Haloalkanes and Haloarenes Test

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Haloalkanes and Haloarenes Test - 63

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Weekly Quiz Competition

Question 1
1 / -0

Which of the following sequence of reactions (reagents) can be used for the conversion of $$Ph{CH}_{2}{CH}_{3}$$ into $$PhCH={CH}_{2}$$?

A
$$SO{Cl}_{2};\ {H}_{2}O$$

B
$${SO}_{2}{Cl}_{2};\ alc. KOH$$

C
$${Cl}_{2}/hv;\ {H}_{2}O$$

D
$$SO{Cl}_{2}; \ alc. KOH$$

Solution

Ethyl benzene can be converted to styrene by using sulphuryl chloride followed by alc. KOH. Sulphuryl chloride is used for the allylic substitution and alc. KOH is used for dehydrochlorination.
Question 2
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Which of the following sequences would yield m-nitro chlorobenzene $$(Z)$$ from benzene?

A
Benzene $$\xrightarrow [ ]{ { Cl }_{ 2 }/Fe{ Cl }_{ 3 } } (X)\xrightarrow [ { H }_{ 2 }{ SO }_{ 4 } ]{ H{ NO }_{ 3 } } (Z)$$

B
Benzene $$\xrightarrow [ ]{ { H }_{ 2 }{ SO }_{ 4 }/H{ NO }_{ 3 } } (Z)$$

C
Benzene $$\xrightarrow [ ]{ { H }_{ 2 }{ SO }_{ 4 }/H{ NO }_{ 3 } } (X)\xrightarrow [ ]{ { Cl }_{ 2 }/Fe{ Cl }_{ 3 } } (Z)$$

D
All of these

Solution

Hence, the answer is (c).
Question 3
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Bottles containing $$PhI$$ and $$Ph{CH}_{2}I$$ lost their original labels. They were labelled as $$(A)$$ and $$(B)$$ for testing. $$(A)$$ and $$(B)$$ were separately taken in test tubes and boiled with $$NaOH$$ solutions. The end solution in each tube was made acidic with dilutie $$H{NO}_{3}$$ and some $$Ag{NO}_{3}$$ solution was added. Substance $$(B)$$ gave a yellow precipitate. Which of the following statements is true for this experiment?

A
Addition of $$H{NO}_{3}$$ was unnecessary

B
$$(A)$$ was $$PhI$$

C
$$(A)$$ was $$Ph{CH}_{2}I$$

D
$$(B)$$ was $$PhI$$

Solution

$$PhI+aq. NaOH\rightarrow No\ reaction$$
$$PhCH_2I+Aq. NaOH\rightarrow PhCH_2OH+I^{\ominus}\xrightarrow {AgNO_3}AgI(Yellow\ ppt.)$$
(A) was iodobenzene $$PhI$$ and (B) was benzyl iodide $$PhCH_2I$$.
Since (A) is aromatic iodide, it will not give free iodide ions. Hence, no precipitate will be obtained with silver nitrate.
(B) is alkyl iodide. It will give free iodide ions. Hence, yellow precipitate will be obtained with silver nitrate.
Question 4
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The product on monobromination of this compound is :

A

B

C

D

Solution

It is electrophilic substitution, so electrophile must be attacked at o/p position due to higher electron density at these positions. Ring $$(A)$$ is activated by $$+R$$ effect of $$(-NH)$$ group, while ring $$(B)$$ has EW $$\left( -\overset { O\\ \parallel }{ C } - \right) $$ group.
So, SE reaction will take place at p-position of ring $$(A)$$ since o-position is blocked.
Question 5
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Aryl halides are less reactive towards nucleophilic substitution reaction as compared to alkyl halides due to:

A
the formation of less stable carbonium ion

B
resonance stabilisation

C
larger carbon-halogen bond

D
inductive effect

Solution

Aryl halides are less reactive towards nucleophilic substitution reaction as compared to alkyl halides due to resonance stabilization.

Due to resonance, $$C-Cl$$ bond acquires partial double bond character and becomes shorter and stronger and cannot be easily replaced by nucleophiles.

Other reasons for the low reactivity of aryl halides are

(1) Difference in the hybridization states of carbon atom in $$C-X$$ bond. In alkyl halides, the carbon atom is $$\displaystyle sp^3$$ hybridized whereas in aryl halides, it is $$\displaystyle sp^2$$ hybridized and hence, more electronegative. Hence, $$C-X$$ bond in aryl halides is more difficult to break.

(2) Polarity of $$C-X$$ bond in aryl halides is lower than that in alkyl halides. Lower is the polarity, lower is the reactivity.
Question 6
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Directions For Questions

Comprehension # 3
Nucleophilic aliphatic substitution reaction is mainly of two types : $$S_{N}1$$ and $$S_{N}2$$. The $$S_{N}1$$ mechanism is a two step process. Reaction velocity of $$S_{N}1$$ reaction depends only on the concentration of the substrate. Since product formation takes place by the formation of carbocation, optically active substrate gives $$(+)$$ and $$(-)$$ forms of the product. In most of the cases the product usually consists of $$5-20\%$$ inverted product and $$80-95\%$$ racemised species. The more stable the carbocation, the greater is the proportion of racemisation. In solvolysis reaction, the more nucleophilic the solvent, the greater is the proportion of inversion.

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Which one of the following compound will give $$S_{N}1$$ reaction predominantly?

A

B

C

D
None of these

Solution

The compound of option (A) $$\displaystyle H_5C_6-C(CH_3)_2Br$$ will give $$\displaystyle S_N1$$ reaction predominately.
It is a tertiary alkyl halide. Also the carbocation intermediate (formed by loss of bromide ion) is stabilized by resonance with the benzene ring.
Question 7
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Directions For Questions

Comprehension # 3
Nucleophilic aliphatic substitution reaction is mainly of two types : $$S_{N}1$$ and $$S_{N}2$$. The $$S_{N}1$$ mechanism is a two step process. Reaction velocity of $$S_{N}1$$ reaction depends only on the concentration of the substrate. Since product formation takes place by the formation of carbocation, optically active substrate gives $$(+)$$ and $$(-)$$ forms of the product. In most of the cases the product usually consists of $$5-20\%$$ inverted product and $$80-95\%$$ racemised species. The more stable the carbocation, the greater is the proportion of racemisation. In solvolysis reaction, the more nucleophilic the solvent, the greater is the proportion of inversion.

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Which of the following compounds will give $$S_{N}1$$ and $$S_{N}2$$ reactions with considerable rate?
I.$$\;C_6H_5-CH_2-Br$$
II.$$\;CH_2=CH-CH_2-Br$$
III.$$\,CH_3-CH(Br)CH_3$$
Select the correct answer from the codes below below.

A
$$\;I,\,II\,and\,III$$

B
$$\;I,\,II\,and\,IV$$

C
$$\;II,\,III\,and\,IV$$

D
$$\;I,\,III\,and\,IV$$

Solution

The following compounds will give $$S_{N}1$$ and $$S_{N}2$$ reactions with a considerable rate.
$$I.\;C_6H_5-CH_2-Br$$
$$II.\;CH_2=CH-CH_2-Br$$
$$III.\,CH_3-CH(Br)CH_3$$
I and II are primary alkyl halides. Hence, they do react by $$S_N2$$ mechanism. However, the carbocation formed is stabilized by resonance with either C=C double bond or benzene ring. Hence, they also react by $$S_N1$$ mechanism.
The compound III is a secondary alkyl bromide. It reacts by $$S_N1$$ and $$S_N2$$ mechanism.
However, the compound IV will undergo $$S_N1$$ reaction faster than $$S_N2$$ reaction as it is sterically hindered and the approach of the nucleophile from the side opposite to that of leaving group is difficult.
Question 8
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Directions For Questions

Comprehension # 2
Nucleophilic substitution reactions generally expressed as
$$Nu^-+R\,-\,L\longrightarrow R-\,Nu+L^-$$
Where $$Nu^-\longrightarrow Nucleophile\:;\:R-L\longrightarrow substrate\:;\:L\longrightarrow leaving\:group$$
The best leaving groups are those that become the most stable ions after they depart. Since most leaving groups leave as a negative ion, the best leaving groups are those ions that stabilize a negative charge most effectively. A good leaving group should be
(a) electron-withdrawing to polarize the carbon
(b) stable once it has left (not a strong base)
(c) polarisable- to maitain partial bonding with the carbon in the transition state (both $$S_{N}1$$ and $$S_{N}2$$). This bonding helps to stabilise the transition state and reduces the activation energy.

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$$Cl^-\;\;\;\;\;\;\;\;\;\;\;\;CH_3O^-\;\;\;\;\;\;\;\;\;\;\;\;CH_3S^-\;\;\;\;\;\;\;\;\;\;\;\;I^-$$
$$(I)\;\;\;\;\;\;\;\;\;\;\;\;(II)\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;(III)\;\;\;\;\;\;\;\;\;\;\;\;(IV)$$
The correct order of increasing leaving group capability of above anions:

A
$$III < IV < II < I$$

B
$$II < III < I < IV$$

C
$$II < IV < III < I$$

D
$$I < III < II < IV$$

Solution

The correct order of increasing leaving group capability of the anions is
$$\displaystyle II (CH_3O^-) < III (CH_3S^-) < I (Cl^-) < IV (I^-) $$
Iodide ion is the best leaving group and methoxide ion is the worst leaving group.
The negative charge on large iodide atom is stabilized. In methoxide ion, the $$+I$$ (electron releasing) effect of methyl group intensifies the negative charge on $$O$$ atom and destablizes the ion.
A good leaving group should be
(a) electron-withdrawing to polarize the carbon
(b) stable once it has left (not a strong base)
(c) polarisable- to maintain partial bonding with the carbon in the transition state (both $$S_N1$$ and $$S_N2$$).
This bonding helps to stabilize the transition state and reduces the activation energy.
Question 9
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Directions For Questions

Comprehension # 2
Nucleophilic substitution reactions generally expressed as
$$Nu^-+R\,-\,L\longrightarrow R-\,Nu+L^-$$
Where $$Nu^-\longrightarrow Nucleophile\:;\:R-L\longrightarrow substrate\:;\:L\longrightarrow leaving\:group$$
The best leaving groups are those that become the most stable ions after they depart. Since most leaving groups leave as a negative ion, the best leaving groups are those ions that stabilize a negative charge most effectively. A good leaving group should be
(a) electron-withdrawing to polarize the carbon
(b) stable once it has left (not a strong base)
(c) polarisable- to maitain partial bonding with the carbon in the transition state (both $$S_{N}1$$ and $$S_{N}2$$). This bonding helps to stabilise the transition state and reduces the activation energy.

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Among the following which is feasible?

A
$$\;X^-+CH_3-CH_2-H\longrightarrow CH_3-CH_2-X+H^-$$

B
$$\;X^-+CH_3-OH\longrightarrow CH_3-X+\bar{O}H$$

C
$$\;{X}\bar+H_3C-\overset { \oplus }{ \underset { \underset { H }{ \mid } }{ O } } H\longrightarrow CH_{ 3 }-X+ H_{ 2 }O$$

D
$$\;X^-+CH_3-CH_3\longrightarrow CH_3-X+\bar{C}H_3$$

Solution

The reaction $$\displaystyle \;{X}\bar+H_3C-\overset { \oplus }{ \underset { \underset { H }{ \mid } }{ O } } H\longrightarrow CH_{ 3 }-X+ H_2O $$ is most feasible as water is the stable, neutral molecule and not a strong base.
Question 10
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Arrange the following in the increasing order of their ability as a leaving group: $$CF_3SO_3^-,\,CH_3SO_3^-\,and\,CH_3COO^-$$

A
$$ CH_3SO_3^- $$ > $$CF_3SO_3^-$$ > $$ CH_3COO^-$$

B
$$CF_3SO_3^-$$ > $$ CH_3SO_3^- $$ > $$ CH_3COO^-$$

C
$$CF_3SO_3^-$$ > $$ CH_3COO^-$$ > $$ CH_3SO_3^- $$

D
$$CF_3SO_3^-$$ < $$ CH_3SO_3^- $$ < $$ CH_3COO^-$$

Solution

The increasing order of their ability as a leaving group is $$CF_3SO_3^-$$ > $$ CH_3SO_3^- $$ > $$ CH_3COO^-$$.
The electronegative $$F$$ atoms show $$-I$$ effect and remove the electron density and stabilise the negative charge. Hence, the leaving group ability of $$CF_3SO_3^-$$ is maximum. Acetate ion has the least ability as a leaving group.