Organic Chemistry Some Basic Principles and Techniques Test

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Organic Chemistry Some Basic Principles and Techniques Test - 37

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Question 1
1 / -0

What is the ratio of (x/y) in the given problem?

A
$$1.33$$

B
$$2$$

C
$$2.5$$

D
$$3$$

Solution

(1) (Refer to Image 1)
(2) (Refer to Image 2)
In reaction $$1$$; $$2$$ equivalents of $$CH_3MgBr$$ are needed to react with $$2$$ equivalent $$-COCl$$ and another two equivalents are needed to react with $$2$$ equivalents of $$-COCH_3$$. That's why total four equivalents of $$CH_3MgBr$$ are needed. But in reaction $$2$$; $$2$$ equivalents of $$CH_3MgBr$$ are needed for one ester group and another one equivalent is needed for $$-CH_2Cl$$ conversion into $$-CH_2-CH_3$$.
$$\therefore x/y= 4/3$$
Question 2
1 / -0

Product (A) is :

A

B

C

D

Solution

As refer to the above image, so carbocation rearrangement takes place to form a stable compound.
Question 3
1 / -0

Product obtained in this reaction is :

A

B

C
$$Ph - C \equiv C - I$$

D
$$I - C \equiv C - H$$

Solution

Refer to image 01

This is a nucleophilic substitution reaction where -I is getting substituted.

Refer to image 02 molecule helps to retain the same configuration due to its bulky nature
Question 4
1 / -0

In which of the following raections 1, 3 - butadiene will be obtained as a major product ?

A
$$Br - CH_2 - CH_2 - CH_2 - CH_2 - Br \frac{(CH_3)_3 COK(2 mole)}{(CH_3)_3COH)}$$

B
$$HO - CH_2- CH_2- CH_2- CH_2 -OH \xrightarrow[]{conc H_2 SO_4 }$$

C
$$H_2C =CH -C=CH\xrightarrow[Ni_2B]{H_2[1moles]}$$

D
All above these

Solution

All reactions produce $$1,3-dibutene$$ as the product

$$Br-CH_2-CH_2-CH_2-CH_2-Br\overset {(CH_3)_3COH}{\longrightarrow} CH_2=CH-CH=CH_2$$

$$HO-CH_2-CH_2-CH_2-CH_2-OH\overset {H_2SO_4}{\longrightarrow} CH_2=CH-CH=CH_2$$

$$H_2C=CH-C\equiv CH\overset {H_2/Ni}{\longrightarrow} CH_2=CH-CH=CH_2$$

Reactions are dehydrohalogenation, dehydration and reduction respectively.
Question 5
1 / -0

Transition state $$2$$ is structurally most likely as :

A
intermediate $$1$$

B
transition state $$3$$

C
intermediate $$2$$

D
product

Solution

Generally transition state tries to bring stability with less energy difference. From the figure it is evident that energy difference between $$TS^2$$ and intermediate 2 is least. So it is most likely that $$TS^2$$ and intermediate 2 have similar structures.
Question 6
1 / -0

Give the structural formula of compound $$(B)$$ :

A
$$C{ H }_{ 3 }-{ \left( C{ H }_{ 2 } \right) }_{ 2 }-C\equiv C-SO_{ 3 }H$$

B
$$C{ H }_{ 3 }-{ \left( C{ H }_{ 2 } \right) }_{ 2 }-C\equiv C-C{ H }_{ 3 }$$

C

D
$$C{ H }_{ 3 }-{ \left( C{ H }_{ 2 } \right) }_{ 2 }-C\equiv C-C{ H }_{ 2 }$$

Solution

Refer to image

$${ LiNH }_{ 2 }$$ strong base attacks terminal hydrogen

$$CH_3CH_2C\equiv CH+LiNH_2\longrightarrow CH_3CH_2CH_2-C\equiv C^{\ominus}$$

Hence, here nucleophilic substitution reaction occurs where an alkyne is formed as terminal alkyne.
$$CH_3CH_2\equiv C^{\ominus}\quad \underrightarrow{(CH_3)_2SO_4}\quad CH_3-CH_2-CH_2-C\equiv C-CH_3$$
Question 7
1 / -0

Product (C) is

A

B

C

D
Both (a) and (b)

Solution

NBS leads to allylic hydrogenation and forms product A and then A react with $$Mg$$ to form lrignard reagent. or reacts with aldehyde to form an adduct B and then B forms diol with $$OsO_4$$
Question 8
1 / -0

In polyence that cantain differently substituted (c=c) doble bond , it is possible to hydrogenate chemeselectively one (c=c) double bond product is:

A

B

C

D

Solution

We know hat the least stable double bond undergoes hydrogenation easily so as to increase its stability.
Hence,
Refer to image
The terminal bond is less stable so, it is hydrogenated.
Question 9
1 / -0

Predict the major product :

A

B

C

D

Solution

In acidic medium, rearrangement occurs until the structure obtained is least energetic and the most stable state.

refer to image.

Hence, carbocation rearrangement also occurs to stabilise the transition state.
Question 10
1 / -0

Product (Z) is:

A

B

C

D

Solution

Alkene on treatment with $$Br_2/hv$$ forms a bromoalkene in accordance with markonikov addition,their product on heating with alcohol $$KOH$$ producesan elimination product ,alkene.Alkene on further exposed to $$Br$$ and peroxide generates a product in accordance with antimarkonikov rule.