Hydrocarbons Test

Result

Hydrocarbons Test - 39

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Weekly Quiz Competition

Question 1
1 / -0

Which of the following used for the conversion of $$2- hexyne$$ into $$2- hexene$$?

A
$$H_{2}/Pd/BaSO_{4}$$

B
$$H_{2}, PtO_{2}$$

C
$$NaBH_{4}$$

D
$$Li-NH_{3}/C_{2}H_{5}OH$$

Solution

The parial reduction of alkynes by active metal in liquid ammonia takes place through trans vinylic anion which ultimately produces trans alkene. 2-hexyne gives trans-2-hexene on treatment $$Li-NH_3/C_2H_5OH$$
Question 2
1 / -0

Which of the following is not the product of hydration of alkene?

A

B

C

D

Solution
Question 3
1 / -0

Which of the following is aromatic?

A
i

B
ii

C
iii

D
iv

Solution
Question 4
1 / -0

A gas decoloursises alkaline $$KMnO_{4}$$ solution but does not give precipitate with silver nitrate. It is:

A
$$CH_{4}$$

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

C
$$C_{2}H_{4}$$

D
$$C_{2}H_{2}$$

Solution

Baeyer's reagent is alkaline $$KMnO_4$$, it reacts with unsaturated hydrocarbon (alkene, alkyne, etc.) and decolourises its pink colour. And silver nitrate reacts with acidic hydrogen of alkyne and makes white precipitate. Among the given options, methane and ethane are alkanes, so they don't react with either solution.
Ethylene is unsaturated but doesn't have acidic hydrogen, so it react with Baeyer's reagent but does not react with silver nitrate. Alkyne reacts with both reagents.
Question 5
1 / -0

Which of the following will produce 4- methyl - 1- pentyne ?

A
ethyne + 1) $$NaNH_{2}$$ 2) 1- iodobutane

B
propyne +1) $$NaNH_{2}$$ 2) 1- iodpropane

C
ethyne +1) $$NaNH_{2}$$ 2) 1- iodo -2 methylprpane

D
propyne +1) $$NaNH_{2}$$ 2) -iodoprepane

Solution
Question 6
1 / -0

Reaction between $${ CH }_{ 2 }=C=O$$ and $${ C }_{ 2 }{ H }_{ 5 }OH$$ forms

A
Methyl acctate

B
Methyl formate

C
Ethyl acctate

D
Acctic acid
Question 7
1 / -0

$$Ph - CH_{2} - CH = CH - CH_{3} \underset {(ii) Alc.KOH \ iii)NaNH_2}{\xrightarrow {(i) Br_{2}}}$$.

A
$$Ph - CH = CH - CH = CH_{2}$$

B
$$Ph - CH_{2} - \underset {\underset {OH}{|}}{C}H - \underset {\underset {OH}{|}}{C}H - CH_{3}$$

C
$$Ph - CH_{2} - C \equiv C - CH_{3}$$

D
$$Ph - C \equiv C - CH_{2} - CH_{3}$$

Solution

$$Ph-C{H}_{2}-CH=CH-C{H}_{3} \xrightarrow{{Br}_{2}} Ph-C{H}_{2}-\overset{\underset{|}{Br}}{CH}-\overset{\underset{|}{Br}}{CH}-C{H}_{3} \xrightarrow[-2 \; HBr]{Alc. \; KOH, NaNH_2} Ph-C{H}_{2}-C\equiv C-C{H}_{3}$$

Option C is correct.
Question 8
1 / -0

$$CH_{3} - CH = CH - CH = N - CH_{3} \xrightarrow {LiAlH_{4}}$$ What is final product?

A
$$CH_{3} - CH_{2} - CH_{2} - CH_{2} - NH - CH_{3}$$

B
$$CH_{3} - CH = CH - CH_{2} - NH - CH_{3}$$

C
$$CH_{3} - CH_{2} - CH_{2} - CH - N - CH_{3}$$

D
$$CH_{3} - CH = CH - CH_{2} - OH$$

Solution
Question 9
1 / -0

Write the product of the given reaction:

A

B

C

D

Solution

In the given reaction, addition of $$DCl$$ and $$DI$$ takes place acorss double bond according to Markownikoff's rule.
Question 10
1 / -0

IUPAC name of given compound is:

A
3, 4 - Dimethyl - 4 - propylhept - 1- en - 6 - yne

B
3, 5 - Dimethyl - 4 - propylhept - 1- en - 6 - yne

C
3, 4 - Dimethyl - 4 - propylhept - 6- en - 1 - yne

D
4 - propyl - 3, 5 - dimethylhept - 6- en - 1 - yne

Solution

In the case of polyfunctional compounds, one of the functional groups is chosen as the principal functional group and the compound is then named on that basis. The remaining functional groups, which are subordinate functional groups, are named as substituents using the appropriate prefixes

Alkene chosen as the principal functional group here.

Hence the correct IUPAC name is:

3, 5 - Dimethyl - 4 - propylhept - 1- en - 6 - yne