Haloalkanes and Haloarenes Test

Result

Haloalkanes and Haloarenes Test - 44

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Weekly Quiz Competition

Question 1
1 / -0

Specific rotation is the rotation produced by $$1$$g of material in $$1$$ml of liquid in a $$1$$ dm tube. Which of the following molecules have largest value of specific rotation? (Hint : Molecular mass increasing).

A

B

C

D

Solution

Would have more specific rotation due to -D groups in excess
Question 2
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The IUPAC name of $$(CH_3)_2CH-CH_2-CH_2Br$$ is ?

A
1-bromopentane

B
1-bromo-3-methylbutane

C
2-methyl-4-bromobutane

D
2-methyl-3-bromopropane

Solution

4 3 2 1
$$CH_{3}-\underset {\underset{\displaystyle {CH_{3}} }{\displaystyle |}}{C}H–CH_{2}–CH_{2}Br$$

Its IUPAC name is 1-bromo-3-methylbutane.

The correct option is B.
Question 3
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Directions For Questions

Dehydrohalogenation involves removal of the halogen atom together with a hydrogen atom from a carbon atom adjacent to the one with a halogen atom. Alcoholic KOH is used for dehydrohalogenation. According to Saytzeff's rule, when two alkenes may be formed, the alkene which is most substituted is the major product.

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Arrange the following alkyl halides in decreasing order of the rate of $$\beta$$- elimination reaction with alcoholic KOH.
$${ CH }_{ 3 }{ CH }_{ 2 }{ CH }_{ 2 }Br\quad { CH }_{ 3 }-\underset { \underset { { \displaystyle \,\,\,\,CH }_{ 3 } }{ | } }{ CH } -{ CH }_{ 2 }Br\quad { CH }_{ 3 }-{ CH }_{ 2 }-\underset { \underset {\displaystyle Br }{ | } }{ CH } -{ CH }_{ 3 }\quad $$
(i) (ii) (iii)

A
$$(ii) > (iii) > (i)$$

B
$$(iii) > (ii) > (i)$$

C
$$(i) > (ii) > (iii)$$

D
$$(ii) > (i) > (iii)$$

Solution

Rate of dehydrohalogenation in alkyl halide is in the order $$\Rightarrow \; 3°>2°>1°$$.
Thus, correct order is $$(iii)>(ii)>(i)$$.
Question 4
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The IUPAC name of tertiary butyl chloride is ?

A
2-chloro-2-methylpropane

B
3-chlorobutane

C
4-chlorobutane

D
1, 2-chloro-3-methylpropane.

Solution

Tertiary butyl chloride is an organochloride with formula $$ (CH_3)_3CCl $$. The longest continuous chain of carbons is 3. It is propane. The remaining groups are chloro and methyl. Both are at position 2. Chlorine comes before methyl in the alphabet, so the IUPAC name of tertiary butyl chloride is 2-chloro-2-methylpropane.
Question 5
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An alkane $$C_6H_{14}$$ gives only two monochloro derivatives on chlorination. Its possible structure is:

A
$$CH_3CH_2CH_2CH_2CH_2CH_3$$

B
$$CH_3-{\underset{\displaystyle CH_3}{\underset{|}CH}}-CH_2CH_2CH_3$$

C
$$CH_3CH_2-{\underset{\displaystyle CH_3}{\underset{|}CH}}

-CH_2CH_3$$

D
$$CH_3-{\overset{\displaystyle CH_3}{\overset{|}CH}}-{\overset{\displaystyle CH_3}{\overset{|}CH}}-CH_3$$

Solution

Monochloro derivation means substitution of single chlorine atom by replacing it with any hydrogen attached to the carbon atom.
(D) will give only two monochloroderivative:
$$CH_3-{\overset{\displaystyle CH_3}{\overset{|}CH}}-{\overset{\displaystyle CH_3}{\overset{|}CH}}-CH_3 \xrightarrow{{Cl}_{2}} CH_3-{\overset{\displaystyle CH_3}{\overset{|}CH}}-\underset{\underset{\displaystyle Cl}|}{{\overset{\displaystyle \,\,\,\,CH_3}{\overset{|}C}}}-CH_3 + Cl-CH_2-{\overset{\displaystyle CH_3}{\overset{|}CH}}-{\overset{\displaystyle CH_3}{\overset{|}CH}}-CH_3$$
Question 6
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Identify $$X$$ and $$Y$$ in the following reaction.
$$H_ 2\underset { \underset {\displaystyle Br }{ | } }{ C } -\underset { \underset { \displaystyle Br }{ | } }{ CH_2}+KOH\xrightarrow [ ]{ alcohol } X\xrightarrow [ ]{ NaNH_ 2 } Y$$

A
$$X: CH_3CHBr, \,\,Y: CH_2=CH_2$$

B
$$X: CH_2OH-CH_2OH, \,\, Y: CH_2=CH_2$$

C
$$X: CH_2=CHBr, \,\, Y: CH\equiv CH$$

D
$$X: CH\equiv CBr, \,\, Y: CH\equiv CH$$

Solution

The reactant is symmetrical alkanes. On reacting with alcoholic KOH, dehydrohalogenation reaction will occur. It will give monobromoalkene.
$$Br{H}_{2}C-C{H}_{2}Br + KOH \xrightarrow{alcohol} C{H}_{2}=CHBr$$
$$C{H}_{2}=CHBr$$ on reaction with $$NaN{H}_{2}$$ releases sodium bromide and ammonia and forms ethyne.
$$C{H}_{2}=CHBr \xrightarrow{NaN{H}_{2}} CH \equiv CH + NaBr + N{H}_{3}$$
Question 7
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$$2-$$$ Bromopentane is treated with alcoholic $$KOH$$ solution. What will be the major product formed in this reaction and what is this type of elimination called?

A
Pent $$-1-$$ ene, $$\beta$$- Elimination

B
Pent $$-2-$$ ene, $$\beta$$- Elimination

C
Pent $$-1-$$ ene, Nucleophilic substitution

D
Pent $$-2-$$ ene, Nucleophilic substitution
Solution
Pent-2-ene is major product known as Saytzeff's Product and it is more stable alkene.
The process is known as $$\beta-$$ elimination as it involves elimination of $$\beta-$$ Hydrogen.
Question 8
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In halogenation of aromatic hydrocarbon, a halogen carrier is used which is generally a Lewis acid. The main function of this reagent is to generate the species :

A
$$ X $$

B
$$ X^- $$

C
$$ X^+ $$

D
$$ X^{\bullet}$$

Solution

Halogen carrier generates an electrophile with the help of halogen. e.g. during chlorination $${Cl}^{+}$$ is generated.
$$Al{X}_{3} + {X}_{2} \longrightarrow {Al{X}_{4}}^{-} + {X}^{+}$$
Question 9
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The correct decreasing order of following alkyl halides according to the rate of $$\beta$$ - elimination reaction with alcoholic KOH will be:
(i)$${ CH }_{ 3 }-\overset { \underset { | }{ H } }{ \underset { \underset { { CH }_{ 3 } }{ | } }{ C } } -{ CH }_{ 2 }Br$$
(ii) $$ CH_3-CH_2-Br$$
(iii) $$CH_3-CH_2-CH_2-Br$$

A
(i) > (ii) > (iii)

B
(iii) > (ii) > (i)

C
(ii) > (iii) > (i)

D
(i) > (iii) > (ii)

Solution

Stability of alkene after $$\beta$$-elimination directly depends on the number of $$\beta$$-substituents.
Since number of $$\beta$$- methyl substituent in (i) is 2, (iii) is 1 and (ii) is 0, thus the correct order of stability is: $$(i)>(iii)>(ii)$$.
Question 10
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Which of the following compounds will react with Na to form 4,5- diethyloctane?

A
$$CH_3CH_2CH_2CH_2Br$$

B
$$CH_3CH_2CH_2-{\underset{CH_3}{\underset{|}C}H}-CH_2CH_2Br$$

C
$$CH_3CH_2CH_2CH_2-{\underset{Br}{\underset{|}C}H}-CH_3$$

D
$$ CH_3CH_2CH_2-{\underset{Br}{\underset{|}C}H}-CH_2CH_3$$

Solution

Answer:-
The reaction is wurtz reaction and proceeds as follows: