Hydrocarbons Test

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Hydrocarbons Test - 30

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Weekly Quiz Competition

Question 1
1 / -0

Among the following free radical bromination reactions, select those in which 2 halide is the major product

A
P, Q, R, S

B
P, R, U

C
P, R, S, T

D
P, Q, R, S, T

Solution

Solution:- (B) P, R, U
Bromine is more selective.
$$Q, S, T$$ form $$3°$$ halide as major and $$P, R, U$$ form $$2°$$ halide as major.
Question 2
1 / -0

What is the product (B) of the following reaction sequence?

A

B

C

D

Solution

In the first reaction the ethyl replaces the bromide ion from the alkane and then that product further reacts with it,which leads to the formation of final product after hydrolysis.
Question 3
1 / -0

For the given Question (1,2,3) consider the following reaction.
How many monohalo derivatives are possible (excluding stereoisomers) ?

A
$$3$$

B
$$4$$

C
$$5$$

D
$$6$$

Solution

Refer to image
There are total 4 products formed.
Question 4
1 / -0

On catalytic reduction $$(H_2/Pd)$$, how many alkenes will give 2-methylbutane?

A
1

B
2

C
3

D
4

Solution

So, 3 alkenes will give 2-methyl butane on reduction with $$H_{2}/Pt$$

The alkenes which will give 2-methylbutane on catalytic reduction are:
1. 3-methylbutene
2. 2-methylbut-2-ene
3. 2-methylbutene
Question 5
1 / -0

Select the chain propagation steps in the free-radical chlorination of methane.
(1)$$Cl_2\, \rightarrow 2Cl^{\cdot}$$
(2) $$Cl^\cdot \, + \, CH_4 \, \xrightarrow \, CH_3C1 \,+ \, H^\cdot$$
(3) $$C1^\cdot \, + \, CH_4 \, \xrightarrow \, CH_3 ^\cdot \, + \, HCI$$
(4) $$H^\cdot \, + \, Cl_2 \, \xrightarrow \, HCl+Cl^\cdot$$
(5) $$CH_3 ^\cdot \, + \, Cl_2 \, \xrightarrow \, CH_3CI+Cl^\cdot$$

A
2,3,5

B
1,3,6

C
3,5

D
2,3,4

Solution

(1) $$Cl_2 \longrightarrow 2Cl^\cdot$$ [Initiation]
(2) $$Cl^\cdot+CH_4 \longrightarrow CH_3Cl+H^\cdot$$
(3) $$Cl^\cdot + CH_4 \longrightarrow CH_3^\cdot+ HCl$$ [Propagation]
(4) $$H^\cdot+Cl_2 \longrightarrow HCl+Cl^\cdot$$
(5) $$CH_3^\cdot+Cl_2 \longrightarrow CH_3Cl+Cl^\cdot$$ [Propagation]
In the first propagation step (3), a $$Cl^\cdot$$ combines with a hydrogen on $$CH_4$$ giving $$HCl$$ and $$^\cdot CH_3$$. In the $$2^{nd}$$ propagation step (5) $$Cl_2$$ combines with $$^\cdot CH_3$$ to form $$CH_3Cl$$ and $$Cl^\cdot$$.
Question 6
1 / -0

$$(A) \, + \, Cl_2 \, \xrightarrow {hv} \, monochloro \, product$$
To maximise the yield of monochloro product in the above reaction?

A
$$Cl_2$$ must be added in excess

B
Reactant (A) must be added in excess

C
Reaction must be carried out in dark

D
Reaction must be carried out with equimolar mixture of $$Cl_2$$ and A

Solution

To maximize the yield of monochloro product the alkane/alkene/alkyne i.e. the hydrocarbon reactant has to be taken in excess. On the other hand, to maximize the yield of polychloro product, chlorine has to be taken in excess.
Question 7
1 / -0

An alkane $$(mol.\ wt.\ = 86)$$ on bromination gives only two monobromo derivatives (excluding stereoisomers). The alkane is:

A
$$CH_3-\underset{CH_3}{\underset{|}{C}}H-CH_2-CH_2-CH_3$$

B
$$CH_3-\overset{CH_3}{\overset{|}{\underset{CH_3}{\underset{|}{C}}}}-CH_2-CH_3$$

C
$$CH_3-\underset{CH_3}{\underset{|}{C}}H-\underset{CH_3}{\underset{|}{C}}H-CH_3$$

D
$$CH_3-\overset{CH_3}{\overset{|}{\underset{CH_3}{\underset{|}{C}}}}-CH_3$$

Solution

$$CH_3-\underset {|}{CH}-CH_2-CH_2-CH_3 \xrightarrow [h\nu]{Br_2} 5$$ monobromoderivatives
$$CH_3$$ $$A$$
$$CH_3$$
$$CH_3-\underset {|}{\overset {|}{C}}-CH_2-CH_3 \xrightarrow [h\nu]{Br_2} 3$$ monobromoderivatives
$$CH_3$$ $$B$$
$$CH_3-\underset {|}{CH}-\underset {|}{CH}-CH_3 \xrightarrow [h\nu]{Br_2} 2$$ monobromoderivatives
$$CH_3$$ $$CH_3$$
Among them $$A,B$$ and $$C$$ have only molecular weight $$86$$. Now $$A$$ has $$5$$ different equivalent $$H$$ atoms and $$B$$ has $$3$$ different equivalent $$H$$ atoms, so $$A$$ and $$B$$ will give $$5$$ and $$3$$ monobromo derivatives respectively. But $$C$$ has $$2$$ different equivalent $$H$$ atoms, so $$C$$ will give $$2$$ monobromo derivatives.
Question 8
1 / -0

$$\xrightarrow[hv]{cl_2}$$ (x). (x) = total number of di-chloro product S-2 chloro hexane.

A
$$7$$

B
$$6$$

C
$$16$$

D
none

Solution

$$Cl$$
$$H_3C-\overset {|}{\underset {|}C}-CH_2-CH_2-CH_2-CH_3 \xrightarrow [h\nu]{Cl_2} (x)$$
$$H$$
The possible product may be
$$Cl$$
(1)$$H_2\underset {|}{C}-\overset {|}{\underset {|}C^*}-CH_2-CH_2-CH_2-CH_3 \longrightarrow 2^1=2$$ stereoisomers
$$Cl$$ $$H$$
$$Cl$$
(2) $$H_3C-\overset {|}{\underset {|}C}-CH_2-CH_2-CH_2-CH_3 \longrightarrow 2^o= 1$$ stereoisomer
$$Cl$$
$$Cl$$ $$Cl$$
(3) $$H_3C-_*\overset {|}{\underset {|}C}-_*\overset {|}{\underset {|}C}-CH_2-CH_2-CH_3= 2^2=4$$ stereoisomers
$$H$$ $$H$$
$$Cl$$ $$Cl$$
(4) $$H_3C-^*\overset {|}{\underset {|}C}-CH_2-\overset {|}{\underset {|}C^*}-CH_2-CH_3=2^2=4$$ stereoisomers
$$H$$ $$H$$
$$Cl$$ $$Cl$$
(5) $$H_3C-_*\overset {|}{\underset {|}C}-CH_2-CH_2-_*\overset {|}{\underset {|}C}-CH_3=$$ as it is symmetrical molecule= $$2^{2-1}+2^{2/2-1}=3$$ stereoisomers
$$H$$ $$H$$
$$Cl$$
(6) $$H_3C-\overset {|}{\underset {|}C}-CH_2-CH_2-CH_2-CH_2-Cl= 2^1=2$$ stereoisomers
$$H$$
So, total number of di chloro product obtained= $$2+1+4+4+3+2=16$$ product
Question 9
1 / -0

Rank the hydrogen atoms $$(H_a, H_b, H_c)$$ in the following molecules according to their acidic strengths:

A
$$a > b > c$$

B
$$b > a > c$$

C
$$b > c > a$$

D
$$a > c > b$$

Solution

Removal of $$H_c$$ creates a negative charge on $$sp^2$$ hybridized carbon which is highly unstable, due to which $$H_c$$ will be least acidic.
Removal of $$H_b$$ creates a negative charge which is stabilized by resonance due to conjugation with double bond, There is no such factor in case of $$H_a$$
Hence the order of acid strength is $$b>a>c$$.
Question 10
1 / -0

In this reaction $$SbF_5$$ acts as:

A
an acid

B
a base

C
a nucleophile

D
an electrophile

Solution

As $$F$$ is a good nucleophile, so it will attack on $$SbF_5$$, which is electrophilic in nature and act as electrophile.