Haloalkanes and Haloarenes Test

Result

Haloalkanes and Haloarenes Test - 41

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

which of the following compound is the best starting material for reaction:

A

B

C

D

Solution

The option (B) represents the correct starting material. It on elimination of a molecule of HBr gives most substituted alkene which is the desired product.
If we start with other options, then isomers of the product will be obtained.
Question 2
1 / -0

$$C_6H_6+Cl_2(excess)\xrightarrow[dark,\, cold]{Anhy. Alcl_3}P$$
Product, $$P$$ is:

A
$${ C }_{ 6 }{ H }_{ 5 }Cl$$

B
$${ C }_{ 6 }{ H }_{ 4 }{ Cl }_{ 2 }$$

C
$${ C }_{ 6 }{ H }_{ 6 }{ Cl }_{ 6 }$$

D
$${ C }_{ 6 }{ Cl }_{ 6 }$$

Solution

Benzene on treatment with excess of chlorine in the presence of anhydrous $$Al{ Cl }_{ 3 }$$ in dark yields hexachlorobenzene.
Question 3
1 / -0

With respect to chlorobenzene, which of the following statements is not correct?

A
$$Cl$$ is ortho-para directing

B
$$Cl$$ exhibits $$+M$$ effect

C
$$Cl$$ is ring deactivating

D
$$Cl$$ is meta directing

Solution

$$Cl$$ exhibits $$-I$$ and $$+M$$ effect.

Due to which it is ortho/para directing but ring deactivating.

Option D is correct.
Question 4
1 / -0

An unknown alkyl halide $$[$$X$$]$$ reacts with alcoholic KOH and produces a hydrocarbon $$(C_4H_8)$$ as the major product. Ozonolysis of this hydrocarbon affords one mole of propanaldehyde and one mole of formaldehyde. Suggest which organic compound among the following has the correct structure of the above alkyl halide $$[$$X$$]$$?

A
$$CH_3CH_2CH_2CH_2Br$$

B
$$CH_3CH(Br)CH_2CH_3$$

C
$$CH_3CH(Br)CH(Br)CH_3$$

D
$$Br(CH_2)_4Br$$

Solution

As the ozonolysis gives propanaldehyde and formaldehyde thus the hydrocarbon is $$\underset{1-butene}{CH_3CH_2CH=CH_2(C_4H_8)}$$
$$1-$$butene can be obtained by $$CH_3-CH_2-CH_2-CH_2-Br\overset{Alc. KOH}{\underset{-HBr}{\rightarrow}}CH_3CH_2CH=CH_2$$.
Question 5
1 / -0

Which of the following is the correct order for strength of $$C - X$$ bond ?

A
$$ CH_3F < CH_3Cl < CH_3Br < CH_3I$$

B
$$CH_3F > CH_3Cl >CH_3Br >CH_3I$$

C
$$CH_3I>CH_3F > CH_3Cl > CH_3Br$$

D
$$CH_3Cl > CH_3Br > CH_3F >CH_3I$$

Solution

As the size of halogen atom increases, the $$C-X $$ bond length increases and hence, the $$C-X$$ bond strength decreases.
Since, the order of size of halogen atom is
$$F < Cl < Br < I$$
Thus, the order of C-X bond strength is
$$C-F > C-Cl > C-Br > C- I$$
Question 6
1 / -0

In alkaline hydrolysis of tertiary halide by aqueous alkali if concentration of alkali doubled, then the reaction:

A
will be doubled

B
will be halved

C
will remain constant

D
None of the above

Solution

In alkaline hydrolysis of tertiary halide by aqueous alkali, if concentration of alkali is doubled, then the reaction will remain constant because t-alkyl halides with aqueous alkali give $${S}_{N}1$$ reaction and the rate of $${S}_{N}1$$ reaction is not based upon the concentration of nucleophile (i.e., alkali).
Question 7
1 / -0

The reaction of $$1-$$bromo$$-3-$$chlorocyclobutane with metallic sodium in dioxane gives:

A

B

C

D

Solution

Here bromides are better leaving group than chloride so reaction occurs by elimination of bromide ion first followed by removal of chloride forming bicyclo compound.
So, the correct option is 'D'.
Question 8
1 / -0

How many optical isomer will the tartaric acid show?

A
$$3$$

B
$$4$$

C
$$0$$

D
$$2$$

Solution

The molecule has an element of symmetry and n is even. The number of d and l forms $$a=2^{n-1}$$

Meso form, $$m=2^{\displaystyle(\frac{n}{2}-1)}$$

Total $$=a+m$$
$$n=2$$

$$a=2^{n-1}=2^{2-1}$$

$$m=2^{1-1}=2^o=1$$

Total $$=2+1=3$$.
Question 9
1 / -0

Which of the following cycloalkane gives open chain compound when it reacts with bromine?

A
Cyclopropane

B
Cyclopentane

C
Cyclohexane

D
Cyclo-octane

Solution

Cyclo propane with $$Br_{2}$$.
The ring is broken because cyclopropane suffers badly from ring strain. The bond angle in the ring are $$60^{\circ}$$ rather than the normal value of about $$109.5^{\circ}$$ when the carbon makes four single bonds.
Question 10
1 / -0

.

A

B

C

D

Solution

The above reaction is the preparation of aniline from chlorobenzene. In $$KNH_{2} - {NH_{2}}^{-}$$ forms and substitutes on Benzene.