Haloalkanes and Haloarenes Test

Result

Haloalkanes and Haloarenes Test - 23

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

The major product obtained in the following reaction is :

A
$$C_{6}H_{5}CH= CHC_{6}H_{5}$$

B
$$(+)C_{6}H_{5}CH\left ( O^{t}Bu \right )CH_{2}C_{6}H_{5}$$

C
$$(-)C_{6}H_{5}CH\left ( O^{t}Bu \right )CH_{2}C_{6}H_{5}$$

D
$$(\pm)C_{6}H_{5}CH\left ( O^{t}Bu \right )CH_{2}C_{6}H_{5}$$

Solution

Alkene is the product as bulky hindered base and secondary alkyl bromide favors elimination over substitution.
Hence, the option (A) is the correct answer.
Question 2
1 / -0

For the compounds $$CH_3Cl, CH_3Br, CH_3I$$ and $$CH_3F$$, the correct order of increasing C-halogen bond length is:

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

B
$$CH_3F < CH_3Br < CH_3Cl < CH_3I$$

C
$$CH_3F < CH_3I < CH_3Br < CH_3Cl$$

D
$$CH_3Cl < CH_3Br < CH_3F < CH_3I$$

Solution

$$C$$-halogen bond length totally depends on the electronegativity difference between Carbon and halogen. The more the difference is, the less is the bond length. The electronegativity order of halogens is $$I < Br < Cl < F$$. Hence, the correct order of increasing $$C$$-halogen bond length is $$CH_3F<CH_3Cl<CH_3Br<CH_3I$$.

Hence, the correct option is $$(A)$$
Question 3
1 / -0

The major product obtained in the photo catalysed bromination of $$2$$-methylbutane is:

A
$$1$$-bromo-$$2$$-methylbutane

B
$$1$$-bromo-$$3$$-methylbutane

C
$$2$$-bromo-$$3$$-methylbutane

D
$$2$$-bromo-$$2$$-methylbutane

Solution

The major product obtained in the photo catalysed bromination of 2-methylbutane is 2-bromo-2-methylbutane.

Hence, the correct option is $$\text{D}$$
Question 4
1 / -0

$$3-$$Methyl-pent$$-2-$$ene on reaction with $$HBr$$ in presence of peroxide forms an addition product.

The number of possible stereoisomers for the products:

A
$$Zero$$

B
$$Two$$

C
$$Four$$

D
$$Six$$

Solution

$$3-Methyl-pent-2-ene$$ on reaction with $$HBr$$ in presence of $$peroxide$$ forms an addition product.

A molecule of $$HBr$$ is added to $$C=C$$ double bond. The addition follows anti-Markowikoff's rule. $$Br$$ is added to less substituted C atom.

The number of possible stereoisomers for the products is four. The product has 2 chiral C atoms and is unsymmetrical.

Hence, 4 stereoisomers are possible.
Question 5
1 / -0

Which one of the following reagents is not suitable for the elimination reaction?

A
$$NaI$$

B
$${NaOEt}/{EtOH}$$

C
$${NaOH}/{{H}_{2}O}$$

D
$${NaOH}/{{H}_{2}O-EtOH}$$

Solution

With $$NaI$$ substitution reaction will take place.
Question 6
1 / -0

The major product of the following reaction is :

A

B

C

D

Solution

When an alkyl bromide is treated with $$\ C_2H_5ONa/ C_2H_5-OH$$, it undergoes dehydrobromintion. A molecule of $$\displaystyle HBr$$ is eliminated and $$\displaystyle C=C$$ double bond is formed. In present example, $$ C=C$$ double bond is conjugated with aromatic ring. This leads to stability.
Question 7
1 / -0

The increasing order of the reactivity of the following halides for the $$S_{N}1$$ reaction is :
$$CH_3\underset{Cl}{\underset{|}{C}}HCH_2CH_3$$ $$CH_3CH_2CH_2Cl$$ $$p-H_3CO-C_6H_4-CH_2Cl$$
(I) (II) (III)

A
$$(II)< (I)< (III)$$

B
$$(I)< (III)< (II)$$

C
$$(II)< (III)< (I)$$

D
$$(III)< (II)< (I)$$

Solution

The increasing order of the reactivity of the halides for the SN1 reaction is $$\displaystyle
(II)<(I)<(III)$$.
$$\displaystyle (II)$$ is primary alkyl halide.
$$\displaystyle (I)$$ is secondary alkyl halide.
Primary alkyl halide is less reactive towards SN1 reaction than secondary alkyl halide.
In $$\displaystyle (III)$$, the carbocation generated is stabilised due to resonance. Hence, $$\displaystyle (III)$$ is most reactive towards SN1 reaction.
Question 8
1 / -0

$$1-$$bromo$$-3-$$chlorocyclobutane when treated with two equivalents of $$Na$$, in the presence of ether which of the following will be formed?

A

B

C

D

Solution

$$1-$$bromo$$-3-$$chlorocyclobutane when treated with two equivalents of $$Na$$, in the presence of dry ether(D.E.) will form the product represented by option D. It is an intramolecular Wurtz reaction and involves formation of $$C-C$$ single bond between two $$C$$ atoms that are attached to halogen atoms.
Question 9
1 / -0

Among the following, the reaction that proceeds through an electrophilic substitution is:

A

B

C

D

Solution

Hence option B is correct.
Question 10
1 / -0

For the following reactions:
Which of the following statements is correct?

A
(a) and (b) are elimination reactions and (c) is addition reaction.

B
(a) is elimination, (b) is substitution and (c) is addition reaction.

C
(a) is elimination, (b) and (c) are substitution reactions.

D
(a) is substitution, (b) and (c) are addition reactions.

Solution

(a) is elimination,
When an alkyl bromide is heated with alcoholic $$KOH$$ solution, a molecule of HBr is eliminated to form $$C=C$$ double bond.
(b) is substitution
When secondary alkyl bromide is heated with aqueous $$KOH$$, bromine atom is substituted with $$-OH$$ group.
(c) is addition reaction
A molecule of bromine is added across $$C=C$$ double bond.