Organic Chemistry Some Basic Principles and Techniques Test

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Organic Chemistry Some Basic Principles and Techniques Test - 47

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Weekly Quiz Competition

Question 1
1 / -0

The incorrect IUPAC name is:

A
$$CH_{3}CH_{2}CONH_{2} \longrightarrow$$ Propanamide

B
$$CH_{3}CH_{2}COOCH_{3} \longrightarrow$$ Methyl Propanoate

C
$$CH_3-\overset{CH_3}{\overset{|}{C}H} - CH = CH - CH_3 \longrightarrow$$ 2-Methyl pent-3-ene

D
$$CH_3CH_2-O - \underset{CH_2-CH_3}{\underset{|}{C}H - C}H_3 \longrightarrow$$ 2-Ethoxy butane

Solution

According to rule, double bound is given priority so labelling is done from the right. The IUPAC name is $$4-Methyl pent-2-ene$$.

$$\underset{5}{C}H_3-\overset{CH_3}{\overset{|}{\underset{4}{C}} -} \underset{3}{C}H = \underset{2}{C}H - \overset{1}{C}H_3$$

Hence, option C is correct.
Question 2
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IUPAC Name of

A
4-Bromo-6-chloro-2-ethyl-1-methylcyclohex-1-ene

B
5-Bromo-1-chloro-3-ethyl-2-methylcyclohex-2-ene

C
5-Bromo-3-chloro-1-ethyl-2-methylcyclohex-1-ene

D
1-Bromo-5-chloro-3-ethyl-4-methylcyclohex-3-ene

Solution
Question 3
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Which of the following carbocation will undergo rearrangement?

A

B

C

D
Question 4
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What is the major elimination product of the given reaction

A

B

C

D
Question 5
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Write the IUPAC names and symbols for the elements having atomic numbers.

A
$$ 102 $$

B
$$ 107 $$

C
$$ 114 $$

D
$$ 119 $$

E
$$ 120 $$
Question 6
1 / -0

Dibromopropane on treatment with X moles of $${ Na }{ NH }_{ 2 }$$ followed by treatment with ethyl bromide gave a pentyne.

The value of X is :

A
One

B
Two

C
Three

D
Four

Solution
Question 7
1 / -0

Which one of the following reagents will not convert ethyl alcohol into ethyl chloride?

A
$$HCl-ZnCl_2$$

B
$$CCl_4$$

C
$$PCl_5$$

D
$$SOCl_2$$

Solution

$$HCl- ZnCl_2$$ will not convert ethyl alcohol into ethyl chloride. As in acidic condition dehydration of alcohol generally takes place.
Question 8
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$$CH_3-\underset{\,\,\,\,CH_3}{\underset{|}{\,\,\,CH}}-CH=CH_2+HBr$$ product which is Prodominate : $$X$$ is

A
$$CH_3\underset{\,\,\,\,CH_3}{\underset{|}{\,\,\,\,CH}}-CH_2CH_2Br$$

B
$$CH_3-\overset{\,\,\,Br}{\overset{|}{\underset{\,\,\,\,CH_3}{\underset{|}{C}}}}-CH_2-CH_3$$

C
$$CH_3-\underset{\,\,\,\,CH_3}{\underset{|}{\,\,\,\,CH}}-\underset{\,\,Br}{\underset{|}{\,\,\,CH}}-CH_3$$

D
$$None\ is\ correct$$

Solution

$$CH_3-\underset{CH_3}{\underset{|}{CH}}-CH=CH_2+HBr\rightarrow CH_3-\underset{CH_3}{\underset{|}{CH}}-C^{\oplus}-CH_3$$
$$\downarrow$$
$$CH_3-\overset{Br}{\overset{|}{\underset{CH_3}{\underset{|}{C}}}}-CH_2-CH_3\leftarrow CH_3-C^{\oplus}-CH_2-CH_3$$
Here, again Carbocation Rearrangement takes place to tertiary carbon.
(Refer to Image)
Question 9
1 / -0

Select the true statement about benzene amongst the following

A
Because of unsaturation benzene easily undergoes addition

B
There are two types of C-C bonds in benzene molecule

C
There is cyclic delocalisation in pi bonds in benzene

D
Mono substitution of benzene gives 3 isomeric products.

Solution

In benzene molecule,all$$ C-C $$ bond length are equal,it does not give addition product but there is a cyclic delocalisation of $$\pi$$ bonds in benzene ring Mono-substitution of benzene gives single isomeric product.
Question 10
1 / -0

Identify the final product in the reaction :

A

B

C

D

Solution

Grignard's addition to aldehyde give alcohol upon reaction with acid, ring expansion takes place followed by carbocation formation.

Hence the correct option is D.