Organic Chemistry Some Basic Principles and Techniques Test

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Organic Chemistry Some Basic Principles and Techniques Test - 60

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Weekly Quiz Competition

Question 1
1 / -0

Which of the following react with $$HBr$$ at faster rate?

A

B

C

D

Solution

The reaction with $$HBr$$ is faster when the carbocation formed is more stable carbocation formed by (Refer to Image) is more stable because the carbocation formed is tertiary carbocation.
Question 2
1 / -0

Product obtained in the reaction is?

A
Diastereomer

B
Racemic

C
Meso

D
Optically pure enantiomer

Solution
Question 3
1 / -0

$$CH_3-\overset{O}{\overset{||}{C}}-H\overset{HCN}{\rightarrow}(A)\overset{H_3O^{\oplus}}{\rightarrow}(B)\underset{\Delta}{\rightarrow}(C)\overset{LiAlH_4}{\rightarrow}(D)\overset{HIO_4}{\rightarrow}HCHO+(E)$$
Compound (C) can show geometrical isomerism. Product (E) of the reaction will be.

A
$$CH_3-\overset{O}{\overset{||}{C}}-CH_3$$

B
$$CH_3-CH_2-\overset{O}{\overset{||}{C}}-H$$

C
$$CH_3-CHO$$

D
HCHO

Solution
Question 4
1 / -0

At which value of n the formation of six membered ring takes place ?

A
n=2

B
n=3

C
n=-5

D
n=6

Solution

(Refer to Image 1)
Thus on compare the product with $$6$$- membered ring
(Refer to Image 2)
So, $$n=3$$ .
Question 5
1 / -0

Identify the major product.

A

B

C

D

Solution

In this reaction, the $$H^+$$ ion from the $$H_2SO_4$$ attacks on $$OH$$ group to form a carbocation on the given compound.
This carbocation is resonance stabilized by the shift of the ethyl group from one carbon to another.
Question 6
1 / -0

A

B

C

D

Solution

$$O$$
$$CH_3-\overset {||}{C}-Cl$$ attacks the electron donating group which has less electron density.
Here, both Nitrogen and oxygen act as electron donating groups. But Nitrogen is less electronegative compared to oxygen.
So, (Refer to Image)
Question 7
1 / -0

Consider the above sequence and answer the following

Reduction $$R-CH_2OH\rightarrow R-CH_3$$ can be done by:

A
$$LiAlH_4$$

B
$$NaBH_4-AlCl_4$$

C
$$H_2-Ni$$

D
Red P+ HI

Solution

Alcohols on reaction with red phsophorus and $$HI$$ reduces it to alkanes.
Mechanism:-
$$R-CH_2-\overset {. .}{O}H\xrightarrow []{HI} R-CH_2-I$$
$$\downarrow Red P,HI$$
$$R-CH_3$$
Question 8
1 / -0

Product (A) of the given reaction is (bromination occur not in the benzene ring).

A

B

C

D

Solution

Bromine attacks at $$2^o$$ carbon which is away from ketone group, so that there will be less electron-withdrawing nature.
(Refer to Image)
Answer is option A.
Question 9
1 / -0

Product $$(D)$$ in the given sequence is:

A

B

C

D

Solution

In first step, Friedel Craft acylation takes place, in which acyl group attached to the benzene ring.
$$\longrightarrow$$ On addition of $$SOCl_2$$ nucleophilic substitution takes place on acid.
$$\longrightarrow$$ On addition of $$NaN_3$$ again nucleophilic substitution takes place, which further eliminate $$N_2$$ and Nitrene formation takes place.
$$\longrightarrow$$ On addition of $$MeOH$$ on Nitrene, Hoffmann Rearrangement takes place in which the alkyl group shifted to electron deficient Nitrogen.
$$\longrightarrow$$ Then $$LiAlH_4$$ reduce both acid & ketone group.
$$\longrightarrow$$ $$NaH$$ abstract proton from $$OH$$ not from $$NH$$ as $$O^{\circleddash}$$ is most stable than $$N^{\circleddash}$$.
$$\longrightarrow O^{\circleddash}$$ will attack on benzene ring and nucleophilic substitution takes place.
$$O$$ substitute $$Cl$$ not $$CF_3$$, because $$C-Cl$$ bond is more polar than $$C-CF_3$$ bond.
Question 10
1 / -0

What is the appropriate reagent to carry out given synthesis, i.e., A, B, C respectively are.

A
$$Br_2/H^+, LiAlH_4, H^{\oplus}$$

B
$$Br_2/H^+, NaBH_4, HO^-$$

C
$$NBS, AlCl_3, HO^-$$

D
$$Br_2/HO^-, BF_3, HO^-$$

Solution

Ketone tautomeries to only one side of carbon in the compound. It substitutes Bromine.
(Refer to Image)
(i) $$Br,H^+$$
(ii) $$NaBH_4$$
(iii) $$O\overset {-}{H}$$