Organic Chemistry : Some Basic Principles & Techniques Test

Result

Organic Chemistry : Some Basic Principles & Techniques Test - 4

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Weekly Quiz Competition

Question 1
4 / -1

How many a and 7C bonds are present in HC = C - CH = CH - CH₃?

A
9σ, 4π

B
10σ, 3π

C
6σ, 6π

D
5σ, 5π

Solution

No. of σ-bonds = 10; No. of π-bonds = 3
Question 2
4 / -1

Which type of hybridisation of each carbon is there in the compound?
CH₃ - CH = CH - CN

A
sp³, sp², sp², sp

B
sp³, sp², sp², sp³

C
sp³,sp²,sp³,sp³

D
sp³,sp²,sp,sp³

Solution

From left 1-sp³, 2-sp², 3-sp², 4-sp
Question 3
4 / -1

Which of the following represents the given sequence of hybridisation of carbon atoms from left to right sp², sp², sp, sp?

A
H₂C = CH - C = CH

B
HC = C - CH = CH₂

C
H₃C-CH = CH-CH₃

D
H₂C = CH - CH = CH₂

Solution
Question 4
4 / -1

Select the correct order of basic nature

A
CH₃CH^-₂ > CH₂ = CH^- > CH ≡ C^- > OH^-

B
CH₃CH^-₂ > CH ≡ C^- > CH₂ = CH^- > OH^-

C
CH₃CH^-₂ > OH^- > CH ≡ C^- > CH₂ = CH^-

D
OH^- > CH ≡ C^- > CH₂ = CH^- > CH₃CH^-₂

Solution

Order of acidic strength is
H₂O > CH = CH > CH₂ = CH₂ > CH₃CH₃
We know that a stronger add has weak conjugate base and viceversa, thus order of basic nature of given compounds is
CH3 - CH^-₂> CH₂ = CH^- > CH ≡ C^- > OH^-
Question 5
4 / -1

What are the hybridization and shapes of the following molecules?
(i) CH₃F
(ii) HC ≡ N

A
(i) sp², trigonal planar; (ii) sp³, tetrahedral

B
(i) sp³, tetrahedral; (ii) sp, linear

C
(i) sp, linear; (ii) sp², trigonal planar

D
(i) sp², trigonal planar, (ii) sp², trigonal planar

Solution

CH₃F - sp³ hybridised carbon, tetrahedral shape
(ii) HC = N - sp hybridised carbon, linear shape.