Organic Chemistry Some Basic Principles and Techniques Test

Result

Organic Chemistry Some Basic Principles and Techniques Test - 42

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

The product(s) is / are:

A
(A) is an alcohol ; (B) is an aldehyde

B
(A) is an aldehyde ; (B) is an alcohol

C
(A) and (B) both are alcohols

D
(A) and (B) both are aldehydes

Solution

A reagent used for both reactions is reducing, it will reduce the carbonyl-chloride to the alcohol group.
Question 2
1 / -0

The third member of the family of alkenynes has the molecular formula?

A
$$C_6H_6$$

B
$$C_5H_6$$

C
$$C_6H_8$$

D
$$C_4H_4$$

Solution

Alkynes are compounds with adjacent ene (double) and yne (triple) bonds.

Now refer to given image for following :

1st member of enyne family= R1, R2, R3, R4 = H
Formula = $$C_4H_4$$

2nd member of enyne family= R1=CH3; R2, R3, R4 = H
Formula = $$C_5H_6$$

3rd member of enyne family= R1, R2=CH3; R3, R4 = H
Formula = $$C_6H_8$$

(C) is the correct answer.
Question 3
1 / -0

In the balanced reaction,
$$xBrO_3^-+ yCr^{+3} + zH_2O \rightarrow pBr_2 +qCrO_4^{-2} + rH^+$$
What are x,y,z respectively?

A
$$6, 10, 11$$

B
$$6, 10, 20$$

C
$$6, 8, 22$$

D
$$6, 10, 22$$

Solution

$$\overset{+5}{B}rO_3^-+Cr^{3+}+H_2O \rightarrow \overset{O}{B}r_2+\overset{6+}{Cr}O_4^{2-}+H^+$$
$$3BrO_3^-+5Cr^{3+}+\dfrac{11}{2}H_2O\rightarrow \dfrac{3}{2}Br_2+5CrO_4^{2-}+22H^+$$
$$6BrO_3^-+10Cr^{3+}+11H_2O\rightarrow 3Br_2+10CrO_4^-+44H^+$$
$$\therefore x=6,y=10,z=11$$
Question 4
1 / -0

Isopropylamine can be obtained by

A
$${\left( {C{H_3}} \right)_2}C = O + N{H_3} \to ?$$

B
$${\left( {C{H_3}} \right)_2}CO + N{H_3}$$

C
$${\left( {C{H_3}} \right)_2}CH - Br + NaN{H_2} \to $$

D
$${\left( {C{H_3}} \right)_2}CH - Br?$$

Solution
Question 5
1 / -0

What are the values of N and M?

A
$$6, 6$$

B
$$6, 4$$

C
$$4, 4$$

D
$$3, 3$$

Solution

Since A has four different types of $$H$$ atoms, so it on chlorinatives gives $$4$$ isomeric products.

Again as four products are chemically different, on fractional distillation they give $$4$$ isomeric products. So $$N=4$$ and $$M=4$$.

Therefore, option C is correct.
Question 6
1 / -0

What is E ?

A

B

C

D

Solution

$$D$$ will be the final product for the following conversion.
Question 7
1 / -0

What is the major product in the given reaction?

A

B

C

D
None of these

Solution

(Refer to Image 1)
As alcoholic $$KOH$$ act as base here thus elimination reaction takes place and most stable alkene form.
(Refer to Image 2)
It follows $$E_2$$ mechanism in which $$^-OH$$ abstract proton and double bond form & $$^-Br$$ leave in a single step.
Question 8
1 / -0

In homlogous series, successive members differ by:

A
$$CH_{3}-$$

B
$$CH_{3}-CH_{2}-$$

C
$$-CH_{2}-$$

D
$$-H$$

Solution

A homologous series is a series of compounds in which each member differs from the next/previous by $$-C{ H }_{ 2 }-$$ or 14 mass units. Physical properties change in members of homologous series but chemical properties remain almost the same because the functional group remains the same.
Question 9
1 / -0

The end product $$(B)$$ of the reaction sequence:

A

B

C

D

Solution

The above reaction is quite similar to the Hinsberg test which is used to identify $$2^o$$ amine.
Question 10
1 / -0

Major product of above reaction is:

A

B

C

D

Solution

The above reaction is bromination reaction. After the addition of bromine, the following carbocation undergoes resonance stabilization. The delocalisation of positive is occur in the ring.