Question 4 1.5 pts Compound D (C28H22) decolorizes a solution of bromine in CCl4

Question

Question 4 1.5 pts Compound D (C28H22) decolorizes a solution of bromine in CCl4. 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 (C28H40). How many rings does compound D possess? Question 5 1.5 pts Which one of the following statements is correct? O The IR absorption band due to a single Be-O bond is more intense that the absorption band due to a single Mg-CI bond O The IR absorption band due to a single Be-O bond is less intense that the absorption band due to a single Mg-Cl bond None of the above

Explanation / Answer

4) degree of unsaturation dn = ( nC x 2 + 2 - nH) / 2

where nC = 28 , nH = 22 , we can find dn

dn = ( 28 x 2 + 2 - 22) / 2 = 18 = number of double bonds + number of rings ( since no triple or phenyl present)

now after reatcion with excess H2 we got C28H40

H added = H40 - H28 = 12 H

H2 added = 6 , hence we have 6 double bonds as 1H2 reacts with 1C=C ,

dn = number of double bonds + rings

18 = 6 + number of rings

number of rings = 12

5) The higher is Electronegative difference the more will be intensity of bond. MgCl has more electronegative difference between atoms ( Mg and Cl) comapred to that of BeO

Hence The IR absorbtion band due to single bond BeO is less intense than absorbtion band due to single bond Mg-Cl band

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