A
the formation of multiple products.

B
reaction is reversible

C
reagent used are impure

D
none of these
Question 8
1 / -0

The order of rate of $$S_N1$$ in following iodo compounds is:

A
I > II > III

B
I > III > II

C
III > II > I

D
II > III > I
Question 9
1 / -0

$$A$$ and $$B$$ are:

A

B

C

D
none of these

Solution

When dibromo alkaline is treated with alc. $$KOH$$ followed by soda-amide and acidification, two molecules of $$HBr$$ are lost and an alkyne is obtained. This alkyne is product (A).
This alkyne then reacts with sodamide followed by ethyl chloride to form product (B). H atom of terminal alkyne is replaced with ethyl group.
Question 10
1 / -0

What is the product of given reaction?

A

B

C

D

Solution

Relation occur two two times triple bond formed