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vova2212 [387]
3 years ago
13

When an unknown amine reacts with an unknown acid chloride, an amide with a molecular mass of 163 g/mol (M = 163 m/z) is formed.

In the infrared spectrum, important absorptions appear at 1661, 750 and 690 cm–1. The 13C NMR and DEPT spectra are provided below. Draw the structure of the product as the resonance contributor lacking any formal charges.

Chemistry
1 answer:
DanielleElmas [232]3 years ago
6 0

Answer:

            As the molecular mass of given amine is 163 g/mol (a odd number) it means that this compound contains a odd number of Nitrogen atoms. We will first apply Rule of Thirteen to get the molecular formula.

Rule of Thirteen:

First divide the parent peak value by 13 as,

                = 163 ÷ 13

                = 12.53

Now, multiply 13 by 12,

                = 13 × 12 (here, 12 specifies number of carbon atoms)

                = 156

Now subtract 156 from 163,

                = 163 - 156

                = 7

Add 7 into 12,

                = 7 + 12

                = 19 (hydrogen atoms)

So, the rough formula we have is,

                                                       C₁₂H₁₉

Now, add one Nitrogen atom to above formula and subtract one Carbon and 2 Hydrogen atoms as these numbers are equal to atomic mass of Nitrogen atom as,

                C₁₂H₁₉   -------N-------->    C₁₁H₁₇N

Also, as shown in ¹³C-NMR there is one peak around 180 ppm and the peak at 1661 cm⁻¹ in IR spectrum is characteristic to carbonyl group hence, we will add one oxygen atom to the chemical formula accordingly. i.e.

                C₁₁H₁₇N   -------O-------->    C₁₀H₁₃NO

Molecular Formula: C₁₀H₁₃NO

Also,

In NMR the the four peaks around 120 ppm are assigned to a mono substituted benzene ring.

The absence of IR peak above 3200 cm⁻³ also confirms that the amine is tertiary in nature and there is no hydrogen attached to the nitrogen atom.

It can be observed that the peaks in upfield are duplicating. This can be due to the presence of rotamers of said compound.

The most plausible structure for given data is shown below, and the resonance structure along with rotamers are also shown.

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Answer:

Correct option is

B

5 liters of CH

4

(g)NO

2

at STP

No. of molecules=

22.4

5

mol=

22.4

5

×N

A

molecules

A) 5ℊ of H

2

(g)

No. of moles=

2

5

mol=

2

5

×N

A

molecules

B) 5l of CH

4

(g)

No. of moles of CH

4

=

22.4

5

mol=

22.4

5

N

A

molecules

C) 5 mol of O

2

=5N

A

O

2

molecules

D) 5×10

23

molecules of CO

2

(g)

Molecules of 5l NO

2

(g) at STP=5l of CH

4

(g) molecules at STP

Therefore, option B is correct.

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