Spectroscopy is a powerful analytical technique that allows us to identify and characterize the chemical structure of organic compounds. In this MCQ 11.3 Spectroscopic identification of organic compounds, you will have the opportunity to test your knowledge of spectroscopic identification.
1.Which of the following is NOT a type of spectroscopy commonly used to identify organic compounds?
A) Infrared spectroscopy
B) Nuclear magnetic resonance spectroscopy
C) Mass spectrometry
D) X-ray crystallography
Answer: D) X-ray crystallography
2.Which of the following is NOT a characteristic feature of infrared spectra?
A) Absorption peaks
B) Emission peaks
C) Sharp lines
D) Continuous spectra
Answer: B) Emission peaks
3.Which of the following statements about proton nuclear magnetic resonance (1H NMR) spectroscopy is NOT true?
A) It can be used to determine the number and types of protons in a compound.
B) It can be used to determine the relative ratio of protons in a compound.
C) It can be used to determine the electronic structure of a compound.
D) It can be used to determine the spatial arrangement of atoms in a compound.
Answer: C) It can be used to determine the electronic structure of a compound.
4.Which of the following statements about mass spectrometry is NOT true?
A) It can be used to determine the molecular weight of a compound.
B) It can be used to determine the molecular formula of a compound.
C) It can be used to determine the spatial arrangement of atoms in a compound.
D) It can be used to determine the fragmentation pattern of a compound.
Answer: C) It can be used to determine the spatial arrangement of atoms in a compound.
5.Which of the following is NOT a characteristic feature of a mass spectrum?
A) Molecular ion peak
B) Fragmentation peaks
C) Absorption peaks
D) Emission peaks
Answer: C) Absorption peaks
6.Which of the following statements about infrared spectroscopy is NOT true?
A) It can be used to determine functional groups in a compound.
B) It can be used to determine the molecular weight of a compound.
C) It is based on the absorption of infrared radiation by a compound.
D) It can be used to determine the spatial arrangement of atoms in a compound.
Answer: B) It can be used to determine the molecular weight of a compound.
7.Which of the following is NOT a type of proton nuclear magnetic resonance (1H NMR) spectroscopy?
A) High-resolution 1H NMR
B) Low-resolution 1H NMR
C) 2D 1H NMR
D) X-ray crystallography
Answer: D) X-ray crystallography
8.Which of the following statements about mass spectrometry is NOT true?
A) It can be used to determine the molecular weight of a compound.
B) It can be used to determine the molecular formula of a compound.
C) It is based on the principle of conservation of mass and charge.
D) It can be used to determine the electronic structure of a compound.
Answer: D) It can be used to determine the electronic structure of a compound.
9.The index of hydrogen deficiency (IHD) is a measure of the degree of unsaturation in a compound, which is determined by the number of multiple bonds or rings present in the compound. Here are some multiple choice questions and answers on the index of hydrogen deficiency:
10.Which of the following compounds has the highest index of hydrogen deficiency?
A) Ethane (C2H6)
B) Ethene (C2H4)
C) Ethyne (C2H2)
D) Benzene (C6H6)
Answer: C) Ethyne (C2H2)
Find more questions on MCQ 11.3 Spectroscopic identification
11.Which of the following compounds has the lowest index of hydrogen deficiency?
A) Ethane (C2H6)
B) Ethene (C2H4)
C) Ethyne (C2H2)
D) Benzene (C6H6)
Answer: A) Ethane (C2H6)
12.The index of hydrogen deficiency of a compound can be calculated by:
A) Adding the number of multiple bonds and rings in the compound
B) Subtracting the number of hydrogen atoms from the number of carbon atoms
C) Dividing the number of carbon atoms by the number of hydrogen atoms
D) Subtracting the number of hydrogen atoms from twice the number of carbon atoms
Answer: D) Subtracting the number of hydrogen atoms from twice the number of carbon atoms
13.Which of the following compounds has an index of hydrogen deficiency of 1?
A) Propene (C3H6)
B) Butene (C4H8)
C) Cyclopentene (C5H8)
C) Benzene (C6H6)
Answer: A) Propene (C3H6)
14.Which of the following compounds has an index of hydrogen deficiency of 2?
A) Propene (C3H6)
B) Butene (C4H8)
C) Cyclopentene (C5H8)
C) Benzene (C6H6)
Answer: C) Cyclopentene (C5H8)
15.Which of the following compounds has an index of hydrogen deficiency of 3?
A) Propene (C3H6)
B) Butene (C4H8)
C) Cyclopentene (C5H8)
C) Benzene (C6H6)
Answer: D) Benzene (C6H6)
16.The index of hydrogen deficiency can be used to determine the number of:
A) Carbon atoms in a compound
B) Hydrogen atoms in a compound
C) Multiple bonds or rings in a compound
D) All of the above
Answer: C) Multiple bonds or rings in a compound
T17.he index of hydrogen deficiency is equal to the sum of the:
A) Number of multiple bonds and rings in the compound
B) Number of carbon atoms and hydrogen atoms in the compound
C) Number of carbon atoms and the molecular weight of the compound
D) None of the above
Answer: A) Number of multiple bonds and rings in the compound
18.Which of the following techniques can be used to determine the electronic structure of a compound?
A. Infrared spectroscopy
B. Nuclear magnetic resonance (NMR) spectroscopy
C. Mass spectrometry
D. X-ray crystallography
Answer: B) Nuclear magnetic resonance (NMR) spectroscopy
19.Which of the following compounds will show a single signal with no splitting pattern (i.e. a singlet) in their high resolution 1H NMR spectrum?
A) Benzene
B) 2,2-dimethylpropane
C) 2-methylpropane
D) Ethanol
Answer: A) Benzene
20.Which of the following compounds will show a signal in its 1H NMR spectrum that is split into a doublet?
A) Bromopropane, CH3CH2CH2Br
B) Propanal, CH3CH2CHO
C) Propanone, CH3COCH3
D) Ethanal, CH3CHO
Answer: A) Bromopropane, CH3CH2CH2Br
21.Which of the following statements about TMS is NOT true?
A) TMS is very unreactive, so it does not interfere with the sample being analyzed.
B) TMS is volatile, so it can easily be removed from the sample after the spectrum has been run.
C) TMS is toxic, so it should be handled with care.
D) TMS has a strong single signal in its 1H NMR spectrum due to all twelve protons being in the same chemical environment.
Answer: C) TMS is toxic, so it should be handled with care.
22.What is the primary reason why TMS is used as a reference material in 1H NMR spectroscopy?
A) TMS absorbs upfield well away from most other protons, so it does not interfere with the signals from the sample.
B) TMS is soluble in water, so it can easily be used to prepare samples for NMR analysis.
C) TMS has a strong single signal in its 1H NMR spectrum due to all twelve protons being in the same chemical environment.
D) TMS is very unreactive, so it does not interfere with the sample being analyzed.
Answer: C) TMS has a strong single signal in its 1H NMR spectrum due to all twelve protons being in the same chemical environment.
23.Which of the following statements about TMS is NOT true?
A) TMS has a strong single signal in its 1H NMR spectrum due to all twelve protons being in the same chemical environment.
B) TMS absorbs upfield well away from most other protons, so it does not interfere with the signals from the sample.
C) TMS is volatile, so it can easily be removed from the sample after
24.Which of the following statements about the 1H NMR spectra for the two isomers of dibromoethane is NOT true?
A) The 1H NMR spectrum for 1,1-dibromoethane shows a single signal that is split into a triplet.
B) The 1H NMR spectrum for 1,2-dibromoethane shows a single signal that is split into a singlet.
C) The 1H NMR spectrum for 1,1-dibromoethane shows two signals that are split into doublets.
D) The 1H NMR spectrum for 1,2-dibromoethane shows two signals that are split into triplets.
Answer: C) The 1H NMR spectrum for 1,1-dibromoethane shows two signals that are split into doublets.
In the 1H NMR spectrum of 1,1-dibromoethane, the two signals are split into triplets, not doublets. The 1H NMR spectrum for 1,2-dibromoethane shows a single signal that is split into a singlet and two signals that are split into triplets.
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