Spectroscopy Question with Answer- CSIR NET, GATE
SPECTROSCOPY-1: Question and Answer
Hello everyone today we will discuss some questions with answers. This is the Spectroscopy Q&A Series-1. Today we will talk about spectroscopy questions and answer for CSIR NET, GATE, IIT JAM, TIFR, and other types of entrance examination.
How to solve the question?
To solve the question -get the knowledge under the whole topic then you have to easy to solve the question. First of all, you have gathered the knowledge, by practicing question answers. Secondly, read the question carefully, you have to find the most common concept which is already solved the question in a practice session.
How to solve the question without wasting any time? Or how to solve the question in a short time?
We are recement to that is we follow, first of all, we are seeing the question paper, raffle read all question. Then choose the most common and easy question, then we have to solve the question without wasting any time. Then we have to read carefully the remaining question in the paper- in this portion solve the question one by one.
How to read NMR Spectroscopy in Organic Chemistry? Or What are the reference books for NMR Spectroscopy?
Nuclear Magnetic Resonance Spectroscopy (NMR Spectroscopy) is the most important topic for a chemist. We easily found any unknown chemical structure, during the pulse read. So, NMR Spectroscopy needs to be in the future. 1H NMR, 13C NMR Spectroscopy is the most valuable and important part of any examination such as CSIR NET, GATE, TIFR, DRDO, IIT JAM, etc.
First of all, we have to require knowledge of the basic concept of NMR spectroscopy, you have to follow the book by C.N Banwell (Fundamental molecular spectroscopy). After clearing the whole portion of fundamental and introduction spectroscopy, you have to require- the concept of organic spectroscopy such as chemical shift value calculation, shielding, and de-shielding effect, population factor, electronegativity effect, hybridization effect, H-bond effect, etc. for these types of concepts are easier to follow the book by Pavia (Introduction to spectroscopy).
Before reading a Book clear your Basic Concept of Spectroscopy. That is very important. We are mentioned in the earlier stage.
Today we will discuss a few questions for CSIR NET, GATE, and IIT JAM: Topic Spectroscopy; structure determination, calculation of the number of signals, etc.
Choose the correct option: Give feedback in the comment section
For the Spectroscopy mixture question with solution, we considered six mixture questions including 1H NMR Spectroscopy, 13C NMR, and IR (Infrared Spectroscopy):
Solve Spectroscopy Question
Q.1. An organic compound (C9H10O3) exhibited the following spectral data:
IR: 3400, 1680 cm –1;
1H NMR: δ 7.8 (1H, d, J = 8 Hz), 7.0 (1 H, d, J = 8Hz), 6.5 (1 H, s), 5.8 (1 H, s, D2O exchangeable), 3.9 (3H, s), 2.3 (3 H, s). The compound is
A. (a)
B. (b)
C. (c)
D. (d)
ANSWER:
Hints:
First find out Double bond equivalent (DBE)= [Number of Carbon] – [Number of monoatomic/2] + [Number of triatomic / 2] +1
Where, the diatomic atom is not included in DBE calculation, such as the oxygen atom.
- The organic compound C9H10O3 DBE = 9 – (10/2) + 1 = 9 – 5 + 1 = 5
One aromatic benzene ring (Three double bonds + one ring = 4) and one double bond (In the case of carbonyl double bond).
- IR: 3400 Cm-1 denote must be present one -OH group, and IR: 1680 Cm-1 denote must be present one carbonyl carbon.
1H NMR Spectroscopy: a total of six signals are found.
Q.2. In the IR spectrum of p-nitrophenyl acetate, the carbonyl absorption band appears at
(A) 1670 cm–1
(B) 1700 cm–1
(C) 1730 cm–1
(D) 1760 cm–1
ANSWER: (B) 1700 cm-1
Hint:
The structure of p-nitrophenyl acetate is given below:
 |
| Infrared spectrum (IR) for p-nitrophenyl acetate carbonyl double bond. |
One carbonyl bond absorption spectra IR: normally 1700-1725 cm-1. We know that NO2 group (-M) effective and (-I) effective, the electron density more withdrawing on the side compared to carbonyl carbon side. That is why the electron delocalization is 40% extra compared to the normal carbonyl carbon. That means decreases the double bond character (up to 40%).
We know the absorption frequency (cm-1) is directly proportional to bond straight.
(V) α (Bond Straight)
So, in the case of bond straight decreases due to delocalization. Therefore, the vibration frequency will be decreased. The absorption frequency: 1700 cm-1.
Q.3. An organic compound exhibited the following 1H NMR spectra data: δ 7.80 (2 H, d, J= 8 Hz), 6.80 (2 H, d, J 8= Hz), 4.10 (2H, q, J =7.2 Hz), 2.4 (3H, s) ,1.25 (3 H, t, J= 7.2 Hz)
The compound, among the choices given below, is,
A. (a)
B. (b)
C. (c)
D. (d)
ANSWER: (B)
Hints: A total of five 1H NMR signals, which are found in (B) compound only.
 |
| Option (B): 1H NMR signal for each hydrogen |
Q.4. The number of signals that appear in the broadband decoupled 13C NMR spectrum of phenanthrene and anthracene, respectively are
(A) Ten and Four
(B) Ten and Ten
(C) Seven and Four
(D) Seven and Seven
ANSWER: (C) Seven and Four
Hints:
 |
| Several 13C NMR Signal: Phenanthrene = seven spectrum signal, and Anthracene =four spectrum signal. |
Phenanthrene and Anthracene 13C NMR Signal: seven and four respectively. Using the mirror plane, find out the equivalent NMR signal.
Q.5. In the 1H NMR spectrum recorded at 293 K, an organic compound (C3H7NO), exhibited signals at δ 7.8 (1H, s), 2.8 (3H, s), and 2.6 (3H, s). The compound is
A. (a)
B. (b)
C. (c)
D. (d)
ANSWER:
Hints:
Organic compound C3H7NO double bond equivalent (DBE) = 3 – (7/2) + (1/2) + 1 = 1
The number of carbon (C) =3, Number of monoatomic atoms (H) = 7, Number of triatomic atoms (N) = 1, and Oxygen do not count in the DBE calculation.
• DBE = 1, Which means one double bond presence in an organic compound.
Q.6. In the 400 MHz 1H NMR spectrum, organic compounds exhibited a doublet. The two lines of the doublet are at δ 2.35 and 2.38 ppm. The coupling constant (J) value is
(A) 3 Hz
(B) 6 Hz
(C) 9 Hz
(D) 12 Hz
ANSWER: (D) 12 Hz
Hints:
We know that the coupling constant (J) = Distance between two spectrum lines in Hz. Given two lines of the doublet are at (δ ppm): 2.35 ppm and 2.38 ppm.
The distance between two spectrum lines = (2.38 – 2.35) = 0.03 ppm = (0.03/400 MHz) =12 Hz
This is the coupling constant (J) = 12 Hz.
Where converted ppm to Hz: chemical shift (δ ppm) divided by the frequency of spectrum machine (400 MHz).
Q.7. Correctly matched structure and carbonyl stretching frequency set is
(1) P-Y, Q-Z, R-X
(2) P-Y, Q-X, R-Z
(3) P-Z, Q-Y, R-X
(4) P-X, Q-Z, R-Y
Answer: (1) P-Y, Q-Z, R-X
Hints:
P: C-O bond 100% delocalized → Bond strength. Carbonyl stretching frequency: 1770 cm-1 (Y).
Q: C-O bond 50% delocalized → Bond strength ↑ (due to decreasing the single bond character and increasing the double bond character of carbonyl bond C=O). Carbonyl stretching frequency: 1800 cm-1 (Z).
R: C-O bond 200% delocalized → Bond strength ↓ (due to increases in the single bond character and decrease in the double bond character of carbonyl bond C=O). Carbonyl stretching frequency: 1750 cm-1 (X).
 |
| Comparing the Stretching frequency of compounds P, Q, and R. |
The Corrected matched structure and carbonyl stretching frequency set are P-Y, Q-Z, and R-X.
Reason:
we know that increasing the delocalization corresponds to decreasing the double bond character of the carbonyl C=O bond (corresponding to increasing the single bond character).
According to vibrational frequency equation: Ï‘ = (1/2Ï€c) √(K/μ) cm-1. Clearly, we see that vibrational frequency is directly proportional to the force constant (K).
Frequency (V) ∝ Force constant (K)
And force constant is directly proportional to Bond strength.
Force constant (K) ∝ Bond strength
That is why we can call bond stretching frequency (vibrational frequency) directly proportional to Bond strength.
Stretching Frequency ∝ Bond strength
We know that the bond strength of the double bond is more than the single bond. Single bond character stretching frequency is lower than double bond character.
Q.8. The number of chemical shift non-equivalent protons expected in the 1H NMR spectrum of α-pinene is
A.7
B. 8
C. 9
D. 10
Answer: (D) 10
Hints: The Structure of α-pinene give a total of 10 1H NMR signal including
 |
| α-pinene give a total of 10 1H NMR signal including. |
Spectroscopy Question with Answer PDF Download free:
CSIR NET, GATE solved questions on spectroscopy.
Conclusion
Test question for spectroscopy-1, this is for entrance examinations such as GATE, and CSIR NET. We are hopeful that this question-answer series helpful for beginners. Thank you for reading. Do not forget to like, share and comment. Give your feedback in below comment section.
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