Question 4 [1.5 pts] Compound D (C2ot22) decolorizes a solution of bromine in CC

Question

Question 4 [1.5 pts] Compound D (C2ot22) decolorizes a solution of bromine in CCla. Independent analysis shows that compound D does not possess any alkynes or phenyl rings. In the presence of Pt and excess H2 compound D undergoes conversion to compound E (C2sH40). How many rings does compound D possess? Question 5 11.5 pts] Which one of the following statements is correct? a The IR absorption band due to a single Be-O bond is more intense that the absorption band due to a single Mg-Cl bond The IR absorption band due to a single Be-O bond is less intense that the absorption band due to a single Mg-CI bond D None of the above Question 6 [1.5 pts] What is the wavenumber of radiation that has a frequency of 1.947 x 10 Hz? Enter just the numerical value expressed as a whole number. Don't be alarmed when Canvas adds trailing zeros to the number you type in. This action will not affect the grading of your answer

Explanation / Answer

Solution

4)

Compound D has the molecular formula C28H22 and it is decolorizes a solution of bromine in CCl4­, it means that the compound D has double bonds.

When compound D is reacted H2/Pt, it converted into compound E having molecular formula C28H40.

Now calculate the degree of unsaturation of compound D

Unsaturation in compound D = 28-(22/2) +1

                                     = 28-11+1 = 18

So there are 18 position of unsaturation and in compound E there are increased H from 22 to 40 means 18 H added by hydrogenation, so there are 9 double bond.

So remaining sit of unsaturation or no. of rings = 18 - 9 = 9

Thus, compound D possesses 9 rings in its structure.

5) The correct statement is as

“The IR absorption band due to a single Be-O bond is more intense than the absorption band due to a single Mg-Cl bond.”

6)

Frequency = 1.947 x 1014 Hz = 5.136 x 10-15 s-1

We know that

Wave length x frequency = velocity of light

Wave length = velocity of light / frequency

= (2.99 x 108 m/s)/ 5.136 x 10-15 s-1

= 0.582165109 x 107 m

= 5821651.09 m

Thus wavelength = 5821651 m (ans)

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