Hydrocarbons Test

Result

Hydrocarbons Test - 61

Score
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Weekly Quiz Competition

Question 1
1 / -0

(x) may be

A

B

C

D

Solution
Question 2
1 / -0

The end product Y of the reaction is:
$$C_2H_5MgBr $$ + $$ S $$ $$ \underrightarrow { \Delta } $$ $$ X $$ $$\underrightarrow { H_3O^+ } $$ $$Y$$

A
$$C_2H_6$$

B
$$C_2H_5SMgBr$$

C
$$C_2H_5SH$$

D
$$C_2H_5OH$$

Solution
Question 3
1 / -0

A hydrocarbon contains $$7.70\%$$ hydrogen $$0.56\ L$$ of the hydrogen weight $$1.95\ g$$ STP. Then the hydrocarbon is an

A
Alkane

B
Alkene

C
Alkyne

D
Arene

Solution
Question 4
1 / -0

Alkanes can be prepared from Grignard reagents by reacting with

A
Alcohols

B
Primary amines

C
Alkynes

D
All of them

Solution

Grignard reagents react rapidly with acidic hydrogen atoms in molecules such as alcohols and water to produce alkanes.

A) Grignard reagents react rapidly with acidic hydrogen atoms in molecules such as alcohols and water. When a Grignard reagent reacts with water, a proton replaces the halogen, and the product is an alkane. The Grignard reagent therefore provides a pathway for converting a haloalkane to an alkane in two steps.

B) Though we typically want them to do nucleophilic addition to something, these powerful bases tend to go for protons -- a faster reaction -- if there is just about anything that might make the proton vulnerable. Attaching the proton to an electronegative atom like nitrogen is enough, so primary and even secondary amines will react like that with Grignard reagents. We just deprotonate the amine, the hydrocarbon part of the Grignard reagent ends up an an alkane .

C) We recall that alkynes are more acidic than alkanes. Therefore, the acid–base reaction of an alkyne with a readily available Grignard reagent gives a Grignard reagent of the alkyne. This alkynide ion of the Grignard reagent reacts with carbonyl compounds.

Hence , option D is correct.
Question 5
1 / -0

Identify the compound which decolourise the brown color of $$Br_{2}$$ dissolved in $$CCI_{4}$$ and gives only two products.

A

B

C

D
Question 6
1 / -0

A
$$2$$

B
$$3$$

C
$$4$$

D
$$5$$

Question 7

1 / -0

The CORRECT match is?

List-$$1$$ Conversion	List-$$2$$ Process involved
A) Benzene $$\rightarrow$$ Cyclohexane	$$1)$$ Controlled hydrogenation
B) Ethylene $$\rightarrow$$ Formaldehyde	$$2)$$ Hydrogenation
C) Acetylene $$\rightarrow$$ Ethylene	$$3)$$ Ozonolysis
D) Benzene $$\rightarrow$$ Toluene	$$4)$$ Alkylation

A-$$3$$, B-$$4$$, C-$$1$$, D-$$2$$

A-$$1$$, B-$$2$$, C-$$3$$, D-$$4$$

A-$$2$$, B-$$3$$, C-$$1$$, D-$$4$$

A-$$4$$, B-$$3$$, C-$$2$$, D-$$1$$

Question 8
1 / -0

Identify the compound which decolorize the brown colour of $$Br_{2}$$ dissolved in $$CCl_{4}$$ and gives only two products?

A

B

C

D
Question 9
1 / -0

Correct statement is:

A
Product is meso in nature

B
Product mixture is optically inactive in nature

C
Product is 2,2-dibromo butane

D
Theoritically 4 stereo-isomers are possible for product.

Solution
Question 10
1 / -0

$$RCH=CH_{2}$$ can be obtained by:

A
$$R - \overset{O}{\overset{||}{C}} - H$$ and $$(C_{6}H_{5})_{3}P=CH_{2}$$

B
By heating $$RCH_{2}CH_{2}CH_{2}OCOCH_{3}$$

C
By heating $$RCH_2 CH_2 \underset{O}{\underset{\downarrow}{N}} (CH_3)_2$$

D
All of these

Solution