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Test: IR Spectroscopy - Chemistry MCQ


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10 Questions MCQ Test Organic Chemistry - Test: IR Spectroscopy

Test: IR Spectroscopy for Chemistry 2024 is part of Organic Chemistry preparation. The Test: IR Spectroscopy questions and answers have been prepared according to the Chemistry exam syllabus.The Test: IR Spectroscopy MCQs are made for Chemistry 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: IR Spectroscopy below.
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Test: IR Spectroscopy - Question 1

 Which compound having molecular formula C5H10 shows absorption at 1380 cm-1?

Detailed Solution for Test: IR Spectroscopy - Question 1

C5H10 → 1380 cm-1. 1380 cm-1 represents a C–C bond band. All have C – C bond. Option Pentyne does not fulfill the H-atom requirement. So, correct option is Cyclopentane.

Test: IR Spectroscopy - Question 2

 What is the effect of ring strain in lactone (cyclic ester) or a lactam (cyclic amide)?

Detailed Solution for Test: IR Spectroscopy - Question 2

Carbonyl stretching order in both the compounds is 6 > 5 > 4 > 3. Ring strain increases, carbonyl frequency increases.

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Test: IR Spectroscopy - Question 3

 What is the relation between wave number of IR absorption and the reduced mass?

Detailed Solution for Test: IR Spectroscopy - Question 3

, frequency is directly proportional to wave number. So, wave number is inversely proportional to reduced mass as shown in the above relation.

Test: IR Spectroscopy - Question 4

Which of the following molecules will not show infrared spectrum?

Detailed Solution for Test: IR Spectroscopy - Question 4

Correct option is H2 as HH2 do not have dynamic dipole moment, so no spectrum will be observed.

Test: IR Spectroscopy - Question 5

Why Monomeric saturated aliphatic carboxylic acids show carbonyl stretching frequency near 1760 cm-1, while saturated aliphatic ketones near 1720 cm-1?

Detailed Solution for Test: IR Spectroscopy - Question 5

Monomeric saturated aliphatic carboxylic acids show carbonyl stretching frequency near 1760 cm-1, while saturated aliphatic ketones near 1720 cm-1 because I effect is dominant in carboxylic acids over the mesomeric effect.

Test: IR Spectroscopy - Question 6

Why in the IR spectrum of Benzoyl chloride, a weak band near 1750 cm-1 is formed?

Detailed Solution for Test: IR Spectroscopy - Question 6

Benzoyl chloride

A weak band near 1750 cm-1. Since, band is of weak intensity, it must be due to fermi Resonance. So correct option is Fermi resonance between C = O band and first overtone.

Test: IR Spectroscopy - Question 7

Why ketenes absorb in IR at a very high frequency (2150 cm-1)?

Detailed Solution for Test: IR Spectroscopy - Question 7

Ketenes absorb in IR at a very high frequency (2150 cm-1) because inner C is sp2 hybridized. Bonds with more s character absorb at a higher frequency.

Test: IR Spectroscopy - Question 8

A compound C8 H6 decolorizes Br2 in CCl4 and gives a white precipitate with Tollen’s reagent. It has sharp band at 3300 cm-l and weak bands at 3085, 2110 cm-l. What is this compound?

Detailed Solution for Test: IR Spectroscopy - Question 8

C8 H6 → Sharp band at 3300 cm-l
Weak band at 3085, 2110 cm-l
Double bond equivalent = C + 1 – H⁄2 – X⁄2 + N⁄2,
(where C→ Carbon, H→ Hydrogen, X→ Halogen, N→ Nitrogen)
= 8 + 1 – 6⁄2 = 6.
One double bond equivalent represents either one bond or one ring.
Now, band at 3300 shows C–H bond, 2110 may be because of C≡C triple bond
Now, we check each option one by one
Option Octene → octene → does not match m.f. C8 H6.


C ≡ C → 2110
D.B.E. → 5πbond
M.f. → C8 H6. Hence, the correct option is Phenyl acetylene.

Test: IR Spectroscopy - Question 9

What is the number of vibrational degrees of freedom in C6H5CH3?

Detailed Solution for Test: IR Spectroscopy - Question 9

C6H5CH3 is nonlinear.
No. of vibration degree of freedom = 3N – 6
= 3(15) – 6 = 39.

Test: IR Spectroscopy - Question 10

The phosphorescence spectrum of the excited species is due to which transition?

Detailed Solution for Test: IR Spectroscopy - Question 10

The phosphorescence spectrum of the excited species is due to triplet to singlet transitions. In an excited singlet state, the electron is promoted in the same spin orientation as it was in the ground state (paired). In a triplet excited stated, the electron that is promoted has the same spin orientation (parallel) to the other unpaired electron.

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