Organic Chemistry Some Basic Principles and Techniques Test

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Organic Chemistry Some Basic Principles and Techniques Test - 33

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Weekly Quiz Competition

Question 1
1 / -0

Which of the following is a 3-methylbutyl group?

A
$$ CH_3 CH_2 CH_2 CH_2- $$

B
$$ ( CH_3 CH_2 )_2 CH-$$

C
$$ (CH_3)_3 CCH_2- $$

D
$$ (CH_3)_2 CHCH_2CH_2-$$

Solution

When a hydrogen is removed from butane, butyl group is obtained.
Hence, the structure of 3-methylbutyl group is obtained, when one hydrogen atom is removed from 3-methylbutane.

i.e. $$ CH_3 - \underset { \overset { | }{ CH_3 } }{ CH }{CH_2 CH_3} $$ after removing one hydrogen $$ (CH_3)_2 CHCH_2CH_2- $$

3-methylbutane 3-methylbutyl
Question 2
1 / -0

The molecular formulae for phosgene and tear gas are .......... and ......... respectively.

A
$$COCl_2$$ and $$CCl_3 NO_2$$

B
$$SOCl_2$$ and $$CCl_3 NO_2$$

C
$$COCl_2$$ and $$CCI_3NO_2$$

D
$$SOCl_2$$ and $$CCI_3 NO_2$$
Question 3
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Among the xylenes, which is thermodynamically most stable?

A

B

C

D
All are equally stable

Solution

Methyl is an ortho-para directing group.
In ortho and para-xylene, electron density becomes high due to +R effect of methyl group, whereas this is not seen in the case of meta-xylene.
Question 4
1 / -0

Identify the correct statement.

A
Reaction mechanisms are studied using isotopic labelling.

B
Isolation of reactive intermediates is a method to establish reaction mechanism.

C
Both $$A$$ and $$B$$ are correct.

D
Neither $$A$$ nor $$B$$ is correct.

Solution

Both $$A$$ and $$B$$ are correct.
$$A$$) Reaction mechanisms are studied using isotopic labeling. One of the atoms of one reactant/ reagent is replaced with its isotope. Then, the product in which this labelled atom is present is determined. From this information, we can decide which atom of reactant goes to which product.
$$ \displaystyle CH_3-CO-O^*H + CH_3-OH \xrightarrow {H^+}CH_3-COO-CH_3+ H_2O^*$$
In the above reaction, O atom of $$ \displaystyle -OH$$ group of acetic acid is labelled. It is then observed that labelled O atom is found in water. Hence, we can conclude that
$$ \displaystyle -OH$$ group of acetic acid combines with H atom of methanol.
$$B$$) Isolation of reactive intermediates is a method to establish reaction mechanism.
Consider the conversion $$ \displaystyle A \rightarrow D$$ .
It can proceed through following two different mechanisms.
$$ \displaystyle A \rightarrow B \rightarrow D$$.
$$ \displaystyle A \rightarrow C \rightarrow D$$.
Suppose that we are able to isolate intermediate B but we cannot isolate intermediate C. Hence, we can say that the reaction mechanism is $$ \displaystyle A \rightarrow B \rightarrow D$$.
Question 5
1 / -0

$$CH_3-CH_2-\underset{CH_2}{\underset{||}{C}-}COOH$$
The correct IUPAC name of the above compound is:

A
$$2$$-methyl butanoic acid

B
$$2$$-Ethylprop-$$2$$-enoic acid

C
$$2$$-carboxy-$$1$$-butene

D
none of these
Solution
According to IUPAC rules, the priority order is
Functional groups $${-COOH}$$
Double bond at 2nd position
No. of carbons in a chain $${3}$$
So the name of the compound is 2-ethylprop-2-enoic acid.
Question 6
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Which of the following is least stable?

A
$$CH_{3} - \overset {(+)}{CH} - CH_{3}$$

B
$$CH_{3}CH_{2} -\overset{(+)} CH_{2}$$

C
$$CH_{3} - \underset {\underset {CH_{3}}{|}}{\overset {(+)}{C}} - CH_{3}$$

D
$$(CH_{3})_{3}C - \underset {\underset {C_{6}H_{5}}{|}}{\overset {(+)}{C}}-H$$

Solution

Stability of carbocation:

$$CH_3^{\oplus} < \underset {CH_3CH_2^{\oplus}}{1^o carbocation} < \underset {CH_3-\underset {CH_3}{\underset {|}{CH^{\oplus}}}}{2^o carbocation} < \underset {CH_3-\underset {CH_3}{\underset {|}{C}}^{\oplus}-CH}{3^o} carbocation$$

$$CH_3-CH_2-CH_2^{\oplus}$$ is $$1^o$$ carbocation. Hence, it is least stable.
Question 7
1 / -0

The IUPAC name of dimethyl ether is :

A
ethoxymethane

B
methoxymethane

C
methoxyethane

D
ethoxyethane

Solution

Dimethyl ether is $$CH_3OCH_3$$.

It's IUPAC name is methoxymethane.

Hence, option $$B$$ is correct.
Question 8
1 / -0

Which of the following can have vacant $$p-$$orbital?

A
$${ _{ }^{ \bullet }{ CH } }_{ 3 }$$

B
$$\overset { \oplus }{ C } { H }_{ 3 }$$

C
$$\upharpoonleft \downharpoonright { CH }_{ 2 }$$

D
Both (b) and (c)

Solution

$$CH^{-}_{3}\rightarrow $$ $$1s^{2}2s^{2} 2p^{4}\rightarrow$$vacant p- orbital not found.
$$CH^{+}_{3}\rightarrow $$ $$1s^{2}2s^{2} 2p^{2}\rightarrow$$vacant p- orbital found.
$$CH_{2}\rightarrow $$ $$1s^{2}2s^{2} 2p^{2}\rightarrow$$vacant p- orbital found.
Question 9
1 / -0

The correct IUPAC name of the compound is:

A
$$1,2,3-triaminobutan-1,3-one$$

B
$$2,4-diamino-3-oxobutanamide$$

C
$$1,3-dioxobutane-1,2,4-triamine$$

D
$$1,3,4-triaminobutane-2,4-one$$

Solution

IUPAC name is $$2,4-diamino$$ $$3-oxobutanamide$$
Question 10
1 / -0

IUPAC name of above compound is:

A
5-ethenyleyclopenta-1,3-diene

B
3-ethenylcyclopenta-1,4-diene

C
1-ethenylcyclopenta-2,4-diene

D
2-ethenylcyclopenta-1,3-diene

Solution

The IUPAC has suggested few rules to be followed during the naming of a compound.
The naming should be in alphabetical order. From $$C_1$$, ethyne group is hanging and the rest is a ring of $$5-C$$.
So, the name goes as 1-ethenylcyclopenta-2,4-diene.