Question

Identify the following compound: C10H10O2: NMR: δ 2.82 (6 H, s), δ 8.13 (4 H, s)...

Identify the following compound:

C10H10O2:
NMR: δ 2.82 (6 H, s), δ 8.13 (4 H, s)

IR: 1681 cm-1, no O-H stretch

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Concepts and reason

1H NMR Spectroscopy:

• Nuclear magnetic resonance(NMR)
spectroscopy is one of the techniques in analytical chemistry that is widely used to determine the purity of sample, and also to predict the structure of the organic compounds.

• The NMR
 spectroscopy is a phenomenon that is observed, when the frequency of nuclei of atoms (sample) resonates with frequency of the rotating magnetic field.

• The ΗΝMR
spectroscopy determines different types of hydrogens (chemically non-equivalent hydrogens) that are present in a molecule.

Infrared Spectroscopy (IR):

• Infrared spectroscopy is an important analytical tool to determine the functional groups in the chemical compounds.

• It is also called as the vibrational spectroscopy.

• Any compound having covalent bonds absorbs a range of frequencies of the electromagnetic radiation in the infrared region of the electromagnetic spectrum.

• The IR radiations lie in the wavelength range of 400 - 800 nm.

Fundamentals

1H NMR Spin-Spin Coupling Patterns:

neighboringNumber
(chemically
Number
of
of
Splitting
Peak
hydrogens
equivalent)
non-
peaks
(n 1)
heights
ratio
Name
n
Singlet

The degree of Unsaturation or index of hydrogen deficiency:

DoU = 2C +N- H-X+2
2
where
Degree of unsaturation is DoU
The number of carbons is C.
The number of nitrogen is N
The number o

The molecular formula represents the total number of atoms that are present in the compound. In Chemistry,

IHD refers to the index of hydrogen deficiency which is also called as the degree of unsaturation. IHD (index of hydrogen deficiency) is a formula, which is used to determine the chemical structure from the molecular formula.

A cyclic structure, which is a double and triple bond in structure, can be determined with the help of the degree of unsaturation values.

For example:

IHD = 0, no unsaturation in the molecule. The structure is saturated hydrocarbon. All bonds are single bonds in the compound.

IHD = 1, the structure may contain 1 double bond or 1 ring.

IHD = 2, the structure may contain 2 double bonds or 1 ring with 1 double bond.

IHD = 3, the structure may contain 3 double bonds or 1 ring with 2 double bonds.

IHD = 4, the structure may contain 4 double bonds or 2 triple bonds, or 1 ring with three double bonds or 2 rings with 2 double bonds or 3 rings with 1 double bond.

The mathematical equation for the IR wavenumber:

k
v =
2пс \и

Where,

Wavenumber of absorption = .

Velocity of light = C
.

Force constant = .

Reduced mass = .

IR frequency and range of absorption:

absorption
of
Range
Functional group
(cm)
Alkane (C-H
2800-2900
Alcohol (O-H)
Alkene (C C)
3400-3600
1640-1680
3020-3100
(C-

Given molecular formula:
DoU 210) +0 -10-0+2
2
DoU= 20- 10+ 2
12
DoU
2
DoU 2

Molecular Formula= C^H1202
IR Frequency= 1681 cm
Ketone Functional Group , RCOR = 1750-1680 cm1
|Hence, the molecule contains

From the given H NMR spectra:
Number of hydrogens
Chemical shift
Multiplicity
singlet
Around 8.13ppm (most deshielded)
4H
Ar

The structure of the compound is:

Н
Н.
ketone group
Н
н
Н
Н
)=c
4H, 8.13, singlet
6H, 2.82, singlet
ketone group
H
T
T
T
T
(о
T

Ans:

The structure for the given chemical formulae is:

CHз
Hас

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