Organic Chemistry Some Basic Principles and Techniques Test

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Organic Chemistry Some Basic Principles and Techniques Test - 41

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Weekly Quiz Competition

Question 1
1 / -0

IUPAC name of the bond line structure is:

A
trimethyl propane

B
pent-1-yne

C
pent-2-ene

D
1, 3-pentadiene

Solution

According to IUPAC convention, numbering should be done such that minimum number should be given on groups or parent chain.
Name is: $$Pentyne$$ or $$Pent-1-yne$$ or $$1-pentyne$$
Question 2
1 / -0

The IUPAC name of the compound is:

A
1-hydroxy-4-methylpentan-3-one

B
2-methyl-5-hydroxypentan-3-one

C
4-methyl-3-oopentan-1-ol

D
hexan-1-ol-3-one

Solution

According to IUPAC convention, numbering is done in such a way that heto group and $$-OH$$ group get less numbering with highest priority .
So, the name is: $$1-hydroxy-4-methylpentan-3-one$$
Question 3
1 / -0

Nitrene is a/an:

A
nucleophile

B
electrophile

C
charged species

D
free atom

Solution

Nucleophiles are the electron rich species.
Nitrene is a nucleophile due to the lone pair electrons of Nitrogen.
Question 4
1 / -0

Which of the following would be expected to be most highly ionised in water?

A
$${\text{C}}{{\text{H}}_{\text{3}}}{\text{C}}{{\text{H}}_{\text{2}}}{\text{C}}{{\text{H}}_{\text{2}}}{\text{COOH}}$$

B
$${\text{C}}{{\text{H}}_{\text{3}}}{\text{CHClC}}{{\text{H}}_{\text{2}}}{\text{COOH}}$$

C
$${\text{C}}{{\text{H}}_{\text{3}}}{\text{C}}{{\text{H}}_2}{\text{CC}}{{\text{l}}_2}{\text{COOH}}$$

D
$${\text{C}}{{\text{H}}_{\text{3}}}{\text{C}}{{\text{H}}_2}{\text{CHClCOOH}}$$

Solution

Fast ionization is shown by strong acids. Chlorine atom has high $$-I$$ effect so when attached to nearly carbon, it increases acidity thus is more acidic.
Therefore option C is correct .
Question 5
1 / -0

Which of the following reacts with an aqueous solution of $$\left[ {{\text{Ag}}\left( {{\text{NH}}_{\text{3}} } \right)_{\text{2}} } \right]^ + {\text{OH}}^{\text{ - }} $$?

A
$$

CH_3 - C \equiv C - CH_3

$$

B
$$

CH_3 - CH = CH_2

$$

C
$$

CH_3 - C \equiv CH

$$

D
None of these

Solution

Only terminal alkyne reacts with tollen's reagent since these are basic reagent require more acidic proton$$(sp-H)$$.
Question 6
1 / -0

Give the major product of the following reaction.

A

B

C

D

Solution

Here, migration of methylide group favors stable carbocation.
Question 7
1 / -0

Which of the reactions shown given will be faster?

A
Reaction $$1$$

B
Reaction $$2$$

C
Both have equal rates of reaction

D
Cannot be predicted

Solution

Since given reaction is $$S_{N2}$$ and we know that $$S_{N2}$$ favoured in presence of polar aprotic solvent like crown-ethers, DMSO, DMF, etc not $$H_2O$$ which is polar protic solvent.
In reaction 2 polar solvent (crown ether) is used.
Thus reaction 2 is faster than reaction 1.
Question 8
1 / -0

The product will be:

A
Cis - 2 - butene

B
Trans - 2 - butene

C
2 - butyne

D
Buta - 1 , 3 - diene

Solution

Since elimination is trans-in nature so meso- compound will give generally trans- alkene as major.
Question 9
1 / -0

$$CH_3-\overset{CH_3}{\overset{|}{\underset{H}{\underset{|}{C}}}}-\overset{CH_3}{\overset{|}{\underset{CH_3}{\underset{|}{C}}}}-\underset{Br}{\underset{|}{CH}}-CH_3\xrightarrow{CH,OH 30^0}C$$
What will be the major product of the following reactions?

A
$$CH_3-\overset{\,\,\,\,\,CH_3}{\overset{|}{\underset{\,\,\,\,\,\,\,OCH_3}{\underset{|}{C}}}}-\overset{\,\,\,\,\,CH_3}{\overset{|}{\underset{H}{\underset{|}{C}}}}-\overset{CH_3}{\overset{|}{\underset{H}{\underset{|}{C}}}}-CH_3$$

B
$$CH_3-\overset{\,\,\,\,\,CH_3}{\overset{|}{C}}=\overset{\,\,\,\,\,CH_3}{\overset{|}{C}}-\overset{\,\,\,\,\,CH_3}{\overset{|}{CH}}-CH_3$$

C
$$CH_3-\overset{\,\,\,\,\,CH_3}{\overset{|}{\underset{H}{\underset{|}{C}}}}-\underset{\,\,\,\,\,\,\,\,OCH_3}{\underset{|}{\overset{\,\,\,\,\,\,\,CH_3}{\overset{|}{C}}}}-\underset{H}{\underset{|}{\overset{\,\,\,\,\,\,CH_3}{\overset{|}{C}}}}-CH_3$$

D
$$CH_3-\overset{\,\,\,\,\,CH_3}{\overset{|}{\underset{H}{\underset{|}{C}}}}-\underset{\,\,\,\,\,\,\,\,CH_3}{\underset{|}{\overset{\,\,\,\,\,\,\,CH_3}{\overset{|}{C}}}}-CH=CH_2$$

Solution
Question 10
1 / -0

Which of the following is a step in the mechanism of the reaction shown?
$$CH_ 2=CH-CH_ 3+HBr \xrightarrow{ Peroxoide} \underset{Br}{\underset{|}{C}}H_2 - CH_2 - CH_3$$

A

B

C

D

Solution

Mechanism:-
(1) Chain initiation
$$R-O-O-R \longrightarrow R\overset {\cdot}O+\overset {\cdot}OR$$
$$R-\overset {\cdot}O+H-Br \longrightarrow R-OH+\overset {\cdot}Br$$
(2) Chain propagation
A secondary radical is formed as it is more stable as shown below.
$$CH_3-CH=CH_2+\overset {\cdot}{Br}\longrightarrow CH_3-\overset {\cdot}CH-CH_2Br$$
$$CH_3-\overset {\cdot}CH-CH_2Br+HBr \longrightarrow CH_3CH_2-CH_2Br+\overset {\cdot}Br$$
(3) Termination
$$2$$ free radicals hit each other to form a molecule.