Organic Chemistry Some Basic Principles and Techniques Test

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Organic Chemistry Some Basic Principles and Techniques Test - 64

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Weekly Quiz Competition

Question 1
1 / -0

Then the correct structure of $$P$$ & $$Q$$ an respectively are:

A

B

C

D

Solution
Question 2
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A and B are respectively:

A

B

C

D

Solution

$$NaBH_4$$ does not reduce double bonds if present. But $$H_2/Pd$$ reduces double bonds also.
A is (Refer to Image 1)
B is (Refer to Image 2)
Question 3
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X is:

A

B

C

D

Solution

It is acid catalyzed esterification. The reaction between an acid and alcohol is called esterification reaction.
Question 4
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In the given reaction:
$$[X]$$ will be:

A
$$HCHO$$

B

C

D
$$HCN$$

Solution

(Only ester will be reduced because ketone is protected)
Question 5
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The major product of the given reaction is:

A

B

C

D

Solution
Question 6
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Which of the following is not the correct relationship

A
$$II$$ and $$IV$$ are metamer

B
$$I$$ and $$II$$ functional isomer

C
$$I$$ and $$III$$ are chain isomer

D
$$I$$ and $$IV$$ are positional isomer

Solution

$$\bf{Explanation-}$$
$$\bullet$$ Metamers: Compounds having the same molecular formula but the different number of carbon atoms (alkyl groups) on either side of the functional group are called metamers.
Hence $$(II)$$ and $$(IV)$$ are metamers as there is the change in the length of the carbon atoms.

$$\bullet$$ Functional isomer: It is the type of isomerism in which the compounds have the same molecular formula but have different functional groups.
Hence, $$(I)$$ and $$(II)$$ are functional isomers as there is alcohol functionality in $$I$$ and there is ether functional group in $$II$$ and have the same molecular formula.

$$\bullet$$ Chain isomer: Two or more compounds that have a similar molecular formula but different arrangements of carbon atoms in straight or branched chains are called chain isomers.
Hence, $$(I)$$ and $$(III)$$ are chain isomers as there is a change in the chain where the alcohol is attached.

$$\bullet$$ Positional isomer: Positional isomers are constitutional isomers that have the same carbon skeleton and the same functional groups but differ from each other in the location of the functional groups in the carbon chain.

Hence, $$(I)$$ and $$(IV)$$ are not positional isomers because in the case of $$i$$ we have an alcohol functional group, and in $$IV$$ we have ether functional group.

$$\bf{Conclusion-}$$ Correct Option $$D$$.
Question 7
1 / -0

Product of the reaction is?

A

B

C

D

Solution

Substitution of $$-Cl$$ in place of $$-OH$$ with the invasion of configuration

Here attaches to the compound forming a chelate compound then oxidation with $$H^+$$ takes place.
Question 8
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The best method for the preparation of 2.2-dimethylbutane is via the reaction of

A
$${Me}_{3}CBr$$ and $$Me{CH}_{2}Br$$ in $$Na/ether$$

B
$${({Me}_{3}C)}_{2}$$ $$CuLi$$ and $$Me{CH}_{2}Br$$

C
$${(Me{CH}_{2})}_{2}$$ $$CuLi$$ and $${Me}_{3}CBr$$

D
$${Me}_{3}CMgI$$ and $$Me {CH}_{2}I$$

Solution
Question 9
1 / -0

The IUPAC name of

A
2-Ethyl-3-menthyl-1-penten-4-yne

B
2-Ethyl-3-methyl-4-pentyne-1-ene

C

2-Methyl-1-penten-4-yne

D
4-Ethyl-3-methy-1-pentyn-4-ene

Solution
Question 10
1 / -0

Possible product is/are:

$$\bigstar = C-{14}$$ isotope

A

B

C

D

Solution