Organic Chemistry Some Basic Principles and Techniques Test

Result

Organic Chemistry Some Basic Principles and Techniques Test - 16

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

In Lassaigne's method organic compound is fused with:

A
sodium metal

B
zinc dust

C
sodium caronate and zinc dust

D
calcium metal

Solution

$$N,S,$$ halogens and phosphorus present in an organic compound are detected by Lassaigne's test, by fusing the compound with sodium $$(N_{a})$$ metal, which converts the elements present in the compound from covalent to ionic form.

$$Na+C+N\overset{\Delta }{\rightarrow}NaCN$$

$$2Na+S\overset{\Delta }{\rightarrow}Na_{2}S$$

$$2Na+X\overset{\Delta }{\rightarrow}NaX\ (X=Cl, \ Br \ or \ I)$$
Question 2
1 / -0

Neo-heptyl alcohol is correctly represented as :

A

B

C

D

Solution

For Neo-heptyl alcohol there should be two methyl branch on the second carbon of the chain along with alcohol group attached to a primary carbon atom. Thus option C is correct.
Question 3
1 / -0

Which of the following is branched chain alkane?

A
n-pentane

B
pyridine

C
iso-butane

D
n-hexane

Solution

Isobutane is branched chain alkane.
n-pentane and n-hexane are straight chain alkanes.
Pyridine is heterocyclic aromatic compound.
Question 4
1 / -0

Common name of $$CH_{3}CH_{2}COOH$$ :

A
Propionic acid

B
Butyric acid

C
Vinegar

D
Methanoic acid

Solution

$$^{3}CH_{3}^{2}CH_{2}^{1}COOH \Rightarrow $$ Propionic acid$$\Rightarrow $$ 3 corbons, so common name is
common name $$\Rightarrow $$ propionic acid, IUAC $$\Rightarrow $$ Propanoic acid
Question 5
1 / -0

Homologue of $$C_{3}H_{4}$$ is:

A
$$C_{4}H_{6}$$

B
$$C_{2}H_{6}$$

C
$$CH_{3} OH$$

D
$$HCHO$$

Solution

A homologous series is a family of hydrocarbons with similar chemical properties that share the same general formula.

The general formula of the compound in the question is $$C_n H_{2n-2}$$ with $$n=3$$.

Only the compound in option $$A$$ satisfies the given general formula with $$n=4$$.

So, homologue of $$C_3H_4$$ is $$C_4H_6$$.
Question 6
1 / -0

$$CH_{2}OHCHOHCH_{2}OH$$ is glycerol.
It doesn't contain ___________ alcoholic group.

A
$$1^{0}$$

B
$$2^{0}$$

C
$$3^{0}$$

D
both $$1^{0}$$ & $$2^{0}$$

Solution

No $$3^{0}$$ -OH group
$$\begin{matrix} 1^{ 0 } & CH_ 2 & - &OH \\ & |& & \\2^0 & CH&-&OH\\&|&&\\3^0&CH_2&-&OH \end{matrix}$$
Question 7
1 / -0

Which of the following compounds has $$-OH$$ as a functional group?

A
Butanone

B
Butanol

C
Butanoic acid

D
Butanal

Solution

In Butanone, the functional group is $$-C=O$$.
In Butanoic acid, the functional group is $$-COOH$$.
In Butanal, the functional group is $$-CHO$$.
In Butanol, the functional group is $$-OH$$.
Question 8
1 / -0

The functional groups present in the following compound is:
$$CH_{3} CH_{2} CH_{2} COOH$$

A
carboxylic acid

B
carbonate

C
alcohol

D
carboxylate

Solution

The functional group present in $$CH_{3}-CH_{2}-CH_{2}-COOH$$ is carboxylic acid, $$-COOH$$.
Question 9
1 / -0

IUPAC name of the compound

$$CH_2=CH - \underset{CH_2CH_3}{\underset{|}{C}H \,\,-}\, \underset{Cl}{\underset{|}{C}} =CH_2$$

A
3-Ethyl-4-chloropenta-1, 4-diene

B
2-Chloro-3-ethylpenta-1, 4-diene

C
4-Chloroethenylpent-1-ene

D
3-Ethenyl-4-chloropent-4-ene

Solution

$$\overset{5}{C}H_2=\overset{4}{C}H - \underset{CH_2CH_3}{\underset{|}{\overset{3}{C}}H \,\,-} \,\, \underset{Cl}{\underset{|}{\overset{2}{C}}} =\overset{1}{C}H_2$$

The correct IUPAC name is 2-Chloro- 3 ethyl-penta 1,4 -diene

Option B is correct.
Question 10
1 / -0

Which of the following does not belong to the same homologous series ?

A
$$CH_{4}$$

B
$$C_{2}H_{6}$$

C
$$C_{3}H_{8}$$

D
$$C_{4}H_{8}$$

Solution

A, B & C are all saturated carbon compounds with the general formula: $$C_{n}H_{2n+2}$$, whereas $$C_{4}H_{8}$$ is an alkene with general formula $$C_{n}H_{2n}$$

Option D is correct.