Haloalkanes and Haloarenes Test

Result

Haloalkanes and Haloarenes Test - 61

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

Directions For Questions

$$S_N$$ reaction is given by these compounds, which have a nucleophilic group and a good leaving EWG. It should be stable after leaving with bonding pair of $$\overline { e } $$'s and it should have high polarisability.
Nucleophilic aliphatic substitution reaction is mainly of two types $${S_N}{1}$$ and $${S_N}{2}.{S_N}{1}$$ mechanism is a two-step process. Reaction velocity of $${S_N}{1}$$ depends only on the concentration of the substrate. It proceeds via the formation of carbocation, optically active substrate gives $$\oplus $$ and $$\ominus $$ forms of the product.
In most of the cases, the product usually consists of $$5-20$$% inverted and $$(95-80)$$% racemised species. The more stable is the carbocation, th egreater is the proportion of racemisation. In solvolysis reaction, the more nucleophilic is the solvent, the greater is the proportion of inversion.

...view full instructions

Which of the following gives $${S_N}{1}$$ reaction?

A

B

C
$$MeBr$$

D
All

Solution

It is $${3}^{o}$$ $$RX$$ and forms stable $$Ph{Me}_{2}{ C }^{ \oplus }$$ ion due to resonance and $$+I$$ effect of methyl groups.
$$2^o$$ and $$ 1^o$$ halide follows $$S_N2$$ mechanism.
Question 2
1 / -0

Directions For Questions

$$S_N$$ reaction is given by these compounds, which have a nucleophilic group and a good leaving EWG. It should be stable after leaving with bonding pair of $$\overline { e } $$'s and it should have high polarisability.
Nucleophilic aliphatic substitution reaction is mainly of two types $${S_N}{1}$$ and $${S_N}{2}.{S_N}{1}$$ mechanism is a two-step process. Reaction velocity of $${S_N}{1}$$ depends only on the concentration of the substrate. It proceeds via the formation of carbocation, optically active substrate gives $$\oplus $$ and $$\ominus $$ forms of the product.
In most of the cases, the product usually consists of $$5-20$$% inverted and $$(95-80)$$% racemised species. The more stable is the carbocation, th egreater is the proportion of racemisation. In solvolysis reaction, the more nucleophilic is the solvent, the greater is the proportion of inversion.

...view full instructions

For the reaction given, which substrate will give maximum racemisation?

A

B

C

D

Solution

$${Ph}_{3}{ C }^{ \oplus }$$ with EDG is more stable and favours $${SN}^{1}$$ reaction and gives maximum racemisation.
Question 3
1 / -0

Explain the relative rate of $$E2$$ reaction of the given compounds.

A
$$ III < I < II $$

B
$$ I < II < III $$

C
$$ III < II < I $$

D
None of these

Solution

The relative rate of $$E_2$$ reaction of the given compounds is $$III<II<I$$.

In the elimination reaction, a molecule of $$HCl$$ is eliminated. Elimination will be fastest when $$H$$ atom and $$Cl$$ atom are trans to each other. In compound $$I$$, both $$H$$ atoms are trans to $$Cl$$. Hence, the rate of elimination for compound $$I$$ is maximum.

In compound $$III$$, both $$H$$ atoms are cis to $$Cl$$. Hence, the rate of elimination for compound $$III$$ is minimum.

In compound $$II$$, only one $$H$$ atom is trans to $$Cl$$. Hence, the rate of elimination for compound $$II$$ is intermediate.

Hence option C is correct
Question 4
1 / -0

Directions For Questions

$$S_N$$ reaction is given by these compounds, which have a nucleophilic group and a good leaving EWG. It should be stable after leaving with bonding pair of $$\overline { e } $$'s and it should have high polarisability.
Nucleophilic aliphatic substitution reaction is mainly of two types $${S_N}{1}$$ and $${S_N}{2}.{S_N}{1}$$ mechanism is a two-step process. Reaction velocity of $${S_N}{1}$$ depends only on the concentration of the substrate. It proceeds via the formation of carbocation, optically active substrate gives $$\oplus $$ and $$\ominus $$ forms of the product.
In most of the cases, the product usually consists of $$5-20$$% inverted and $$(95-80)$$% racemised species. The more stable is the carbocation, th egreater is the proportion of racemisation. In solvolysis reaction, the more nucleophilic is the solvent, the greater is the proportion of inversion.

...view full instructions

Fugacity power of which group will be maximum?

A
$${ CH }_{ 3 }-O-\overset { O\\ \parallel }{ \underset { \parallel \\ O }{ S } } -{ C }_{ 2 }{ H }_{ 5 }$$

B
$${ CH }_{ 3 }-O-\overset { O\\ \parallel }{ \underset { \parallel \\ O }{ S } } -{ CH }_{ 3 }$$

C

D
$${ CH }_{ 3 }-O-\overset { O\\ \parallel }{ C } -{ CH }_{ 3 }$$

Solution

Stronger the acid or weaker the base, better is the leaving group. In (c), due to $$-I$$ effect of three $$F$$-atoms, it makes stronger acid.
Therefore, the order of leaving group:$$(c)>(b)>(a)>(d)$$
Question 5
1 / -0

Directions For Questions

$$S_N$$ reaction is given by these compounds, which have a nucleophilic group and a good leaving EWG. It should be stable after leaving with bonding pair of $$\overline { e } $$'s and it should have high polarisability.
Nucleophilic aliphatic substitution reaction is mainly of two types $${S_N}{1}$$ and $${S_N}{2}.{S_N}{1}$$ mechanism is a two-step process. Reaction velocity of $${S_N}{1}$$ depends only on the concentration of the substrate. It proceeds via the formation of carbocation, optically active substrate gives $$\oplus $$ and $$\ominus $$ forms of the product.
In most of the cases, the product usually consists of $$5-20$$% inverted and $$(95-80)$$% racemised species. The more stable is the carbocation, th egreater is the proportion of racemisation. In solvolysis reaction, the more nucleophilic is the solvent, the greater is the proportion of inversion.

...view full instructions

Which of the following will give $$S_N$$ reaction?

A
$$R-Br$$

B
$$R-{N}_{3}$$

C
$$R-\overset { \oplus }{ { OH }_{ 2 } } $$

D
All

Solution

As all of these compounds contains one stable nucleophile so they all will give $$S_N$$ reaction.
Question 6
1 / -0

Compound $$(A)$$ on reaction with $${NH}_{2}{NH}_{2}+OH$$ gives:

A

B

C

D

Solution
Question 7
1 / -0

Directions For Questions

$$S_N$$ reaction is given by these compounds, which have a nucleophilic group and a good leaving EWG. It should be stable after leaving with bonding pair of $$\overline { e } $$'s and it should have high polarisability.
Nucleophilic aliphatic substitution reaction is mainly of two types $${S_N}{1}$$ and $${S_N}{2}.{S_N}{1}$$ mechanism is a two-step process. Reaction velocity of $${S_N}{1}$$ depends only on the concentration of the substrate. It proceeds via the formation of carbocation, optically active substrate gives $$\oplus $$ and $$\ominus $$ forms of the product.
In most of the cases, the product usually consists of $$5-20$$% inverted and $$(95-80)$$% racemised species. The more stable is the carbocation, th egreater is the proportion of racemisation. In solvolysis reaction, the more nucleophilic is the solvent, the greater is the proportion of inversion.

...view full instructions

Which of the following give $${S_N}{1}$$ reaction?
Select the correct answer.

A
$$(I),(II)$$ and $$(III)$$

B
$$(I)$$ and $$(II)$$

C
$$(II),(III)$$ and $$(IV)$$

D
$$(I),(III)$$ and $$(IV)$$

Solution

Stability of carbocation:Benzyl $${ C }^{ \oplus }>$$ Allyl $${ C }^{ \oplus }> {3}^{o}{ C }^{ \oplus }(I)>(II)$$
$$(IV)$$ gives $${Me}_{3}-\overset { \oplus }{ { CH }_{ 2 } } $$(neopentyl $${ C }^{ \oplus }$$) (sterically hendered). It undergoes neither $${SN}^{1}$$ nor $${SN}^{2}$$.
Question 8
1 / -0

$$\beta$$-Elimination or anti-elimination reaction is carried out with base $$({ B }^{ \overset { .. }{ \ominus } })$$ as shown.
The following bases are used.
$$(I) \overset { \ominus \quad }{ OH } $$ $$(II) R{ O }^{ \ominus }$$ $$(III) RCO{ O }^{ \ominus }$$ $$(IV) \overset { \ominus }{ C } N$$ $$(V) N{ O }_{ 3 }^{ \ominus }$$
The decreasing order of reactivity for the above elimination is :

A
$$(II)>(I)>(IV)>(III)>(V)$$

B
$$(V)>(III)>(IV)>(I)>(II)$$

C
$$(II)>(I)>(III)>(IV)>(V)$$

D
$$(I)>(II)>(III)>(IV)>(V)$$

Solution

The reagent should be a strong bronsted base.
Acidic order: $$H{NO}_{3}> RCOOH> HCN> {H}_{2}O> ROH$$
Basic order : $${NO}_{3}^{\ominus }< RCO{O}^{\ominus }< {CN}^{\ominus }< \overset { \ominus }{ OH } < {RO}^ { \ominus }$$
Decreasing order of basicities and hence $$\beta$$-elimination is : $$(II)> (I)> (IV)> (V)$$.
Question 9
1 / -0

Consider the given reactions:
Which statement(s) is/are wrong?

A
The product by path $$(I)$$ is $$Me-CH={CH}_{2}(I)$$

B
The product by path $$(II)$$ is $$Me-CH(OEt)Me(II)$$

C
The products are mixture of $$(I)$$ and $$(II)$$ by both paths

D
Path $$(I)$$ proceeds via E2 mechanism, while path $$II$$ proceeds via $${S_N}{1}$$ mechanism

Solution

Statement (C) is wrong
i. In path $$(I)$$, $$Et{ O }^{ \ominus }$$ is a strong base and with $${2}^{o}$$ $$RX$$ groups. The $$E_2$$ product predominates over the $${S_N}{2}$$ product to give $$(Me-CH={CH}_{2})$$
ii. In path $$(II)$$, $$EtOH$$ is a weak base, but a better nucleophile. so $${S_N}{1}$$ reaction is favoured to give $$MeCH(OEt)Me$$
Question 10
1 / -0

Directions For Questions

Isopropyl bromide was treated separately with sodium t-butoxide and sodium ethoxide under two different conditions.
Reaction I:
Treatment of isopropyl bromide with $$({Me}_{3}CONa) $$ at $${40}^{o}C$$ gave almost exlusively compound $$(A) ({C}_{3}{H}_{6})$$
Reaction II
Treatment of $$(i-PrBr)$$ with $$NaO{C}_{2}{H}_{5}$$ at $${30}^{o}C$$ yielded compound $$(A)({C}_{3}{H}_{6})$$ along with a small amount of an ether $$(B)({C}_{5}{H}_{12}O)$$.

Compound $$(A)$$ was readily oxidised by a neutral solution of cold dil. $$KMn{O}_{4}$$ to give a brown precipitate.

...view full instructions

Which of the following represents the intermediate T.S for the formation of compound $$(B)$$?

A

B

C

D

Solution

Hence the answer is (c)