Haloalkanes and Haloarenes Test

Result

Haloalkanes and Haloarenes Test - 60

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Weekly Quiz Competition

Question 1
1 / -0

The compound $$A$$ is :

A

B

C

D
none of the above

Solution

$$E_2$$ reaction is stereospecific reaction and involves trans -elimination. Thus, the product obtained is cis-1,2-diphenyl-1-propene. Thus, the option A is the correct answer.
Question 2
1 / -0

The product of the reaction $$P$$ is:

A

B

C

D
None of the above

Solution

Here, due to possibility of a cyclic stable intermediate, $$KOH$$ reacts with reactant and $$\alpha - C$$ w.r.t. to $$O$$, forms a five-membered ring product.
Question 3
1 / -0

In the given reaction, the products are:

A

B

C

D

Solution

Due to the presence of a strong base, it will prefer elimination reaction and alkene will be formed as major product and SN2 product will be formed as a secondary product.

Hence, the correct option is $$B$$
Question 4
1 / -0

In the following pair of compounds, which of the following relation is correct for nucleophilicity in a polar-protic solvent?

A
$${ CH }_{ 3 }S{ CH }_{ 3 } >{ CH }_{ 3 }O{ CH }_{ 3 }$$

B
$${ CH }_{ 3 }S{ CH }_{ 3 } <{ CH }_{ 3 }O{ CH }_{ 3 }$$

C
$${ CH }_{ 3 }S{ CH }_{ 3 } ={ CH }_{ 3 }O{ CH }_{ 3 }$$

D
None of these

Solution

In a polar-protic solvent, the order of nucleophilicity is $$\displaystyle { CH }_{ 3 }S{ CH }_{ 3 } >{ CH }_{ 3 }O{ CH }_{ 3 } $$
as $$O$$ is more electronegative than $$S$$. Hence, the lone pair of electrons is more strongly held by $$O$$ than by $$S$$.
Hence, it cannot be easily donated by $$O$$.
Hence, dimethyl ether is less nucleophilic than dimethyl sulphide.
Question 5
1 / -0

$$CH_3-\!\!\overset{\,\,CH_3}{\overset{|}{\underset{\,\,\,CH_3}{\underset{|}{C}}}\!\!\!-}\,CH=CH_2\xrightarrow{dil.H_2SO_4}\,\,\xrightarrow[\Delta]{conc.H_3PO_4}\,\,\xrightarrow[(ii)\, Zn / H_2O]{(i)\, O_3}$$
Final product is / are:

A

$$\underset{CH_3}{\underset{|}{C}=}\,O$$

B
$$HCHO + CH_3-\overset{O}{\overset{||}{C}}-\underset{CH_3}{\underset{|}{C}H}-CH_3$$

C
$$CH_3CHO +$$ $$CH_3-\overset{O}{\overset{||}{C}}-CH_3$$

D
$$CH_{3}CHO+HCHO$$

Solution

$$CH_3-\!\!\overset{\,\,CH_3}{\overset{|}{\underset{\,\,\,CH_3}{\underset{|}{C}}}\!\!\!-}\,CH=CH_2\xrightarrow{dil.H_2SO_4}CH_3-\!\!\overset{\,\,CH_3}{\overset{|}{\underset{\,\,\,CH_3}{\underset{|}{C}}}\!\!\!-}\,\underset{OH}{\underset{|}{C}H}-CH_3 \xrightarrow[\Delta]{conc.H_3PO_4} \,\,CH_3-\underset{CH_3}{\underset{|}{C}=}\,\underset{CH_3}{\underset{|}{C}-}\,CH_3$$

$$\xrightarrow[(ii)\, Zn / H_2O]{(i)\, O_3}CH_3-\underset{CH_3}{\underset{|}{C}=}\,O$$

Here first $$2^0$$ alcohol forms as a product which forms symmetrical alkene as secondary product.
Then this alkene in presence of ozone forms ketone as a product.
Question 6
1 / -0

Suppose the double bonded methylene group of $$1$$-bromo-$$2$$-propene is labelled with $$_{ }^{ 13 }{ C },{ H }_{ 2 }{ }^{ 13 }{ C }=CH-{ CH }_{ 2 }Br$$. What product will be formed by methanolysis of this substrate?

A
$${ H }_{ 2 }{ }^{ 13 }{ C }=CH-{ CH }_{ 2 }-O{ CH }_{ 3 }$$ and $${ H }_{ 2 }C=CH-_{ }^{ 13 }{ { CH } }-O{ CH }_{ 3 }$$

B
$${ H }_{ 2 }C=_{ }^{ 13 }{ { CH } }-{ CH }_{ 2 }-O{ CH }_{ 3 }$$

C
$${ H }_{ 2 }{ }^{ 13 }{ { C } }=CH-{ CH }_{ 2 }-O{ CH }_{ 3 }$$

D
$${ H }_{ 2 }C=CH-_{ }^{ 13 }{ { CH }_{ 2 } }-O{ CH }_{ 3 }$$

Solution

Two products are obtained. In the first product, the bromine atom is replaced by the methoxy group. In the second product, the Br atom is replaced by the methoxy group and the double bond is also rearranged.
$$^{13}CH_2=CHCH_2Br \xrightarrow [-HBr] {CH_2OH} ^{13}CH_2=CHCH_2OCH_3+CH_2=CH^{13}CH_2OCH_3$$
Question 7
1 / -0

Which of the given pair of $${ S }_{ N }1$$ reaction would you expect to proceed faster?

A
$$HBr > NaBr $$

B
$$HBr < NaBr $$

C
$$HBr = NaBr $$

D
None of these

Solution

The reaction with HBr is faster than the reaction with NaBr. HBr protonates the oxygen atom of tert butyl alohol and helps in the formation of tert butyl carbocation (by loss of molecule of water) which is the rate determining step.
Question 8
1 / -0

Which alkene is formed as the major product in the $$E_2$$ reaction of the given compound?

A
$$3$$-methylcyclonex-($$1$$)-ene

B
$$1$$-deutero-$$3$$-methylcyclohex-($$1$$)-ene

C
$$1$$-methyl-$$3$$-deuterocyclohex-(1)ene

D
both A and B

Solution

$$3$$-methylcyclohex-$$1$$-ene is formed as the major product in the $$E_2$$ reaction of the given compound.
Question 9
1 / -0

In an Aprotic solvent, which relative ordering best describes the nucleophilicity of the halide ions?

A
Iodide> Bromide> Chloride>> Flouride

B
Iodide> Bromide> Chloride> Flouride

C
Iodide< Bromide< Chloride< Flouride

D
Iodide< Bromide> Chloride> Flouride

Solution

In an aprotic solvent, the increasing ordering for the nucleophilicity of the halide ions
is iodide< bromide< chloride< flouride .
A polar aprotic solvent does not hydrogen bond to nucleophiles to a significant extent, meaning that the nucleophiles have greater freedom in solution. Under these conditions, nucleophilicity correlates well with basicity and fluoride ion, being the most unstable of the halide ions, reacts fastest with electrophiles.
Question 10
1 / -0

What is the best term to describe the rearrangement that causes the conversion of $$3$$-bromo-$$2,2$$-dimethylbutane to $$2,3$$-dimethyl-$$2$$-butanol?

A
$$1,2$$-methyl shift

B
$$1,3$$- hydride shift

C
$$1,3$$-methyl shift

D
$$1,2$$- hydride shift

Solution

$$1,2$$-methyl shift is the best term to describe the rearrangement that causes the conversion of $$3$$-bromo-$$2,2$$-dimethylbutane to $$2,3$$-dimethyl-$$2$$-butanol.