Organic Chemistry Some Basic Principles and Techniques Test

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Organic Chemistry Some Basic Principles and Techniques Test - 39

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Question 1
1 / -0

$$CH_3-\underset{Br}{\underset{|}{\overset{CH_3}{\overset{|}C}}}-CH_3, \, CH_3-\underset{Br}{\underset{|}{C}}-CH_3, \, CH_3-CH-Br$$
Compare rate of $$E_2$$ reaction :

A
c > b > a

B
a > b > c

C
b > a > c

D
c > a > b

Solution

Forming a stable product affects the rate of reaction. Rate is very high for most stable product, hence
Refer to image
Product are
$$CH_2=Ch(Me)_2>CH_2=CHCH_3>CH_2=CH$$
Question 2
1 / -0

Consider the following statements :
(1) Bridgehead halides are inert towards both $$S{_{N}}^{1}$$ and $$S{_{N}}^{2}$$ reactions (till one of the ring size is eight member ring)
(2) The first step in both $$S{_{N}}^{1}$$ and $$E_1$$
(3) $$S{_{N}}^{2}$$ reaction proceed with total retention of configuration
(4) $$E_2$$ elemnations are by use of asolvent of low polarity and high concentration of a strong base
Which og the above statements are correct?

A
1, 2, and 4

B
1 and 3

C
2, 3, and 4

D
1, 2, 3 and 4

Solution

Bridge head halide can't form a stable intermediate hence they can't undergo with both $$SN^1$$ and $$Sn^2$$. The $$SN^1$$ and $$E^1$$ reactions have some first step, leaving of leaving group and $$E_2$$ requires high concentration of bases to occur.
Question 3
1 / -0

identify the rate of reaction of given compounds in $$E_2$$ reaction:

A
$$a > b > c > d$$

B
$$a > c > b > d$$

C
$$b > a > c > d$$

D
$$b > d > a > c$$

Solution

The rate of reaction depends on the leaving ability of halide ions.
Leaving Ability: $$I^->Br^->Cl^->F^-$$ (Acidic nature)
Strong acids act as a strong leaving groups.
Hence rate of reaction is $$a>b>c>d$$
Question 4
1 / -0

In solvolysis of 1,2-dimethyl propyl p-tolunce sulfonate in acetic acid at $$75^{\circ}C$$, how many (alkane + substitution) products will be formed ?

A
2

B
3

C
4

D
5

Solution

Refer to image
Due to the strong donating nature, many substitution occur which are equal to the number of carbon in alkyl chain. Hence five products are formed.
Question 5
1 / -0

$$1.9 \times 10^{9}\ gm$$ of the metal having density $$19\ gm/ml$$ is dispersed in one litre of water to give a sol having spherical metal particles of raidus $$10\ nm.$$ The approximate number of metal sol particles per $$cm^{3}$$ of the sol is given by :

A
$$2.35 \times 10^{9}$$

B
$$54 \times 10^{10}$$

C
$$1.9 \times 10^{9}$$

D
$$2.8 \times 10^{9}$$

Solution
Question 6
1 / -0

Above conversion can be done by :

A
$$NaBH_{4}$$

B
$$LiAlH_{4}$$

C
$$PCC$$

D
$$KMnO_{4}$$

Solution

Lithium aluminium hydride can reduce esters, carboxylic acids, ketones, aldehydes into alcohols.
So, (Refer to Image)
Question 7
1 / -0

In the given conversion best yield will obtained with.

A
$$A=CH_3-\overset{O}{\overset{||}{C}}-Cl, AlCl_3, B=Zn(Hg), HCl$$

B
$$A=Zn(Hg), HCl, B=CH_3-\overset{O}{\overset{||}{C}}-Cl, AlCl_3$$

C
$$A=CH_3-CH_2-Cl, AlCl_3, B=Zn(Hg), HCl$$

D
$$A=NH_2-NH_2/HO^-, D, B=CH_3-CH_2-Cl, AlCl_3$$

Solution

It is a friedel crafts acetylation reaction.The $$Zn(Hg)$$ and $$HCl$$ reduces the $$-CO-$$ group to $$-CH_2$$ group.
Question 8
1 / -0

Product (E) is?

A
Nylon $$66$$

B
Nylon $$6$$

C
Styrene

D
Polystyrene

Solution
Question 9
1 / -0

$$CH_3CHO \overset{10 \, percent \,\, NaOH}{\underset{5^o C }{\longrightarrow}} \underset{\Delta}{\longrightarrow} \overset{H_2O}{\underset{Ni}{\longrightarrow}} (A) $$;
Product (A) of the reaction is:

A
propanol

B
ethanol

C
butanol

D
pentanol

Solution
Question 10
1 / -0

Yield of each step as actually carried out in the laboratory is given. What is overall yield of reaction?

A
$$42\%$$

B
$$31\%$$

C
$$21\%$$

D
$$60\%$$

Solution

The overall yield of reaction= $$x$$
$$\Rightarrow x= \left( \cfrac {yield-1}{100}\right)\times \left(\cfrac {yield-2}{100}\right)\times \left(\cfrac {yield-3}{100}\right)$$
$$=\left(\cfrac {58}{100}\right)\times \left(\cfrac {54}{100}\right)\times\left(\cfrac {68}{100}\right)$$
$$=0.2129$$
$$\Rightarrow$$ % yield= $$21.3$$%
So, overall yield= $$21$$%